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Chapter 2  第二章

Wave-Particle Duality, Probability, and the Schrödinger Equation
波粒二象性、概率和薛定谔方程

Abstract 摘要

The developments outlined in Chapter 1 are often described as the Old Quantum Theory. The rules devised were all ad hoc, and the connection between various separate discoveries, such as the particle nature of radiation, the wave nature of electrons and the Bohr atom (as well as other rules not discussed in our brief survey) did not rest on any firm foundation. Quantum mechanics was discovered twice: first, by Werner Heisenberg in 1925 as matrix mechanics, and then again by Erwin Schrödinger in 1926 as wave mechanics. The two forms were soon found to be identical in content, but wave mechanics became a more useful tool because the mathematics of waves were familiar to many physicists. In this chapter we begin our study of quantum mechanics, and we follow, in spirit, the path laid out by Schrödinger.
第一章概述的發展通常被稱為舊量子理論。所設計的規則都是臨時性的，而且各種不同發現之間的關係，例如輻射的粒子性質、電子的波浪性質和玻爾原子（以及我們的簡短調查中沒有討論的其他規則），都沒有任何堅實的基礎。量子力學被發現了兩次：第一次是由海森堡 (Werner Heisenberg) 在 1925 年發現的矩陣力學，第二次則是由薛定谔 (Erwin Schrödinger) 在 1926 年發現的波動力學。這兩種形式很快就被發現在內容上是相同的，但是波動力學成為了更有用的工具，因為許多物理學家都熟悉波的數學。在本章中，我們將開始研究量子力學，在精神上，我們遵循薛定谔所鋪設的路徑。

2-1 RADIATION AS PARTICLES, ELECTRONS AS WAVES
2-1 輻射為粒子，電子為波

The fact that radiation and electrons exhibit both particle and wave properties raises deep conceptual difficulties, as can be seen from the following considerations: There is no doubt that light consists of individual particles, called photons, which carry energy and momentum, as was first unequivocally demonstrated by the Compton effect. The human eye cannot detect individual photons, but it is fairly close to being a photon counter, since under optimal conditions it takes only 5 − 10 5 − 10 5-105-10 photons to activate the darknessadapted eye. There are devices, known as photomultipliers, that can easily detect individual photons.
事實上，輻射和電子同時展現出粒子和波的特性，這引起了深層次的概念上的困難，這可以從以下的考量中看出：毫無疑問，光是由個別的粒子（稱為光子）所組成，這些粒子帶有能量和動量，康普頓效應（Compton effect）首次明確地證明了這一點。人眼無法偵測到個別的光子，但它相當接近光子計數器，因為在最佳條件下，只需要 5 − 10 5 − 10 5-105-10 光子就能啟動適應黑暗的眼睛。有一種稱為光電倍增管的裝置，可以輕易偵測到個別光子。
An interesting thought experiment is discussed in Dirac’s wonderful book on quantum mechanics. When light of a certain polarization is used to produce electrons (as in the photoelectric effect), the latter are emitted with an angular distribution that depends on the direction of the polarization of the incident photon beam. Since in the photoelectric effect single photons eject single electrons, this implies that individual photons, in addition to carrying energy and momentum, also have polarization properties. Suppose we now send a beam of polarized light with an initial intensity I 0 I 0 I_(0)I_{0} through a crystal, which has the property that only the component of light polarized along a particular axis can pass through it. Thus if the polarization of the initial beam is in the direction of the axis, then the emerging beam will have intensity I 0 I 0 I_(0)I_{0}. If the polarization vector makes an angle θ θ theta\theta with the axis, then the intensity of the emerging beam is I 0 cos 2 ⁡ θ I 0 cos 2 ⁡ θ I_(0)cos^(2)thetaI_{0} \cos ^{2} \theta. Let us look at this result in terms of individual photons. If the beam is totally polarized along the direction of the axis, then all
Dirac 在量子力學的精彩著作中討論了一個有趣的思想實驗。當某種偏振光被用來產生電子時（如光電效應），後者的出射角度分佈取決於入射光子光束的偏振方向。由於在光電效應中，單個光子會射出單個電子，這意味著單個光子除了攜帶能量和動量之外，還具有偏振特性。假設我們現在傳送一束初始強度為 I 0 I 0 I_(0)I_{0} 的偏振光穿過一顆晶體，這顆晶體的特性是只有沿著特定軸偏振的光分量才能穿過它。因此，如果初始光束的偏振方向與軸向一致，則新出現的光束將具有 I 0 I 0 I_(0)I_{0} 的強度。如果偏振向量與軸成 θ θ theta\theta 角，則新出現光束的強度為 I 0 cos 2 ⁡ θ I 0 cos 2 ⁡ θ I_(0)cos^(2)thetaI_{0} \cos ^{2} \theta 。讓我們從單個光子的角度來看看這個結果。如果光束沿著軸的方向完全偏振，那麼所有的光子都會被偏振。
of the photons that make up the beam must have been polarized in that direction. For a beam polarized in a different direction, the intensity is reduced by a factor cos 2 ⁡ θ cos 2 ⁡ θ cos^(2)theta\cos ^{2} \theta. This implies that only this fraction of the photons passes through the crystal. However, photons cannot be split into pieces, so that a given photon will either pass through the crystal, or it will not. We have no way of predicting whether an individual photon will pass through. All we can say is that for N N NN incident photons, N cos 2 ⁡ θ N cos 2 ⁡ θ Ncos^(2)thetaN \cos ^{2} \theta will get through, so that the odds, or probability, that a particular photon will get through is cos 2 ⁡ θ cos 2 ⁡ θ cos^(2)theta\cos ^{2} \theta.
組成光束的光子一定是在該方向偏振的。對於偏振方向不同的光束，強度會降低 cos 2 ⁡ θ cos 2 ⁡ θ cos^(2)theta\cos ^{2} \theta 倍。這意味著只有這部分的光子會穿過晶體。然而，光子無法分割成不同的部分，因此特定的光子要麼會穿過晶體，要麼不會。我們無法預測單個光子是否會穿過。我們只能說，對於 N N NN 入射光子， N cos 2 ⁡ θ N cos 2 ⁡ θ Ncos^(2)thetaN \cos ^{2} \theta 會穿透，因此特定光子穿透的機率或概率是 cos 2 ⁡ θ cos 2 ⁡ θ cos^(2)theta\cos ^{2} \theta 。
We also know from classical optics that a beam of light consisting of many photons will exhibit wavelike properties-that is, diffraction and interference. An experiment carried out by G. I. Taylor in 1909 was the first to show that a beam of light gave rise to a diffraction pattern around a needle even when the intensity of the light was so low that only one photon at a time passed by the needle. Since then, many more experiments showed that the interference and diffraction properties cannot be due to the collective effect of the many photons in a beam. This raises new problems. Consider a thought experiment, which is a variant of the Taylor experiment, in which a very low intensity beam of light is directed at a screen with two slits in it. The photons are then detected at a second screen (Fig. 2-1). The intensity is such that at a given time no more than one photon passes through the two-slit screen. After very many photons have passed by, we see the classically expected diffraction pattern. Classically this is well understood: If the electric fields at a particular point r r r\mathbf{r} on the detecting screen due to electromagnetic waves crossing slits 1 and 2 are E 1 ( r , t ) E 1 ( r , t ) E_(1)(r,t)\mathbf{E}_{1}(\mathbf{r}, t) and E 2 ( r , t ) E 2 ( r , t ) E_(2)(r,t)\mathbf{E}_{2}(\mathbf{r}, t) respectively, then the total field at the point r r r\mathbf{r} at the time t t tt is the sum of the fields. This is a consequence of the superposition rules for electric fields, which in turn is a consequence of the fact that Maxwell’s equations for the electromagnetic fields are linear. The intensity at the screen is proportional to the square of the total electric field, and thus to ( E 1 ( r , t ) + E 2 ( r , t ) ) 2 E 1 ( r , t ) + E 2 ( r , t ) 2 (E_(1)(r,t)+E_(2)(r,t))^(2)\left(\mathbf{E}_{1}(\mathbf{r}, t)+\mathbf{E}_{2}(\mathbf{r}, t)\right)^{2}. The interference pattern is due to the presence of the E 1 ( r , t ) ⋅ E 2 ( r , t ) E 1 ( r , t ) ⋅ E 2 ( r , t ) E_(1)(r,t)*E_(2)(r,t)\mathbf{E}_{1}(\mathbf{r}, t) \cdot \mathbf{E}_{2}(\mathbf{r}, t) cross term in the square of the sum of the fields. If only slit 1 were open, the intensity would be proportional to E 1 ( r , t ) 2 E 1 ( r , t ) 2 E_(1)(r,t)^(2)\mathbf{E}_{1}(\mathbf{r}, t)^{2}, and if only slit 2 were open, the intensity would be proportional to E 2 ( r , t ) 2 E 2 ( r , t ) 2 E_(2)(r,t)^(2)\mathbf{E}_{2}(\mathbf{r}, t)^{2}. If we now translate intensity into probability, as suggested by our discussion about polarization, we find that if only slit 1 is open, the probability of finding a photon at r r r\mathbf{r} is P 1 ( r , t ) P 1 ( r , t ) P_(1)(r,t)P_{1}(\mathbf{r}, t), and if only slit 2 is open, the probability of finding a photon at r r r\mathbf{r} is P 2 ( r , t ) P 2 ( r , t ) P_(2)(r,t)P_{2}(\mathbf{r}, t). However, if both slits are open, the probability is not the sum of the probabilities associated with each slit.
我們也從古典光學中得知，由許多光子組成的光束會呈現出波浪般的特性，也就是衍射與干涉。G. I. Taylor 在 1909 年所做的實驗首次顯示，即使光束的強度很低，每次只有一個光子經過針頭，針頭周圍也會產生衍射圖案。自此之後，更多的實驗顯示干涉和衍射的特性不可能是由於光束中許多光子的集體效應。這引起了新的問題。考慮一個思想實驗，它是泰勒實驗的變體，在這個實驗中，一束強度很低的光束會射向一個有兩道細縫的螢幕。光子會在第二個螢幕上被偵測到 (圖 2-1)。光束的強度是這樣的：在給定時間內，不會有超過一個光子通過雙縫螢幕。當非常多的光子通過後，我們會看到經典預期的衍射圖案。這在經典上是很好理解的：如果在偵測螢幕上的特定點 r r r\mathbf{r} 上，由於電磁波穿越狭縫 1 和 2 而產生的電場分別是 E 1 ( r , t ) E 1 ( r , t ) E_(1)(r,t)\mathbf{E}_{1}(\mathbf{r}, t) 和 E 2 ( r , t ) E 2 ( r , t ) E_(2)(r,t)\mathbf{E}_{2}(\mathbf{r}, t) ，那麼在 t t tt 時，點 r r r\mathbf{r} 上的總電場就是這兩個電場的總和。這是電場的疊加規則的結果，而疊加規則又是 Maxwell 電磁場方程式是線性的結果。螢幕上的強度與總電場的平方成正比，因此也與 ( E 1 ( r , t ) + E 2 ( r , t ) ) 2 E 1 ( r , t ) + E 2 ( r , t ) 2 (E_(1)(r,t)+E_(2)(r,t))^(2)\left(\mathbf{E}_{1}(\mathbf{r}, t)+\mathbf{E}_{2}(\mathbf{r}, t)\right)^{2} 成正比。干擾圖案是由於在電場總和的平方中存在 E 1 ( r , t ) ⋅ E 2 ( r , t ) E 1 ( r , t ) ⋅ E 2 ( r , t ) E_(1)(r,t)*E_(2)(r,t)\mathbf{E}_{1}(\mathbf{r}, t) \cdot \mathbf{E}_{2}(\mathbf{r}, t) 交叉項所致。 如果只有狭缝 1 打开，强度将与 E 1 ( r , t ) 2 E 1 ( r , t ) 2 E_(1)(r,t)^(2)\mathbf{E}_{1}(\mathbf{r}, t)^{2} 成比例，如果只有狭缝 2 打开，强度将与 E 2 ( r , t ) 2 E 2 ( r , t ) 2 E_(2)(r,t)^(2)\mathbf{E}_{2}(\mathbf{r}, t)^{2} 成比例。如果我們現在將強度轉換成概率，就像我們在討論偏振時所建議的，我們會發現如果只有狭縫 1 打開，在 r r r\mathbf{r} 處找到一個光子的概率是 P 1 ( r , t ) P 1 ( r , t ) P_(1)(r,t)P_{1}(\mathbf{r}, t) ，而如果只有狭縫 2 打開，在 r r r\mathbf{r} 處找到一個光子的概率是 P 2 ( r , t ) P 2 ( r , t ) P_(2)(r,t)P_{2}(\mathbf{r}, t) 。但是，如果兩個狭縫都打開，概率就不是與每個狭縫相關的概率之和了。
The only way to resolve these difficulties is to assume that each photon interferes with itself. This can be handled by assuming that each photon is described by its own electric field, e ( r , t ) e ( r , t ) e(r,t)\mathbf{e}(\mathbf{r}, t), and that in the presence of two slits, the photon field at the detector is the sum of two terms. These are associated with the presence of two slits, so that
解決這些難題的唯一方法是假設每個光子都會自相干擾。假設每個光子都有自己的電場 e ( r , t ) e ( r , t ) e(r,t)\mathbf{e}(\mathbf{r}, t) 來描述，而且在有兩個縫隙的情況下，偵測器上的光子場是兩個項的總和，就可以處理這個問題。這兩項與兩道狹縫的存在有關，因此
e ( r , t ) = e 1 ( r , t ) + e 2 ( r , t ) e ( r , t ) = e 1 ( r , t ) + e 2 ( r , t ) e(r,t)=e_(1)(r,t)+e_(2)(r,t)\mathbf{e}(\mathbf{r}, t)=\mathbf{e}_{1}(\mathbf{r}, t)+\mathbf{e}_{2}(\mathbf{r}, t)
Figure 2-1 Interference pattern resulting from the passage of a beam of photons through a screen with two open slits.
圖 2-1 光子束穿過有兩個開縫的螢幕時所產生的干涉圖案。
just as for a classical light wave. Note that we are still talking about a single photon. The only real requirements are (1) that the field e ( r , t e ( r , t e(r,t\mathbf{e}(\mathbf{r}, t ) obeys a linear equation and that (2) in the classical limit a large collection of photons acts in accordance with Maxwell’s equations. The actual formulation of a quantum theory of photons is somewhat complicated, and we leave the discussion of this to Supplement 18-A.
就像古典光波一樣。請注意，我們討論的仍是單一光子。唯一真正的要求是：(1) 場 e ( r , t e ( r , t e(r,t\mathbf{e}(\mathbf{r}, t ) 遵循一個線性等式；(2) 在經典極限中，一大堆光子依照麥克斯韋方程行事。光子量子理論的實際公式有點複雜，我們留待補篇 18-A 再討論。
At this point we turn our attention to electrons. From the time of their discovery, electrons were described as particles. They appear to travel along trajectories determined by the electric and magnetic forces acting on them, they have mass, and they carry energy and momentum. Nevertheless, they have wavelike properties, as first determined in the diffraction of electron beams by crystals. As demonstrated by the beautiful two-slit experiment of A. Tonomura presented in Chapter 1 (see Fig. 1-9) the pattern of the hits of electrons on the absorbing screen slowly builds up to a wavelike interference pattern. This experiment is a realization of the thought experiments we discussed above in connection with photons. Individual electrons appear to hit the screen at random. As the number of electrons increases, the expected interference pattern emerges. Again, the place where a single electron hits cannot be affected by the fact that other electrons came before it, or will come after it, so that the emergence of the pattern must lie in the property of each electron. By analogy of our conjectures in connection with the photon, we may expect that the properties of a single electron are described by an analog of the one-photon electric field e ( r , t ) e ( r , t ) e(r,t)\mathbf{e}(\mathbf{r}, \boldsymbol{t}). We thus expect that an electron will be described by a wave function traditionally denoted by ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t). In order to get an electron to interfere with itself we must insist that ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) obeys a linear equation. In that case, a sum of two wave functions is also a wave function, so that the superposition rules apply. Furthermore, we expect that the predictability of where an electron hits (as if it were a classical particle) will be replaced by a statement, involving the wave function ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t), about the probability that an electron arrives at r r r\mathbf{r}. The rest of this chapter is devoted to arguments that lead us to the correct form of the linear equation obeyed by ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(r, t), the Schrödinger equation, the probability for finding an electron at r r r\mathbf{r} at a time t t tt in terms of ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) and other general properties of the wave function. We shall approach this by constructing waves that might simulate the properties of particles. The study of wave packets will be helpful in this matter, even though the idea that there are real waves that act like particles is not correct.
此時，我們將目光轉向電子。從發現電子開始，電子就被描述成粒子。它們看起來是沿著由作用在它們身上的電力和磁力所決定的軌跡移動，它們有質量，並且帶有能量和動量。儘管如此，電子仍具有波狀特性，這在電子束被水晶衍射的過程中首次被確定。正如第一章所介紹的 A. Tonomura 美麗的雙縫實驗（見圖 1-9）所證明的，電子在吸收螢幕上的撞擊圖案慢慢地形成波浪狀的干涉圖案。這個實驗實現了我們上面討論過的與光子有關的思想實驗。個別電子看起來是隨機撞擊螢幕。隨著電子數量的增加，預期的干涉圖案就會出現。同樣地，單個電子撞擊的位置不會受到在它之前或之後的其他電子的影響，所以圖案的出現一定是每個電子的特性。根據我們對光子的猜想，我們可以預期單一電子的屬性是由單光子電場 e ( r , t ) e ( r , t ) e(r,t)\mathbf{e}(\mathbf{r}, \boldsymbol{t}) 的類比所描述的。因此，我們預期電子將由傳統上以 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) 表示的波函数來描述。為了讓電子自相干擾，我們必須堅持 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) 遵循一個線性等式。在這種情況下，兩個波函数之和也是一個波函数，因此疊加規則適用。 此外，我們預期電子撞擊位置的可預測性 (就像它是一個經典粒子一樣) 將被一個涉及波函数 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) 的陳述所取代，這個陳述是關於電子到達 r r r\mathbf{r} 的概率。本章的其餘部分將用於論證，引導我們找到 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(r, t) 所遵從的線性等式的正確形式、薛定谔等式、在 t t tt 的時間 r r r\mathbf{r} 找到電子的概率，以及波函数的 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) 和其他一般屬性。我們將透過建構可能模擬粒子特性的波來處理這個問題。波包的研究在這個問題上會很有幫助，儘管有真正的波會像粒子一樣行動的想法並不正確。

2-2 PLANE WAVES AND WAVE PACKETS
2-2 平面波和波包

A harmonic wave propagating in the positive x x xx-direction with wave number k k kk has the form
沿波數 k k kk 正向 x x xx 傳播的諧波的形式為
ψ k ( x , t ) = A 1 cos ⁡ ( k x − ω t ) + A 2 sin ⁡ ( k x − ω t ) ψ k ( x , t ) = A 1 cos ⁡ ( k x − ω t ) + A 2 sin ⁡ ( k x − ω t ) psi_(k)(x,t)=A_(1)cos(kx-omega t)+A_(2)sin(kx-omega t)\psi_{k}(x, t)=A_{1} \cos (k x-\omega t)+A_{2} \sin (k x-\omega t)
or equivalently 或相當於
ψ k ( x , t ) = A e i ( k x − ω t ) + B e − i ( k x − ω t ) ψ k ( x , t ) = A e i ( k x − ω t ) + B e − i ( k x − ω t ) psi_(k)(x,t)=Ae^(i(kx-omega t))+Be^(-i(kx-omega t))\psi_{k}(x, t)=A e^{i(k x-\omega t)}+B e^{-i(k x-\omega t)}
The wave number k k kk is related to the wavelength by
波數 k k kk 與波長的關係為
k = 2 π / λ k = 2 π / λ k=2pi//lambdak=2 \pi / \lambda
and the angular frequency ω ω omega\omega is related to the period T T TT by
而角頻率 ω ω omega\omega 與週期 T T TT 的關係為
ω = 2 π / T ω = 2 π / T omega=2pi//T\omega=2 \pi / T
and thus to the frequency ν = 1 / T ν = 1 / T nu=1//T\nu=1 / T by
因此，頻率 ν = 1 / T ν = 1 / T nu=1//T\nu=1 / T 由
ω = 2 π ν ω = 2 π ν omega=2pi nu\omega=2 \pi \nu
In general ω ω omega\omega will be related to k k kk in some way. For example in the case of light propagating in a vacuum, ν = c / λ ν = c / λ nu=c//lambda\nu=c / \lambda where c c cc is thgspetd of light. Hence ω = k c ω = k c omega=kc\omega=k c. This relationship is rot tree for light in a dispersive medium. There ν = c / n λ ν = c / n λ nu=c//n lambda\nu=c / n \lambda, where n n nn is the refractive index of the medium, and n n nn is generally a function of the wavelength n = n = n=n= n ( λ ) n ( λ ) n(lambda)n(\lambda). We shall determine below the relation between ω ω omega\omega and k k kk for the waves of interest to us.
一般而言， ω ω omega\omega 會以某種方式與 k k kk 相關。例如，在光在真空中傳播的情況下， ν = c / λ ν = c / λ nu=c//lambda\nu=c / \lambda 其中 c c cc 是光的thgspetd。因此 ω = k c ω = k c omega=kc\omega=k c 。對於在色散介質中的光，這個關係是rot tree。在這裡， ν = c / n λ ν = c / n λ nu=c//n lambda\nu=c / n \lambda ，其中 n n nn 是介質的折射率，而 n n nn 通常是波長 n = n = n=n= n ( λ ) n ( λ ) n(lambda)n(\lambda) 的函數。下面我們將針對我們感興趣的波來確定 ω ω omega\omega 和 k k kk 之間的關係。
Since ψ k ( x , t ) ψ k ( x , t ) psi_(k)(x,t)\psi_{k}(x, t) does not depend on y y yy or z z zz, it takes on the same value everywhere on the y − z y − z y-zy-z plane, and is thus called a plane wave. ψ k ( x , t ) ψ k ( x , t ) psi_(k)(x,t)\psi_{k}(x, t) is a plane wave for all possible values of k k kk, and it is therefore possible to take a superposition of plane waves with different amplitudes A ( k ) A ( k ) A(k)A(k) and B ( k ) B ( k ) B(k)B(k). There is nothing in what we have done so far to require A ( k ) A ( k ) A(k)A(k) and B ( k ) B ( k ) B(k)B(k) to be real, and, in fact, we shall see that in general we must allow ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) to be complex. Let us consider the wave A ( k ) e i ( k x − ω t ) A ( k ) e i ( k x − ω t ) A(k)e^(i(kx-omega t))A(k) e^{i(k x-\omega t)} and add such waves for a variety of values of k . 1 k . 1 k.^(1)k .{ }^{1} The superposition is called a wave packet, and it takes the form
由於 ψ k ( x , t ) ψ k ( x , t ) psi_(k)(x,t)\psi_{k}(x, t) 並不取決於 y y yy 或 z z zz ，它在 y − z y − z y-zy-z 平面上的任何地方都取相同的值，因此稱為平面波。 ψ k ( x , t ) ψ k ( x , t ) psi_(k)(x,t)\psi_{k}(x, t) 對於所有可能的 k k kk 值都是平面波，因此可以取振幅不同的 A ( k ) A ( k ) A(k)A(k) 和 B ( k ) B ( k ) B(k)B(k) 平面波的疊加。在我們目前所做的工作中，沒有任何地方需要 A ( k ) A ( k ) A(k)A(k) 和 B ( k ) B ( k ) B(k)B(k) 是實數，事實上，我們將會看到，在一般情況下，我們必須允許 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 是複數。讓我們考慮波 A ( k ) e i ( k x − ω t ) A ( k ) e i ( k x − ω t ) A(k)e^(i(kx-omega t))A(k) e^{i(k x-\omega t)} ，並為 k . 1 k . 1 k.^(1)k .{ }^{1} 的各種值加上這樣的波，這個疊加被稱為波包，它的形式是
ψ ( x , t ) = ∫ − ∞ ∞ d k A ( k ) e i ( k x − ω t ) ψ ( x , t ) = ∫ − ∞ ∞   d k A ( k ) e i ( k x − ω t ) psi(x,t)=int_(-oo)^(oo)dkA(k)e^(i(kx-omega t))\psi(x, t)=\int_{-\infty}^{\infty} d k A(k) e^{i(k x-\omega t)}
We begin by considering the wave packet at time t = 0 t = 0 t=0t=0,
我們首先考慮時間 t = 0 t = 0 t=0t=0 時的波包、
ψ ( x , 0 ) = ∫ − ∞ ∞ d k A ( k ) e i k x ψ ( x , 0 ) = ∫ − ∞ ∞   d k A ( k ) e i k x psi(x,0)=int_(-oo)^(oo)dkA(k)e^(ikx)\psi(x, 0)=\int_{-\infty}^{\infty} d k A(k) e^{i k x}
and illustrate it by considering a special form, called the gaussian form
並考慮一種稱為高斯形式的特殊形式來加以說明
A ( k ) = e − α ( k − k 0 ) 2 / 2 A ( k ) = e − α k − k 0 2 / 2 A(k)=e^(-alpha(k-k_(0))^(2)//2)A(k)=e^{-\alpha\left(k-k_{0}\right)^{2} / 2}
This function is centered about k 0 k 0 k_(0)k_{0}, and it falls off ranidly away from that center. We shall see that it is the-width of the square of this function that is of interest. The square falls to 1 / 3 1 / 3 1//31 / 3 of its peak value when α ( k − k 0 ) 2 ≅ 1 α k − k 0 2 ≅ 1 alpha(k-k_(0))^(2)~=1\alpha\left(k-k_{0}\right)^{2} \cong 1. This means that the width can be taken to be Δ k = 2 / α Δ k = 2 / α Delta k=2//sqrtalpha\Delta k=2 / \sqrt{\alpha}. The integral in (2-5) can now be done in steps. We first make a change of variables to q ′ = k − k 0 q ′ = k − k 0 q^(')=k-k_(0)q^{\prime}=k-k_{0}, and we end up with
這個函數的中心點是 k 0 k 0 k_(0)k_{0} ，而它會從中心點離開。我們將會看到，這個函數的平方寬度才是我們感興趣的地方。當 α ( k − k 0 ) 2 ≅ 1 α k − k 0 2 ≅ 1 alpha(k-k_(0))^(2)~=1\alpha\left(k-k_{0}\right)^{2} \cong 1 時，平方會下降到其峰值的 1 / 3 1 / 3 1//31 / 3 。這表示寬度可以取為 Δ k = 2 / α Δ k = 2 / α Delta k=2//sqrtalpha\Delta k=2 / \sqrt{\alpha} 。現在可以分步完成 (2-5) 中的積分。我們先變更變數為 q ′ = k − k 0 q ′ = k − k 0 q^(')=k-k_(0)q^{\prime}=k-k_{0} ，最後就會得到
= 2 π α e i k 0 x e − x 2 / 2 α = 2 π α e i k 0 x e − x 2 / 2 α {:=sqrt((2pi)/(alpha))e^(ik_(0)x)e^(-x^(2)//2alpha):}\begin{aligned} & =\sqrt{\frac{2 \pi}{\alpha}} e^{i k_{0} x} e^{-\boldsymbol{x}^{2} / 2 \alpha} \end{aligned}
Aside from the scale factor, which could have been absorbed by slightly modifying A ( k ) A ( k ) A(k)A(k), we end up with a plane-wave factor characterized by the wave number k 0 k 0 k_(0)k_{0} and a modulating function that acts to localize the packet about x = 0 x = 0 x=0x=0. The width of that packet, which is also gaussian, is defined in the same way as before: We square the function and see where it drops off to about 1 / 3 1 / 3 1//31 / 3. This provides the width Δ x = 2 α Δ x = 2 α Delta x=2sqrtalpha\Delta x=2 \sqrt{\alpha}. We see that there is a reciprocal relation between the width of the function A ( k ) A ( k ) A(k)A(k) that determines the shape of the wave packet, and the width of the wave packet. In fact, the product of the two widths has the property that Δ k Δ x = 4 Δ k Δ x = 4 Delta k Delta x=4\Delta k \Delta x=4 here. The specific value of the number on the right side is not important. A slightly different definition of the widths would have given us a different an-
除了尺度因數 (它可以透過稍微修改 A ( k ) A ( k ) A(k)A(k) 而被吸收)，我們最後得到一個平面波因數，它的特徵是波數 k 0 k 0 k_(0)k_{0} 和一個調變函數，它的作用是將 x = 0 x = 0 x=0x=0 的封包局部化。該封包的寬度也是高斯，定義方式與之前相同：我們將函數平方，並查看其下降至 1 / 3 1 / 3 1//31 / 3 左右的位置。這提供了寬度 Δ x = 2 α Δ x = 2 α Delta x=2sqrtalpha\Delta x=2 \sqrt{\alpha} 。我們可以看到，決定波包形狀的函數 A ( k ) A ( k ) A(k)A(k) 的寬度與波包的寬度之間存在著對等關係。事實上，這兩個寬度的乘積具有 Δ k Δ x = 4 Δ k Δ x = 4 Delta k Delta x=4\Delta k \Delta x=4 此處的特性。右邊數字的具體值並不重要。如果寬度的定義稍有不同，我們就會得到不同的---------------。
swer. What is quite general is that the product is independent of α α alpha\alpha It is actually a general result of Fourier integrals that
swer。一般而言，乘積與 α α alpha\alpha 無關。
Δ k Δ x > 1 2 Δ k Δ x > 1 2 Delta k Delta x > (1)/(2)\Delta k \Delta x>\frac{1}{2}
so that this reciprocal relation is true for wave packets in general. We illustrate this in the example that follows.
所以一般而言，這個互易關係對於波包是真實的。我們在下面的範例中說明這一點。

