Question

Suppose you add a constant${\mathbf{V}}_{\mathbf{0}}$ to the potential energy (by “constant” I mean independent ofxas 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:$\mathbf{exp}\mathbf{(}\mathbf{-}{\mathbf{iV}}_{\mathbf{0}}\mathbf{t}\mathbf{/}\sout{\mathbf{h}}\mathbf{)}$. What effect does this have on the expectation value of a dynamical variable?

Short answer

This constant will give the wave function a time-dependent phase factor. However, it will have no effect on the dynamical variables' expected values.

Step-by-step solution

The Schrödinger equation

The Schrödinger equation is given by,

$\mathbf{i}\sout{\mathbf{h}}\frac{\mathbf{\partial }\mathbf{\psi }}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\frac{\mathbf{-}\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{\psi }}{\mathbf{\partial }{\mathbf{X}}^{\mathbf{2}}}\mathbf{+}\mathbf{V\psi }$ …… (1)

This equation is without a constant V₀.

Schrödinger equation after adding constant V0

If a constant is added to the potential energy, the Schrödingerequation will be,

$\mathrm{i}\sout{\mathrm{h}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=\frac{-\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{X}}^2}+(\mathrm{V}+{\mathrm{V}}_0)\mathrm{\psi }$ …… (2)

A time-dependent phase factor will be picked up by the wave function that is ${\mathrm{e}}^{-{\mathrm{N}}_0\mathrm{t}/\sout{\mathrm{h}}}$. For doing the calculation a new wave function $\varphi$will be taken that is equal to the old wavefunction which is multiplied with the phase factor.

$\mathrm{\Phi }={\mathrm{\psi e}}^{-{\mathrm{iv}}_0\mathrm{t}/\sout{\mathrm{h}}}$

Replacement of first equation with wave function

Now, equation (1) will be replaced with this new wave function (the originalSchrödingerequation before adding the constant to the potential energy). So, the equation will be:

$$\begin{gathered} i\sout{h}\frac{\partial \Phi }{\partial t}=\frac{-\sout{h}}{2m}\frac{{\partial }^2\Phi }{\partial X^2}+V\Phi \\ i\sout{h}\frac{\partial \Phi }{\partial t}\left(\psi e^{-i{v}_{0}t/\sout{h}}\right)=\frac{-\sout{h}}{2m}\frac{{\partial }^2}{\partial X^2}\left(\psi e^{-i{v}_{0}t/\sout{h}}\right)+V\left(\psi e^{-i{v}_{0}t/\sout{h}}\right) \\ i\sout{h}\frac{\partial \psi }{\partial t}{e}^{-i{v}_{0}t/\sout{h}}+\psi V_0{e}^{-i{v}_{0}t/\sout{h}}=\frac{-\sout{h}}{2m}\frac{{\partial }^2\psi }{\partial X^2}{e}^{-i{v}_{0}t/\sout{h}}+V\psi e^{-i{v}_{0}t/\sout{h}} \\ i\sout{h}\frac{\partial \psi }{\partial t}=\frac{-\sout{h}}{2m}\frac{{\partial }^2\psi }{\partial X^2}+V\psi -\psi V_0 \\ i\sout{h}\frac{\partial \psi }{\partial t}=\frac{-\sout{h}}{2m}\frac{{\partial }^2\psi }{\partial X^2}+(V+V_0)\psi \end{gathered}$$

Calculating the expectation value of a particle using the new potential energy

The negative sign is absorbed into the constant in this case. The extra phase factor will also have no effect on the dynamical variables' expected values.

Because the dynamical variables do not depend on time (just on position), the extra phase component will have no effect on the dynamical variables' expected value.

Hence, it’s the same as the old potential energy.