Time partial derivative of a wave function

1. Oct 2, 2012

NeroKid

the probability of finding particle is a constant with time <ψ|$\partialψ$/$\partial(t)$> = -<$\partialψ$/$\partial(t)$|ψ> , the equation holds for all ψ so the time derivative operator is an anti-hermitian operator, but then consider any hermitian operator A, the rate of change of A is d(<ψ|Aψ>)/dt = <$\partialψ$/$\partial(t)$|Aψ>+<ψ|$\partial(Aψ)$/$\partial(t)$> , interchange the anti hermitian operator for the first one the equation equals to 0 which means every observable is conserved in all situation and it's clearly wrong, so where did I think wrong? Please show me

Last edited: Oct 2, 2012
2. Oct 3, 2012

dextercioby

What your derivation is trying to say is that the time derivative of the expectation value of an observable in an arbitray pure state is time independent iff the commutator between the hamiltonian and the observable in that pure state is 0, thing which is a pretty standard result of the theory.

3. Oct 3, 2012

NeroKid

yes but that is what after you change frome time derivative to hamiltonian , what i was trying to say is that because the time derivative is anti-hermitian it has the form < f|Ag> = - <Af|g> so any observable is conserved

4. Oct 4, 2012

5. Oct 4, 2012

Demystifier

NeroKid,
In your attempt to prove that time derivative is an anti-hermitian operator on the Hilbert space, you tacitly take for granted that time derivative is at least an operator on the Hilbert space. However, that is wrong. The time derivative is not an operator on the Hilbert space. This is because states in the Hilbert space are functions of x, not functions of x and t. The parameter t is an external parameter, not associated with the Hilbert space.

If you are still confused, try to work out your argument with x instead of t. Do you see a difference?

Last edited: Oct 4, 2012
6. Oct 4, 2012

NeroKid

ok I see now tks a lot