EXAMPLE 2-1 範例 2-1

Consider a wave packet for which
考慮一個波包，其
A ( k ) = N − K ≤ k ≤ K = 0 elsewhere A ( k ) = N − K ≤ k ≤ K = 0  elsewhere  {:[A(k)=N-K <= k <= K],[=0" elsewhere "]:}\begin{aligned} A(k) & =N & & -K \leq k \leq K \\ & =0 & & \text { elsewhere } \end{aligned}
Calculate ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0), and use some reasonable definition of the width to show that (2-8) is satisfied.
計算 ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) ，並使用一些合理的寬度定義來證明 (2-8) 滿足。
SOLUTION We have 解法 我們有
ψ ( x , 0 ) = ∫ − K K d k N e i k x = N i x ( e i K x − e − K x ) = 2 N sin ⁡ K x x ψ ( x , 0 ) = ∫ − K K   d k N e i k x = N i x e i K x − e − K x = 2 N sin ⁡ K x x psi(x,0)=int_(-K)^(K)dkNe^(ikx)=(N)/(ix)(e^(iKx)-e^(-Kx))=2N(sin Kx)/(x)\psi(x, 0)=\int_{-K}^{K} d k N e^{i k x}=\frac{N}{i x}\left(e^{i K x}-e^{-K x}\right)=2 N \frac{\sin K x}{x}
The definition of A ( k ) A ( k ) A(k)A(k) easily shows that Δ k = 2 K Δ k = 2 K Delta k=2K\Delta k=2 K. A reasonable definition of Δ x Δ x Delta x\Delta x might be the distance between the two points at which ψ ( x ) ψ ( x ) psi(x)\psi(x) first vanishes as it gets away from x = 0 x = 0 x=0x=0. This happens when K x = ± π K x = ± π Kx=+-piK x= \pm \pi, so that Δ x = 2 π / K Δ x = 2 π / K Delta x=2pi//K\Delta x=2 \pi / K. It follows that
A ( k ) A ( k ) A(k)A(k) 的定義很容易說明 Δ k = 2 K Δ k = 2 K Delta k=2K\Delta k=2 K 。 Δ x Δ x Delta x\Delta x 的合理定義可能是 ψ ( x ) ψ ( x ) psi(x)\psi(x) 離開 x = 0 x = 0 x=0x=0 時首先消失的兩點之間的距離。這發生在 K x = ± π K x = ± π Kx=+-piK x= \pm \pi 時，因此 Δ x = 2 π / K Δ x = 2 π / K Delta x=2pi//K\Delta x=2 \pi / K 。由此可知
Δ k Δ x = 4 π Δ k Δ x = 4 π Delta k Delta x=4pi\Delta k \Delta x=4 \pi
which certainly satisfies (2-8).
這當然滿足 (2-8)。
The mathematical description of how wave packets move is a little messy, so we set the material off in a subsection. The important result is that for a wave packet for which the spread about a particular value of k k kk-say, k 0 k 0 k_(0)k_{0}-is small, so that it looks a lot like a plane wave modulated by a very wide function in x x xx, then the wave packet moves with the group velocity
波包如何移動的數學描述有點亂，所以我們把材料放在一個小節中。重要的結果是，對於 k k kk 的特定值 (例如 k 0 k 0 k_(0)k_{0} )，波包的擴散很小，因此它看起來很像由 x x xx 中一個很寬的函數所調變的平面波，那麼波包的移動速度就是群速度
v g = ( ∂ ω ( k ) ∂ k ) k = k 0 v g = ∂ ω ( k ) ∂ k k = k 0 v_(g)=((del omega(k))/(del k))_(k=k_(0))v_{g}=\left(\frac{\partial \omega(k)}{\partial k}\right)_{k=k_{0}}
and its width in x x xx spreads as a function of time. In the following subsection we see this in detail for our Gaussian wave packet.
的寬度會隨著時間而擴散。在下面的小節中，我們會看到高斯波包的詳細情況。

*How Wave Packets Move *波包如何移動

As indicated in (2-4), the motion of the wave package can be obtained by integration, provided we know how ω ω omega\omega depends on k k kk. Let us assume that A ( k ) A ( k ) A(k)A(k) is sharply peaked about the value of k = k 0 k = k 0 k=k_(0)k=k_{0}. We may make the approximation
如 (2-4) 所示，只要我們知道 ω ω omega\omega 如何取決於 k k kk ，就可以透過積分得到波包的運動。讓我們假設 A ( k ) A ( k ) A(k)A(k) 在 k = k 0 k = k 0 k=k_(0)k=k_{0} 的值附近有尖銳的峰值。我們可以做近似
= ω ( k 0 ) + ( k − k 0 ) ( ∂ ω ∂ k ) k = k 0 + 1 2 ( k − k 0 ) 2 ( ∂ 2 ω ∂ k 2 ) k = k 0 = ω k 0 + k − k 0 ∂ ω ∂ k k = k 0 + 1 2 k − k 0 2 ∂ 2 ω ∂ k 2 k = k 0 quad=omega(k_(0))+(k-k_(0))((del omega)/(del k))_(k=k_(0))+(1)/(2)(k-k_(0))^(2)((del^(2)omega)/(delk^(2)))_(k=k_(0))\quad=\omega\left(k_{0}\right)+\left(k-k_{0}\right)\left(\frac{\partial \omega}{\partial k}\right)_{k=k_{0}}+\frac{1}{2}\left(k-k_{0}\right)^{2}\left(\frac{\partial^{2} \omega}{\partial k^{2}}\right)_{k=k_{0}}
Taylor expansion 泰勒擴展
With this, 有了這個
( k x − ω t ) = ( k 0 x − ω ( k 0 ) t ) + ( k − k 0 ) [ x − ( ∂ ω ∂ k ) k 0 t ] − 1 2 ( k − k 0 ) 2 ( ∂ 2 ω ∂ k 2 ) k 0 t ( k x − ω t ) = k 0 x − ω k 0 t + k − k 0 x − ∂ ω ∂ k k 0 t − 1 2 k − k 0 2 ∂ 2 ω ∂ k 2 k 0 t (kx-omega t)=(k_(0)x-omega(k_(0))t)+(k-k_(0))[x-((del omega)/(del k))_(k_(0))t]-(1)/(2)(k-k_(0))^(2)((del^(2)omega)/(delk^(2)))_(k_(0))t(k x-\omega t)=\left(k_{0} x-\omega\left(k_{0}\right) t\right)+\left(k-k_{0}\right)\left[x-\left(\frac{\partial \omega}{\partial k}\right)_{k_{0}} t\right]-\frac{1}{2}\left(k-k_{0}\right)^{2}\left(\frac{\partial^{2} \omega}{\partial k^{2}}\right)_{k_{0}} t
When this is inserted into (2-4) we get, after changing variables to q = k − k 0 q = k − k 0 q=k-k_(0)q=k-k_{0}, the following:
將此插入 (2-4) 之後，將變數變更為 q = k − k 0 q = k − k 0 q=k-k_(0)q=k-k_{0} 之後，我們會得到以下結果：
ψ ( x , t ) = e i ( k 0 x − ω ( k 0 ) t ) ∫ − ∞ ∞ d q A ( q + k 0 ) e i q ( x − v s t ) e − i q 2 ( ∂ 2 ω / ∂ k 2 ) 0 t 2 ψ ( x , t ) = e i k 0 x − ω k 0 t ∫ − ∞ ∞   d q A q + k 0 e i q x − v s t e − i q 2 ∂ 2 ω / ∂ k 2 0 t 2 psi(x,t)=e^(i(k_(0)x-omega(k_(0))t))int_(-oo)^(oo)dqA(q+k_(0))e^(iq(x-v_(s)t))e^(-iq^(2)(del^(2)omega//delk^(2))_(0)t2)\psi(x, t)=e^{i\left(k_{0} x-\omega\left(k_{0}\right) t\right)} \int_{-\infty}^{\infty} d q A\left(q+k_{0}\right) e^{i q\left(x-v_{s} t\right)} e^{-i q^{2}\left(\partial^{2} \omega / \partial k^{2}\right)_{0} t 2}
If the last factor in the integrand were absent, as is the case when ω = k ω = k omega=k\omega=\boldsymbol{k} for example, then the integral becomes a function of ( x − v g t ) x − v g t (x-v_(g)t)\left(x-v_{g} t\right). If this function peaks at x = 0 x = 0 x=0x=0 at time t = 0 t = 0 t=0t=0, then it will peak at x = v g t x = v g t x=v_(g)tx=v_{g} t at a later time t t tt. The velocity with which the packet moves is the group velocity. The second term in the integrand, when it does not vanish, modifies the amplitude A ( q + k 0 ) A q + k 0 A(q+k_(0))A\left(q+k_{0}\right) and thus the shape of the wave packet; this is best seen in terms of our gaussian packet. The integral is worked out just like the one that leads to (2-7), since the extra term is also gaussian, as the exponent is quadratic in q 2 q 2 q^(2)q^{2}. With the notation,
如果在積分中沒有最後一個因數，例如 ω = k ω = k omega=k\omega=\boldsymbol{k} 時的情況，那麼積分就會變成 ( x − v g t ) x − v g t (x-v_(g)t)\left(x-v_{g} t\right) 的函數。如果這個函數在 t = 0 t = 0 t=0t=0 時於 x = 0 x = 0 x=0x=0 達到峰值，那麼它會在稍後的 t t tt 時於 x = v g t x = v g t x=v_(g)tx=v_{g} t 達到峰值。封包移動的速度就是群速度。積分項中的第二項，當它不消失時，會改變振幅 A ( q + k 0 ) A q + k 0 A(q+k_(0))A\left(q+k_{0}\right) ，進而改變波包的形狀；這在我們的高斯波包中看得最清楚。積分的運算方式與 (2-7) 的運算方式相同，因為額外的項也是高斯的，因為指數是 q 2 q 2 q^(2)q^{2} 的二次方。使用記號、
( ∂ 2 ω ∂ k 2 ) k = k 0 = β ∂ 2 ω ∂ k 2 k = k 0 = β ((del^(2)omega)/(delk^(2)))_(k=k_(0))=beta\left(\frac{\partial^{2} \omega}{\partial k^{2}}\right)_{k=k_{0}}=\beta
We see that in addition to changing x x xx to x − v g t x − v g t x-v_(g)tx-v_{g} t, we change α α alpha\alpha to α + 2 i β t α + 2 i β t alpha+2i beta t\alpha+2 i \beta t. We therefore get
我們看到，除了將 x x xx 改為 x − v g t x − v g t x-v_(g)tx-v_{g} t 之外，我們還將 α α alpha\alpha 改為 α + 2 i β t α + 2 i β t alpha+2i beta t\alpha+2 i \beta t 。因此我們得到
ψ ( x , t ) = 2 π α + 2 i β t e i ( k 0 x − ω 0 t ) e − ( α − v t t ) 2 2 α + 4 i β t ψ ( x , t ) = 2 π α + 2 i β t e i k 0 x − ω 0 t e − α − v t t 2 2 α + 4 i β t psi(x,t)=sqrt((2pi)/(alpha+2i beta t))e^(i(k_(0)x-omega_(0)t))e^(-((alpha-v_(t)t)^(2))/(2alpha+4i beta t))\psi(x, t)=\sqrt{\frac{2 \pi}{\alpha+2 i \beta t}} e^{i\left(k_{0} x-\omega_{0} t\right)} e^{-\frac{\left(\alpha-v_{t} t\right)^{2}}{2 \alpha+4 i \beta t}}
This is a rather untransparent expression, involving a complex function of x x xx and t t tt, but as we shall see soon, it’s absolute magnitude has a physical meaning. We therefore calculate
這是一個相當不透明的表達方式，涉及到 x x xx 和 t t tt 的複雜函數，但正如我們即將看到的，它的絕對大小具有物理意義。因此，我們可以計算出
| ψ ( x , t ) | 2 = ψ ∗ ( x , t ) ψ ( x , t ) | ψ ( x , t ) | 2 = ψ ∗ ( x , t ) ψ ( x , t ) |psi(x,t)|^(2)=psi^(**)(x,t)psi(x,t)|\psi(x, t)|^{2}=\psi^{*}(x, t) \psi(x, t)
The pre-factor is easily handled; the phase factor has magnitude 1, and in the second exponential, the exponent has to be added to its complex conjugate. When all of this is done, we get
前因數很容易處理；相位因數的大小為 1，在第二個指數中，指數必須加上它的複共轭。完成所有這些後，我們得到
| ψ ( x , t ) | 2 = 2 π α 2 + 4 β 2 t 2 e − α ( x − v k t ) 2 α 2 + 4 β 2 t 2 | ψ ( x , t ) | 2 = 2 π α 2 + 4 β 2 t 2 e − α x − v k t 2 α 2 + 4 β 2 t 2 |psi(x,t)|^(2)=(2pi)/(sqrt(alpha^(2)+4beta^(2)t^(2)))e-(alpha(x-v_(k)t)^(2))/(alpha^(2)+4beta^(2)t^(2))|\psi(x, t)|^{2}=\frac{2 \pi}{\sqrt{\alpha^{2}+4 \beta^{2} t^{2}}} e-\frac{\alpha\left(x-v_{k} t\right)^{2}}{\alpha^{2}+4 \beta^{2} t^{2}}
Comparison with the value at t = 0 t = 0 t=0t=0 shows that the width, initially given by 2 α 2 α 2sqrtalpha2 \sqrt{\alpha}, now becomes 2 α + 4 β 2 t 2 / α = 2 α 1 + 4 β 2 t 2 / α 2 2 α + 4 β 2 t 2 / α = 2 α 1 + 4 β 2 t 2 / α 2 2sqrt(alpha+4beta^(2)t^(2)//alpha)=2sqrtalphasqrt(1+4beta^(2)t^(2)//alpha^(2))2 \sqrt{\alpha+4 \beta^{2} t^{2} / \alpha}=2 \sqrt{\alpha} \sqrt{1+4 \beta^{2} t^{2} / \alpha^{2}}. The result is that the wave packet traveling with speed v g v g v_(g)v_{g} spreads as time increases. When α α alpha\alpha is large, so that the packet is broad, the spreading is small. Nevertheless it is there, and this stands in the way of interpreting the wave packet as describing the particle itself.
與 t = 0 t = 0 t=0t=0 處的值比較，可以發現最初以 2 α 2 α 2sqrtalpha2 \sqrt{\alpha} 表示的寬度，現在變成 2 α + 4 β 2 t 2 / α = 2 α 1 + 4 β 2 t 2 / α 2 2 α + 4 β 2 t 2 / α = 2 α 1 + 4 β 2 t 2 / α 2 2sqrt(alpha+4beta^(2)t^(2)//alpha)=2sqrtalphasqrt(1+4beta^(2)t^(2)//alpha^(2))2 \sqrt{\alpha+4 \beta^{2} t^{2} / \alpha}=2 \sqrt{\alpha} \sqrt{1+4 \beta^{2} t^{2} / \alpha^{2}} 。結果是以速度 v g v g v_(g)v_{g} 行走的波包會隨著時間增加而擴散。當 α α alpha\alpha 較大時，波包較寬，擴散較小。儘管如此，它還是存在，而且這會妨礙我們把波包詮釋為描述粒子本身。

2-3 THE PROBABILITY INTERPRETATION OF THE WAVE FUNCTION
2-3 波函数的概率解释

At this point we recall that in the case of photons, the intensity, proportional to [ e ( r , t ) ] 2 [ e ( r , t ) ] 2 [e(r,t)]^(2)[\mathbf{e}(\mathbf{r}, t)]^{2}, was interpreted as being proportional to the probability of finding a photon in the vicinity of r r r\mathbf{r} at the time t t tt. Since we were led to the conclusion that ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) had to be complex, we assume that it is | ψ ( r , t ) | 2 | ψ ( r , t ) | 2 |psi(r,t)|^(2)|\psi(\mathbf{r}, t)|^{2} that is related to the corresponding probability of finding an electron in the vicinity of r r r\mathbf{r} at time t t tt. For simplicity we deal with motion in one dimension (though the generalization is straightforward) and assert
在這一點上，我們回想一下，在光子的情況下，與 [ e ( r , t ) ] 2 [ e ( r , t ) ] 2 [e(r,t)]^(2)[\mathbf{e}(\mathbf{r}, t)]^{2} 成比例的強度被解釋為與 t t tt 時在 r r r\mathbf{r} 附近找到光子的概率成比例。由於我們得到的結論是 ψ ( r , t ) ψ ( r , t ) psi(r,t)\psi(\mathbf{r}, t) 必須是複雜的，因此我們假設是 | ψ ( r , t ) | 2 | ψ ( r , t ) | 2 |psi(r,t)|^(2)|\psi(\mathbf{r}, t)|^{2} 與時間 t t tt 時在 r r r\mathbf{r} 附近找到電子的相對概率有關。為了簡單起見，我們處理單維的運動（雖然廣泛化是很直接的），並斷言
P ( x , t ) d x = | ψ ( x , t ) | 2 d x P ( x , t ) d x = | ψ ( x , t ) | 2 d x P(x,t)dx=|psi(x,t)|^(2)dxP(x, t) d x=|\psi(x, t)|^{2} d x
The probability interpretation is due to Max Born who, shortly after the discovery of the Schrödinger equation, studied the scattering of a beam of electrons by a target and was led to the above form.
概率詮釋來自 Max Born，他在發現薛定谔方程後不久，研究了電子束被目標散射的情況，並引出了上述形式。
With this interpretation of | ψ ( x , t ) | 2 | ψ ( x , t ) | 2 |psi(x,t)|^(2)|\psi(x, t)|^{2} the spreading of the wave packet presents no problems. All it implies is that an electron, known to be in a certain region with some probability distribution, will, with increasing time, have an increasing probability of being found outside that region.
在這種 | ψ ( x , t ) | 2 | ψ ( x , t ) | 2 |psi(x,t)|^(2)|\psi(x, t)|^{2} 的詮釋下，波包的擴散不會產生任何問題。它所暗示的只是，一個已知在某個區域內的電子，隨著時間的增加，在該區域外被發現的機率也會增加。
The appearance of probability in quantum mechanics differs from its appearance in classical physics. Here it is not a statement of ignorance about what is “really” going on, as is the case when we speak of the probability of a coin-toss leading to heads or tails, but it is a basic limitation on what we can know when the wave function is known. The mathematical implications of this interpretation of ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) will be discussed at length in the next chapter.
量子力學中概率的出現不同於其在經典物理學中的出現。在這裡，它並不是對「真正」發生了什麼事的無知聲明，就像我們談到擲硬幣導致正面或反面的概率一樣，而是當波函數已知時，我們所能知道的事的基本限制。下一章將詳細討論 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 這種詮釋的數學含意。
The probability interpretation allows us to understand electron interference. As a consequence of the linearity of the equation for ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t), a wave function of the form
概率詮釋讓我們了解電子干擾。由於 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 等式的線性關係，波形為
ψ ( x , t ) = N ( ψ 1 ( x , t ) + ψ 2 ( x , t ) ) ψ ( x , t ) = N ψ 1 ( x , t ) + ψ 2 ( x , t ) psi(x,t)=N(psi_(1)(x,t)+psi_(2)(x,t))\psi(x, t)=N\left(\psi_{1}(x, t)+\psi_{2}(x, t)\right)
is a solution, if both ψ 1 ψ 1 psi_(1)\psi_{1} and ψ 2 ψ 2 psi_(2)\psi_{2} are solutions. Let ψ 1 ψ 1 psi_(1)\psi_{1} be the wave function of the electron that describes the system with slit 2 closed. This wave function is then definitely associated with passage through slit 1 . Similarly, if ψ 2 ψ 2 psi_(2)\psi_{2} is the wave function with slit 1 closed, then the wave function with both slits open will then be the sum of the wave functions ψ 1 ψ 1 psi_(1)\psi_{1} and ψ 2 ψ 2 psi_(2)\psi_{2}. Consequently, the probability density of finding an electron at a point x x xx on the photographic plate behind the slits is proportional to | ψ 1 ( x , t ) + ψ 2 ( x , t ) | 2 ψ 1 ( x , t ) + ψ 2 ( x , t ) 2 |psi_(1)(x,t)+psi_(2)(x,t)|^(2)\left|\psi_{1}(x, t)+\psi_{2}(x, t)\right|^{2}.
是一個解，如果 ψ 1 ψ 1 psi_(1)\psi_{1} 和 ψ 2 ψ 2 psi_(2)\psi_{2} 都是解。讓 ψ 1 ψ 1 psi_(1)\psi_{1} 成為電子的波函数，描述狭縫 2 閉合時的系統。這個波函数肯定與通過狭縫 1 有關。同樣地，如果 ψ 2 ψ 2 psi_(2)\psi_{2} 是狹縫 1 閉合時的波函数，那麼兩個狹縫都打開的波函数就是 ψ 1 ψ 1 psi_(1)\psi_{1} 和 ψ 2 ψ 2 psi_(2)\psi_{2} 的和。因此，在狹縫後方感光板上的 x x xx 點找到電子的概率密度與 | ψ 1 ( x , t ) + ψ 2 ( x , t ) | 2 ψ 1 ( x , t ) + ψ 2 ( x , t ) 2 |psi_(1)(x,t)+psi_(2)(x,t)|^(2)\left|\psi_{1}(x, t)+\psi_{2}(x, t)\right|^{2} 成正比。
We have 我們有
| ψ 1 ( x , t ) + ψ 2 ( x , t ) | 2 = | ψ 1 ( x , t ) | 2 + | ψ 2 ( x , t ) | 2 + 2 Re ( ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) ψ 1 ( x , t ) + ψ 2 ( x , t ) 2 = ψ 1 ( x , t ) 2 + ψ 2 ( x , t ) 2 + 2 Re ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) |psi_(1)(x,t)+psi_(2)(x,t)|^(2)=|psi_(1)(x,t)|^(2)+|psi_(2)(x,t)|^(2)+2Re(psi_(1)(x,t)psi_(2)^(**)(x,t):}\left|\psi_{1}(x, t)+\psi_{2}(x, t)\right|^{2}=\left|\psi_{1}(x, t)\right|^{2}+\left|\psi_{2}(x, t)\right|^{2}+2 \operatorname{Re}\left(\psi_{1}(x, t) \psi_{2}^{*}(x, t)\right.
and the third term clearly exhibits the interference. This effect requires that there be a single electron source, so that the phase difference of the two wave functions ψ 1 ψ 1 psi_(1)\psi_{1} and ψ 2 ψ 2 psi_(2)\psi_{2} does not vary randomly. If the phase difference varied unpredictably, then the probability would be determined by
而第三項明顯地展現了干擾。這種效果需要有單一電子源，因此兩個波函数 ψ 1 ψ 1 psi_(1)\psi_{1} 和 ψ 2 ψ 2 psi_(2)\psi_{2} 的相位差不會隨機變化。如果相位差無法預測地變化，那麼概率將由以下公式決定
P ( x , t ) = | ψ 1 ( x , t ) | 2 + | ψ 2 ( x , t ) | 2 P ( x , t ) = ψ 1 ( x , t ) 2 + ψ 2 ( x , t ) 2 P(x,t)=|psi_(1)(x,t)|^(2)+|psi_(2)(x,t)|^(2)P(x, t)=\left|\psi_{1}(x, t)\right|^{2}+\left|\psi_{2}(x, t)\right|^{2}

EXAMPLE 2-2 範例 2-2

Consider a two-slit experiment, in which the wave function at slit 1 acquires an arbitrary random phase that is to be averaged over, so that the total wave function at the screen is ψ ( x , t ) = ψ ( x , t ) = psi(x,t)=\psi(x, t)= e t ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) e t ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) e^(t phi)psi_(1)(x,t)+psi_(2)(x,t)e^{t \phi} \psi_{1}(x, t)+\psi_{2}(x, t). (Such a situation might arise if there were t t tt wo incoherent electron sources, one at each slit). Show that under these circumstances the interference averages out.
考慮一個雙縫實驗，其中縫 1 處的波函数會獲得一個任意的隨機相位，這個相位會被平均，因此螢幕上的總波函数是 ψ ( x , t ) = ψ ( x , t ) = psi(x,t)=\psi(x, t)= e t ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) e t ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) e^(t phi)psi_(1)(x,t)+psi_(2)(x,t)e^{t \phi} \psi_{1}(x, t)+\psi_{2}(x, t) 。(如果有 t t tt 兩個非相干電子源，每個狭缝有一個，就可能會出現這種情況）。證明在這些情況下，干擾會平均化。
SOLUTION We need to calculate
解答 我們需要計算

