# The Time Independence of Normalization

1. Jun 8, 2014

### kq6up

On page 13 of Griffith's "Introduction to Quantum Mechanics 2nd ed" David goes into a long (relatively speaking) proof of why a normalized pair of quantum state vectors will not at some time later become "un-normalized". It seems like just putting the Psi's in a braket the e^(-it) "time dependence" term would just cancel out -- showing that normalization is not time dependent. Could anyone take a shot at why this is not the case -- other than the fact he has not developed braket notation or shown that the time dependence a separate exponent factor? Maybe I just answered my own question :D Is there anything wrong with showing it this way?

Thanks,
Chris Maness

2. Jun 8, 2014

### aim1732

The general wave function is built of a linear combination of the eigenfunctions. When you do this, the exponential time-dependent phases do not cancel out. When he shows that the normalization remains constant in time, he shows it for a general wave function that is a solution of just the Schrodinger equation, and not the eigenvalue equation in general.

3. Jun 8, 2014

4. Jun 9, 2014

5. Jun 9, 2014

### stevendaryl

Staff Emeritus
Well, it's enough to know that

$\int \psi^* H \psi dx$

is real. That's because Schrodinger's equation implies that:

$\dfrac{d}{dt} \int \psi^* \psi dx = \dfrac{2}{\hbar} Im(\int \psi^* H \psi dx)$

where $Im$ means the imaginary part.