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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}