# Wavefunction Normalization at Different Times

1. Feb 2, 2014

### user3

In Introduction to Quantum Mechanics by Griffith, when he is normalizing a wave function that's dependent on both x and t, he lets t=0 , and solves for the constant (A). But if the integration of ψ^2 at any time t is 1, then is it correct to let t = 2, for instance, instead of 0 and solve for A? If yes, that would mean that A could be an infinite number of different A's, and that would be wrong because we would get different values for expectation values of position for every constant A. And if No, explain why please.

2. Feb 2, 2014

### stevendaryl

Staff Emeritus
Since the normalization is a constant over time, it doesn't matter which moment you choose to normalize it -- you'll get the same normalization constant.

3. Feb 2, 2014

### user3

If you normalize this wave function at t=0 and at t=2, you don't get the same constant:
ψ=Ae^[ -a[(mx^2)/hbar + i t] ] where m, a ,and A are positive real constants.

at t = 0 , you get A=(2am/hbar pi) ^1/4 (solution)

at t=2 , I got A = e^4ai * (2am/hbar pi) ^1/4

if i hadn't done a stupid mistake in the calculations.

4. Feb 2, 2014

### stevendaryl

Staff Emeritus
The only thing that matters for normalization constant $A$ is the absolute value, $|A|$. The phase can be chosen arbitrarily.

5. Feb 2, 2014

### Staff: Mentor

This is correct.

This is incorrect. Hint: write your wave function in the form
$$\Psi(x,t) = Ae^{-amx^2/\hbar}e^{-iat}$$
which might make things slightly more obvious.

When you calculate $|\Psi|^2 = \Psi^*\Psi$, what happens to the factor that contains t? Remember that $\Psi^*$ means complex conjugate of $\Psi$.

Last edited: Feb 2, 2014
6. Feb 2, 2014

### user3

but what happens if the "i" was not there: ψ=Ae^[ -a[(mx^2)/hbar + t] ] ?

7. Feb 2, 2014

### Dr Transport

That is not an eigen-function of the Schrodinger equation....

8. Feb 2, 2014

Staff Emeritus
Then it's a mistake. (It's not a phase)

9. Feb 2, 2014

### atyy

To choose the normalization, fix the time (any time will do). Even at any fixed time, there are an infinite number of possible normalizations, just different by $\exp(i\theta)$, where $\theta$ is a constant. These different normalizations are physically equivalent, just like in electrostatics the zero of potential is arbitrary. So just pick one.

Schroedinger's equation will take care that the wave function at other times remains correct.

Last edited: Feb 2, 2014