Question 1.8: Suppose you add a constant V0 to the potential energy (by “c...

Suppose you add a constant V_0 to the potential energy (by “constant” I mean independent of x as well as t). In classical mechanics this doesn’t change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor. \exp(-iV_0t/\hbar ) .What effect does this have on the expectation value of a dynamical variable?

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Suppose Ψ satisfies the Schrödinger equation without V_0 : i \hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\partial x^{2}}+V \Psi . We want to find the solution Ψ_0 with V_0 : i \hbar \frac{\partial \Psi_0}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi_0}{\partial x^{2}}+(V+V_0) \Psi_0 .

Claim: Ψ_0 = Ψ e^{-i V_0 t/\hbar} .

Proof: i \hbar \frac{\partial \Psi_{0}}{\partial t}=i \hbar \frac{\partial \Psi}{\partial t} e^{-i V_{0} t / \hbar}+i \hbar \Psi\left(-\frac{i V_{0}}{\hbar}\right) e^{-i V_{0} t / \hbar}=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\partial x^{2}}+V \Psi\right] e^{-i V_{0} t / \hbar}+V_{0} \Psi e^{-i V_{0} t / \hbar}

= -\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi_{0}}{\partial x^{2}}+\left(V+V_{0}\right) \Psi_{0} .  QED

This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq. 1.36.

\langle Q(x, p))=\int \Psi^{*}[Q(x,-i \hbar \partial / \partial x)] \Psi d x     (1.36).

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