随机相位角差 随机相位差

| ψ ( x , t ) | 2 = ( e i ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) ) ( e − i ϕ ψ 1 ∗ ( x , t ) + ψ 2 ∗ ( x , t ) ) = | ψ 1 ( x , t ) | 2 + | ψ 2 ( x , t ) | 2 + e i ϕ ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) + e − i ϕ ψ 2 ( x , t ) ψ 1 ∗ ( x , t ) = | ψ 1 ( x , t ) | 2 + | ψ 2 ( x , t ) | 2 + 2 cos ⁡ ϕ Re ( ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) ) + 2 sin ⁡ ϕ Im ( ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) ) | ψ ( x , t ) | 2 = e i ϕ ψ 1 ( x , t ) + ψ 2 ( x , t ) e − i ϕ ψ 1 ∗ ( x , t ) + ψ 2 ∗ ( x , t ) = ψ 1 ( x , t ) 2 + ψ 2 ( x , t ) 2 + e i ϕ ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) + e − i ϕ ψ 2 ( x , t ) ψ 1 ∗ ( x , t ) = ψ 1 ( x , t ) 2 + ψ 2 ( x , t ) 2 + 2 cos ⁡ ϕ Re ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) + 2 sin ⁡ ϕ Im ψ 1 ( x , t ) ψ 2 ∗ ( x , t ) {:[|psi(x","t)|^(2)=(e^(i phi)psi_(1)(x,t)+psi_(2)(x,t))(e^(-i phi)psi_(1)^(**)(x,t)+psi_(2)^(**)(x,t))],[=|psi_(1)(x,t)|^(2)+|psi_(2)(x,t)|^(2)+e^(i phi)psi_(1)(x","t)psi_(2)^(**)(x","t)+e^(-i phi)psi_(2)(x","t)psi_(1)^(**)(x","t)],[=|psi_(1)(x,t)|^(2)+|psi_(2)(x,t)|^(2)+2cos phi Re(psi_(1)(x,t)psi_(2)^(**)(x,t))+2sin phi Im(psi_(1)(x,t)psi_(2)^(**)(x,t))]:}\begin{aligned} |\psi(x, t)|^{2} & =\left(e^{i \phi} \psi_{1}(x, t)+\psi_{2}(x, t)\right)\left(e^{-i \phi} \psi_{1}^{*}(x, t)+\psi_{2}^{*}(x, t)\right) \\ & =\left|\psi_{1}(x, t)\right|^{2}+\left|\psi_{2}(x, t)\right|^{2}+e^{i \phi} \psi_{1}(x, t) \psi_{2}^{*}(x, t)+e^{-i \phi} \psi_{2}(x, t) \psi_{1}^{*}(x, t) \\ & =\left|\psi_{1}(x, t)\right|^{2}+\left|\psi_{2}(x, t)\right|^{2}+2 \cos \phi \operatorname{Re}\left(\psi_{1}(x, t) \psi_{2}^{*}(x, t)\right)+2 \sin \phi \operatorname{Im}\left(\psi_{1}(x, t) \psi_{2}^{*}(x, t)\right) \end{aligned}
The angle ϕ ϕ phi\phi varies randomly from electron to electron, so that in the pattern of dots, the terms involving this angle average out to zero.
角度 ϕ ϕ phi\phi 會隨電子的不同而隨機變化，因此在點的圖案中，涉及此角度的項平均為零。
There is a potential difficulty with the probability interpretation. Consider a beam of electrons passing through a screen with a double slit. Suppose we could determine in some
概率解釋有潛在的困難。考慮一束電子通過一個有雙縫的螢幕。假設我們可以在某些情況下確定

30 Chapter 2 Wave-Particle Duality, Probability, and the Schrödinger Equation
30 第 2 章 波粒二象性、概率和薛定谔方程

way through which slit each given electron passes. If the electrons are far apart, then as far as any given electron is concerned, we might as well have closed the other slit. If such a detection means were available to us, we could divide all the electrons, arriving at the screen on which they are detected, into two classes: those that went through slit 1 and those that went through slit 2 . In that case, however, we would get the distribution (2-17). We are forced to the conclusion that we get an interference pattern only if the experiment does not allow us to determine which slit the electrons go through. If we somehow arrange to find out what the slit of passage was, then the interference pattern disappears and the probability is just the sum of the individual probabilities. The rule is simple:
每個特定的電子通過哪一條細縫。如果電子之間的距離很遠，那麼就任何一個電子而言，我們都可以關閉另一條細縫。如果我們有這樣的檢測方法，我們就可以把所有到達檢測螢幕的電子分成兩類：一類是通過狭缝 1 的電子，另一類是通過狭缝 2 的電子。然而，在這種情況下，我們會得到分佈 (2-17)。只有在實驗無法讓我們確定電子經過哪個狭縫時，我們才會被迫得出干涉圖案的結論。如果我們以某種方式安排找出通過的裂縫，那麼干擾圖案就會消失，而概率只是個別概率的總和。這個規則很簡單：
If the paths are not determined, add the wave functions and square; if the paths are determined, square the wave functions and add.
如果路徑未定，則將波函数相加並平方；如果路徑已定，則將波函数平方並相加。
If we look at the discussion in Example 2-2, we see that somehow the acquisition of knowledge in which way the electron went must introduce a random phase into the component of the wave function that is being “looked at.” For further discussion see Chapter 20.
如果我們看一下例 2-2 的討論，就會發現不知何故，在獲得電子往哪個方向走的知識時，一定會在被 「觀察 」的波函数分量中引入隨機相位。進一步的討論請參閱第 20 章。

2-4 THE SCHRÖDINGER EQUATION
2-4 薛定谔方程

We have constructed a wave function that may be used to describe satisfactorily the probability of finding a freely traveling electron at x x xx, at a time t t tt. We make the connection with physics by first recalling that according to de Broglie k = p / ℏ k = p / ℏ k=p//ℏk=p / \hbar and as suggested by the Planck relation ω = E / ℏ ω = E / ℏ omega=E//ℏ\omega=E / \hbar Thus the wave packet may be rewritten in the form
我們建構了一個波函数，可用來滿意地描述在 t t tt 時，在 x x xx 處找到一個自由移動電子的概率。我們首先回想一下，根據 de Broglie 的 k = p / ℏ k = p / ℏ k=p//ℏk=p / \hbar 和普朗克關係 ω = E / ℏ ω = E / ℏ omega=E//ℏ\omega=E / \hbar 的建議，波包可以改寫成以下形式，從而與物理學建立起聯繫
ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) e i ( p x − E t ) ℏ ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) e i ( p x − E t ) ℏ psi(x,t)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)e^(i(px-Et)ℏ)\psi(x, t)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) e^{i(p x-E t) \hbar}
We now take E ( p ) = p 2 / 2 m E ( p ) = p 2 / 2 m E(p)=p^(2)//2mE(p)=p^{2} / 2 m for a free particle. We see that the group velocity is
現在我們取 E ( p ) = p 2 / 2 m E ( p ) = p 2 / 2 m E(p)=p^(2)//2mE(p)=p^{2} / 2 m 為自由粒子。我們看到群速度為
ν g = ∂ ω ∂ k = ∂ E ∂ p = p m Relation between E and G.velocity ν g = ∂ ω ∂ k = ∂ E ∂ p = p m  Relation between E and G.velocity  nu_(g)=(del omega)/(del k)=(del E)/(del p)=(p)/(m)quad" Relation between E and G.velocity "\nu_{g}=\frac{\partial \omega}{\partial k}=\frac{\partial E}{\partial p}=\frac{p}{m} \quad \text { Relation between E and G.velocity }
which confirms our association of ℏ ω ℏ ω ℏomega\hbar \omega with the energy. Here ϕ ( p ) / 2 π ℏ ϕ ( p ) / 2 π ℏ phi(p)//sqrt(2piℏ)\phi(p) / \sqrt{2 \pi \hbar} plays the role played by A ( k ) A ( k ) A(k)A(k) in eq. (2-5).
這證實了 ℏ ω ℏ ω ℏomega\hbar \omega 與能量的關係。這裡 ϕ ( p ) / 2 π ℏ ϕ ( p ) / 2 π ℏ phi(p)//sqrt(2piℏ)\phi(p) / \sqrt{2 \pi \hbar} 扮演了公式 (2-5) 中 A ( k ) A ( k ) A(k)A(k) 所扮演的角色。
Now suppose that the particle under consideration is not free, so that instead of E = p 2 / 2 m E = p 2 / 2 m E=p^(2)//2mE=p^{2} / 2 m, we have
現在假設考慮中的粒子不是自由的，因此我們有 E = p 2 / 2 m E = p 2 / 2 m E=p^(2)//2mE=p^{2} / 2 m ，而不是 E = p 2 / 2 m E = p 2 / 2 m E=p^(2)//2mE=p^{2} / 2 m 。
Generally, particles can be bounded, so
一般而言，粒子是有邊界的，所以
E = p 2 / 2 m + V ( x ) E = p 2 / 2 m + V ( x ) E=p^(2)//2m+V(x)E=p^{2} / 2 m+V(x)
If we were to mindlessly insert this into the exponent in eq. (2-18), the wave function would be changed rather trivially: the new wave function ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) is just the product of the free particle wave function, and a factor Now this factor is a pure phase factor, whose absolute square is 1 . This would mean that the addition of a potential to the energy in This is patently wrong since in a potential, the velocity of a particle changes from place to place What we need to do is to find the equation that ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) is a solution of, and then modify that equation to take into account the presence of a potential V ( x ) V ( x ) V(x)V(x).
如果我們不假思索地將其插入公式 (2-18) 中的指數，波函数就會發生相當微不足道的變化：新的波函数 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 只是自由粒子波函数與一個因子的乘積，而這個因子是純相位因子，其絕對平方為 1。我們需要做的是找到 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的方程，然後修訂這個方程，以考慮到電勢 V ( x ) V ( x ) V(x)V(x) 的存在。
We proceed as follows: If we differentiate with respect to time, we find that
我們的步驟如下：如果我們對時間進行微分，我們會發現
i ℏ ∂ ψ ( x , t ) ∂ t = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) E ( p ) e i ( p x − E t ) ℏ ℏ = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) p 2 2 m e i ( p x − E t ) ℏ i ℏ ∂ ψ ( x , t ) ∂ t = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) E ( p ) e i ( p x − E t ) ℏ ℏ = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) p 2 2 m e i ( p x − E t ) ℏ {:[iℏ(del psi(x,t))/(del t)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)E(p)e^(i(px-Et)ℏℏ)],[=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)(p^(2))/(2m)e^(i(px-Et)ℏ)]:}\begin{aligned} i \hbar \frac{\partial \psi(x, t)}{\partial t} & =\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) E(p) e^{i(p x-E t) \hbar \hbar} \\ & =\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) \frac{p^{2}}{2 m} e^{i(p x-E t) \hbar} \end{aligned}
On the other hand, 另一方面、
ℏ i ∂ ∂ x ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) p e i ( p x − E t ) ℏ ℏ i ∂ ∂ x ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) p e i ( p x − E t ) ℏ (ℏ)/(i)(del)/(del x)psi(x,t)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)pe^(i(px-Et)ℏ)\frac{\hbar}{i} \frac{\partial}{\partial x} \psi(x, t)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) p e^{i(p x-E t) \hbar}
and hence 因此
( ℏ i ∂ ∂ x ) 2 ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) p 2 e i ( p x − E t ) / ℏ ℏ i ∂ ∂ x 2 ψ ( x , t ) = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) p 2 e i ( p x − E t ) / ℏ ((ℏ)/(i)(del)/(del x))^(2)psi(x,t)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)p^(2)e^(i(px-Et)//ℏ)\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right)^{2} \psi(x, t)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) p^{2} e^{i(p x-E t) / \hbar}
We may combine these two results to get an equation for ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) that is solved by (2-18). This is
我們可以結合這兩個結果，得到 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的方程式，由 (2-18) 解得。這就是
i ℏ σ ψ ( x , t ) r t = − π 2 2 m r 2 ( x , t ) γ x 2 i ℏ σ ψ ( x , t ) r t = − π 2 2 m r 2 ( x , t ) γ x 2 iℏ(sigma psi(x,t))/(rt)=-(pi^(2))/(2m)(r^(2)(x,t))/(gammax^(2))i \hbar \frac{\sigma \psi(x, t)}{r t}=-\frac{\pi^{2}}{2 m} \frac{r^{2}(x, t)}{\gamma x^{2}}
This is the 這是
Although we started from a solution of (2-22), the equation takes precedence over the solution. It is easy to see that starting from the equation, a solution of the form e i ( k x + ω t ) e i ( k x + ω t ) e^(i(kx+omega t))e^{i(k x+\omega t)} would correspond to a negative kinetic energy and that A cos ⁡ ( k x − ω t ) + B sin ⁡ ( k x − ω t ) A cos ⁡ ( k x − ω t ) + B sin ⁡ ( k x − ω t ) A cos(kx-omega t)+B sin(kx-omega t)A \cos (k x-\omega t)+B \sin (k x-\omega t) will only be a solution if B = i A B = i A B=iAB=i A. To the extent that (2-22) is a translation of E = p 2 / 2 m E = p 2 / 2 m E=p^(2)//2mE=p^{2} / 2 m, with E E EE being replaced by i ℏ ∂ ∂ t i ℏ ∂ ∂ t iℏ(del)/(del t)i \hbar \frac{\partial}{\partial t} and p p pp by − i ℏ ∂ ∂ x − i ℏ ∂ ∂ x -iℏ(del)/(del x)-i \hbar \frac{\partial}{\partial x}, we can generalize the energy equation in the presence of a potential V ( x ) V ( x ) V(x)V(x),
雖然我們從 (2-22) 的解開始，但等式優先於解。很容易看出，從等式開始， e i ( k x + ω t ) e i ( k x + ω t ) e^(i(kx+omega t))e^{i(k x+\omega t)} 形式的解將對應於負動能，而 A cos ⁡ ( k x − ω t ) + B sin ⁡ ( k x − ω t ) A cos ⁡ ( k x − ω t ) + B sin ⁡ ( k x − ω t ) A cos(kx-omega t)+B sin(kx-omega t)A \cos (k x-\omega t)+B \sin (k x-\omega t) 只有在 B = i A B = i A B=iAB=i A 的情況下才是解。只要 (2-22) 是 E = p 2 / 2 m E = p 2 / 2 m E=p^(2)//2mE=p^{2} / 2 m 的轉換，而 E E EE 被 i ℏ ∂ ∂ t i ℏ ∂ ∂ t iℏ(del)/(del t)i \hbar \frac{\partial}{\partial t} 取代， p p pp 被 − i ℏ ∂ ∂ x − i ℏ ∂ ∂ x -iℏ(del)/(del x)-i \hbar \frac{\partial}{\partial x} 取代，我們就可以在電勢 V ( x ) V ( x ) V(x)V(x) 存在的情況下概括能量方程式、
E = p 2 2 m + V ( x ) E = p 2 2 m + V ( x ) E=(p^(2))/(2m)+V(x)E=\frac{p^{2}}{2 m}+V(x)
to the general Schrödinger equation
到一般薛定谔方程
2 t ≡ i s ( x t ) ℏ 2 t ≡ i s ( x t ) ℏ 2t-=(is(xt))/(ℏ)2 t \equiv \frac{i s(x t)}{\hbar}
i ℏ ∂ ψ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 ψ ( x , t ) ∂ x 2 + V ( x ) ψ ( x , t ) i ℏ ∂ ψ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 ψ ( x , t ) ∂ x 2 + V ( x ) ψ ( x , t ) iℏ(del psi(x,t))/(del t)=-(ℏ^(2))/(2m)(del^(2)psi(x,t))/(delx^(2))+V(x)psi(x,t)i \hbar \frac{\partial \psi(x, t)}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi(x, t)}{\partial x^{2}}+V(x) \psi(x, t)
This is the basic equation that we will be working with in much of this book.
這是我們在這本書的大部分內容中都會用到的基本等式。
Let us return to the free particle case. The most general solution of that equation depends on the form of ϕ ( p ) ϕ ( p ) phi(p)\phi(p), and we shall now show that this is determined by the initial condition-that is, by 2 ( x , 0 ) 2 ( x , 0 ) 2(x,0)2(x, 0) In fact, if we t = 0 t = 0 t=0t=0 in the solution (2-18), we get
讓我們回到自由粒子的情況。該方程式最一般的解取決於 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 的形式，我們現在要證明這取決於初始條件，也就是 2 ( x , 0 ) 2 ( x , 0 ) 2(x,0)2(x, 0) 事實上，如果我們在解 (2-18) 中 t = 0 t = 0 t=0t=0 ，我們會得到
ψ ( x , 0 ) = 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) e ℓ p x / ℏ ψ ( x , 0 ) = 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) e ℓ p x / ℏ psi(x,0)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)e^(ℓpx//ℏ)\psi(x, 0)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) e^{\ell p x / \hbar}
and this equation determines ϕ ( p ) ϕ ( p ) phi(p)\phi(p). The solution of (2-23) is also determined by ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0). This is in contrast to the familiar wave equation
而這個等式決定了 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 。(2-23) 的解也由 ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) 決定。這與我們熟悉的波方程式
∂ 2 f ( x , t ) ∂ t 2 = u 2 ∂ 2 f ( x , t ) ∂ x 2 ∂ 2 f ( x , t ) ∂ t 2 = u 2 ∂ 2 f ( x , t ) ∂ x 2 (del^(2)f(x,t))/(delt^(2))=u^(2)(del^(2)f(x,t))/(delx^(2))\frac{\partial^{2} f(x, t)}{\partial t^{2}}=u^{2} \frac{\partial^{2} f(x, t)}{\partial x^{2}}
in which both f ( x , 0 ) f ( x , 0 ) f(x,0)f(x, 0) and ( ∂ f ( x , t ) / ∂ t ) t = 0 ( ∂ f ( x , t ) / ∂ t ) t = 0 (del f(x,t)//del t)_(t=0)(\partial f(x, t) / \partial t)_{t=0} have to be specified. The difference is a consequence of the fact that the Schrödinger equation is of first order in t t tt. We shall see that this is closely related to the probability interpretation of ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t).
其中 f ( x , 0 ) f ( x , 0 ) f(x,0)f(x, 0) 和 ( ∂ f ( x , t ) / ∂ t ) t = 0 ( ∂ f ( x , t ) / ∂ t ) t = 0 (del f(x,t)//del t)_(t=0)(\partial f(x, t) / \partial t)_{t=0} 都必須指定。這個差異是薛定谔方程在 t t tt 中屬於一階的結果。我們將會看到這與 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的概率詮釋密切相關。

*The Relation between ϕ ( p ) ϕ ( p ) phi(p)\phi(p) and ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0)
ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 和 ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) 之間的關係

The relation between ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) and ϕ ( p ) ϕ ( p ) phi(p)\phi(p) is obtained by noting that (2-23) is a Fourier integral and thus can be inverted. Here we make use of the properties of Fourier integrals outlined in Supplement 2-A. [www.wiley.com/college/gasiorowicz]
ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) 與 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 之間的關係可由 (2-23) 得到，注意 (2-23) 是傅立葉積分，因此可以倒轉。在此，我們使用補充 2-A 中概述的傅立葉積分特性。[www.wiley.com/college/gasiorowicz] 。
32 Chapter 2 Wave-Particle Duality, Probability, and the Schrödinger Equation
32 第 2 章 波粒二象性、概率和薛定谔方程
We multiply ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) by e − i p ′ x / ℏ e − i p ′ x / ℏ e^(-ip^(')x//ℏ)e^{-i p^{\prime} x / \hbar} and integrate over all x x xx. This leads to
我們將 ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) 乘以 e − i p ′ x / ℏ e − i p ′ x / ℏ e^(-ip^(')x//ℏ)e^{-i p^{\prime} x / \hbar} 並對所有 x x xx 進行整合。這導致
∫ − ∞ ∞ d x ψ ( x , 0 ) e − i p ′ x / ℏ = 1 2 π ℏ ∫ − ∞ ∞ d x ∫ − ∞ ∞ d p ϕ ( p ) e i ( p − p ′ x / ℏ ∫ − ∞ ∞   d x ψ ( x , 0 ) e − i p ′ x / ℏ = 1 2 π ℏ ∫ − ∞ ∞   d x ∫ − ∞ ∞   d p ϕ ( p ) e i p − p ′ x / ℏ int_(-oo)^(oo)dx psi(x,0)e^(-ip^(')x//ℏ)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dxint_(-oo)^(oo)dp phi(p)e^(i(p-p^(')x//ℏ:})\int_{-\infty}^{\infty} d x \psi(x, 0) e^{-i p^{\prime} x / \hbar}=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x \int_{-\infty}^{\infty} d p \phi(p) e^{i\left(p-p^{\prime} x / \hbar\right.}
We now use ∫ − ∞ ∞ d x e i ( p − p ′ ) x / ℏ = 2 π ℏ δ ( p − p ′ ) ∫ − ∞ ∞   d x e i p − p ′ x / ℏ = 2 π ℏ δ p − p ′ int_(-oo)^(oo)dxe^(i(p-p^('))x//ℏ)=2piℏdelta(p-p^('))\int_{-\infty}^{\infty} d x e^{i\left(p-p^{\prime}\right) x / \hbar}=2 \pi \hbar \delta\left(p-p^{\prime}\right) to get 2 π ℏ ϕ ( p ′ ) 2 π ℏ ϕ p ′ sqrt(2piℏ)phi(p^('))\sqrt{2 \pi \hbar} \phi\left(p^{\prime}\right) on the right-hand side, so that
現在我們使用 ∫ − ∞ ∞ d x e i ( p − p ′ ) x / ℏ = 2 π ℏ δ ( p − p ′ ) ∫ − ∞ ∞   d x e i p − p ′ x / ℏ = 2 π ℏ δ p − p ′ int_(-oo)^(oo)dxe^(i(p-p^('))x//ℏ)=2piℏdelta(p-p^('))\int_{-\infty}^{\infty} d x e^{i\left(p-p^{\prime}\right) x / \hbar}=2 \pi \hbar \delta\left(p-p^{\prime}\right) 來得到右側的 2 π ℏ ϕ ( p ′ ) 2 π ℏ ϕ p ′ sqrt(2piℏ)phi(p^('))\sqrt{2 \pi \hbar} \phi\left(p^{\prime}\right) ，因此
ϕ ( p ) = 1 2 π ℏ ∫ − ∞ ∞ d x ψ ( x , 0 ) e − i p x / ℏ ϕ ( p ) = 1 2 π ℏ ∫ − ∞ ∞   d x ψ ( x , 0 ) e − i p x / ℏ phi(p)=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dx psi(x,0)e^(-ipx//ℏ)\phi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x \psi(x, 0) e^{-i p x / \hbar}

2-5 THE HEISENBERG UNCERTAINTY RELATIONS
2-5 海森堡不確定性關係

Let us return to the reciprocal relation obtained in eq. (2-8),
讓我們回到公式 (2-8) 所得到的倒易關係、
Δ k Δ x > 1 / 2 Δ k Δ x > 1 / 2 Delta k Delta x > 1//2\Delta k \Delta x>1 / 2
Recalling the identification of ℏ k ℏ k ℏk\hbar k with the momentum, this relation takes the form
回想 ℏ k ℏ k ℏk\hbar k 與動量的關係，此關係的形式為
This is called the Heisenberg uncertainty relation, or the Heisenberg indeterminacy relation. It arose in the context of our discussion of the wave packets, but as we now see, it is a statement about the wave function. We saw that ψ ( x ) ψ ( x ) psi(x)\psi(x) cannot describe a particle that is both well-localized in space and has a sharp momentum. This is in great contrast to classical mechanics. What the relation states is that there is a quantitative limitation on the accuracy with which we can describe a system using our familiar, classical notions of position and momentum. Position and momentum are said to be complementary variables. We can illustrate this limitation by a couple of examples. 2 2 ^(2){ }^{2}
這就是所謂的海森堡不確定性關係 (Heisenberg uncertainty relation)，或是海森堡不確定性關係 (Heisenberg indeterminacy relation)。它是在我們討論波包時產生的，但我們現在看到，它是關於波函数的聲明。我們看到 ψ ( x ) ψ ( x ) psi(x)\psi(x) 無法描述一個既在空間中定位良好，又有銳利動量的粒子。這與經典力學形成很大的對比。這個關係說明，我們使用熟悉的、經典的位置和動量概念來描述一個系統時，在精確度上有量的限制。位置和動量被稱為互補變數。我們可以用幾個例子來說明這個限制。 2 2 ^(2){ }^{2}

Diffraction of a Photon Beam
光子光束的衍射

Consider a beam of photons, passing through a slit of width a a aa (Fig. 2-2). When the beam is treated as an electromagnetic wave, one can show that the beam is diffracted with an angular spread of magnitude
考慮一束光子，通過寬度為 a a aa 的細縫（圖 2-2）。當光束被視為電磁波時，我們可以顯示光束的衍射角度擴散幅度為
θ ≈ λ a θ ≈ λ a theta~~(lambda )/(a)\theta \approx \frac{\lambda}{a}
where λ λ lambda\lambda is the wavelength of the light. The particle-wave duality of quantum mechanics allows us to describe the beam as a sequence of photons passing through the slit. Under those circumstances, one would not expect a spreading of the beam. The uncertainty relation rescues us, so to speak, from a paradox. The photon momentum in the x x xx-direction is given by p x = h / λ p x = h / λ p_(x)=h//lambdap_{x}=h / \lambda. The y y yy-coordinate of the photon is indeterminate to the extent that
其中 λ λ lambda\lambda 是光的波長。量子力學的粒子-波二重性讓我們可以將光束描述為一連串光子通過狭縫。在這種情況下，我們不會預期光束會擴散。可以說，不確定性關係將我們從悖論中拯救出來。 x x xx 方向的光子動量由 p x = h / λ p x = h / λ p_(x)=h//lambdap_{x}=h / \lambda 給出。光子的 y y yy 坐標在以下範圍內是不確定的
Δ y ≤ a Δ y ≤ a Delta y <= a\Delta y \leq a
2 2 ^(2){ }^{2} Many examples are discussed in W. Heisenberg, The Physical Principles of the Quantum Theory, Dover Publications, New York, 1930, and in most books of quantum mechanics listed in the bibliography. See also J.
2 2 ^(2){ }^{2} W. Heisenberg，The Physical Principles of the Quantum Theory，Dover Publications，New York，1930，以及書目中列出的大多數量子力學書籍中都討論了許多例子。另請參閱 J.
Bernstein, P. M. Fishbane, and S. Gasiorowicz, loc.cit.
Figure 2-2 Diffraction of a beam of photons that pass through a slit in a screen.
圖 2-2 一束光子通過螢幕上的細縫時產生的衍射。
This implies that the momentum of the photon in the y y yy-direction is uncertain to the extent that
這意味著光子在 y y yy 方向上的動量是不確定的，其程度為
Δ p y ≥ h Δ y > h a Δ p y ≥ h Δ y > h a Deltap_(y) >= (h)/(Delta y) > (h)/(a)\Delta p_{y} \geq \frac{h}{\Delta y}>\frac{h}{a}
This, however corresponds to a spread of the beam of the order of
然而，這相當於光束的擴散量為
θ ≈ Δ p y p x ≈ h / a h / λ ≈ λ a θ ≈ Δ p y p x ≈ h / a h / λ ≈ λ a theta~~(Deltap_(y))/(p_(x))~~(h//a)/(h//lambda)~~(lambda )/(a)\theta \approx \frac{\Delta p_{y}}{p_{x}} \approx \frac{h / a}{h / \lambda} \approx \frac{\lambda}{a}
in agreement with the “wave description” of the incoming beam. 3 3 ^(3){ }^{3}
與入射光束的「波描述」一致。 3 3 ^(3){ }^{3}

