A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states: Ψ(x,0) = A[ψ1(x) + ψ2(x)]. (a) Normalize Ψ(x,0) .(That is, find A. This is very easy, if you exploit the orthonormality of ψ1 and ψ2. Recall that, having normalized Ψ at t = 0,

Question

A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states: [latex] Ψ(x,0) = A[ψ_1(x) + ψ_2(x)] [/latex].

(a) Normalize Ψ(x,0) .(That is, find A. This is very easy, if you exploit the orthonormality of [latex] ψ_1 [/latex] and [latex] ψ_2 [/latex]. Recall that, having normalized Ψ at t = 0,you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).)

(b) Find Ψ(x ,t) and [latex] \left|Ψ(x,t)\right| ^2 [/latex]. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let [latex] ω ≡ \pi ^2\hbar /2ma^2 [/latex].

(c) Compute [latex] \left\langle x\right\rangle [/latex]. Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than a/2 , go directly to jail.)

(d) Compute [latex] \left\langle p\right\rangle [/latex]. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”)

(e) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of H. How does it compare with [latex] E_1 [/latex] and [latex] E_2 [/latex]?

Accepted Answer

(a)

[latex] |\Psi|^{2}=\Psi^{*} \Psi=|A|^{2}\left(\psi_{1}^{*}+\psi_{2}^{*} ight)\left(\psi_{1}+\psi_{2} ight)=|A|^{2}\left[\psi_{1}^{*} \psi_{1}+\psi_{1}^{*} \psi_{2}+\psi_{2}^{*} \psi_{1}+\psi_{2}^{*} \psi_{2} ight] [/latex].

[latex] 1=\int|\Psi|^{2} d x=|A|^{2} \int\left[\left|\psi_{1} ight|^{2}+\psi_{1}^{*} \psi_{2}+\psi_{2}^{*} \psi_{1}+\left|\psi_{2} ight|^{2} ight] d x=2|A|^{2} \Rightarrow A= 1/\sqrt{2} [/latex].

(b)

[latex] \Psi(x, t)=\frac{1}{\sqrt{2}}\left[\psi_{1} e^{-i E_{1} t / \hbar}+\psi_{2} e^{-i E_{2} t / \hbar} ight] [/latex] (but [latex] \frac{E_n}{\hbar} = n^2 ω) [/latex].

[latex] =\frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}\left[\sin \left(\frac{\pi}{a} x ight) e^{-i \omega t}+\sin \left(\frac{2 \pi}{a} x ight) e^{-i 4 \omega t} ight] = \frac{1}{\sqrt{a}} e^{-i \omega t}\left[\sin \left(\frac{\pi}{a} x ight)+\sin \left(\frac{2 \pi}{a} x ight) e^{-3 i \omega t} ight] [/latex].

[latex] |\Psi(x, t)|^{2}=\frac{1}{a}\left[\sin ^{2}\left(\frac{\pi}{a} x ight)+\sin \left(\frac{\pi}{a} x ight) \sin \left(\frac{2 \pi}{a} x ight)\left(e^{-3 i \omega t}+e^{3 i \omega t} ight)+\sin ^{2}\left(\frac{2 \pi}{a} x ight) ight] [/latex].

= [latex] \frac{1}{a}\left[\sin ^{2}\left(\frac{\pi}{a} x ight)+\sin ^{2}\left(\frac{2 \pi}{a} x ight)+2 \sin \left(\frac{\pi}{a} x ight) \sin \left(\frac{2 \pi}{a} x ight) \cos (3 \omega t) ight] [/latex].

(c)

[latex] \langle x angle=\int x|\Psi(x, t)|^{2} d x [/latex].

[latex] =\frac{1}{a} \int_{0}^{a} x\left[\sin ^{2}\left(\frac{\pi}{a} x ight)+\sin ^{2}\left(\frac{2 \pi}{a} x ight)+2 \sin \left(\frac{\pi}{a} x ight) \sin \left(\frac{2 \pi}{a} x ight) \cos (3 \omega t) ight] d x [/latex].

[latex] \int_{0}^{a} x \sin ^{2}\left(\frac{\pi}{a} x ight) d x=\left.\left[\frac{x^{2}}{4}-\frac{x \sin \left(\frac{2 \pi}{a} x ight)}{4 \pi / a}-\frac{\cos \left(\frac{2 \pi}{a} x ight)}{8(\pi / a)^{2}} ight] ight|_{0} ^{a}=\frac{a^{2}}{4}=\int_{0}^{a} x \sin ^{2}\left(\frac{2 \pi}{a} x ight) d x [/latex].

[latex] \int_{0}^{a} x \sin \left(\frac{\pi}{a} x ight) \sin \left(\frac{2 \pi}{a} x ight) d x=\frac{1}{2} \int_{0}^{a} x\left[\cos \left(\frac{\pi}{a} x ight)-\cos \left(\frac{3 \pi}{a} x ight) ight] d x [/latex].

[latex] =\frac{1}{2}\left[\frac{a^{2}}{\pi^{2}} \cos \left(\frac{\pi}{a} x ight)+\frac{a x}{\pi} \sin \left(\frac{\pi}{a} x ight)-\frac{a^{2}}{9 \pi^{2}} \cos \left(\frac{3 \pi}{a} x ight)-\frac{a x}{3 \pi} \sin \left(\frac{3 \pi}{a} x ight) ight]_{0}^{a} [/latex].

[latex] =\frac{1}{2}\left[\frac{a^{2}}{\pi^{2}}(\cos (\pi)-\cos (0))-\frac{a^{2}}{9 \pi^{2}}(\cos (3 \pi)-\cos (0)) ight]=-\frac{a^{2}}{\pi^{2}}\left(1-\frac{1}{9} ight)=-\frac{8 a^{2}}{9 \pi^{2}} [/latex].

[latex] herefore\langle x angle=\frac{1}{a}\left[\frac{a^{2}}{4}+\frac{a^{2}}{4}-\frac{16 a^{2}}{9 \pi^{2}} \cos (3 \omega t) ight] = \frac{a}{2}\left[1-\frac{32}{9 \pi^{2}} \cos (3 \omega t) ight] [/latex].

Amplitude: [latex] \frac{32}{9 \pi^{2}}\left(\frac{a}{2} ight)=0.3603(a / 2) [/latex]; angular frequency: [latex] 3 \omega=\frac{3 \pi^{2} \hbar}{2 m a^{2}} [/latex].

(d)

[latex] \langle p angle =m \frac{d\langle x angle}{d t}=m\left(\frac{a}{2} ight)\left(-\frac{32}{9 \pi^{2}} ight)(-3 \omega) \sin (3 \omega t)= \frac{8 \hbar}{3 a} \sin (3 \omega t) [/latex].

(e) You could get either [latex] E_{1}=\pi^{2} \hbar^{2} / 2 m a^{2} [/latex] or [latex] E_{2} = 2 \pi^{2} \hbar^{2} / m a^{2} [/latex], with equal probability [latex] P_1 = P_2 = 1/2 [/latex].

So [latex] \langle H angle = [latex] \frac{1}{2}\left(E_{1}+E_{2} ight)=\frac{5 \pi^{2} \hbar^{2}}{4 m a^{2}} [/latex] ; it's the average of [latex] E_1 [/latex] and [latex] E_2 [/latex].

Question 2.5: A particle in the infinite square well has as its initial wa...

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