Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of and in Problem 2.5: Ψ(x,0)=

Question

Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of and in Problem 2.5:

[latex] \Psi(x, t)=\sum_{n=1}^{\infty} c_{n} \psi_{n}(x) e^{-i E_{n} t / \hbar}=\sum_{n=1}^{\infty} c_{u} \Psi_{n}(x, t) [/latex] (2.17).

[latex] \Psi(x, 0)=A\left[\psi_{1}(x)+e^{i \phi} \psi_{2}(x)\right] [/latex],

where ϕ is some constant. Find Ψ(x,t) , [latex] \left|Ψ(x,t) \right| ^2[/latex] and [latex] \left\langle x\right\rangle [/latex] and compare your results with what you got before. Study the special cases ϕ = 2 π and ϕ = π. (For a graphical exploration of this problem see the applet in footnote 9 of this chapter.)

Accepted Answer

From Problem 2.5, we see that

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

[latex] \left|Ψ(x,t) \right| ^2 = \frac{1}{a}\left[\sin ^{2}\left(\frac{\pi}{a} x\right)+\sin ^{2}\left(\frac{2 \pi}{a} x\right)+2 \sin \left(\frac{\pi}{a} x\right) \sin \left(\frac{2 \pi}{a} x\right) \cos (3 \omega t-\phi)\right] [/latex].

and hence [latex] \langle x\rangle=\frac{a}{2}\left[1-\frac{32}{9 \pi^{2}} \cos (3 \omega t-\phi)\right] [/latex].

This amounts physically to starting the clock at a different time (i.e., shifting the t = 0 point).

If [latex] ϕ = \frac{π}{2} [/latex], so [latex] \Psi(x, 0)=A\left[\psi_{1}(x)+i \psi_{2}(x)\right] [/latex], then [latex] \cos (3 \omega t-\phi)=\sin (3 \omega t) [/latex]; [latex] \left\langle x\right\rangle [/latex] starts at [latex] \frac{a}{2} [/latex].

If ϕ = π, so [latex] \Psi(x, 0)=A\left[\psi_{1}(x)-\psi_{2}(x)\right] [/latex], then [latex] \cos (3 \omega t-\phi)=-\cos (3 \omega t) [/latex]; [latex] \left\langle x\right\rangle [/latex] starts at [latex] \frac{a}{2}\left(1+\frac{32}{9 \pi^{2}}\right) [/latex].

Question 2.6: Although the overall phase constant of the wave function is ...

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