Inability to Localize Bohr Orbits
無法定位玻爾軌道

Quantum mechanics does not allow us to talk about classical orbits of an electron in the Coulomb field of a proton. Suppose we want to conduct an experiment to study the location of the electron in an orbit. Let us limit ourselves to circular orbits. We will want to distinguish between an orbit of radius characterized by the quantum number n n nn and a neighboring orbit. If we use a photon beam to do this, we need a beam of wavelength such that
量子力學不允許我們談論電子在質子的庫倫場中的經典軌道。假設我們想要進行實驗來研究電子在軌道中的位置。讓我們僅限於圓形軌道。我們將要區分半徑以量子數 n n nn 為特徵的軌道和鄰近的軌道。如果我們使用光子光束來做這件事，我們需要一束波長如下的光束
λ ≪ r n + 1 − r n H原子乹道养 = ℏ m e c α [ ( n + 1 ) 2 − n 2 ] ≈ h m e c α n λ ≪ r n + 1 − r n  H原子乹道养  = ℏ m e c α ( n + 1 ) 2 − n 2 ≈ h m e c α n lambda≪(r_(n+1)-r_(n))/(" H原子乹道养 ")=(ℏ)/(m_(e)c alpha)[(n+1)^(2)-n^(2)]~~(h)/(m_(e)c alpha)n\lambda \ll \frac{r_{n+1}-r_{n}}{\text { H原子乹道养 }}=\frac{\hbar}{m_{e} c \boldsymbol{\alpha}}\left[(n+1)^{2}-n^{2}\right] \approx \frac{h}{m_{e} c \boldsymbol{\alpha}} n原子乹道养
(where we have again neglected factors of the order of 2 π 2 π 2pi2 \pi ). Such a photon, by particlewave duality, will transfer momentum of the order of
(其中我們再次忽略了 2 π 2 π 2pi2 \pi 的數量級因子）。這樣的光子，根據粒子波二重性，將傳遞以下數量級的動量
p γ ≈ h λ ≫ m e c α n p γ ≈ h λ ≫ m e c α n p_(gamma)~~(h)/( lambda)≫(m_(e)c alpha)/(n)p_{\gamma} \approx \frac{h}{\lambda} \gg \frac{m_{e} c \alpha}{n}
to the electron. This means that the energy transfer to the electron is
到電子。這意味著電子的能量轉移是
Δ E ≈ p Δ p m e ≈ p p γ m e ≫ m e ( c α ) 2 n 2 Δ E ≈ p Δ p m e ≈ p p γ m e ≫ m e ( c α ) 2 n 2 Delta E~~(p Delta p)/(m_(e))~~(pp_(gamma))/(m_(e))≫(m_(e)(c alpha)^(2))/(n^(2))\Delta E \approx \frac{p \Delta p}{m_{e}} \approx \frac{p p_{\gamma}}{m_{e}} \gg \frac{m_{e}(c \alpha)^{2}}{n^{2}}
which is large enough to kick the electron out of its “orbit.” Again, the attempt to localize the electron must be accompanied by an uncontrollable momentum transfer to it.
大到足以將電子踢出其 「軌道」。同樣地，將電子局部化的嘗試必須伴隨著不可控制的動量轉移給它。
One more comment: The heading of this section included Relations. We shall see in Chapter 5 that the uncertainty relation we used above follows formally from the
還有一個意見：本節的標題包括了關係。我們將在第五章看到，我們在上面使用的不確定性關係，正式來自於

34 Chapter 2 Wave-Particle Duality, Probability, and the Schrödinger Equation
34 第 2 章 波粒二象性、概率和薛定谔方程

Schrödinger equation, provided we use a suitable definition of the meaning of Δ x Δ x Delta x\Delta x and Δ p Δ p Delta p\Delta p. There is another relation that is not so directly derivable, but it is nevertheless applicable. This is a relation that is familiar in classical wave theory. A wave train that is limited in time, so that it lasts a time Δ t Δ t Delta t\Delta t, must have a spread in frequencies given by Δ ν ≥ 1 Δ t Δ ν ≥ 1 Δ t Delta nu >= (1)/(Delta t)\Delta \nu \geq \frac{1}{\Delta t}, which translates into
Schrödinger 方程，只要我們對 Δ x Δ x Delta x\Delta x 和 Δ p Δ p Delta p\Delta p 的意義使用適當的定義。還有一種關係不是那麼直接可以推導出來，但它仍然適用。這是經典波理論所熟悉的關係。一列波在時間上是有限的，因此它持續的時間 Δ t Δ t Delta t\Delta t ，在頻率上一定會有 Δ ν ≥ 1 Δ t Δ ν ≥ 1 Δ t Delta nu >= (1)/(Delta t)\Delta \nu \geq \frac{1}{\Delta t} 所給出的擴散，換句話說就是
Δ E Δ t > h Δ E Δ t > h Delta E Delta t > h\Delta E \Delta t>h
We shall see an example of this when we discuss the lifetime of an electron in an excited state and the associated width of the spectral line in Chapter 17.
當我們在第 17 章討論電子在激發狀態下的壽命以及相關的光譜線寬度時，我們會看到一個例子。

INTERIM SUMMARY 中期摘要

The particle nature of radiation and electrons (and other particles such as neutrons, helium nuclei, and even complex molecules) is incompatible with the observed wave properties that these particles manifest when suitable “wave-detection” experiments are performed. The suggested solution is that the particles are described by wave functions that obey a superposition principle. These must obey a linear equation (not necessarily the familiar “wave equation”), they must be complex in general, and the wave functions must be interpreted as yielding only probability statements about the behavior of individual particles. Some good guesses lead to the linear Schrödinger equation (2-23) for the wave function ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) that describes a particle in a potential V ( x ) V ( x ) V(x)V(x). We also stated the content of the Born interpretation-namely, that the wave function ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) is a probability amplitude, and its absolute square yields a probability.
輻射和電子（以及其他粒子，例如中子、氦核，甚至複雜的分子）的粒子性質與所觀察到的波的特性是不相容的，這些粒子在進行適當的「波偵測」實驗時會顯示出波的特性。建議的解決方案是，粒子是由遵守疊加原理的波函数來描述的。這些波函數必須遵從一個線性等式（不一定是我們熟悉的「波等式」），它們在一般情況下必須是複雜的，而且波函數必須被詮釋為只產生關於個別粒子行為的概率陳述。一些好的猜測引導出描述位勢 V ( x ) V ( x ) V(x)V(x) 中一個粒子的波函数 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的線性薛定谔方程 (2-23)。我們也說明了 Born 詮釋的內容 - 即波函數 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 是一個概率振幅，它的絕對平方會產生一個概率。

2-6 MORE ON THE PROBABILITY INTERPRETATION
2-6 更多概率解釋

For the interpretation given in (2-15) to hold, we must require that
為了使 (2-15) 所給出的解釋成立，我們必須要求
∫ − ∞ ∞ d x P ( x , t ) = ∫ − ∞ ∞ d x | ψ ( x , t ) | 2 = 1 ∫ − ∞ ∞   d x P ( x , t ) = ∫ − ∞ ∞   d x | ψ ( x , t ) | 2 = 1 int_(-oo)^(oo)dxP(x,t)=int_(-oo)^(oo)dx|psi(x,t)|^(2)=1\int_{-\infty}^{\infty} d x P(x, t)=\int_{-\infty}^{\infty} d x|\psi(x, t)|^{2}=1
since the particle must be somewhere in the range − ∞ < x < ∞ − ∞ < x < ∞ -oo < x < oo-\infty<x<\infty. Suppose we find a solution of the Schrödinger equation for which the integral in (2-29) is not equal to unity. As a consequence of the linearity of the Schrödinger equation, if ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) is a solution, so is A ψ ( x , t ) A ψ ( x , t ) A psi(x,t)A \psi(x, t), and with a proper choice of A A AA one can normalize the wave function ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) so it does indeed satisfy (2-29). We shall see below that all we need is that
因為粒子必須在 − ∞ < x < ∞ − ∞ < x < ∞ -oo < x < oo-\infty<x<\infty 範圍內的某處。假設我們找到一個薛定谔方程的解，其 (2-29) 中的積分不等於 unity。由於薛定谔方程的线性，如果 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 是一个解，那么 A ψ ( x , t ) A ψ ( x , t ) A psi(x,t)A \psi(x, t) 也是一个解，只要适当选择 A A AA ，就可以将波函数 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 归一化，使它确实满足 (2-29)。下面我們會看到，我們只需要
∫ − ∞ ∞ d x | ψ ( x , 0 ) | 2 < ∞ ∫ − ∞ ∞   d x | ψ ( x , 0 ) | 2 < ∞ int_(-oo)^(oo)dx|psi(x,0)|^(2) < oo\int_{-\infty}^{\infty} d x|\psi(x, 0)|^{2}<\infty
that is, the initial state wave functions must be square integrable. Since we may need to deal with integrals of the type
也就是說，初始狀態的波函数必須是可平方積分的。由於我們可能需要處理以下類型的積分
∫ − ∞ ∞ d x ψ ∗ ( x , t ) x n ψ ( x , t ) ∫ − ∞ ∞   d x ψ ∗ ( x , t ) x n ψ ( x , t ) int_(-oo)^(oo)dxpsi^(**)(x,t)x^(n)psi(x,t)\int_{-\infty}^{\infty} d x \psi^{*}(x, t) x^{n} \psi(x, t)
and 和
∫ − ∞ ∞ d x ψ ∗ ( x , t ) ( ∂ ∂ x ) n ψ ( x , t ) ∫ − ∞ ∞   d x ψ ∗ ( x , t ) ∂ ∂ x n ψ ( x , t ) int_(-oo)^(oo)dxpsi^(**)(x,t)((del)/(del x))^(n)psi(x,t)\int_{-\infty}^{\infty} d x \psi^{*}(x, t)\left(\frac{\partial}{\partial x}\right)^{n} \psi(x, t)
we will require that the wave functions ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) go to zero rapidly as x → ± ∞ x → ± ∞ x rarr+-oox \rightarrow \pm \infty, often faster than any power of x x xx. We shall also require that the wave functions ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) be continuous in x x xx.
我們會要求波函数 ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0) 以 x → ± ∞ x → ± ∞ x rarr+-oox \rightarrow \pm \infty 的速度快速歸零，通常比 x x xx 的任何幂還要快。我們也會要求波函数 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 在 x x xx 中是連續的。

The Importance of Phases 階段的重要性

The emphasis on | ψ ( x , t ) | 2 | ψ ( x , t ) | 2 |psi(x,t)|^(2)|\psi(x, t)|^{2} as the physically relevant quantity might lead to the impression that the phase of the wave function is of no importance. If we write ψ = R e i θ ψ = R e i θ psi=Re^(i theta)\psi=R e^{i \theta}, then indeed | ψ | 2 = R 2 | ψ | 2 = R 2 |psi|^(2)=R^(2)|\psi|^{2}=R^{2} independent of θ θ theta\theta. However, the linearity of the equation allows us to add solutions, as in our discussion of the electron interference pattern with two slits. We see that
強調 | ψ ( x , t ) | 2 | ψ ( x , t ) | 2 |psi(x,t)|^(2)|\psi(x, t)|^{2} 為物理相關量，可能會讓人覺得波函数的相位並不重要。如果我們寫 ψ = R e i θ ψ = R e i θ psi=Re^(i theta)\psi=R e^{i \theta} ，那麼事實上 | ψ | 2 = R 2 | ψ | 2 = R 2 |psi|^(2)=R^(2)|\psi|^{2}=R^{2} 與 θ θ theta\theta 無關。然而，等式的線性允許我們增加解，就像我們討論兩個狹縫的電子干涉圖樣一樣。我們可以看到
| R 1 e i θ 1 + R 2 e i θ 2 | 2 = R 1 2 + R 2 2 + 2 R 1 R 2 cos ⁡ ( θ 1 − θ 2 ) R 1 e i θ 1 + R 2 e i θ 2 2 = R 1 2 + R 2 2 + 2 R 1 R 2 cos ⁡ θ 1 − θ 2 |R_(1)e^(itheta_(1))+R_(2)e^(itheta_(2))|^(2)=R_(1)^(2)+R_(2)^(2)+2R_(1)R_(2)cos(theta_(1)-theta_(2))\left|R_{1} e^{i \theta_{1}}+R_{2} e^{i \theta_{2}}\right|^{2}=R_{1}^{2}+R_{2}^{2}+2 R_{1} R_{2} \cos \left(\theta_{1}-\theta_{2}\right)
depends on the relative phase. An overall phase in the total wave function can be ignored, or chosen arbitrarily for convenience.
取決於相位。總波函數中的整體相位可以忽略，或為方便起見任意選擇。

The Probability Current 概率電流

The ability to normalize the wave function is only possible if the constant A A AA mentioned above is a constant, independent of time. We now show that this is the case. We will use (2-23) and its complex conjugate, with the explicit assumption that the potential energy V ( x ) V ( x ) V(x)V(x) is real. Under these circumstances we have
只有當上面提到的常數 A A AA 是獨立於時間的常數時，才有可能將波函數正常化。我們現在要證明這一點。我們將使用 (2-23) 及其複共轭，並明確假定势能 V ( x ) V ( x ) V(x)V(x) 是實數。在這種情況下，我們有
− i ℏ ∂ ψ ∗ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 ψ ∗ ( x , t ) ∂ x 2 + V ( x ) ψ ∗ ( x , t ) − i ℏ ∂ ψ ∗ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 ψ ∗ ( x , t ) ∂ x 2 + V ( x ) ψ ∗ ( x , t ) -iℏ(delpsi^(**)(x,t))/(del t)=-(ℏ^(2))/(2m)(del^(2)psi^(**)(x,t))/(delx^(2))+V(x)psi^(**)(x,t)-i \hbar \frac{\partial \psi^{*}(x, t)}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi^{*}(x, t)}{\partial x^{2}}+V(x) \psi^{*}(x, t)
We may use this to calculate
我們可以用它來計算
∂ ∂ t P ( x , t ) = ∂ ψ ∗ ∂ t ψ + ψ ∗ ∂ ψ ∂ t = 1 i ℏ ( ℏ 2 2 m ) ( ∂ 2 ψ ∗ ∂ x 2 ψ − ψ ∗ ∂ 2 ψ ∂ x 2 ) = − ∂ ∂ x [ ℏ 2 i m ( ψ ∗ ∂ ψ ∂ x − ∂ ψ ∗ ∂ x ψ ) ] ∂ ∂ t P ( x , t ) = ∂ ψ ∗ ∂ t ψ + ψ ∗ ∂ ψ ∂ t = 1 i ℏ ℏ 2 2 m ∂ 2 ψ ∗ ∂ x 2 ψ − ψ ∗ ∂ 2 ψ ∂ x 2 = − ∂ ∂ x ℏ 2 i m ψ ∗ ∂ ψ ∂ x − ∂ ψ ∗ ∂ x ψ {:[(del)/(del t)P(x","t)=(delpsi^(**))/(del t)psi+psi^(**)(del psi)/(del t)],[=(1)/(iℏ)((ℏ^(2))/(2m))((del^(2)psi^(**))/(delx^(2))psi-psi^(**)(del^(2)psi)/(delx^(2)))],[=-(del)/(del x)[(ℏ)/(2im)(psi^(**)(del psi)/(del x)-(delpsi^(**))/(del x)psi)]]:}\begin{aligned} \frac{\partial}{\partial t} P(x, t) & =\frac{\partial \psi^{*}}{\partial t} \psi+\psi^{*} \frac{\partial \psi}{\partial t} \\ & =\frac{1}{i \hbar}\left(\frac{\hbar^{2}}{2 m}\right)\left(\frac{\partial^{2} \psi^{*}}{\partial x^{2}} \psi-\psi^{*} \frac{\partial^{2} \psi}{\partial x^{2}}\right) \\ & =-\frac{\partial}{\partial x}\left[\frac{\hbar}{2 i m}\left(\psi^{*} \frac{\partial \psi}{\partial x}-\frac{\partial \psi^{*}}{\partial x} \psi\right)\right] \end{aligned}
If we now define the f l u x f l u x fluxf l u x or, equivalently, the probability current by
如果我們現在定義 f l u x f l u x fluxf l u x ，或等同於以下列方式定義概率電流
j ( x , t ) = ℏ 2 i m ( ψ ∗ ∂ ψ ∂ x − ∂ ψ ∗ ∂ x ψ ) j ( x , t ) = ℏ 2 i m ψ ∗ ∂ ψ ∂ x − ∂ ψ ∗ ∂ x ψ j(x,t)=(ℏ)/(2im)(psi^(**)(del psi)/(del x)-(delpsi^(**))/(del x)psi)j(x, t)=\frac{\hbar}{2 i m}\left(\psi^{*} \frac{\partial \psi}{\partial x}-\frac{\partial \psi^{*}}{\partial x} \psi\right)
we get 我們得到
∂ ∂ t P ( x , t ) + ∂ ∂ x j ( x , t ) = 0 ∂ ∂ t P ( x , t ) + ∂ ∂ x j ( x , t ) = 0 (del)/(del t)P(x,t)+(del)/(del x)j(x,t)=0\frac{\partial}{\partial t} P(x, t)+\frac{\partial}{\partial x} j(x, t)=0
When this is integrated over all space, we find that
當這個值整合到所有空間時，我們發現
∂ ∂ t ∫ − ∞ ∞ d x P ( x , t ) = − ∫ − ∞ ∞ d x ∂ ∂ x j ( x , t ) = 0 ∂ ∂ t ∫ − ∞ ∞   d x P ( x , t ) = − ∫ − ∞ ∞   d x ∂ ∂ x j ( x , t ) = 0 (del)/(del t)int_(-oo)^(oo)dxP(x,t)=-int_(-oo)^(oo)dx(del)/(del x)j(x,t)=0\frac{\partial}{\partial t} \int_{-\infty}^{\infty} d x P(x, t)=-\int_{-\infty}^{\infty} d x \frac{\partial}{\partial x} j(x, t)=0
The last step follows from the fact that for square integrable functions j ( x , t ) j ( x , t ) j(x,t)j(x, t) vanishes at ± ∞ ± ∞ +-oo\pm \infty. Incidentally, had we allowed for discontinuities in ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) we would have been led to delta functions in the flux, and hence in the probability density, which is unacceptable in a physically observed quantity. [Eq. (2-34) then implies that A A AA is constant.]
最後一個步驟是因為對於平方可整函數 j ( x , t ) j ( x , t ) j(x,t)j(x, t) 在 ± ∞ ± ∞ +-oo\pm \infty 消失。順便一提，如果我們允許 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的不連續性，我們就會在通量中產生三角函數，進而在概率密度中產生三角函數，這在物理觀察量中是不可接受的。[公式 (2-34) 意味著 A A AA 是常數。］
Equation (2-33) is a conservation law analogous to the charge conservation equation in classical electrodynamics. It expresses the fact that a change in the density in a region-say, a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq b-is compensated by a net change in flux into that region
等式 (2-33) 是一個守恆定律，類似於經典電動力學的電荷守恆等式。它表達了一個事實，就是在一個區域中，例如 a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq b ，密度的改變會被流入該區域的流量淨改變所補償。
d d t ∫ a b d x P ( x , t ) = − ∫ a b d x ∂ ∂ x j ( x , t ) = j ( a , t ) − j ( b , t ) d d t ∫ a b   d x P ( x , t ) = − ∫ a b   d x ∂ ∂ x j ( x , t ) = j ( a , t ) − j ( b , t ) (d)/(dt)int_(a)^(b)dxP(x,t)=-int_(a)^(b)dx(del)/(del x)j(x,t)=j(a,t)-j(b,t)\frac{d}{d t} \int_{a}^{b} d x P(x, t)=-\int_{a}^{b} d x \frac{\partial}{\partial x} j(x, t)=j(a, t)-j(b, t)

EXAMPLE 2-3 範例 2-3

What is the probability current for a wave function of the form
對於形式為
(a) A e i k x + B − i k x A e i k x + B − i k x Ae^(ikx)+B^(-ikx)A e^{i k x}+B^{-i k x}, (b) A e − α x A e − α x Ae^(-alpha x)A e^{-\alpha x}, © R ( x ) e i S ( x ) / ℏ R ( x ) e i S ( x ) / ℏ R(x)e^(iS(x)//ℏ)R(x) e^{i S(x) / \hbar} ?
(a) A e i k x + B − i k x A e i k x + B − i k x Ae^(ikx)+B^(-ikx)A e^{i k x}+B^{-i k x} , (b) A e − α x A e − α x Ae^(-alpha x)A e^{-\alpha x} , © R ( x ) e i S ( x ) / ℏ R ( x ) e i S ( x ) / ℏ R(x)e^(iS(x)//ℏ)R(x) e^{i S(x) / \hbar} ?

SOLUTION 解決方案

(a) With ψ ( x ) = A e i k x + B e − i k x ψ ( x ) = A e i k x + B e − i k x psi(x)=Ae^(ikx)+Be^(-ikx)\psi(x)=A e^{i k x}+B e^{-i k x}, we have d ψ / d x = i k ( A i k x − B e − i k x ) d ψ / d x = i k A i k x − B e − i k x d psi//dx=ik(A^(ikx)-Be^(-ikx))d \psi / d x=i k\left(A^{i k x}-B e^{-i k x}\right) and ψ ∗ = A ∗ e − i k x + B ∗ e i k x ψ ∗ = A ∗ e − i k x + B ∗ e i k x psi^(**)=A^(**)e^(-ikx)+B^(**)e^(ikx)\psi^{*}=A^{*} e^{-i k x}+B^{*} e^{i k x}, so that
(a) 有了 ψ ( x ) = A e i k x + B e − i k x ψ ( x ) = A e i k x + B e − i k x psi(x)=Ae^(ikx)+Be^(-ikx)\psi(x)=A e^{i k x}+B e^{-i k x} ，我們就有 d ψ / d x = i k ( A i k x − B e − i k x ) d ψ / d x = i k A i k x − B e − i k x d psi//dx=ik(A^(ikx)-Be^(-ikx))d \psi / d x=i k\left(A^{i k x}-B e^{-i k x}\right) 和 ψ ∗ = A ∗ e − i k x + B ∗ e i k x ψ ∗ = A ∗ e − i k x + B ∗ e i k x psi^(**)=A^(**)e^(-ikx)+B^(**)e^(ikx)\psi^{*}=A^{*} e^{-i k x}+B^{*} e^{i k x} ，因此
ℏ 2 i m ( ψ ∗ d ψ d x − d ψ ∗ d x ψ ) = ℏ 2 i m [ ( A ∗ e − i k x + B ∗ e i k x ) i k ( A e i k x − B e − i k x ) − c . c ] = ℏ k m ( | A | 2 − | B | 2 ) = p m ( | A | 2 − | B | 2 ) ℏ 2 i m ψ ∗ d ψ d x − d ψ ∗ d x ψ = ℏ 2 i m A ∗ e − i k x + B ∗ e i k x i k A e i k x − B e − i k x − c . c = ℏ k m | A | 2 − | B | 2 = p m | A | 2 − | B | 2 {:[(ℏ)/(2im)(psi^(**)(d psi)/(dx)-(dpsi^(**))/(dx)psi)=(ℏ)/(2im)[(A^(**)e^(-ikx)+B^(**)e^(ikx))ik(Ae^(ikx)-Be^(-ikx))-c.c]],[=(ℏk)/(m)(|A|^(2)-|B|^(2))=(p)/(m)(|A|^(2)-|B|^(2))]:}\begin{aligned} \frac{\hbar}{2 i m}\left(\psi^{*} \frac{d \psi}{d x}-\frac{d \psi^{*}}{d x} \psi\right) & =\frac{\hbar}{2 i m}\left[\left(A^{*} e^{-i k x}+B^{*} e^{i k x}\right) i k\left(A e^{i k x}-B e^{-i k x}\right)-\mathrm{c} . \mathrm{c}\right] \\ & =\frac{\hbar k}{m}\left(|A|^{2}-|B|^{2}\right)=\frac{p}{m}\left(|A|^{2}-|B|^{2}\right) \end{aligned}
(b) Here ψ ( x ) ψ ( x ) psi(x)\psi(x) is real, which means that j ( x , 0 ) j ( x , 0 ) j(x,0)j(x, 0) is manifestly zero. Note that this result is true for any ψ ψ psi\psi of the form A A AA (real function of x ) where A A AA may be a complex number.
(b) 此處 ψ ( x ) ψ ( x ) psi(x)\psi(x) 為實數，這表示 j ( x , 0 ) j ( x , 0 ) j(x,0)j(x, 0) 顯然為零。請注意，這個結果對於任何形式 A A AA (x 的實數函數 ) 的 ψ ψ psi\psi 都是真實的，其中 A A AA 可以是複數。
© Here © 此處
ℏ 2 i m ( R e − i S / ℏ ( d R d x e i S / ℏ + i R 1 ℏ d S d x e i S / ℏ ) − c.c. ) = 1 m R 2 d S d x ℏ 2 i m R e − i S / ℏ d R d x e i S / ℏ + i R 1 ℏ d S d x e i S / ℏ −  c.c.  = 1 m R 2 d S d x (ℏ)/(2im)(Re^(-iS//ℏ)((dR)/(dx)e^(iS//ℏ)+iR(1)/(ℏ)(dS)/(dx)e^(iS//ℏ))-" c.c. ")=(1)/(m)R^(2)(dS)/(dx)\frac{\hbar}{2 i m}\left(R e^{-i S / \hbar}\left(\frac{d R}{d x} e^{i S / \hbar}+i R \frac{1}{\hbar} \frac{d S}{d x} e^{i S / \hbar}\right)-\text { c.c. }\right)=\frac{1}{m} R^{2} \frac{d S}{d x}
These comments are sufficient to proceed with out discussion of the consequences of the probability interpretation in quantum mechanics.
這些評論足以讓我們繼續討論量子力學中概率解釋的結果。

