# Hermitian adjoint of the time derivative?

1. Feb 11, 2008

### leright

So I had a QM test today and I needed to show that the energy operator is hermitian. This was easy to show provided that the the adjoint of d/dt is -d/dt. I know this is the case for the spatial derivative but is it the case with the time derivative? The bra-ket is an integral over x not time so the proof isn't clear to me.

2. Feb 11, 2008

### jostpuur

The time derivative isn't an operator in the same sense as many other operators. If you have some fixed state vector $|\psi\rangle$, you cannot calculate $\partial_t|\psi\rangle$. It's not operator like that. I think, in the exam, you should have shown that the Hamiltonian operator is Hermitian.

3. Feb 11, 2008

### leright

wel, that's what the exam asked....it asked to show that ihbar(d/dt) is hermitian. To show that you need to assume d/dt is anti hermitian, right? oh, I see....the hamiltonian operator can take a time derivative form and a spatial derivative form...ha

4. Feb 11, 2008

### jostpuur

So was is the energy operator, or the $i\hbar\partial_t$? The energy operator would have been the Hamiltonian $H$, that depends on the system.

5. Feb 11, 2008

### cks

Can I prove this way?

$$<\phi|H|\phi> = <\phi|E|\phi> = <\phi|E^*|\phi> = <\phi|H^+|\phi>$$

So

$$H=H^+$$

6. Feb 11, 2008

### leright

that is the energy operator. the hamiltonian can take the form of a time derivative or a second spatial derivative....I think.

7. Feb 11, 2008

### jostpuur

When $|\psi(t)\rangle$, a state vector that is a function of time, satisfies the Schrödinger's equation

$$i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle,$$

then, by definition, the operators $i\hbar\partial_t$ and $H$ will have the same effect on any such vector, but in general these are different operators. They must be of course, since if they were the same, the SE would have no content about the time evolution of the state!

In other words, if $|\xi(t)\rangle$ is some arbitrary parametrization of a path in the state space, then you can have

$$i\hbar\partial_t|\xi(t)\rangle \neq H|\xi(t)\rangle.$$

So, they are different operators, and it is correct to write

$$i\hbar\partial_t \neq H.$$

Last edited: Feb 11, 2008
8. Aug 6, 2008

### Domnu

Sorry to revive an old thread, but would this proof work?

We know that

$$\hat{H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)$$

Now, clearly, $$V(x)$$ is Hermitian. In addition, we can ignore all multiplicative constants in the problem (a multiple of a Hermitian operator is still Hermitian). But we know that $$d/dx$$ is anti-Hermitian, so this means that $$(d^2 / d x^2)^\dagger = d^2/dx^2$$, implying that the Hamiltonian operator is Hermitian.