Inner product of two wave functions is constant in time

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.16.

The fact that the normalization of the wave function is constant over time is actually a special case of a more general theorem, which is

\displaystyle  \frac{d}{dt}\int_{-\infty}^{\infty}\Psi_{1}^*\Psi_{2}dx=0 \ \ \ \ \ (1)

for any two normalizable solutions to the Schrödinger equation (with the same potential). The proof of this follows a similar derivation to that in section 1.4 of Griffiths’s book.

The derivative in the integrand is (where we’re using a subscript {t} or {x} to denote a derivative with respect that variable):

\displaystyle   \frac{\partial}{\partial t}\left(\Psi_{1}^*\Psi_{2}\right) \displaystyle  = \displaystyle  \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t} \ \ \ \ \ (2)

From the Schrödinger equation

\displaystyle   \Psi_{2t} \displaystyle  = \displaystyle  i\frac{\hbar}{2m}\Psi_{2xx}-\frac{i}{\hbar}V\Psi_{2}\ \ \ \ \ (3)
\displaystyle  \Psi_{1t}^* \displaystyle  = \displaystyle  -i\frac{\hbar}{2m}\Psi_{1xx}^*+\frac{i}{\hbar}V\Psi_{1}^*\ \ \ \ \ (4)
\displaystyle  \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t} \displaystyle  = \displaystyle  i\frac{\hbar}{2m}\left(-\Psi_{1xx}^*\Psi_{2}+\Psi_{2xx}\Psi_{1}^*\right)+\frac{i}{\hbar}V\left(\Psi_{1}^*\Psi_{2}-\Psi_{1}^*\Psi_{2}\right)\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  i\frac{\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi_{2x}\Psi_{1}^*-\Psi_{1x}^*\Psi_{2}\right) \ \ \ \ \ (6)

Inserting this into 1 and integrating gives zero because all wave functions go to zero at infinity. [Of course, the theorem doesn’t hold if {\Psi_{1}} and {\Psi_{2}} are solutions for different potentials, because in that case the potential term wouldn’t cancel out in 5.]

One thought on “Inner product of two wave functions is constant in time

  1. Kamila

    Hello Sir, I really appreciate your work, but could you please tell something about the physical meaning of this theorem? I cannot find it anywhere…

    Reply

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