2-7 EXPECTATION VALUES AND THE MOMENTUM IN WAVE MECHANICS
2-7 波動力學的期望值與動量

Given the probability density P ( x , t ) P ( x , t ) P(x,t)P(x, t), we can calculate the expectation value of x , x 2 x , x 2 x,x^(2)x, x^{2}, or, for that matter, any function of x x xx. (See Supplement 2-B for a discussion of expectation values) [www.wiley.com/college/gasiorowicz]. We haye
給定概率密度 P ( x , t ) P ( x , t ) P(x,t)P(x, t) ，我們可以計算出 x , x 2 x , x 2 x,x^(2)x, x^{2} 的期望值，或者， x x xx 的任何函數。(有關期望值的討論，請參閱補充 2-B）[www.wiley.com/college/gasiorowicz]。我們
表示成這種次序
因為 f ( x ) f ( x ) f(x)f(x) 有洔和微分有関 因為 f ( x ) f ( x ) f(x)f(x) 有洔和微分有関
Given our assumptions about the behavior of ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) as x → ± ∞ x → ± ∞ x rarr+-oox \rightarrow \pm \infty, there is no problem with the convergence of the integral. Note that we inserted f ( x ) f ( x ) f(x)f(x) between ψ ∗ ( x , t ) ψ ∗ ( x , t ) psi^(**)(x,t)\psi^{*}(x, t) and ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t). Clearly the order in which we put the three functions does not matter when we integrate. We will soon find that sometimes f ( x ) f ( x ) f(x)f(x) involves derivatives with respect to x x xx, and then the order matters. Note that ⟨ f ( x ) ⟩ ⟨ f ( x ) ⟩ (:f(x):)\langle f(x)\rangle depends on time, even if it does not have an explicit time dependence, because in its definition the wave function ψ ψ psi\psi may have a nontrivial time dependence.
鑑於我們假設 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 的行為為 x → ± ∞ x → ± ∞ x rarr+-oox \rightarrow \pm \infty ，積分的收斂沒有問題。請注意，我們在 ψ ∗ ( x , t ) ψ ∗ ( x , t ) psi^(**)(x,t)\psi^{*}(x, t) 和 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 之間插入了 f ( x ) f ( x ) f(x)f(x) 。很明顯，當我們進行整合時，這三個函數的順序並不重要。我們很快就會發現，有時候 f ( x ) f ( x ) f(x)f(x) 會涉及相對於 x x xx 的導數，這時順序就很重要了。請注意， ⟨ f ( x ) ⟩ ⟨ f ( x ) ⟩ (:f(x):)\langle f(x)\rangle 取決於時間，即使它沒有明確的時間依賴性，因為在它的定義中，波函数 ψ ψ psi\psi 可能有非小數的時間依賴性。
The above expression does not tell us how to calculate the average of the momentum, or a function of the momentum, because we don’t know what to insert between ψ ∗ ψ ∗ psi^(**)\psi^{*} and ψ ψ psi\psi when calculating ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle, for example.
上面的表達式並沒有告訴我們如何計算動量的平均值，或是動量的函數，因為我們不知道例如在計算 ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle 時，應該在 ψ ∗ ψ ∗ psi^(**)\psi^{*} 和 ψ ψ psi\psi 之間插入什麼。
The Momentum in Wave Mechanics
波浪力學的動量

計算动量的期望值 計算動量的期望值

We approach the calculation of ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle by noting that classically
我們在計算 ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle 時注意到，經典地
p = m v = m d x d t p = m v = m d x d t p=mv=m(dx)/(dt)p=m v=m \frac{d x}{d t}
γ 2 t ∗ γ t = − 1 i ℏ ( − ℏ 2 2 m γ 2 ψ ∗ ∂ x 2 + V v ∗ ) x ∂ 4 ∂ t = 1 i ℏ ( − ℏ 2 2 m γ 2 ψ γ x 2 + V 2 ) γ 2 t ∗ γ t = − 1 i ℏ − ℏ 2 2 m γ 2 ψ ∗ ∂ x 2 + V v ∗ x ∂ 4 ∂ t = 1 i ℏ − ℏ 2 2 m γ 2 ψ γ x 2 + V 2 {:[(gamma^(2)t^(**))/(gamma t)=-(1)/(iℏ)(-(ℏ^(2))/(2m)(gamma^(2)psi^(**))/(delx^(2))+Vv^(**))^(x)],[(del^(4))/(del t)=(1)/(iℏ)(-(ℏ^(2))/(2m)(gamma^(2)psi)/(gammax^(2))+V2)]:}\begin{aligned} & \frac{\gamma^{2} t^{*}}{\gamma t}=-\frac{1}{i \hbar}\left(-\frac{\hbar^{2}}{2 m} \frac{\gamma^{2} \psi^{*}}{\partial x^{2}}+V v^{*}\right)^{x} \\ & \frac{\partial^{4}}{\partial t}=\frac{1}{i \hbar}\left(-\frac{\hbar^{2}}{2 m} \frac{\gamma^{2} \psi}{\gamma x^{2}}+V 2\right) \end{aligned}
so we try 所以我們嘗試
⟨ p ⟩ = m d d t ⟨ x ⟩ = m d d t ∫ − ∞ ∞ d x ψ ∗ ( x , t ) x ψ ( x , t ) = m ∫ − ∞ ∞ d x ( ∂ ψ ∗ ( x , t ) ∂ t x ψ ( x , t ) + ψ ∗ ( x , t ) x ∂ ψ ( x , t ) ∂ t ) ⟨ p ⟩ = m d d t ⟨ x ⟩ = m d d t ∫ − ∞ ∞   d x ψ ∗ ( x , t ) x ψ ( x , t ) = m ∫ − ∞ ∞   d x ∂ ψ ∗ ( x , t ) ∂ t x ψ ( x , t ) + ψ ∗ ( x , t ) x ∂ ψ ( x , t ) ∂ t {:[(:p:)=m(d)/(dt)(:x:)=m(d)/(dt)int_(-oo)^(oo)dxpsi^(**)(x","t)x psi(x","t)],[=mint_(-oo)^(oo)dx((delpsi^(**)(x,t))/(del t)x psi(x,t)+psi^(**)(x,t)x(del psi(x,t))/(del t))]:}\begin{aligned} \langle p\rangle & =m \frac{d}{d t}\langle x\rangle=m \frac{d}{d t} \int_{-\infty}^{\infty} d x \psi^{*}(x, t) x \psi(x, t) \\ & =m \int_{-\infty}^{\infty} d x\left(\frac{\partial \psi^{*}(x, t)}{\partial t} x \psi(x, t)+\psi^{*}(x, t) x \frac{\partial \psi(x, t)}{\partial t}\right) \end{aligned}
Note that there is no d x / d t d x / d t dx//dtd x / d t under the integral sign. The only quantity that varies with time is ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t), and it is this variation that gives rise to a change in ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle with time. We again use the Schrödinger equation and its complex conjugate to evaluate the above. We end up with
請注意，在積分符號下沒有 d x / d t d x / d t dx//dtd x / d t 。唯一會隨時間變化的量是 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) ，而正是這個變化引起了 ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle 隨時間的變化。我們再次使用薛定谔方程和它的複共轭来评估上面的内容。最後我們得到
⟨ p ⟩ = ℏ 2 i ∫ − ∞ ∞ d x ( ∂ 2 ψ ∗ ∂ x 2 x ψ − ψ ∗ x ∂ 2 ψ ∂ x 2 ) ⟨ p ⟩ = ℏ 2 i ∫ − ∞ ∞   d x ∂ 2 ψ ∗ ∂ x 2 x ψ − ψ ∗ x ∂ 2 ψ ∂ x 2 (:p:)=(ℏ)/(2i)int_(-oo)^(oo)dx((del^(2)psi^(**))/(delx^(2))x psi-psi^(**)x(del^(2)psi)/(delx^(2)))\langle p\rangle=\frac{\hbar}{2 i} \int_{-\infty}^{\infty} d x\left(\frac{\partial^{2} \psi^{*}}{\partial x^{2}} x \psi-\psi^{*} x \frac{\partial^{2} \psi}{\partial x^{2}}\right)
Now 現在
∂ 2 ψ ∗ ∂ x 2 x ψ = ∂ ∂ x [ ∂ ψ ∗ ∂ x x ψ ] − ∂ ψ ∗ ∂ x ψ − ∂ ψ ∗ ∂ x x ∂ ψ ∂ x = ∂ ∂ x [ ∂ ψ ∗ ∂ x x ψ ] − ∂ ∂ x ( ψ ∗ ψ ) + ψ ∗ ∂ ψ ∂ x − ∂ ∂ x [ ψ ∗ x ∂ ψ ∂ x ] + ψ ∗ ∂ ψ ∂ x + ψ ∗ x ∂ 2 ψ ∂ x 2 ∂ 2 ψ ∗ ∂ x 2 x ψ = ∂ ∂ x ∂ ψ ∗ ∂ x x ψ − ∂ ψ ∗ ∂ x ψ − ∂ ψ ∗ ∂ x x ∂ ψ ∂ x = ∂ ∂ x ∂ ψ ∗ ∂ x x ψ − ∂ ∂ x ψ ∗ ψ + ψ ∗ ∂ ψ ∂ x − ∂ ∂ x ψ ∗ x ∂ ψ ∂ x + ψ ∗ ∂ ψ ∂ x + ψ ∗ x ∂ 2 ψ ∂ x 2 {:[(del^(2)psi^(**))/(delx^(2))x psi=(del)/(del x)[(delpsi^(**))/(del x)x psi]-(delpsi^(**))/(del x)psi-(delpsi^(**))/(del x)x(del psi)/(del x)],[=(del)/(del x)[(delpsi^(**))/(del x)x psi]-(del)/(del x)(psi^(**)psi)+psi^(**)(del psi)/(del x)-(del)/(del x)[psi^(**)x(del psi)/(del x)]+psi^(**)(del psi)/(del x)+psi^(**)x(del^(2)psi)/(delx^(2))]:}\begin{aligned} \frac{\partial^{2} \psi^{*}}{\partial x^{2}} x \psi & =\frac{\partial}{\partial x}\left[\frac{\partial \psi^{*}}{\partial x} x \psi\right]-\frac{\partial \psi^{*}}{\partial x} \psi-\frac{\partial \psi^{*}}{\partial x} x \frac{\partial \psi}{\partial x} \\ & =\frac{\partial}{\partial x}\left[\frac{\partial \psi^{*}}{\partial x} x \psi\right]-\frac{\partial}{\partial x}\left(\psi^{*} \psi\right)+\psi^{*} \frac{\partial \psi}{\partial x}-\frac{\partial}{\partial x}\left[\psi^{*} x \frac{\partial \psi}{\partial x}\right]+\psi^{*} \frac{\partial \psi}{\partial x}+\psi^{*} x \frac{\partial^{2} \psi}{\partial x^{2}} \end{aligned}
This means that the integrand in (2-38) has the form
這表示在 (2-38) 中的積分有以下形式
∂ ∂ x ( ∂ ψ ∗ ∂ x x ψ − ψ ∗ x ∂ ψ ∂ x − ψ ∗ ψ + 2 ψ ∗ ∂ ψ ∂ x ∂ ∂ x ∂ ψ ∗ ∂ x x ψ − ψ ∗ x ∂ ψ ∂ x − ψ ∗ ψ + 2 ψ ∗ ∂ ψ ∂ x (del)/(del x)((delpsi^(**))/(del x)x psi-psi^(**)x(del psi)/(del x)-psi^(**)psi+2psi^(**)(del psi)/(del x):}\frac{\partial}{\partial x}\left(\frac{\partial \psi^{*}}{\partial x} x \psi-\psi^{*} x \frac{\partial \psi}{\partial x}-\psi^{*} \psi+2 \psi^{*} \frac{\partial \psi}{\partial x}\right.
Because the wave functions vanish at infinity, the first term does not contribute, and the integral gives
由於波函数在无穷远处消失，因此第一项没有贡献，积分得到
⟨ p ⟩ = ∫ − ∞ ∞ d x ψ ∗ ( x , t ) ( ℏ i ∂ ∂ x ) ψ ( x , t ) ⟨ p ⟩ = ∫ − ∞ ∞   d x ψ ∗ ( x , t ) ℏ i ∂ ∂ x ψ ( x , t ) (:p:)=int_(-oo)^(oo)dxpsi^(**)(x,t)((ℏ)/(i)(del)/(del x))psi(x,t)\langle p\rangle=\int_{-\infty}^{\infty} d x \psi^{*}(x, t)\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right) \psi(x, t)
This suggests that the momentum be represented by the differential operator
這意味著動量可以用微分算子來表示
p o p = ℏ i ∂ ∂ x p o p = ℏ i ∂ ∂ x p_(op)=(ℏ)/(i)(del)/(del x)p_{o p}=\frac{\hbar}{i} \frac{\partial}{\partial x}
铁氜督各 鐵氜督各
Once we accept this, we can easily calculate
一旦我們接受這一點，我們就可以很容易地計算出
⟨ p 2 ⟩ = ∫ − ∞ ∞ d x ψ ∗ ( x , t ) ( ℏ i ∂ ∂ x ) 2 ψ ( x , t ) = − ℏ 2 ∫ − ∞ ∞ d x ψ ∗ ( x , t ) ∂ 2 ψ ( x , t ) ∂ x 2 p 2 = ∫ − ∞ ∞   d x ψ ∗ ( x , t ) ℏ i ∂ ∂ x 2 ψ ( x , t ) = − ℏ 2 ∫ − ∞ ∞   d x ψ ∗ ( x , t ) ∂ 2 ψ ( x , t ) ∂ x 2 (:p^(2):)=int_(-oo)^(oo)dxpsi^(**)(x,t)((ℏ)/(i)(del)/(del x))^(2)psi(x,t)=-ℏ^(2)int_(-oo)^(oo)dxpsi^(**)(x,t)(del^(2)psi(x,t))/(delx^(2))\left\langle p^{2}\right\rangle=\int_{-\infty}^{\infty} d x \psi^{*}(x, t)\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right)^{2} \psi(x, t)=-\hbar^{2} \int_{-\infty}^{\infty} d x \psi^{*}(x, t) \frac{\partial^{2} \psi(x, t)}{\partial x^{2}}
and more generally 以及更廣泛地
⟨ f ( p ) ⟩ = ∫ − ∞ ∞ d x ψ ∗ ( x , t ) f ( ℏ i ∂ ∂ x ) ψ ( x , t ) ⟨ f ( p ) ⟩ = ∫ − ∞ ∞   d x ψ ∗ ( x , t ) f ℏ i ∂ ∂ x ψ ( x , t ) (:f(p):)=int_(-oo)^(oo)dxpsi^(**)(x,t)f((ℏ)/(i)(del)/(del x))psi(x,t)\langle f(p)\rangle=\int_{-\infty}^{\infty} d x \psi^{*}(x, t) f\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right) \psi(x, t)
The form of the momentum operator in (2-40) raises the question whether the expectation value of the momentum in some states could be imaginary. We can, in fact, show that the expectation value of p p pp is always real. We write
動量算子在 (2-40) 中的形式提出了一個問題：在某些狀態下，動量的期望值是否可能是虛的。事實上，我們可以證明 p p pp 的期望值永遠是實數。我們寫成
⟨ p ⟩ − ⟨ p ⟩ ∗ = ∫ − ∞ ∞ d x [ ψ ∗ ( ℏ i ) ∂ ψ ∂ x − ψ ( − ℏ i ) ∂ ψ ∗ ∂ x ] = ℏ i ∫ − ∞ ∞ d x [ ψ ∗ ∂ ψ ∂ x + ψ ∂ ψ ∗ ∂ x ] = ℏ i ∫ − ∞ ∞ d x ∂ ∂ x ( ψ ∗ ψ ) = 0 ⟨ p ⟩ − ⟨ p ⟩ ∗ = ∫ − ∞ ∞   d x ψ ∗ ℏ i ∂ ψ ∂ x − ψ − ℏ i ∂ ψ ∗ ∂ x = ℏ i ∫ − ∞ ∞   d x ψ ∗ ∂ ψ ∂ x + ψ ∂ ψ ∗ ∂ x = ℏ i ∫ − ∞ ∞   d x ∂ ∂ x ψ ∗ ψ = 0 {:[(:p:)-(:p:)^(**)=int_(-oo)^(oo)dx[psi^(**)((ℏ)/(i))(del psi)/(del x)-psi(-(ℏ)/(i))(delpsi^(**))/(del x)]],[=(ℏ)/(i)int_(-oo)^(oo)dx[psi^(**)(del psi)/(del x)+psi(delpsi^(**))/(del x)]=(ℏ)/(i)int_(-oo)^(oo)dx(del)/(del x)(psi^(**)psi)=0]:}\begin{aligned} \langle p\rangle-\langle p\rangle^{*} & =\int_{-\infty}^{\infty} d x\left[\psi^{*}\left(\frac{\hbar}{i}\right) \frac{\partial \psi}{\partial x}-\psi\left(-\frac{\hbar}{i}\right) \frac{\partial \psi^{*}}{\partial x}\right] \\ & =\frac{\hbar}{i} \int_{-\infty}^{\infty} d x\left[\psi^{*} \frac{\partial \psi}{\partial x}+\psi \frac{\partial \psi^{*}}{\partial x}\right]=\frac{\hbar}{i} \int_{-\infty}^{\infty} d x \frac{\partial}{\partial x}\left(\psi^{*} \psi\right)=0 \end{aligned}
The last step follows from the square integrability of the wave functions. Sometimes one has occasion to use functions that are not square integrable but that obey periodic boundary conditions, such as
最後一個步驟來自於波函数的平方可整性。有時候，我們需要使用不具備平方可整性，但符合週期性邊界條件的函數，例如
ψ ( x ) = ψ ( x + L ) ψ ( x ) = ψ ( x + L ) psi(x)=psi(x+L)\psi(x)=\psi(x+L)
Also, under these circumstances
此外，在這些情況下
⟨ p ⟩ − ⟨ p ⟩ ∗ = ℏ i ∫ 0 L d x ∂ ∂ x ( ψ ∗ ψ ) = ℏ i ( | ψ ( L ) | 2 − | ψ ( 0 ) | 2 ) = 0 ⟨ p ⟩ − ⟨ p ⟩ ∗ = ℏ i ∫ 0 L   d x ∂ ∂ x ψ ∗ ψ = ℏ i | ψ ( L ) | 2 − | ψ ( 0 ) | 2 = 0 (:p:)-(:p:)^(**)=(ℏ)/(i)int_(0)^(L)dx(del)/(del x)(psi^(**)psi)=(ℏ)/(i)(|psi(L)|^(2)-|psi(0)|^(2))=0\langle p\rangle-\langle p\rangle^{*}=\frac{\hbar}{i} \int_{0}^{L} d x \frac{\partial}{\partial x}\left(\psi^{*} \psi\right)=\frac{\hbar}{i}\left(|\psi(L)|^{2}-|\psi(0)|^{2}\right)=0
An operator whose expectation value for all admissible wave functions is real is called a hermitian operator. We see that the momentum operator is hermitian.
對於所有可接受的波函数，其期望值都是實數的算子稱為荷米子算子。我們可以看到動量算子是全等的。
The form of the momentum operator allows us to write the Schrödinger equation in the form
動量算子的形式允許我們以下列形式寫出薛定谔方程
the Schrödinger equation in − ℏ 2 m γ x , t ) 2 γ x 2 + V ( x ) ψ ( x , t ) = 1 2 m ( h λ γ γ x ) 2 ψ ( x , t ) + V ( x ) ψ ( x , t = p 2 2 m + V ( x ) ψ ( x , t ) = H ψ ( x , t ) transcription of the energy  the Schrödinger equation in  − ℏ 2 m γ x , t ) 2 γ x 2 + V ( x ) ψ ( x , t ) = 1 2 m h λ γ γ x 2 ψ ( x , t ) + V ( x ) ψ ( x , t = p 2 2 m + V ( x ) ψ ( x , t ) = H ψ ( x , t )  transcription of the energy  {:[" the Schrödinger equation in "],[-(ℏ)/(2m)(gamma x,t)^(2))/(gammax^(2))+V(x)psi(x","t)],[=(1)/(2m)((h)/( lambda)(gamma)/(gamma x))^(2)psi(x","t)+V(x)psi(x","t],[=(p^(2))/(2m)+V(x)psi(x","t)=H psi(x","t)],[" transcription of the energy "]:}\begin{aligned} & \text { the Schrödinger equation in } \\ & -\frac{\hbar}{2 m} \frac{\gamma x, t)^{2}}{\gamma x^{2}}+V(x) \psi(x, t) \\ & =\frac{1}{2 m}\left(\frac{h}{\lambda} \frac{\gamma}{\gamma x}\right)^{2} \psi(x, t)+V(x) \psi(x, t \\ & =\frac{p^{2}}{2 m}+V(x) \psi(x, t)=H \psi(x, t) \\ & \text { transcription of the energy } \end{aligned}ö
代入 schroidinger > i ℏ ∂ ψ ( x , t ) ∂ t = H ψ ( x , t ) = 1 2 m ( h i γ γ x ) 2 ψ ( x , t ) + V ( x ) ψ ( x , t )  代入 schroidinger  > i ℏ ∂ ψ ( x , t ) ∂ t = H ψ ( x , t ) = 1 2 m h i γ γ x 2 ψ ( x , t ) + V ( x ) ψ ( x , t ) " 代入 schroidinger " > iℏ(del psi(x,t))/(del t)=H psi(x,t)=(1)/(2m)((h)/(i)(gamma)/(gamma x))^(2)psi(x,t)+V(x)psi(x,t)\text { 代入 schroidinger }>i \hbar \frac{\partial \psi(x, t)}{\partial t}=H \psi(x, t)=\frac{1}{2 m}\left(\frac{h}{i} \frac{\gamma}{\gamma x}\right)^{2} \psi(x, t)+V(x) \psi(x, t)代入
where the operator H H HH, called the Hamiltonian operator, is a transcription of the energy
其中算子 H H HH ，稱為漢密爾頓算子，是能量的轉換
H = p o p 2 2 m + V ( x ) H = p o p 2 2 m + V ( x ) H=(p_(op)^(2))/(2m)+V(x)H=\frac{p_{o p}^{2}}{2 m}+V(x)
into operator form, using the operator form of p p pp given in (2-38). The Hamiltonian operator is particularly significant in quantum mechanics, and we shall analyze it in detail in many examples.
轉換成算子形式，使用 p p pp 的算子形式在 (2-38) 中給出。哈密爾頓算子在量子力學中特別重要，我們將在許多範例中詳細分析它。

动昌場
Wave Function in Momentum Space
動昌場 動量空間中的波函数

Now that we have obtaned a way of representing the momentum, we can discuss the physical meaning of ϕ ( p ) ϕ ( p ) phi(p)\phi(p), which was found to have the form
既然我們已經得到表示動量的方法，我們就可以討論 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 的物理含義，它的形式是

动量星波出现的概率 動量星波出現的概率

ϕ ( p ) = 1 2 π ℏ ∫ − ∞ d x ψ ( x , 0 ) e − i p x / ℏ ϕ ( p ) = 1 2 π ℏ ∫ − ∞   d x ψ ( x , 0 ) e − i p x / ℏ phi(p)=(1)/(sqrt(2piℏ))int_(-oo)dx psi(x,0)e^(-ipx//ℏ)\phi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty} d x \psi(x, 0) e^{-i p x / \hbar}
We can calculate 我們可以計算出
∫ − ∞ ∞ d p ϕ ∗ ( p ) ϕ ( p ) = ∫ − ∞ ∞ d p ϕ ∗ ( p ) 1 2 π ℏ ∫ − ∞ ∞ d x ψ ( x ) e − i p x / h = 1 2 π ℏ ∫ − ∞ ∞ d x ψ ( x ) ∫ − ∞ ∞ d p ϕ ∗ ( p ) e − i p x / h = ∫ − ∞ ∞ d x ψ ( x ) ψ ∗ ( x ) = 1 点点? ? ∫ − ∞ ∞   d p ϕ ∗ ( p ) ϕ ( p ) = ∫ − ∞ ∞   d p ϕ ∗ ( p ) 1 2 π ℏ ∫ − ∞ ∞   d x ψ ( x ) e − i p x / h = 1 2 π ℏ ∫ − ∞ ∞   d x ψ ( x ) ∫ − ∞ ∞   d p ϕ ∗ ( p ) e − i p x / h = ∫ − ∞ ∞   d x ψ ( x ) ψ ∗ ( x ) = 1  点点? ?  {:[int_(-oo)^(oo)dpphi^(**)(p)phi(p)=int_(-oo)^(oo)dpphi^(**)(p)(1)/(sqrt(2piℏ))int_(-oo)^(oo)dx psi(x)e^(-ipx//h)],[=(1)/(sqrt(2piℏ))int_(-oo)^(oo)dx psi(x)int_(-oo)^(oo)dpphi^(**)(p)e^(-ipx//h)],[=int_(-oo)^(oo)dx psi(x)psi^(**)(x)=1" 点点? ? "]:}\begin{aligned} \int_{-\infty}^{\infty} d p \phi^{*}(p) \phi(p) & =\int_{-\infty}^{\infty} d p \phi^{*}(p) \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x \psi(x) e^{-i p x / h} \\ & =\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x \psi(x) \int_{-\infty}^{\infty} d p \phi^{*}(p) e^{-i p x / h} \\ & =\int_{-\infty}^{\infty} d x \psi(x) \psi^{*}(x)=1 \text { 点点? ? } \end{aligned}点点
This result is known as Parseval’s theorem in the mathematicaltiterature. It states that if a
這個結果在數學文獻中被稱為 Parseval 定理。它說明如果一個

Next consider 下一步考慮

x空間 x 空間

⟨ p ⟩ = ∫ − ∞ ∞ d x ψ ∗ ( x ) ℏ i ∂ ψ ( x ) ∂ x = ∫ − ∞ ∞ d x ψ ∗ ( x ) ℏ i d d x 1 2 π ℏ ∫ − ∞ ∞ d p ϕ ( p ) e i p x / h = ∫ − ∞ ∞ d p ϕ ( p ) p 1 2 π ℏ ∫ − ∞ ∞ d x ψ ∗ ( x ) e i p x / ℏ = ∫ − ∞ ∞ d p ϕ ∗ ( p ) p ϕ ( p ) ⟨ p ⟩ = ∫ − ∞ ∞   d x ψ ∗ ( x ) ℏ i ∂ ψ ( x ) ∂ x = ∫ − ∞ ∞   d x ψ ∗ ( x ) ℏ i d d x 1 2 π ℏ ∫ − ∞ ∞   d p ϕ ( p ) e i p x / h = ∫ − ∞ ∞   d p ϕ ( p ) p 1 2 π ℏ ∫ − ∞ ∞   d x ψ ∗ ( x ) e i p x / ℏ = ∫ − ∞ ∞   d p ϕ ∗ ( p ) p ϕ ( p ) {:[(:p:)=int_(-oo)^(oo)dxpsi^(**)(x)(ℏ)/(i)(del psi(x))/(del x)=int_(-oo)^(oo)dxpsi^(**)(x)(ℏ)/(i)(d)/(dx)(1)/(sqrt(2piℏ))int_(-oo)^(oo)dp phi(p)e^(ipx//h)],[=int_(-oo)^(oo)dp phi(p)p(1)/(sqrt(2piℏ))int_(-oo)^(oo)dxpsi^(**)(x)e^(ipx//ℏ)],[=int_(-oo)^(oo)dpphi^(**)(p)p phi(p)]:}\begin{aligned} \langle p\rangle & =\int_{-\infty}^{\infty} d x \psi^{*}(x) \frac{\hbar}{i} \frac{\partial \psi(x)}{\partial x}=\int_{-\infty}^{\infty} d x \psi^{*}(x) \frac{\hbar}{i} \frac{d}{d x} \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d p \phi(p) e^{i p x / h} \\ & =\int_{-\infty}^{\infty} d p \phi(p) p \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty} d x \psi^{*}(x) e^{i p x / \hbar} \\ & =\int_{-\infty}^{\infty} d p \phi^{*}(p) p \phi(p) \end{aligned}
This result, together with (2-47), strongly suggests that ϕ ( p ) ϕ ( p ) phi(p)\phi(p) should be interpreted as the wave function in momentum space, with | ϕ ( p ) | 2 | ϕ ( p ) | 2 |phi(p)|^(2)|\phi(p)|^{2} yielding the probability density for finding the particle with momentum p p pp.
這個結果加上 (2-47)，強烈建議 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 應該被解釋為動量空間中的波函数，而 | ϕ ( p ) | 2 | ϕ ( p ) | 2 |phi(p)|^(2)|\phi(p)|^{2} 則產生找到動量為 p p pp 的粒子的概率密度。
Lest the reader think that in spite of this symmetry between x x xx - and p p pp-space, p = p = p=p= − i ℏ ( ∂ / ∂ x ) − i ℏ ( ∂ / ∂ x ) -iℏ(del//del x)-i \hbar(\partial / \partial x) is an operator and x x xx is not, we note that x x xx is, in fact, an operator too. It happens to have a simple (multiplicative) form in x x xx-space, but if we want to calculate ⟨ f ( x ) ⟩ ⟨ f ( x ) ⟩ (:f(x):)\langle f(x)\rangle in momentum space, then we can show by methods very similar to the ones used
為了避免讀者以為儘管 x x xx - 和 p p pp 空間之間存在對稱，但 p = p = p=p= − i ℏ ( ∂ / ∂ x ) − i ℏ ( ∂ / ∂ x ) -iℏ(del//del x)-i \hbar(\partial / \partial x) 是一個算子，而 x x xx 不是，我們注意到 x x xx 事實上也是一個算子。它在 x x xx 空間中碰巧有一個簡單 (乘法) 的形式，但是如果我們想要計算動量空間中的 ⟨ f ( x ) ⟩ ⟨ f ( x ) ⟩ (:f(x):)\langle f(x)\rangle ，那麼我們可以用非常類似的方法來證明

p → p + 2 ⧸ p → p + 2 ⧸ p rarr p+2""⧸""p \rightarrow p+2 \not 之周的机率  p → p + 2 ⧸ p → p + 2 ⧸ p rarr p+2""⧸""p \rightarrow p+2 \not 之周的機率

