The Time Independence of Normalization

[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

 

[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.

 

[atyy]

For a time-independent Hamiltonian, it is legitimate to just demonstrate that exp(-iHt) is unitary, where H is the Hamiltonian operator.

http://ocw.mit.edu/courses/nuclear-...ng-2012/lecture-notes/MIT22_02S12_lec_ch6.pdf (see section 6.1.2)

 

[Jorriss]

For a time-independent Hamiltonian, it is legitimate to just demonstrate that exp(-iHt) is unitary, where H is the Hamiltonian operator.

http://ocw.mit.edu/courses/nuclear-...ng-2012/lecture-notes/MIT22_02S12_lec_ch6.pdf (see section 6.1.2)

While this is true, Griffiths does not introduce unitary operators, the time evolution operator, etc I believe and he certainly does not do it by page 13.

 

[stevendaryl]

While this is true, Griffiths does not introduce unitary operators, the time evolution operator, etc I believe and he certainly does not do it by page 13.

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.