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.

6 thoughts on “Adding a constant to the potential introduces a phase factor

  1. Cindy

    you solve this problem by using ‘time-independent Schrödinger equation’
    but in this question the author didn’t point out that V is indep. of time
    is that a precondition that the author didn’t point out??

    1. Chris

      Magic happens. Take for example a simple case in which we have a potential of the form V(x,t) = V(x) + D*B(x)*C(t), where D is a “sufficiently small constant”, where C(t) is 0 until time = 0, at which point it becomes any arbitrary function of time. At time = 0 your wave function can be expressed as a linear combination of coefficients times energy eigenstates of the hamiltonian with V(x) (since at that point the time dependence has just turned on and has had no time to act). You can use the magic of linear algebra to show that these coefficients become coefficients of time, in otherwords time dependence in this case causes the coefficients to vary in time. The differential equation that governs this time dependence is an INFINITE DIMENSIONAL array of first order differential equations, with each equation involving all the other coefficients, so it is coupled to infinite order! This is often simplified though since multiplied by each coefficient in each differential equation is a corresponding matrix element. For example I believe the rate of change in the first coefficient for the 1st energy eigenstate goes something like C1(t)*e^i(E1-E1)t*c(t)* + C2(t)*e^i(E2-E1)t*c(t)* + C3(t)*e^(i(E3-E1)t)*c(t)*…. etc. This often reduces to an infinite number of differential equations with a finite number of terms governing each ones time evolution, since the matrix elements are often 0 for the terms.

  2. John

    You solved the question using content covered in chapter 2, but this question is in chapter 1 – is there a way to solve it using only the content from chapter 1?

    1. gwrowe Post author

      Since Griffiths essentially gives the answer in the question, you could just plug this into the Schrodinger equation and check that it works.


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