⇒ | ϕ ( p ) | 2 = ⇒ | ϕ ( p ) | 2 = =>|phi(p)|^(2)=\Rightarrow|\phi(p)|^{2}= protability density in moneritum apace
⇒ | ϕ ( p ) | 2 = ⇒ | ϕ ( p ) | 2 = =>|phi(p)|^(2)=\Rightarrow|\phi(p)|^{2}= 在moneritum apace中的可保護性密度

⇒ ϕ ( p ) = ⇒ ϕ ( p ) = =>phi(p)=\Rightarrow \phi(p)= wave function and more generally
⇒ ϕ ( p ) = ⇒ ϕ ( p ) = =>phi(p)=\Rightarrow \phi(p)= 波函数和更一般的
in momentum space 在動量空間
⟨ x ⟩ = ∫ − ∞ ∞ d p ϕ ∗ ( p ) ( i ℏ ∂ ∂ p ) ϕ ( p ) ⟨ x ⟩ = ∫ − ∞ ∞   d p ϕ ∗ ( p ) i ℏ ∂ ∂ p ϕ ( p ) (:x:)=int_(-oo)^(oo)dpphi^(**)(p)(iℏ(del)/(del p))phi(p)\langle x\rangle=\int_{-\infty}^{\infty} d p \phi^{*}(p)\left(i \hbar \frac{\partial}{\partial p}\right) \phi(p)
波迫伐用 : p p quad p\quad p 篓 ⇒ ϕ ( p ) ⇒ ϕ ( p ) =>phi(p)\Rightarrow \phi(p)
波迫伐用 : p p quad p\quad p 篓 ⇒ ϕ ( p ) ⇒ ϕ ( p ) =>phi(p)\Rightarrow \phi(p)
⟨ f ( x ) ⟩ = ∫ − ∞ ∞ d p ϕ ∗ ( p ) f ( i ℏ ∂ ∂ p ) ϕ ( p ) ⟨ f ( x ) ⟩ = ∫ − ∞ ∞   d p ϕ ∗ ( p ) f i ℏ ∂ ∂ p ϕ ( p ) (:f(x):)=int_(-oo)^(oo)dpphi^(**)(p)f(iℏ(del)/(del p))phi(p)\langle f(x)\rangle=\int_{-\infty}^{\infty} d p \phi^{*}(p) f\left(i \hbar \frac{\partial}{\partial p}\right) \phi(p)
In other words, the operator x x xx has the representation
換句話說，運算符 x x xx 的表示為
x = i ℏ ∂ ∂ p x = i ℏ ∂ ∂ p x=iℏ(del)/(del p)x=i \hbar \frac{\partial}{\partial p}

in momentum space. 在動量空間中。
40 Chapter 2 Wave-Particle Duality, Probability, and the Schrödinger Equation
40 第 2 章 波粒二象性、概率和薛定谔方程
The following example illustrates some computations for a specific wave function ψ ( x ) ψ ( x ) psi(x)\psi(x).
以下範例說明特定波函数 ψ ( x ) ψ ( x ) psi(x)\psi(x) 的一些計算。

EXAMPLE 2-4 範例 2-4

Consider a particle whose normalized wave function is
考慮一個粒子，其歸一化波函数為
ψ ( x ) = 2 α α x e − α x x > 0 = 0 x < 0 ψ ( x ) = 2 α α x e − α x x > 0 = 0 x < 0 {:[psi(x)=2alphasqrtalphaxe^(-alpha x)x > 0],[=0x < 0]:}\begin{aligned} \psi(x) & =2 \alpha \sqrt{\alpha} x e^{-\alpha x} & & x>0 \\ & =0 & & x<0 \end{aligned}
(a) For what value of x x xx does P ( x ) = | ψ ( x ) | 2 P ( x ) = | ψ ( x ) | 2 P(x)=|psi(x)|^(2)P(x)=|\psi(x)|^{2} peak?
(a) x x xx 的峰值是 P ( x ) = | ψ ( x ) | 2 P ( x ) = | ψ ( x ) | 2 P(x)=|psi(x)|^(2)P(x)=|\psi(x)|^{2} 的什麼值？
(b) Calculate ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle and ⟨ x 2 ⟩ x 2 (:x^(2):)\left\langle x^{2}\right\rangle.
(b) 計算 ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle 和 ⟨ x 2 ⟩ x 2 (:x^(2):)\left\langle x^{2}\right\rangle 。
© What is the probability that the particle is found between x = 0 x = 0 x=0x=0 and x = 1 / α x = 1 / α x=1//alphax=1 / \alpha ?
© 在 x = 0 x = 0 x=0x=0 和 x = 1 / α x = 1 / α x=1//alphax=1 / \alpha 之間發現粒子的概率是多少？
(d) Calculate ϕ ( p ) ϕ ( p ) phi(p)\phi(p) and use this to calculate ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle and ⟨ p 2 ⟩ p 2 (:p^(2):)\left\langle p^{2}\right\rangle.
(d) 計算 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) ，並以此計算 ⟨ p ⟩ ⟨ p ⟩ (:p:)\langle p\rangle 和 ⟨ p 2 ⟩ p 2 (:p^(2):)\left\langle p^{2}\right\rangle 。

SOLUTION 解決方案

(a) The peak in P ( x ) P ( x ) P(x)P(x) occurs when d P ( x ) / d x = 0 d P ( x ) / d x = 0 dP(x)//dx=0d P(x) / d x=0-that is, when
(a) P ( x ) P ( x ) P(x)P(x) 的峰值出現在 d P ( x ) / d x = 0 d P ( x ) / d x = 0 dP(x)//dx=0d P(x) / d x=0 時，也就是當
which is at x = 1 / α x = 1 / α x=1//alphax=1 / \alpha. 位於 x = 1 / α x = 1 / α x=1//alphax=1 / \alpha 。
⟨ x ⟩ ∫ − ∞ d d x ( x 2 a − 2 α x 2 d x × ( 2 x ( 1 − α x ) e − 2 α x = 0 a ) ν ⟨ x ⟩ ∫ − ∞ d d x x 2 a − 2 α x 2   d x × 2 x ( 1 − α x ) e − 2 α x = 0 a ν (:x:)int_(-oo)^((d)/(dx)(x^(2)(a^(-2alpha x))/(2):})dx xx((2x(1-alpha x)e^(-2alpha x)=0)/(sqrta))^(nu)\langle x\rangle \int_{-\infty}^{\frac{d}{d x}\left(x^{2} \frac{a^{-2 \alpha x}}{2}\right.} d x \times\left(\frac{2 x(1-\alpha x) e^{-2 \alpha x}=0}{\sqrt{a}}\right)^{\nu}
八 = π p ⟨ x ⟩ = ∫ 0 ∞ d x x ( 4 α 3 x 2 e − 2 α x ) = 1 4 α ∫ 0 ∞ d y y 3 e − y = 3 ! 4 α = 3 2 α ⟨ x ⟩ 2 = ∫ 0 ∞ d x x 2 ( 4 α 3 x 2 e − 2 α x ) = 4 ! 8 α 2 = 3 α 2  八  = π p ⟨ x ⟩ = ∫ 0 ∞   d x x 4 α 3 x 2 e − 2 α x = 1 4 α ∫ 0 ∞   d y y 3 e − y = 3 ! 4 α = 3 2 α ⟨ x ⟩ 2 = ∫ 0 ∞   d x x 2 4 α 3 x 2 e − 2 α x = 4 ! 8 α 2 = 3 α 2 {:[" 八 "=(pi )/(p)],[(:x:)=int_(0)^(oo)dxx(4alpha^(3)x^(2)e^(-2alpha x))=(1)/(4alpha)int_(0)^(oo)dyy^(3)e^(-y)=(3!)/(4alpha)=(3)/(2alpha)],[(:x:)^(2)=int_(0)^(oo)dxx^(2)(4alpha^(3)x^(2)e^(-2alpha x))=(4!)/(8alpha^(2))=(3)/(alpha^(2))]:}\begin{aligned} & \text { 八 }=\frac{\pi}{p} \\ & \langle x\rangle=\int_{0}^{\infty} d x x\left(4 \alpha^{3} x^{2} e^{-2 \alpha x}\right)=\frac{1}{4 \alpha} \int_{0}^{\infty} d y y^{3} e^{-y}=\frac{3!}{4 \alpha}=\frac{3}{2 \alpha} \\ & \langle x\rangle^{2}=\int_{0}^{\infty} d x x^{2}\left(4 \alpha^{3} x^{2} e^{-2 \alpha x}\right)=\frac{4!}{8 \alpha^{2}}=\frac{3}{\alpha^{2}} \end{aligned}八
© The desired probability is
© 所需的概率為
P = ∫ 0 1 / α d x ( 4 α 3 ) x 2 e − 2 α x = 1 2 ∫ 0 2 d y y 2 e − y = 0.32 P = ∫ 0 1 / α   d x 4 α 3 x 2 e − 2 α x = 1 2 ∫ 0 2   d y y 2 e − y = 0.32 P=int_(0)^(1//alpha)dx(4alpha^(3))x^(2)e^(-2alpha x)=(1)/(2)int_(0)^(2)dyy^(2)e^(-y)=0.32P=\int_{0}^{1 / \alpha} d x\left(4 \alpha^{3}\right) x^{2} e^{-2 \alpha x}=\frac{1}{2} \int_{0}^{2} d y y^{2} e^{-y}=0.32
(d)
ϕ ( p ) = 1 2 π ℏ ∫ 0 ∞ d x e − i p x / ℏ ( 2 α α ) x e − α x = 4 α 3 2 π ℏ d d α ∫ 0 ∞ d x e − ( α + i p / ℏ ) x = − 4 α 3 2 π ℏ 1 ( α + i p / ℏ ) 2 ϕ ( p ) = 1 2 π ℏ ∫ 0 ∞   d x e − i p x / ℏ ( 2 α α ) x e − α x = 4 α 3 2 π ℏ d d α ∫ 0 ∞   d x e − ( α + i p / ℏ ) x = − 4 α 3 2 π ℏ 1 ( α + i p / ℏ ) 2 {:[phi(p)=(1)/(sqrt(2piℏ))int_(0)^(oo)dxe^(-ipx//ℏ)(2alphasqrtalpha)xe^(-alpha x)],[=sqrt((4alpha^(3))/(2piℏ))(d)/(d alpha)int_(0)^(oo)dxe^(-(alpha+ip//ℏ)x)=-sqrt((4alpha^(3))/(2piℏ))(1)/((alpha+ip//ℏ)^(2))]:}\begin{aligned} \phi(p) & =\frac{1}{\sqrt{2 \pi \hbar}} \int_{0}^{\infty} d x e^{-i p x / \hbar}(2 \alpha \sqrt{\alpha}) x e^{-\alpha x} \\ & =\sqrt{\frac{4 \alpha^{3}}{2 \pi \hbar}} \frac{d}{d \alpha} \int_{0}^{\infty} d x e^{-(\alpha+i p / \hbar) x}=-\sqrt{\frac{4 \alpha^{3}}{2 \pi \hbar}} \frac{1}{(\alpha+i p / \hbar)^{2}} \end{aligned}
From this we can calculate
由此我們可以計算出
⟨ p ⟩ = ∫ − ∞ ∞ d p p | ϕ ( p ) | 2 = 4 α 3 2 π ℏ ∫ − ∞ ∞ d p p ( α 2 + p 2 / ℏ 2 ) 2 = 0 ⟨ p ⟩ 2 = 4 α 3 2 π ℏ ∫ − ∞ ∞ d p p 2 ( α 2 + p 2 / ℏ 2 ) 2 = 8 α 3 2 π ℏ ∫ 0 ∞ d p p 2 ( α 2 + p 2 / ℏ 2 ) 2 ⟨ p ⟩ = ∫ − ∞ ∞   d p p | ϕ ( p ) | 2 = 4 α 3 2 π ℏ ∫ − ∞ ∞   d p p α 2 + p 2 / ℏ 2 2 = 0 ⟨ p ⟩ 2 = 4 α 3 2 π ℏ ∫ − ∞ ∞   d p p 2 α 2 + p 2 / ℏ 2 2 = 8 α 3 2 π ℏ ∫ 0 ∞   d p p 2 α 2 + p 2 / ℏ 2 2 {:[(:p:)=int_(-oo)^(oo)dpp|phi(p)|^(2)=(4alpha^(3))/(2piℏ)int_(-oo)^(oo)dp(p)/((alpha^(2)+p^(2)//ℏ^(2))^(2))=0],[(:p:)^(2)=(4alpha^(3))/(2piℏ)int_(-oo)^(oo)dp(p^(2))/((alpha^(2)+p^(2)//ℏ^(2))^(2))=(8alpha^(3))/(2piℏ)int_(0)^(oo)dp(p^(2))/((alpha^(2)+p^(2)//ℏ^(2))^(2))]:}\begin{aligned} & \langle p\rangle=\int_{-\infty}^{\infty} d p p|\phi(p)|^{2}=\frac{4 \alpha^{3}}{2 \pi \hbar} \int_{-\infty}^{\infty} d p \frac{p}{\left(\alpha^{2}+p^{2} / \hbar^{2}\right)^{2}}=0 \\ & \langle p\rangle^{2}=\frac{4 \alpha^{3}}{2 \pi \hbar} \int_{-\infty}^{\infty} d p \frac{p^{2}}{\left(\alpha^{2}+p^{2} / \hbar^{2}\right)^{2}}=\frac{8 \alpha^{3}}{2 \pi \hbar} \int_{0}^{\infty} d p \frac{p^{2}}{\left(\alpha^{2}+p^{2} / \hbar^{2}\right)^{2}} \end{aligned}
With the change of variables p = ℏ α tan ⁡ θ p = ℏ α tan ⁡ θ p=ℏalpha tan thetap=\hbar \alpha \tan \theta, we get
變數 p = ℏ α tan ⁡ θ p = ℏ α tan ⁡ θ p=ℏalpha tan thetap=\hbar \alpha \tan \theta 變更後，我們得到
⟨ p 2 ⟩ = 4 α 2 ℏ 2 π ∫ 0 π / 2 d θ sin 2 ⁡ θ = α 2 ℏ 2 p 2 = 4 α 2 ℏ 2 π ∫ 0 π / 2   d θ sin 2 ⁡ θ = α 2 ℏ 2 (:p^(2):)=(4alpha^(2)ℏ^(2))/(pi)int_(0)^(pi//2)d thetasin^(2)theta=alpha^(2)ℏ^(2)\left\langle p^{2}\right\rangle=\frac{4 \alpha^{2} \hbar^{2}}{\pi} \int_{0}^{\pi / 2} d \theta \sin ^{2} \theta=\alpha^{2} \hbar^{2}
The introduction of operators brings in a new concern. Products of operators need careful definition, because the order in which they act is important. Consider for example
運算符號的引入帶來了新的問題。運算符號的乘積需要仔細定義，因為運算符號的作用順序非常重要。例如
x p ψ ( x ) = − i ℏ x d ψ ( x ) d x x p ψ ( x ) = − i ℏ x d ψ ( x ) d x xp psi(x)=-iℏx(d psi(x))/(dx)x p \psi(x)=-i \hbar x \frac{d \psi(x)}{d x}
whereas 鉴于
p x ψ ( x ) = − i ℏ d d x ( x ψ ( x ) ) = − i ℏ x d ψ ( x ) d x − i ℏ ψ ( x ) p x ψ ( x ) = − i ℏ d d x ( x ψ ( x ) ) = − i ℏ x d ψ ( x ) d x − i ℏ ψ ( x ) px psi(x)=-iℏ(d)/(dx)(x psi(x))=-iℏx(d psi(x))/(dx)-iℏpsi(x)p x \psi(x)=-i \hbar \frac{d}{d x}(x \psi(x))=-i \hbar x \frac{d \psi(x)}{d x}-i \hbar \psi(x)
which is different from (2-52). Note that we can deduce
這與 (2-52) 不同。請注意，我們可以推導出
[ p , x ] ψ ( x ) ≡ ( p x − x p ) ψ ( x ) = − i ℏ ψ ( x ) [ p , x ] ψ ( x ) ≡ ( p x − x p ) ψ ( x ) = − i ℏ ψ ( x ) [p,x]psi(x)-=(px-xp)psi(x)=-iℏpsi(x)[p, x] \psi(x) \equiv(p x-x p) \psi(x)=-i \hbar \psi(x)
Since this is true for all ψ ( x ) ψ ( x ) psi(x)\psi(x), we conclude that we have an operator relation, which reads
由於這對所有 ψ ( x ) ψ ( x ) psi(x)\psi(x) 都是真實的，因此我們得出結論，我們有一個算子關係，其內容如下
[ p , x ] = − i ℏ [ p , x ] = − i ℏ [p,x]=-iℏ[p, x]=-i \hbar
This is a commutation relation, and it is interesting because it is a relation between operators, independent of what wave function this acts on. The difference between classical physics and quantum mechanics lies in that physical variables are described by operators and these do not necessarily commute. The commutator (2-55) will be seen to lie at the heart of the Heisenberg uncertainty relation.
這就是換向關係，它的有趣之處在於它是運算符之間的關係，與作用於何種波函数無關。經典物理與量子力學的差異在於物理變數是由算子來描述的，而這些算子不一定是換向的。換向器 (2-55) 將被視為海森堡不確定性關係的核心。

