# Adding a constant to the potential introduces a phase factor

Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers

Required physics: Schrödinger equation, probability density

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.8.

The time-independent Schrödinger equation in one dimension can be separated into two equations as follows:

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$ $\displaystyle =$ $\displaystyle E\psi(x)\ \ \ \ \ (1)$ $\displaystyle i\hbar\frac{d\Xi(t)}{dt}$ $\displaystyle =$ $\displaystyle E\Xi(t) \ \ \ \ \ (2)$

and the general solution is

$\displaystyle \Psi\left(x,t\right)=\psi\left(x\right)\Xi\left(t\right) \ \ \ \ \ (3)$

The time component can be solved as

$\displaystyle \Xi\left(t\right)=Ce^{-iEt/\hbar} \ \ \ \ \ (4)$

where ${C}$ is the constant of integration.

If we add a constant (in both space and time) ${V_{0}}$ to the potential, then the original Schrödinger equation becomes

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\Psi}{dx^{2}}+V(x)\Psi+V_{0}\Psi$ $\displaystyle =$ $\displaystyle i\hbar\frac{\partial\Psi}{\partial t}\ \ \ \ \ (5)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\Psi}{dx^{2}}+V(x)\Psi$ $\displaystyle =$ $\displaystyle i\hbar\frac{\partial\Psi}{\partial t}-V_{0}\Psi \ \ \ \ \ (6)$

Applying separation of variables gives us

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{1}{\psi(x)}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)$ $\displaystyle =$ $\displaystyle E\ \ \ \ \ (7)$ $\displaystyle i\hbar\frac{1}{\Xi(t)}\frac{\partial\Xi}{\partial t}-V_{0}$ $\displaystyle =$ $\displaystyle E \ \ \ \ \ (8)$

[Since ${V_{0}}$ is independent of both ${x}$ and ${t}$, we could put it in either the ${\psi\left(x\right)}$ or the ${\Xi\left(t\right)}$ equation, but putting it in the ${\Xi}$ equation eliminates it from the more complex ${\psi}$ equation, so we’ll do that.]

The solution to 8 is now

$\displaystyle \Xi\left(t\right)=Ce^{-i\left(E+V_{0}\right)t/\hbar} \ \ \ \ \ (9)$

so we’ve introduced a phase factor ${e^{-iV_{0}t/\hbar}}$ into the overall wave function ${\Psi}$. For the time-independent Schrödinger equation, all quantities of physical interest involve multiplying the complex conjugate ${\Psi^*}$ by some operator ${\hat{Q}\left(x\right)}$ that depends only on ${x}$, operating on ${\Psi}$. That is, we’re interested only in quantities of the form

 $\displaystyle \Psi^*\left[\hat{Q}\left(x\right)\Psi\right]$ $\displaystyle =$ $\displaystyle \left|C\right|^{2}e^{+i\left(E+V_{0}\right)t/\hbar}e^{-i\left(E+V_{0}\right)t/\hbar}\psi^*\left[\hat{Q}\left(x\right)\psi\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|C\right|^{2}\psi^*\left[\hat{Q}\left(x\right)\psi\right] \ \ \ \ \ (11)$

Thus the phase factor disappears when calculating any physical quantity.