Hermitian Operator in Inner Product

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Homework Statement

\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>

Homework 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?

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.

 

[Unto]

Help?

 

[vela]

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

 

[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?

 

[vela]

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}\rangleto \langle\psi_1\vert\hat{A}\vert\psi_2\rangle?

 

[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.

 

[vela]

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*}

 

[Unto]

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

EDIT:

Done, thanks a lot, 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.