PROBLEMS 問題

  1. Given that A ( k ) = N / ( k 2 + α 2 ) A ( k ) = N / k 2 + α 2 A(k)=N//(k^(2)+alpha^(2))A(k)=N /\left(k^{2}+\alpha^{2}\right), calculate ψ ( x ) ψ ( x ) psi(x)\psi(x). Plot A ( k ) A ( k ) A(k)A(k) and ψ ( x ) ψ ( x ) psi(x)\psi(x) and show that Δ k Δ x > 1 Δ k Δ x > 1 Delta k Delta x > 1\Delta k \Delta x>1, independent of the choice of α α alpha\alpha.
    給定 A ( k ) = N / ( k 2 + α 2 ) A ( k ) = N / k 2 + α 2 A(k)=N//(k^(2)+alpha^(2))A(k)=N /\left(k^{2}+\alpha^{2}\right) ，計算 ψ ( x ) ψ ( x ) psi(x)\psi(x) 。描繪 A ( k ) A ( k ) A(k)A(k) 和 ψ ( x ) ψ ( x ) psi(x)\psi(x) 並顯示 Δ k Δ x > 1 Δ k Δ x > 1 Delta k Delta x > 1\Delta k \Delta x>1 ，與 α α alpha\alpha 的選擇無關。
  2. The relation between the wavelength λ λ lambda\lambda and the frequency ν ν nu\nu in a wave guide is given by
    波導中波長 λ λ lambda\lambda 與頻率 ν ν nu\nu 的關係為
λ = c ν 2 − ν 0 2 λ = c ν 2 − ν 0 2 lambda=(c)/(sqrt(nu^(2)-nu_(0)^(2)))\lambda=\frac{c}{\sqrt{\nu^{2}-\nu_{0}^{2}}}
What is the group of velocity of such waves?
這種波的速度群是多少？
3. For surface tension waves in shallow water, the relation between frequency and wavelength is given by
3.對於淺水中的表面張力波，頻率與波長的關係為
ν = 2 π T ρ λ 3 ν = 2 π T ρ λ 3 nu=sqrt((2pi T)/(rholambda^(3)))\nu=\sqrt{\frac{2 \pi T}{\rho \lambda^{3}}}
where T T TT is the surface tension and ρ ρ rho\rho is the density. What is the group velocity of the waves?
其中 T T TT 是表面張力， ρ ρ rho\rho 是密度。波的群速度是多少？
4. For deep water gravity waves, the relation between frequency and wavelength is given by
4.對於深水重力波，頻率與波長的關係為
ν = g 2 π λ ν = g 2 π λ nu=sqrt((g)/(2pi lambda))\nu=\sqrt{\frac{g}{2 \pi \lambda}}
What is the group velocity of such waves?
這種波的群速度是多少？
5. Consider the problem of spreading of a gaussian wave packet describing a free particle, with
5.考慮描述一個自由粒子的高斯波包的擴散問題，其中
ω = k 2 ℏ 2 m ω = k 2 ℏ 2 m omega=(k^(2)ℏ)/(2m)\omega=\frac{k^{2} \hbar}{2 m}
Calculate the fractional change in the size of the wave packet in one second if
計算一秒鐘內波包大小變化的分數，如果
(a) The packet represents an electron of mass 0.9 × 10 − 30 kg 0.9 × 10 − 30 kg 0.9 xx10^(-30)kg0.9 \times 10^{-30} \mathrm{~kg}, with the wave packet having dimensions of 10 − 6 m ; 10 − 10 m 10 − 6 m ; 10 − 10 m 10^(-6)m;10^(-10)m10^{-6} \mathrm{~m} ; 10^{-10} \mathrm{~m}.
(a) 波包代表質量 0.9 × 10 − 30 kg 0.9 × 10 − 30 kg 0.9 xx10^(-30)kg0.9 \times 10^{-30} \mathrm{~kg} 的電子，波包的尺寸為 10 − 6 m ; 10 − 10 m 10 − 6 m ; 10 − 10 m 10^(-6)m;10^(-10)m10^{-6} \mathrm{~m} ; 10^{-10} \mathrm{~m} 。
(b) The packet represents an object of mass 10 − 3 kg 10 − 3 kg 10^(-3)kg10^{-3} \mathrm{~kg} and size 0.01 m . [It will be convenient to express the width in units of ℏ / m c ℏ / m c ℏ//mc\hbar / m c, where m m mm is the mass of the particle represented by the packet.]
(b) 該封包代表質量為 10 − 3 kg 10 − 3 kg 10^(-3)kg10^{-3} \mathrm{~kg} 、大小為 0.01 m 的物體。[方便的做法是以 ℏ / m c ℏ / m c ℏ//mc\hbar / m c 的單位來表示寬度，其中 m m mm 是封包所代表的粒子的質量。］
6. A beam of electrons is to be fired over a distance of 10 4 km 10 4 km 10^(4)km10^{4} \mathrm{~km}. If the size of the initial packet is 10 − 3 10 − 3 10^(-3)10^{-3} m , what will be its size upon arrival, if its kinetic energy is (a) 13.6 eV ; (b) 100 MeV ? [Caution: The relation between kinetic energy (K. E.) and momentum is not always K. E. = p 2 / 2 m ! ] = p 2 / 2 m ! {:=p^(2)//2m!]\left.=p^{2} / 2 m!\right]
6.一束電子將射出 10 4 km 10 4 km 10^(4)km10^{4} \mathrm{~km} 的距離。如果初始電子包的大小是 10 − 3 10 − 3 10^(-3)10^{-3} m，當它到達時，如果它的動能是 (a) 13.6 eV ； (b) 100 MeV，它的大小會是多少？[注意：注意：動能 (K. E.) 與動量之間的關係不一定是 K. E. = p 2 / 2 m ! ] = p 2 / 2 m ! {:=p^(2)//2m!]\left.=p^{2} / 2 m!\right] .
7. Consider a wave packet for neutrinos, which are massless to a very good approximation, so that E = p c E = p c E=pcE=p c. Show that such a wave packet does not spread.
7.考慮一個中微子的波包，中微子在很好的近似情況下是無質量的，因此 E = p c E = p c E=pcE=p c 。證明這樣的波包不會擴散。
8. Consider a wave function of the form
8.考慮一個形式為
ψ ( x ) = A e − μ | x | ψ ( x ) = A e − μ | x | psi(x)=Ae^(-mu|x|)\psi(x)=A e^{-\mu|x|}
Calculate the wave function in momentum space ϕ ( p ) ϕ ( p ) phi(p)\phi(p).
計算動量空間 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 中的波函数。
9. Consider the example in Problem 8. Calculate A A AA so that ψ ( x ) ψ ( x ) psi(x)\psi(x) is properly normalized.
9.考慮問題 8 中的範例。計算 A A AA ，使 ψ ( x ) ψ ( x ) psi(x)\psi(x) 適當地規範化。
10. Show that the conservation law (2-33) holds when ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) obeys eq. (2-23), but only if V ( x ) V ( x ) V(x)V(x) is real.
10.證明當 ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t) 遵循公式 (2-23) 時，守恆定律 (2-33) 成立，但前提是 V ( x ) V ( x ) V(x)V(x) 是實數。
11. Suppose the V ( x ) V ( x ) V(x)V(x) is complex. Obtain an expression for
11.假設 V ( x ) V ( x ) V(x)V(x) 是複數。取得
∂ P ( x , t ) ∂ t and d d t ∫ − ∞ ∞ d x P ( x , t ) ∂ P ( x , t ) ∂ t  and  d d t ∫ − ∞ ∞   d x P ( x , t ) (del P(x,t))/(del t)" and "(d)/(dt)int_(-oo)^(oo)dxP(x,t)\frac{\partial P(x, t)}{\partial t} \text { and } \frac{d}{d t} \int_{-\infty}^{\infty} d x P(x, t)
For absorption of particles the last quantity must be negative (since particles disappear, the probability of their being anywhere decreases). What does this tell us about the imaginary part of V ( x ) V ( x ) V(x)V(x) ?
對於微粒的吸收，最後一個量必須是負數（因為微粒消失了，它們在任何地方的機率就會降低）。這對 V ( x ) V ( x ) V(x)V(x) 的假想部分有何啟示？
12. Consider the distribution of grades in a class of 60 students, given by
12.考慮一個有 60 名學生的班級的成績分佈，如下所示
Grades 等級 60 55 50 45 40 35 30 25 20 15 10 5
# students # 學生 1 2 7 9 16 13 3 6 2 0 1 0
Grades 60 55 50 45 40 35 30 25 20 15 10 5 # students 1 2 7 9 16 13 3 6 2 0 1 0| Grades | 60 | 55 | 50 | 45 | 40 | 35 | 30 | 25 | 20 | 15 | 10 | 5 | | :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | | # students | 1 | 2 | 7 | 9 | 16 | 13 | 3 | 6 | 2 | 0 | 1 | 0 |
(a) Plot a histogram of the distribution.
(a) 繪製分佈直方圖。
(b) Calculate the class average.
(b) 計算班級平均值。
© Calculate ( Δ g ) 2 = ( ⟨ g 2 ⟩ − ⟨ g ⟩ 2 ) ( Δ g ) 2 = g 2 − ⟨ g ⟩ 2 (Delta g)^(2)=((:g^(2):)-(:g:)^(2))(\Delta g)^{2}=\left(\left\langle g^{2}\right\rangle-\langle g\rangle^{2}\right). © 計算 ( Δ g ) 2 = ( ⟨ g 2 ⟩ − ⟨ g ⟩ 2 ) ( Δ g ) 2 = g 2 − ⟨ g ⟩ 2 (Delta g)^(2)=((:g^(2):)-(:g:)^(2))(\Delta g)^{2}=\left(\left\langle g^{2}\right\rangle-\langle g\rangle^{2}\right) 。
13. Compare your histogram with a distribution of the form
13.將您的直方圖與形式為
N ( g ) = C e − ( g − ( g ) ) 2 ( Δ g ) 2 N ( g ) = C e − ( g − ( g ) ) 2 ( Δ g ) 2 N(g)=Ce^(-(g-(g))^(2)(Delta g)^(2))N(g)=C e^{-(g-(g))^{2}(\Delta g)^{2}}
with C C CC chosen that ∑ g N ( g ) = 60 ∑ g   N ( g ) = 60 sum_(g)N(g)=60\sum_{g} N(g)=60.
選擇 C C CC ，即 ∑ g N ( g ) = 60 ∑ g   N ( g ) = 60 sum_(g)N(g)=60\sum_{g} N(g)=60 。
14. Show that a grade distribution of the form
14.證明以下形式的等級分布
Grades 等級 60 50 10
# students # 學生 7 34 19
Grades 60 50 10 # students 7 34 19| Grades | 60 | 50 | 10 | | :--- | ---: | ---: | ---: | | # students | 7 | 34 | 19 |
leads to the same average grade, but that the dispersion Δ g Δ g Delta g\Delta g is different. What is it?
導致相同的平均成績，但分散性 Δ g Δ g Delta g\Delta g 不同。這是什麼？
15. Consider the wave function obtained in the first example, 2 N / ( sin ⁡ k x ) / x 2 N / ( sin ⁡ k x ) / x 2N//(sin kx)//x2 N /(\sin k x) / x. For what value of N N NN will it be normalized? [Hint: A useful integral is ∫ − ∞ ∞ d t ( sin ⁡ t t ) 2 = π ∫ − ∞ ∞   d t sin ⁡ t t 2 = π int_(-oo)^(oo)dt((sin t)/(t))^(2)=pi\int_{-\infty}^{\infty} d t\left(\frac{\sin t}{t}\right)^{2}=\pi.]
15.考慮第一個範例中得到的波函数 2 N / ( sin ⁡ k x ) / x 2 N / ( sin ⁡ k x ) / x 2N//(sin kx)//x2 N /(\sin k x) / x 。對於 N N NN 的什麼值，它會被規範化？[提示：一個有用的積分是 ∫ − ∞ ∞ d t ( sin ⁡ t t ) 2 = π ∫ − ∞ ∞   d t sin ⁡ t t 2 = π int_(-oo)^(oo)dt((sin t)/(t))^(2)=pi\int_{-\infty}^{\infty} d t\left(\frac{\sin t}{t}\right)^{2}=\pi 。］
16. Consider the wave function
16.考慮波函数
ψ ( x ) = ( α / π ) 1 / 4 exp ⁡ ( − α x 2 / 2 ) ψ ( x ) = ( α / π ) 1 / 4 exp ⁡ − α x 2 / 2 psi(x)=(alpha//pi)^(1//4)exp(-alphax^(2)//2)\psi(x)=(\alpha / \pi)^{1 / 4} \exp \left(-\alpha x^{2} / 2\right)
Calculate ⟨ x ′ ′ ⟩ x ′ ′ (:x^(''):)\left\langle x^{\prime \prime}\right\rangle for n = 1 , 2 n = 1 , 2 n=1,2n=1,2. Can you quickly write down the result for ⟨ x 17 ⟩ x 17 (:x^(17):)\left\langle x^{17}\right\rangle ?
為 n = 1 , 2 n = 1 , 2 n=1,2n=1,2 計算 ⟨ x ′ ′ ⟩ x ′ ′ (:x^(''):)\left\langle x^{\prime \prime}\right\rangle 。您可以快速寫下 ⟨ x 17 ⟩ x 17 (:x^(17):)\left\langle x^{17}\right\rangle 的結果嗎？
17. Calculate ϕ ( p ) ϕ ( p ) phi(p)\phi(p) for the wave function in problem 16. Calculate ⟨ p n ⟩ p n (:p^(n):)\left\langle p^{n}\right\rangle for n = 1 , 2 n = 1 , 2 n=1,2n=1,2.
17.計算問題 16 中波函數的 ϕ ( p ) ϕ ( p ) phi(p)\phi(p) 。計算 n = 1 , 2 n = 1 , 2 n=1,2n=1,2 的 ⟨ p n ⟩ p n (:p^(n):)\left\langle p^{n}\right\rangle 。
18. Use the definitions ( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 ( Δ x ) 2 = x 2 − ⟨ x ⟩ 2 (Delta x)^(2)=(:x^(2):)-(:x:)^(2)(\Delta x)^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2} and ( Δ p ⟩ 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 ( Δ p ⟩ 2 = p 2 − ⟨ p ⟩ 2 (Delta p:)^(2)=(:p^(2):)-(:p:)^(2)(\Delta p\rangle^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2} with the results of Problems 16 and 17 to show that Δ p Δ x > ℏ / 2 Δ p Δ x > ℏ / 2 Delta p Delta x > ℏ//2\Delta p \Delta x>\hbar / 2.
18.使用定義 ( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 ( Δ x ) 2 = x 2 − ⟨ x ⟩ 2 (Delta x)^(2)=(:x^(2):)-(:x:)^(2)(\Delta x)^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2} 和 ( Δ p ⟩ 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 ( Δ p ⟩ 2 = p 2 − ⟨ p ⟩ 2 (Delta p:)^(2)=(:p^(2):)-(:p:)^(2)(\Delta p\rangle^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2} 與問題 16 和 17 的結果來證明 Δ p Δ x > ℏ / 2 Δ p Δ x > ℏ / 2 Delta p Delta x > ℏ//2\Delta p \Delta x>\hbar / 2 。
19. Make an estimate of the strength of the nuclear potential energy given the following fact: The “size” of the box that roughly describes the nuclear potential is 10 − 15 m 10 − 15 m 10^(-15)m10^{-15} \mathrm{~m}, and it takes 8 MeV to eject a particle from this potential well.
19.根據以下事實，估計一下核勢能的強度：粗略描述核勢能的方框「大小」為 10 − 15 m 10 − 15 m 10^(-15)m10^{-15} \mathrm{~m} ，從這個勢能井射出一個粒子需要 8 MeV。
(a) Use the uncertainty principle to estimate ⟨ p 2 ⟩ p 2 (:p^(2):)\left\langle p^{2}\right\rangle for a nucleon in the box, and given the fact that the mass of the nucleon is M = 1.67 × 10 − 27 kg M = 1.67 × 10 − 27 kg M=1.67 xx10^(-27)kgM=1.67 \times 10^{-27} \mathrm{~kg}, estimate the kinetic energy of the nucleon.
(a) 使用不確定性原理來估計盒子中一個核子的 ⟨ p 2 ⟩ p 2 (:p^(2):)\left\langle p^{2}\right\rangle ，並根據核子的質量是 M = 1.67 × 10 − 27 kg M = 1.67 × 10 − 27 kg M=1.67 xx10^(-27)kgM=1.67 \times 10^{-27} \mathrm{~kg} ，估計核子的動能。
(b) Since the potential that gives rise to the binding must more than compensate for this, what is the negative potential energy?
(b) 由於產生結合的電勢必須補償有餘，那麼負電勢能是多少？
20. Monochromatic light passes through a shutter that opens for a time Δ t = 10 − 10 sec Δ t = 10 − 10 sec Delta t=10^(-10)sec\Delta t=10^{-10} \mathrm{sec}. What is the spread in frequencies caused by the shutter?
20.單色光通過快門，快門打開的時間為 Δ t = 10 − 10 sec Δ t = 10 − 10 sec Delta t=10^(-10)sec\Delta t=10^{-10} \mathrm{sec} 。快門造成的頻率差是多少？
21. Nuclei typically of size 10 − 14 m 10 − 14 m 10^(-14)m10^{-14} \mathrm{~m} frequently emit electrons with energies in the range 1 − 10 MeV 1 − 10 MeV 1-10MeV1-10 \mathrm{MeV}. In the early days of nuclear physics, people believed that electrons “lived” inside the nuclei. Use the uncertainty relation to show that electrons of such energies could not be contained inside the nucleus.
21.通常大小為 10 − 14 m 10 − 14 m 10^(-14)m10^{-14} \mathrm{~m} 的原子核經常會釋放出能量在 1 − 10 MeV 1 − 10 MeV 1-10MeV1-10 \mathrm{MeV} 範圍內的電子。在早期的核物理學中，人們相信電子「住在」原子核內。使用不確定性關係來說明這樣能量的電子不可能包含在原子核內。
22. Show that eq. (2-49) holds.
22.證明公式 (2-49) 成立。

Supplement 2-A 補編 2-A

The Fourier Integral and Delta Functions
傅立葉整數和三角函數

Consider a function f ( x ) f ( x ) f(x)f(x) that is periodic, with period 2 L 2 L 2L2 L, so that
考慮一個週期為 2 L 2 L 2L2 L 的函數 f ( x ) f ( x ) f(x)f(x) ，如此
f ( x ) = f ( x + 2 L ) f ( x ) = f ( x + 2 L ) f(x)=f(x+2L)f(x)=f(x+2 L)
Such a function can be expanded in a Fourier series in the interval ( − L , L ) ( − L , L ) (-L,L)(-L, L), and the series has the form
這樣的函數可以在 ( − L , L ) ( − L , L ) (-L,L)(-L, L) 區間內的傅立葉數列中展開，該數列的形式為
f ( x ) = ∑ n = 0 ∞ A n cos ⁡ n π x L + ∑ n = 1 ∞ B n sin ⁡ n π x L f ( x ) = ∑ n = 0 ∞   A n cos ⁡ n π x L + ∑ n = 1 ∞   B n sin ⁡ n π x L f(x)=sum_(n=0)^(oo)A_(n)cos((n pi x)/(L))+sum_(n=1)^(oo)B_(n)sin((n pi x)/(L))f(x)=\sum_{n=0}^{\infty} A_{n} \cos \frac{n \pi x}{L}+\sum_{n=1}^{\infty} B_{n} \sin \frac{n \pi x}{L}
We can rewrite the series in the form
我們可以將此數列重寫為
f ( x ) = ∑ n = − ∞ ∞ a n e i n π x / L f ( x ) = ∑ n = − ∞ ∞   a n e i n π x / L f(x)=sum_(n=-oo)^(oo)a_(n)e^(in pi x//L)f(x)=\sum_{n=-\infty}^{\infty} a_{n} e^{i n \pi x / L}
which is certainly possible, since
這當然是可能的，因為
cos ⁡ n π x L = 1 2 ( e i n π x / L + e − i n π x / L ) sin ⁡ n π x L = 1 2 i ( e i n π x / L − e − i n π x / L ) cos ⁡ n π x L = 1 2 e i n π x / L + e − i n π x / L sin ⁡ n π x L = 1 2 i e i n π x / L − e − i n π x / L {:[cos((n pi x)/(L))=(1)/(2)(e^(in pi x//L)+e^(-in pi x//L))],[sin((n pi x)/(L))=(1)/(2i)(e^(in pi x//L)-e^(-in pi x//L))]:}\begin{aligned} & \cos \frac{n \pi x}{L}=\frac{1}{2}\left(e^{i n \pi x / L}+e^{-i n \pi x / L}\right) \\ & \sin \frac{n \pi x}{L}=\frac{1}{2 i}\left(e^{i n \pi x / L}-e^{-i n \pi x / L}\right) \end{aligned}
The coefficients can be determined with the help of the orthonormality relation
這些係數可以在正交關係的幫助下確定
1 2 L ∫ − L L d x e i n π x / L e − i m π x / L = δ m n = { 1 m = n 0 m ≠ n 1 2 L ∫ − L L   d x e i n π x / L e − i m π x / L = δ m n = 1 m = n 0 m ≠ n (1)/(2L)int_(-L)^(L)dxe^(in pi x//L)e^(-im pi x//L)=delta_(mn)={[1,m=n],[0,m!=n]:}\frac{1}{2 L} \int_{-L}^{L} d x e^{i n \pi x / L} e^{-i m \pi x / L}=\delta_{m n}= \begin{cases}1 & m=n \\ 0 & m \neq n\end{cases}
Thus 因此
a n = 1 2 L ∫ − L L d x f ( x ) e − i n π x / L a n = 1 2 L ∫ − L L   d x f ( x ) e − i n π x / L a_(n)=(1)/(2L)int_(-L)^(L)dxf(x)e^(-in pi x//L)a_{n}=\frac{1}{2 L} \int_{-L}^{L} d x f(x) e^{-i n \pi x / L}
Let us now rewrite (2A-3) by introducing Δ n Δ n Delta n\Delta n, the difference between two successive integers. Since this is unity, we have
現在讓我們引入 Δ n Δ n Delta n\Delta n ，即兩個連續整數之間的差，來重寫 (2A-3)。由於它是單一，我們有
f ( x ) = ∑ n a n e i n π x / L Δ n = L π ∑ n a n e i n π x / L π Δ n L f ( x ) = ∑ n   a n e i n π x / L Δ n = L π ∑ n   a n e i n π x / L π Δ n L {:[f(x)=sum_(n)a_(n)e^(in pi x//L)Delta n],[=(L)/( pi)sum_(n)a_(n)e^(in pi x//L)(pi Delta n)/(L)]:}\begin{aligned} f(x) & =\sum_{n} a_{n} e^{i n \pi x / L} \Delta n \\ & =\frac{L}{\pi} \sum_{n} a_{n} e^{i n \pi x / L} \frac{\pi \Delta n}{L} \end{aligned}
Let us change the notation by writing
讓我們改變記號，寫成
π n L = k π n L = k (pi n)/(L)=k\frac{\pi n}{L}=k
and 和
π Δ n L = Δ k π Δ n L = Δ k (pi Delta n)/(L)=Delta k\frac{\pi \Delta n}{L}=\Delta k
We also write 我們也寫
L a n π = A ( k ) 2 π L a n π = A ( k ) 2 π (La_(n))/(pi)=(A(k))/(sqrt(2pi))\frac{L a_{n}}{\pi}=\frac{A(k)}{\sqrt{2 \pi}}
Hence (2A-6) becomes 因此 (2A-6) 變成
f ( x ) = ∑ A ( k ) 2 π e i k x Δ k f ( x ) = ∑ A ( k ) 2 π e i k x Δ k f(x)=sum(A(k))/(sqrt(2pi))e^(ikx)Delta kf(x)=\sum \frac{A(k)}{\sqrt{2 \pi}} e^{i k x} \Delta k
If we now let L → ∞ L → ∞ L rarr ooL \rightarrow \infty, then k k kk approaches a continuous variable, since Δ k Δ k Delta k\Delta k becomes infinitesimally small. If we recall the Riemann definition of an integral, we see that in the limit ( 2 A − 10 ) ( 2 A − 10 ) (2A-10)(2 \mathrm{~A}-10) can be written in the form
如果我們現在讓 L → ∞ L → ∞ L rarr ooL \rightarrow \infty ，那麼 k k kk 就會接近連續變數，因為 Δ k Δ k Delta k\Delta k 變得無限小。如果我們回想一下積分的黎曼定義，我們會發現在極限中 ( 2 A − 10 ) ( 2 A − 10 ) (2A-10)(2 \mathrm{~A}-10) 可以寫成以下的形式
f ( x ) = 1 2 π ∫ − ∞ ∞ d k A ( k ) e i k x f ( x ) = 1 2 π ∫ − ∞ ∞   d k A ( k ) e i k x f(x)=(1)/(sqrt(2pi))int_(-oo)^(oo)dkA(k)e^(ikx)f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} d k A(k) e^{i k x}
The coefficient A ( k ) A ( k ) A(k)A(k) is given by
係數 A ( k ) A ( k ) A(k)A(k) 由以下公式給出
A ( k ) = 2 π L π ⋅ 1 2 L ∫ − L L d x f ( x ) e − i n π x / L → 1 2 π ∫ − ∞ ∞ d x f ( x ) e − i k x A ( k ) = 2 π L π ⋅ 1 2 L ∫ − L L   d x f ( x ) e − i n π x / L → 1 2 π ∫ − ∞ ∞   d x f ( x ) e − i k x {:[A(k)=sqrt(2pi)(L)/( pi)*(1)/(2L)int_(-L)^(L)dxf(x)e^(-in pi x//L)],[ rarr(1)/(sqrt(2pi))int_(-oo)^(oo)dxf(x)e^(-ikx)]:}\begin{aligned} A(k) & =\sqrt{2 \pi} \frac{L}{\pi} \cdot \frac{1}{2 L} \int_{-L}^{L} d x f(x) e^{-i n \pi x / L} \\ & \rightarrow \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} d x f(x) e^{-i k x} \end{aligned}
Equations (2A-11) and (2A-12) define the Fourier integral transformations. If we insert the second equation into the first we get
等式 (2A-11) 和 (2A-12) 定義了傅立葉積分轉換。如果我們將第二個等式插入第一個等式，我們會得到
f ( x ) = 1 2 π ∫ − ∞ ∞ d k e i k x ∫ − ∞ ∞ d y f ( y ) e − i k y f ( x ) = 1 2 π ∫ − ∞ ∞   d k e i k x ∫ − ∞ ∞   d y f ( y ) e − i k y f(x)=(1)/(2pi)int_(-oo)^(oo)dke^(ikx)int_(-oo)^(oo)dyf(y)e^(-iky)f(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d k e^{i k x} \int_{-\infty}^{\infty} d y f(y) e^{-i k y}
Suppose now that we interchange, without question, the order of integrations. We then get
現在假設我們毫無疑問地交換積分的順序。我們會得到
f ( x ) = ∫ − ∞ ∞ d y f ( y ) [ 1 2 π ∫ − ∞ ∞ d k e i k ( x − y ) ] f ( x ) = ∫ − ∞ ∞   d y f ( y ) 1 2 π ∫ − ∞ ∞   d k e i k ( x − y ) f(x)=int_(-oo)^(oo)dyf(y)[(1)/(2pi)int_(-oo)^(oo)dke^(ik(x-y))]f(x)=\int_{-\infty}^{\infty} d y f(y)\left[\frac{1}{2 \pi} \int_{-\infty}^{\infty} d k e^{i k(x-y)}\right]
For this to be true, the quantity δ ( x − y ) δ ( x − y ) delta(x-y)\delta(x-y) defined by
若要證明這一點，由以下公式定義的量 δ ( x − y ) δ ( x − y ) delta(x-y)\delta(x-y)
δ ( x − y ) = 1 2 π ∫ − ∞ ∞ d k e i k ( x − y ) δ ( x − y ) = 1 2 π ∫ − ∞ ∞   d k e i k ( x − y ) delta(x-y)=(1)/(2pi)int_(-oo)^(oo)dke^(ik(x-y))\delta(x-y)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d k e^{i k(x-y)}
and called the Dirac delta function must be a very peculiar kind of function; it must vanish when x ≠ y x ≠ y x!=yx \neq y, and it must tend to infinity in an appropriate way when x − y = 0 x − y = 0 x-y=0x-y=0, since the range of integration is infinitesimally small. It is therefore not a function of the usual
並稱為 Dirac delta 函數的函數必須是一種非常特殊的函數；當 x ≠ y x ≠ y x!=yx \neq y 時，它必須消失；當 x − y = 0 x − y = 0 x-y=0x-y=0 時，它必須以適當的方式趨向無窮大，因為積分的範圍是無限小的。因此，它不是一般的
mathematical sense, but it is rather a “generalized function” or a “distribution.” It does not have any meaning by itself, but it can be defined provided it always appears in the form
在數學意義上，它是一種「泛函數」或「分佈」。它本身沒有任何意義，但它可以被定義，只要它總是以下列形式出現
∫ d x f ( x ) δ ( x − a ) ∫ d x f ( x ) δ ( x − a ) int dxf(x)delta(x-a)\int d x f(x) \delta(x-a)
with the function f ( x ) f ( x ) f(x)f(x) sufficiently smooth in the range of values that the argument of the delta function takes. We will take that for granted and manipulate the delta function by itself, with the understanding that at the end all the relations that we write down only occur under the integral sign.
在 delta 函數的參數取值範圍內，函數 f ( x ) f ( x ) f(x)f(x) 充分平滑。我們將視此為理所當然，並自行處理 delta 函數，但要了解，最後我們寫下的所有關係只會在積分符號下出現。
The following properties of the delta function can be demonstrated:
可以證明 delta 函數的下列屬性：
(i)
δ ( a x ) = 1 | a | δ ( x ) δ ( a x ) = 1 | a | δ ( x ) delta(ax)=(1)/(|a|)delta(x)\delta(a x)=\frac{1}{|a|} \delta(x)
This can be seen to follow from
這可以從
f ( x ) = ∫ d y f ( y ) δ ( x − y ) f ( x ) = ∫ d y f ( y ) δ ( x − y ) f(x)=int dyf(y)delta(x-y)f(x)=\int d y f(y) \delta(x-y)
If we write x = a ξ x = a ξ x=a xix=a \xi and y y yy and a η a η a etaa \eta, then this reads
如果我們寫 x = a ξ x = a ξ x=a xix=a \xi 、 y y yy 和 a η a η a etaa \eta ，則讀成
f ( a ξ ) = | a | ∫ d η f ( a η ) δ [ a ( ξ − η ) ] f ( a ξ ) = | a | ∫ d η f ( a η ) δ [ a ( ξ − η ) ] f(a xi)=|a|int d eta f(a eta)delta[a(xi-eta)]f(a \xi)=|a| \int d \eta f(a \eta) \delta[a(\xi-\eta)]
On the other hand, 另一方面、
f ( a ξ ) = ∫ d η f ( a η ) δ ( ξ − η ) f ( a ξ ) = ∫ d η f ( a η ) δ ( ξ − η ) f(a xi)=int d eta f(a eta)delta(xi-eta)f(a \xi)=\int d \eta f(a \eta) \delta(\xi-\eta)
which implies our result.
這意味著我們的結果。
(ii) A relation that follows from ( 2 A − 16 ) ( 2 A − 16 ) (2A-16)(2 \mathrm{~A}-16) is
(ii) 由 ( 2 A − 16 ) ( 2 A − 16 ) (2A-16)(2 \mathrm{~A}-16) 所產生的關係是
δ ( x 2 − a 2 ) = 1 2 | a | [ δ ( x − a ) + δ ( x + a ) ] δ x 2 − a 2 = 1 2 | a | [ δ ( x − a ) + δ ( x + a ) ] delta(x^(2)-a^(2))=(1)/(2|a|)[delta(x-a)+delta(x+a)]\delta\left(x^{2}-a^{2}\right)=\frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]
This follows from the fact that the argument of the delta function vanishes at x = a x = a x=ax=a and x = − a x = − a x=-ax=-a. Thus there are two contributions:
這是由於 delta 函數的參數在 x = a x = a x=ax=a 和 x = − a x = − a x=-ax=-a 處消失。因此有兩個貢獻：
δ ( x 2 − a 2 ) = δ [ ( x − a ) ( x + a ) ] = 1 | x + a | δ ( x − a ) + 1 | x − a | δ ( x + a ) = 1 2 | a | [ δ ( x − a ) + δ ( x + a ) ] δ x 2 − a 2 = δ [ ( x − a ) ( x + a ) ] = 1 | x + a | δ ( x − a ) + 1 | x − a | δ ( x + a ) = 1 2 | a | [ δ ( x − a ) + δ ( x + a ) ] {:[delta(x^(2)-a^(2))=delta[(x-a)(x+a)]],[=(1)/(|x+a|)delta(x-a)+(1)/(|x-a|)delta(x+a)],[=(1)/(2|a|)[delta(x-a)+delta(x+a)]]:}\begin{aligned} \delta\left(x^{2}-a^{2}\right) & =\delta[(x-a)(x+a)] \\ & =\frac{1}{|x+a|} \delta(x-a)+\frac{1}{|x-a|} \delta(x+a) \\ & =\frac{1}{2|a|}[\delta(x-a)+\delta(x+a)] \end{aligned}
More generally, one can show that
一般而言，我們可以證明
δ [ f ( x ) ] = ∑ i δ ( x − x i ) | d f / d x | x = x i δ [ f ( x ) ] = ∑ i   δ x − x i | d f / d x | x = x i delta[f(x)]=sum_(i)(delta(x-x_(i)))/(|df//dx|_(x=x_(i)))\delta[f(x)]=\sum_{i} \frac{\delta\left(x-x_{i}\right)}{|d f / d x|_{x=x_{i}}}
where the x i x i x_(i)x_{i} are the roots of f ( x ) f ( x ) f(x)f(x) in the interval of integration.
其中 x i x i x_(i)x_{i} 是 f ( x ) f ( x ) f(x)f(x) 在整合區間內的根。
1 1 ^(1){ }^{1} The theory of distributions was developed by the mathematician Laurent Schwartz. An introductory treatment may be found in M. J. Lighthill, Introduction to Fourier Analysis and Generalized Functions, Cambridge University Press, Cambridge, England, 1958.
1 1 ^(1){ }^{1} 分佈理論是由數學家勞倫?Lighthill, Introduction to Fourier Analysis and Generalized Functions, Cambridge University Press, Cambridge, England, 1958。
In addition to the representation (2A-15) of the delta function, there are other representations that may prove useful. We discuss several of them.
除了 delta 函數的表示法 (2A-15)，還有其他可能有用的表示法。我們將討論其中幾種。
(a) Consider the form (2A-15), which we write in the form
(a) 考慮形式 (2A-15)，我們寫成以下形式
δ ( x ) = 1 2 π Lim L → ∞ ∫ − L L d k e i k x δ ( x ) = 1 2 π Lim L → ∞ ∫ − L L   d k e i k x delta(x)=(1)/(2pi)Lim_(L rarr oo)int_(-L)^(L)dke^(ikx)\delta(x)=\frac{1}{2 \pi} \operatorname{Lim}_{L \rightarrow \infty} \int_{-L}^{L} d k e^{i k x}
The integral can be done, and we get
可以進行積分，我們得到
δ ( x ) = Lim L → ∞ 1 2 π e i L x − e − i L x i x = Lim L → ∞ sin ⁡ L x π x δ ( x ) = Lim L → ∞ 1 2 π e i L x − e − i L x i x = Lim L → ∞ sin ⁡ L x π x {:[delta(x)=Lim_(L rarr oo)(1)/(2pi)(e^(iLx)-e^(-iLx))/(ix)],[=Lim_(L rarr oo)(sin Lx)/(pi x)]:}\begin{aligned} \delta(x) & =\operatorname{Lim}_{L \rightarrow \infty} \frac{1}{2 \pi} \frac{e^{i L x}-e^{-i L x}}{i x} \\ & =\operatorname{Lim}_{L \rightarrow \infty} \frac{\sin L x}{\pi x} \end{aligned}
(b) Consider the function Δ ( x , a ) Δ ( x , a ) Delta(x,a)\Delta(x, a) defined by
(b) 考慮函數 Δ ( x , a ) Δ ( x , a ) Delta(x,a)\Delta(x, a) 定義如下
Δ ( x , a ) = 0 x < − a = 1 2 a − a < x < a = 0 a < x Δ ( x , a ) = 0 x < − a = 1 2 a − a < x < a = 0 a < x {:[Delta(x","a)=0x < -a],[=(1)/(2a)-a < x < a],[=0a < x]:}\begin{aligned} \Delta(x, a) & =0 & & x<-a \\ & =\frac{1}{2 a} & -a & <x<a \\ & =0 & & a<x \end{aligned}
Then 那麼
δ ( x ) = Lim a → 0 Δ ( x , a ) δ ( x ) = Lim a → 0 Δ ( x , a ) delta(x)=Lim_(a rarr0)Delta(x,a)\delta(x)=\operatorname{Lim}_{a \rightarrow 0} \Delta(x, a)
It is clear that an integral of a product of Δ ( x , a ) Δ ( x , a ) Delta(x,a)\Delta(x, a) and a function f ( x ) f ( x ) f(x)f(x) that is smooth near the origin will pick out the value at the origin
很明顯， Δ ( x , a ) Δ ( x , a ) Delta(x,a)\Delta(x, a) 和函數 f ( x ) f ( x ) f(x)f(x) 的乘積的積分，如果在原點附近是平滑的，則會找出原點的值
Lim a → 0 ∫ d x f ( x ) Δ ( x , a ) = f ( 0 ) Lim a → 0 ∫ d x Δ ( x , a ) = f ( 0 ) Lim a → 0 ∫ d x f ( x ) Δ ( x , a ) = f ( 0 ) Lim a → 0 ∫ d x Δ ( x , a ) = f ( 0 ) {:[Lim_(a rarr0)int dxf(x)Delta(x","a)=f(0)Lim_(a rarr0)int dx Delta(x","a)],[=f(0)]:}\begin{aligned} \operatorname{Lim}_{a \rightarrow 0} \int d x f(x) \Delta(x, a) & =f(0) \operatorname{Lim}_{a \rightarrow 0} \int d x \Delta(x, a) \\ & =f(0) \end{aligned}
© By the same token, any peaked function, normalized to unit area under it, will approach a delta function in the limit that the width of the peak goes to zero. We will leave it to the reader to show that the following are representations of the delta function:
同樣的道理，任何峰值函數，以其下的單位面積規範化後，都會在峰值寬度歸零的極限內接近 delta 函數。我們將讓讀者自行證明以下是 delta 函數的表示：
δ ( x ) = Lim a → 0 1 π a x 2 + a 2 δ ( x ) = Lim a → 0 1 π a x 2 + a 2 delta(x)=Lim_(a rarr0)(1)/(pi)(a)/(x^(2)+a^(2))\delta(x)=\operatorname{Lim}_{a \rightarrow 0} \frac{1}{\pi} \frac{a}{x^{2}+a^{2}}
and 和
δ ( x ) = Lim a → ∞ α π e − α 2 x 2 δ ( x ) = Lim a → ∞ α π e − α 2 x 2 delta(x)=Lim_(a rarr oo)(alpha)/(sqrtpi)e^(-alpha^(2)x^(2))\delta(x)=\operatorname{Lim}_{a \rightarrow \infty} \frac{\alpha}{\sqrt{\pi}} e^{-\alpha^{2} x^{2}}
(d) We will have occasion to deal with orthonormal polynomials, which we denote by the general symbol P n ( x ) P n ( x ) P_(n)(x)P_{n}(x). These have the property that
(d) 我們將有機會處理正交多项式，我們以一般符號 P n ( x ) P n ( x ) P_(n)(x)P_{n}(x) 表示。它們的特性是
∫ d x P m ( x ) P n ( x ) w ( x ) = δ m n ∫ d x P m ( x ) P n ( x ) w ( x ) = δ m n int dxP_(m)(x)P_(n)(x)w(x)=delta_(mn)\int d x P_{m}(x) P_{n}(x) w(x)=\delta_{m n}
where w ( x ) w ( x ) w(x)w(x) may be unity or some simple function, called the weight function. For functions that may be expanded in a series of these orthogonal polynomials, we can write
其中 w ( x ) w ( x ) w(x)w(x) 可以是單一數或一些簡單的函數，稱為權值函數。對於可以在一系列這些正交多項式中展開的函數，我們可以寫成
f ( x ) = ∑ n a n P n ( x ) f ( x ) = ∑ n   a n P n ( x ) f(x)=sum_(n)a_(n)P_(n)(x)f(x)=\sum_{n} a_{n} P_{n}(x)
W-10 Supplement 2-A The Fourier Integral and Delta Functions
W-10 補充 2-A 傅立葉整數和三角函數
If we multiply both sides by w ( x ) P m ( x ) w ( x ) P m ( x ) w(x)P_(m)(x)w(x) P_{m}(x) and integrate over x x xx, we find that
如果我們將兩邊乘以 w ( x ) P m ( x ) w ( x ) P m ( x ) w(x)P_(m)(x)w(x) P_{m}(x) ，並對 x x xx 進行整合，我們會發現
a m = ∫ d y w ( y ) f ( y ) P m ( y ) a m = ∫ d y w ( y ) f ( y ) P m ( y ) a_(m)=int dyw(y)f(y)P_(m)(y)a_{m}=\int d y w(y) f(y) P_{m}(y)
We can insert this into (2A-27) and, prepared to deal with “generalized functions,” we freely interchange sum and integral. We get
我們可以將此插入 (2A-27)，準備處理「一般化函數」，我們可以自由交換總和及積分。我們得到
f ( x ) = ∑ n P n ( x ) ∫ d y w ( y ) f ( y ) P n ( y ) = ∫ d y f ( y ) ( ∑ n P n ( x ) w ( y ) P n ( y ) ) f ( x ) = ∑ n   P n ( x ) ∫ d y w ( y ) f ( y ) P n ( y ) = ∫ d y f ( y ) ∑ n   P n ( x ) w ( y ) P n ( y ) {:[f(x)=sum_(n)P_(n)(x)int dyw(y)f(y)P_(n)(y)],[=int dyf(y)(sum_(n)P_(n)(x)w(y)P_(n)(y))]:}\begin{aligned} f(x) & =\sum_{n} P_{n}(x) \int d y w(y) f(y) P_{n}(y) \\ & =\int d y f(y)\left(\sum_{n} P_{n}(x) w(y) P_{n}(y)\right) \end{aligned}
Thus we get still another representation of the delta function. Examples of the P n ( x ) P n ( x ) P_(n)(x)P_{n}(x) are Legendre polynomials, Hermite polynomials, and Laguerre polynomials, all of which make their appearance in quantum mechanical problems.
因此，我們得到了 delta 函數的另一種表示方法。 P n ( x ) P n ( x ) P_(n)(x)P_{n}(x) 的例子有 Legendre 多段式、Hermite 多段式和 Laguerre 多段式，這些都會在量子力學問題中出現。
Since the delta function always appears multiplied by a smooth function under an integral sign, we can give meaning to its derivatives. For example,
由於 delta 函數總是與平滑函數相乘出現在積分符號下，我們可以賦予其導數意義。舉例來說
∫ − ε ε d x f ( x ) d d x δ ( x ) = ∫ − ε ε d x d d x [ f ( x ) δ ( x ) ] − ∫ − ε ε d x d f ( x ) d x δ ( x ) = − ∫ − ε ε d x d f ( x ) d x δ ( x ) = − ( d f d x ) x = 0 ∫ − ε ε   d x f ( x ) d d x δ ( x ) = ∫ − ε ε   d x d d x [ f ( x ) δ ( x ) ] − ∫ − ε ε   d x d f ( x ) d x δ ( x ) = − ∫ − ε ε   d x d f ( x ) d x δ ( x ) = − d f d x x = 0 {:[int_(-epsi)^(epsi)dxf(x)(d)/(dx)delta(x)=int_(-epsi)^(epsi)dx(d)/(dx)[f(x)delta(x)]-int_(-epsi)^(epsi)dx(df(x))/(dx)delta(x)],[=-int_(-epsi)^(epsi)dx(df(x))/(dx)delta(x)],[=-((df)/(dx))_(x=0)]:}\begin{aligned} \int_{-\varepsilon}^{\varepsilon} d x f(x) \frac{d}{d x} \delta(x) & =\int_{-\varepsilon}^{\varepsilon} d x \frac{d}{d x}[f(x) \delta(x)]-\int_{-\varepsilon}^{\varepsilon} d x \frac{d f(x)}{d x} \delta(x) \\ & =-\int_{-\varepsilon}^{\varepsilon} d x \frac{d f(x)}{d x} \delta(x) \\ & =-\left(\frac{d f}{d x}\right)_{x=0} \end{aligned}
and so on. The delta function is an extremely useful tool, and the student will encounter it in every part of mathematical physics.
等等。delta 函數是非常有用的工具，學生在數學物理的每個部分都會遇到它。
The integral of a delta function is
三角函數的積分為
∫ − ∞ x d y δ ( y − a ) = 0 x < a = 1 x > a ≡ θ ( x − a ) ∫ − ∞ x   d y δ ( y − a ) = 0 x < a = 1 x > a ≡ θ ( x − a ) {:[int_(-oo)^(x)dy delta(y-a),=0,x < a],[,=1,x > a],[,-=theta(x-a)]:}\begin{array}{rlr} \int_{-\infty}^{x} d y \delta(y-a) & =0 & x<a \\ & =1 & x>a \\ & \equiv \theta(x-a) \end{array}
which is the standard notation for this discontinuous function. Conversely, the derivative of the so-called step function is the Dirac delta function:
是這個不連續函數的標準符號。相反地，所謂階梯函數的導數就是 Dirac delta 函數：
d d x θ ( x − a ) = δ ( x − a ) d d x θ ( x − a ) = δ ( x − a ) (d)/(dx)theta(x-a)=delta(x-a)\frac{d}{d x} \theta(x-a)=\delta(x-a)

