# Hermitian Operator in Inner Product

1. Oct 16, 2011

### Unto

1. The problem statement, all variables and given/known data

$\int d^{3} \vec{r} ψ_{1} \hat{A} ψ_{2}$ = $\int d^{3} \vec{r} ψ_{2} \hat{A}* ψ_{1}$

Hermitian operator A, show that this condition is equivalent to requiring $<v|\hat{A}u>$ = $< \hat{A}v|u>$

2. Relevant equations

I changed the definitions of ψ into their bra-ket forms first of all.

Hints say something about the Identity operator, but I don't have any bra's in my equation, what do I do?

3. The attempt at a solution

After changing the ψ into their bra-ket forms and substituting, I am stumped. Any pointers please?

I have introduction into adjoint hermitian operators but I cannot see how this would fit into it.

I would write more about what I did, but your equation editor is very difficult to use, is it possible that you could make a code that mirror's the equation editor on Microsoft Word 2007 - 2011?

Please help me, I'm doing this problem WAY ahead of time and I just want to be good at this stuff. Please don't ignore me, just give me a push in the right direction and I promise I will do all the rest and show you what I did.

2. Oct 16, 2011

### Unto

Help?

3. Oct 17, 2011

### vela

Staff Emeritus
It would help if you showed what you did rather than just describe it.

4. Oct 17, 2011

### Unto

Latex takes forever to use... Well here goes:

We are supposed to assume that $ψ_{1}*$ is a wavefunction, so we use a trick to show just that, whilst the conjugate is now something wierd

$ψ_{1}* = < \vec{r}|ψ_{1} >$

$ψ_{1} = < ψ_{1}|\vec{r} >$

So subst into

$\int d^{3} \vec{r} < ψ_{1}|\vec{r} > \hat{A} < \vec{r}|ψ_{2} >$ = $\int d^{3} \vec{r} < \vec{r}|ψ_{2} > \hat{A}* < ψ_{1}|\vec{r} >$

($| \vec{r} > < \vec{r} |$

Left side can have the r thingy rearranged in terms of identity bra-ket which is 1, so we get:

$< ψ_{1}|\hat{A}|ψ_{2} >$ = $\int d^{3} \vec{r} < \vec{r}|ψ_{2} > \hat{A}* < ψ_{1}|\vec{r} >$

Right side becomes:

$| ψ_{2}>\hat{A}*< ψ_{1}|$

Swap around the functions on the right side to get:

$< ψ_{1}|\hat{A}|ψ_{2} >$ = $< ψ_{2}|\hat{A} *|ψ_{1} >$

Remembering that $\hat{A} = \hat{A}*$ for an hermitian function. Dunno whether this is correct.

Latex is so hard to use, anything easier?

Last edited: Oct 17, 2011
5. Oct 17, 2011

### vela

Staff Emeritus
Can you explain how you went from
$$\int d^3\vec{r}\,\langle\psi_1\vert\vec{r}\rangle\hat{A}\langle\psi_2\vert\vec{r}\rangle$$to $\langle\psi_1\vert\hat{A}\vert\psi_2\rangle$?

6. Oct 17, 2011

### Unto

That's not what I did..

I went from $\int d^{3} \vec{r} < ψ_{1}|\vec{r} > \hat{A} < \vec{r}|ψ_{2} >$ to $< ψ_{1}|\hat{A}|ψ_{2} >$

I used that identity relation with the bra of the r facing the ket of the r... it equals 1.

7. Oct 17, 2011

### vela

Staff Emeritus
Sorry, that was a typo. The thing I don't like is you have the operator sandwiched between the $|\vec{r}\rangle$ and $\langle \vec{r}|$. It's not clear that you can just then eliminate the pair from the expression. It would be better if you wrote it as follows:
\begin{align*}
\int d^3\vec{r}\, \psi_1^* \hat{A} \psi_2 &= \int d^3\vec{r}\, \psi_1^* (\hat{A} \psi_2) \\
&= \int d^3\vec{r}\, \langle\psi_1 | \,\vec{r}\rangle\langle\vec{r}\, |\, \hat{A}\psi_2\rangle \\
&= \langle \psi_1 |\, \hat{A}\psi_2 \rangle
\end{align*}

8. Oct 18, 2011

### Unto

Nicely done broslice, I assume I do the same thing for the right side?

EDIT:

Done, thanks alot, since it's hermitian, the conjugate remains the same as normal and the condition $<v|\hat{A}u>$ = $< \hat{A}v|u>$ is met.

Can you help me on my other topic? Just a point in what to do with the probabilities, thank you in advance.

Last edited: Oct 18, 2011