Supplement 2-B 補編 2-B

A Brief Tutorial on Probability
概率簡易教程

In this supplement we give a brief discussion of probability. For simplicity we start with discrete events. Consider the toss of a six-faced die. If the die is not perfectly symmetric, the outcome for face n ( n = 1 , 2 , … , 6 ) n ( n = 1 , 2 , … , 6 ) n(n=1,2,dots,6)n(n=1,2, \ldots, 6) has a probability p n p n p_(n)p_{n}. In a large number N N NN of tosses, the face " n n nn " turns up a n a n a_(n)a_{n} times, and we say that the probability of getting the face n n nn is
在本補充篇中，我們將簡單討論概率。為了簡單起見，我們從離散事件開始。考慮擲六面骰子。如果骰子不是完全對稱，面 n ( n = 1 , 2 , … , 6 ) n ( n = 1 , 2 , … , 6 ) n(n=1,2,dots,6)n(n=1,2, \ldots, 6) 的結果有一個概率 p n p n p_(n)p_{n} 。在大量 N N NN 的擲骰中，面 " n n nn " 會出現 a n a n a_(n)a_{n} 次，我們說得到面 n n nn 的概率是
p n = a n N p n = a n N p_(n)=(a_(n))/(N)p_{n}=\frac{a_{n}}{N}
Since ∑ a n = N ∑ a n = N suma_(n)=N\sum a_{n}=N, it follows that
由於 ∑ a n = N ∑ a n = N suma_(n)=N\sum a_{n}=N ，因此可得
Σ p n = 1 Σ p n = 1 Sigmap_(n)=1\Sigma p_{n}=1
Let us now assign to each face a certain “payoff.” We assign points in the following manner: 1 point when face 1 turns up, 2 points when face 2 turns up, and so on. In N N NN tosses we get n n nn points a n a n a_(n)a_{n} times, so that the total number of points is Σ n a n Σ n a n Sigma na_(n)\Sigma n a_{n}. This, of course, grows with N N NN. We thus focus on the average value (points per toss), so that
現在讓我們為每個面孔分配一定的「報酬」。我們以下列方式分配點數：面 1 出現時得 1 分，面 2 出現時得 2 分，依此類推。在 N N NN 次擲中，我們得到 n n nn 分 a n a n a_(n)a_{n} 次，因此總分數是 Σ n a n Σ n a n Sigma na_(n)\Sigma n a_{n} 。這當然會隨著 N N NN 而成長。因此，我們將焦點放在平均值（每次擲的點數）上，所以
⟨ n ⟩ = 1 N ∑ n n a n = ∑ n n p n ⟨ n ⟩ = 1 N ∑ n   n a n = ∑ n   n p n (:n:)=(1)/(N)sum_(n)na_(n)=sum_(n)np_(n)\langle n\rangle=\frac{1}{N} \sum_{n} n a_{n}=\sum_{n} n p_{n}
We may be interested in an average of n 2 n 2 n^(2)n^{2}, say, and in that case we calculate
我們可能對 n 2 n 2 n^(2)n^{2} 的平均值感興趣，比方說，在這種情況下，我們計算
⟨ n 2 ⟩ = 1 N ∑ n n 2 a n = ∑ n n 2 p n n 2 = 1 N ∑ n   n 2 a n = ∑ n   n 2 p n (:n^(2):)=(1)/(N)sum_(n)n^(2)a_(n)=sum_(n)n^(2)p_(n)\left\langle n^{2}\right\rangle=\frac{1}{N} \sum_{n} n^{2} a_{n}=\sum_{n} n^{2} p_{n}
and so on. 等等。
When we do not have a discrete outcome, we must deal with densities. To be specific, consider a quantity that varies continuously, for example the height of a population of students. We can make this discrete by making a histogram (Fig. 2B-1) plotting the height by listing people’s heights in 10 − cm 10 − cm 10-cm10-\mathrm{cm} intervals. Nobody will be exactly 180 cm or 190 cm tall, so we just group people, and somehow round things up at the edges. We may want a finer detail, and list the heights in 1-cm intervals or 1-mm intervals, which will continue to make things discrete, but as the intervals become smaller, the histogram resembles more and more a continuous curve. Let us take some interval, d x d x dxd x, and treat it as infinitesimal. The number of people whose height lies between x x xx and x + d x x + d x x+dxx+d x is n ( x ) d x n ( x ) d x n(x)dxn(x) d x. The proportionality to d x d x dxd x is obvious: twice as many people will fall into the interval 2 d x 2 d x 2dx2 d x as fall into d x d x dxd x. It is here that the infinitesimal character of d x d x dxd x comes in: we do not need to decide whether n ( x ) d x n ( x ) d x n(x)dxn(x) d x or n ( x + d x / 2 ) d x n ( x + d x / 2 ) d x n(x+dx//2)dxn(x+d x / 2) d x is to be taken, since we treat ( d x ) 2 ( d x ) 2 (dx)^(2)(d x)^{2} as vanishingly small. If the total population has N N NN members, we can speak of the probability of falling into the particular interval as
當我們沒有離散的結果時，我們必須處理密度。具體來說，考慮一個連續變化的量，例如一群學生的身高。我們可以透過製作直方圖 (圖 2B-1)，以 10 − cm 10 − cm 10-cm10-\mathrm{cm} 間隔列出人們的身高，來繪製身高圖，從而使其離散化。沒有人的身高會正好是 180 公分或 190 公分，因此我們只需將人們分組，然後以某種方式將邊緣四捨五入。我們可能想要更細緻的細節，所以會以 1 公分或 1 公釐的間隔列出身高，這會繼續讓事情離散，但隨著間隔變小，直方圖就會越來越像連續曲線。讓我們取一些間隔， d x d x dxd x ，並將其視為無限小。身高介於 x x xx 和 x + d x x + d x x+dxx+d x 之間的人數是 n ( x ) d x n ( x ) d x n(x)dxn(x) d x 。與 d x d x dxd x 的比例關係顯而易見：落入 2 d x 2 d x 2dx2 d x 區間的人數是落入 d x d x dxd x 區間的人數的兩倍。 d x d x dxd x 的無限小特性就在此出現：我們無需決定應採用 n ( x ) d x n ( x ) d x n(x)dxn(x) d x 還是 n ( x + d x / 2 ) d x n ( x + d x / 2 ) d x n(x+dx//2)dxn(x+d x / 2) d x ，因為我們將 ( d x ) 2 ( d x ) 2 (dx)^(2)(d x)^{2} 視為極小。如果總人口有 N N NN 個成員，我們可以把落入特定區間的概率說成
1 N n ( x ) d x = p ( x ) d x 1 N n ( x ) d x = p ( x ) d x (1)/(N)n(x)dx=p(x)dx\frac{1}{N} n(x) d x=p(x) d x
W-12 Supplement 2-B A Brief Tutorial on Probability
W-12 補充 2-B 概率簡易教程
p ( x ) p ( x ) p(x)p(x) is called a probability density. Since the total number of students is N N NN, we have
p ( x ) p ( x ) p(x)p(x) 稱為概率密度。由於學生總數為 N N NN ，我們有
∫ d x n ( x ) = N ∫ d x n ( x ) = N int dxn(x)=N\int d x n(x)=N
or 或
∫ d x p ( x ) = 1 ∫ d x p ( x ) = 1 int dxp(x)=1\int d x p(x)=1
If we want the probability of finding a student of height between a a aa (say, 150 cm ) and b b bb (say, 180 cm ), we calculate
如果我們想要找到身高介於 a a aa (例如 150 cm) 和 b b bb (例如 180 cm) 之間的學生的概率，我們可以計算出
P ( a ≤ x ≤ b ) = ∫ a b d x p ( x ) P ( a ≤ x ≤ b ) = ∫ a b   d x p ( x ) P(a <= x <= b)=int_(a)^(b)dxp(x)P(a \leq x \leq b)=\int_{a}^{b} d x p(x)
The average height is calculated in the same way as for the discrete case
平均高度的計算方式與離散情況相同
⟨ x ⟩ = ∫ x p ( x ) d x ⟨ x ⟩ = ∫ x p ( x ) d x (:x:)=int xp(x)dx\langle x\rangle=\int x p(x) d x
and we can calculate other quantities such as
我們可以計算其他量，例如
⟨ x 2 ⟩ = ∫ x 2 p ( x ) d x x 2 = ∫ x 2 p ( x ) d x (:x^(2):)=intx^(2)p(x)dx\left\langle x^{2}\right\rangle=\int x^{2} p(x) d x
and so on. Instead of calling these quantities averages, we call them expectation values, a terminology that goes back to the roots of probability theory in gambling theory.
等等。我們不稱這些量為平均值，而是稱它們為期望值，這個名詞可以追溯到賭博理論中概率論的根源。
We are often interested in a quantity that gives a measure of how the heights, say, are distributed about the average. The deviations from the average must add up to zero. Formally it is clear that
我們通常會對一個數量感興趣，這個數量可以量度身高在平均值附近的分佈情況。與平均值的偏差加起來必須為零。形式上很明顯
⟨ x − ⟨ x ⟩ ⟩ = 0 ⟨ x − ⟨ x ⟩ ⟩ = 0 (:x-(:x:):)=0\langle x-\langle x\rangle\rangle=0
since the average of any number, including ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle, is just that number. We can, however, calculate the average value of the square of the deviation, and this quantity will not be zero. In fact,
因為任何數的平均值，包括 ⟨ x ⟩ ⟨ x ⟩ (:x:)\langle x\rangle 在內，都只是這個數。不過，我們可以計算偏差平方的平均值，這個數量不會是零。事實上
⟨ ( x − ⟨ x ⟩ ) 2 = ⟨ x 2 − 2 x ⟨ x ⟩ + ⟨ x ⟩ 2 ⟩ = ⟨ x 2 ⟩ − 2 ⟨ x ⟩ ⟨ x ⟩ + ⟨ x ⟩ 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = ( Δ x ) 2 ( x − ⟨ x ⟩ ) 2 = x 2 − 2 x ⟨ x ⟩ + ⟨ x ⟩ 2 = x 2 − 2 ⟨ x ⟩ ⟨ x ⟩ + ⟨ x ⟩ 2 = x 2 − ⟨ x ⟩ 2 = ( Δ x ) 2 {:[(:(x-(:x:))^(2):}=(:x^(2)-2x(:x:)+(:x:)^(2):)=(:x^(2):)-2(:x:)(:x:)+(:x:)^(2)],[=(:x^(2):)-(:x:)^(2)=(Delta x)^(2)]:}\begin{aligned} \left\langle(x-\langle x\rangle)^{2}\right. & =\left\langle x^{2}-2 x\langle x\rangle+\langle x\rangle^{2}\right\rangle=\left\langle x^{2}\right\rangle-2\langle x\rangle\langle x\rangle+\langle x\rangle^{2} \\ & =\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=(\Delta x)^{2} \end{aligned}
This quantity is called the dispersion.
此量稱為色散。
Figure 2B-1 圖 2B-1
Example of a histogram. 直方圖範例。
The dispersion is used in quantum mechanics to define the uncertainty, so that in the Heisenberg uncertainty relations proper definitions of ( Δ x ) 2 ( Δ x ) 2 (Delta x)^(2)(\Delta x)^{2} and ( Δ p ) 2 ( Δ p ) 2 (Delta p)^(2)(\Delta p)^{2} are given by
在量子力學中，分散被用來定義不確定性，因此在海森堡不確定性關係中， ( Δ x ) 2 ( Δ x ) 2 (Delta x)^(2)(\Delta x)^{2} 和 ( Δ p ) 2 ( Δ p ) 2 (Delta p)^(2)(\Delta p)^{2} 的適當定義為
( Δ x ) 2 = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 ( Δ x ) 2 = x 2 − ⟨ x ⟩ 2 (Delta x)^(2)=(:x^(2):)-(:x:)^(2)(\Delta x)^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}
and 和
( Δ p ) 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 ( Δ p ) 2 = p 2 − ⟨ p ⟩ 2 (Delta p)^(2)=(:p^(2):)-(:p:)^(2)(\Delta p)^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}
There is one more point to be made that is directly relevant to quantum physics. We may illustrate it by going back to the population of students. Let us ask for the probability that a randomly chosen student’s height lies between 160 cm and 170 cm , and that his or her telephone number ends in an even integer. Assuming that there is nothing perverse about how telephone numbers are distributed in the community, half the students will fall into the even category, and thus the desired probability is P ( 160 ≤ x ≤ 170 ) × ( 1 / 2 ) P ( 160 ≤ x ≤ 170 ) × ( 1 / 2 ) P(160 <= x <= 170)xx(1//2)P(160 \leq x \leq 170) \times(1 / 2). This is just an example of a general rule that the probability of two (or more) uncorrelated events is the product of the individual probabilities.
還有一點與量子物理直接相關。我們可以回到學生人口來說明。讓我們詢問隨機選出的學生身高介於 160 cm 和 170 cm 之間，以及他或她的電話號碼以偶數整數結尾的概率。假設電話號碼在社區中的分佈沒有任何不正常的地方，一半的學生會屬於偶數類別，因此期望的概率是 P ( 160 ≤ x ≤ 170 ) × ( 1 / 2 ) P ( 160 ≤ x ≤ 170 ) × ( 1 / 2 ) P(160 <= x <= 170)xx(1//2)P(160 \leq x \leq 170) \times(1 / 2) 。這只是一般規則的一個例子，即兩個（或更多）不相關事件的概率是個別概率的乘積。
Thus, if the probability that a particle is in some state " n n nn " in our laboratory is P ( n ) P ( n ) P(n)P(n), and the probability that a different particle in a laboratory across the country is in a state " m m mm " is P ( m ) P ( m ) P(m)P(m), then the joint probability that one finds " n n nn " in the first laboratory and " m m mm " in the second laboratory is
因此，如果在我們的實驗室中，粒子處於某種狀態 " n n nn 「 的概率是 P ( n ) P ( n ) P(n)P(n) ，而在國外實驗室中，不同粒子處於某種狀態 」 m m mm 「 的概率是 P ( m ) P ( m ) P(m)P(m) ，那麼在第一個實驗室中發現 」 n n nn 「 和在第二個實驗室中發現 」 m m mm " 的共同概率是
P ( m , n ) = P ( n ) P ( m ) P ( m , n ) = P ( n ) P ( m ) P(m,n)=P(n)P(m)P(m, n)=P(n) P(m)
We will find that a similar result holds for probability amplitudes-that is, for wave functions.
我們會發現，類似的結果也適用於概率振幅，也就是波函数。
  1. 1 1 ^(1){ }^{1} We do not consider the wave form B ( k ) exp ⁡ [ − i ( k x − ω t ) ] B ( k ) exp ⁡ [ − i ( k x − ω t ) ] B(k)exp[-i(kx-omega t)]B(k) \exp [-i(k x-\omega t)] because a term proportional to exp ⁡ [ i ω t ] exp ⁡ [ i ω t ] exp[i omega t]\exp [i \omega t] will be seen to be associated with a negative kinetic energy.
    1 1 ^(1){ }^{1} 我們不考慮波形 B ( k ) exp ⁡ [ − i ( k x − ω t ) ] B ( k ) exp ⁡ [ − i ( k x − ω t ) ] B(k)exp[-i(kx-omega t)]B(k) \exp [-i(k x-\omega t)] ，因為與 exp ⁡ [ i ω t ] exp ⁡ [ i ω t ] exp[i omega t]\exp [i \omega t] 成比例的項目會被視為與負動能相關。
  2. 3 3 ^(3){ }^{3} In this example, we have worked with orders of magnitude, so that we have ignored factors of 2 π 2 π 2pi2 \pi in the difference between h h hh and ℏ ℏ ℏ\hbar.
    3 3 ^(3){ }^{3} 在這個範例中，我們使用的是數量級，因此在 h h hh 和 ℏ ℏ ℏ\hbar 之間的差異中，我們忽略了 2 π 2 π 2pi2 \pi 的因數。