1. ## Hermitian operators

So I'm writing an assignment in which I have to define, what is understood by a hermitian operator.
My teacher has told me to definere it as:
<ϕm|A|ϕn> = <ϕn|A|ϕm>* , where lϕn> and lϕm> is the n'th and m'th basis vector.
And using this i then have to proof the more general definition:
<a|A|b> = <b|A|a>*, la> and lb> being arbitrary vectors.
I've tried to do so but I have yet not succeeded:
What I've done is to say: Take a hermitian operator. Since it's hermitian it must satisfy:
A = ∑_(m,n)|ϕm>amn <ϕn| = ∑(m,n)|ϕm> anm*<ϕm|
Which when dotted with 2 arbitrary vectors|ψ> and <φ|equals to:
<φ|A|ψ> = ∑(m,n) <φ|ϕm>amn<ϕn|ψ> = ∑(m,n) <ϕm|φ>*amn <ψ|ϕn>* =
∑(m,n) <ψ|ϕn>* amn <ϕm|φ>*
Since A is hermitian this equals to:
∑(m,n) <ψ|ϕn>* amn <ϕm|φ>* = ∑(m,n) <ψ|ϕn>*anm*<ϕm|φ>* = [∑(m,n) <ψ|ϕn>anm<ϕm|φ>]*
Now the proof would work if |ϕn>anm<ϕm| = |ϕm>amn<ϕn|, but that's not right is it?
Can anyone help me how to proove this? I know it's simple, but I'm still finding it a little hard.

2. I'm re-writing your post for readability. Please check that this is what you meant to say.

So I'm writing an assignment in which I have to define what is understood by a Hermitian operator. My teacher has told me to define it as:

$\displaystyle \langle\phi_{m}|A\phi_{n}\rangle=\langle\phi_{n}|A \phi_{m}\rangle^{*},$ where $\displaystyle |\phi_{n}\rangle$ and $\displaystyle \phi_{m}\rangle$ are the $\displaystyle n$th and $\displaystyle m$th basis vectors, respectively. Using this I then have to prove the more general definition:

$\displaystyle \langle a|Ab\rangle=\langle b|Aa\rangle^{*},$ with $\displaystyle |a\rangle$ and $\displaystyle |b\rangle$ being arbitrary vectors.

I've tried to do so but I have not yet succeeded. What I've done is to say: take a Hermitian operator $\displaystyle A.$ Since it's Hermitian it must satisfy

$\displaystyle \displaystyle A=\sum_{m,n}|\phi_{m}\rangle A_{mn}\langle\phi_{n}|=\sum_{m,n}|\phi_{m}\rangle A_{nm}^{*}\langle\phi_{m}|,$

which, when dotted with an arbitrary ket $\displaystyle |\psi\rangle$ and bra $\displaystyle \langle\varphi|$ produces

$\displaystyle \displaystyle \langle\phi|A\psi\rangle=\sum_{m,n}\langle\varphi| \phi_{m}\rangle A_{mn}\langle\phi_{n}|\psi\rangle=\sum_{mn}\langle \phi_{m}|\varphi\rangle^{*}A_{mn}\langle\psi|\phi_ {n}\rangle^{*}=\sum_{mn}\langle\psi|\phi_{n}\rangl e^{*}A_{mn}\langle\phi_{m}|\varphi\rangle^{*}.$

Since $\displaystyle A$ is Hermitian this yields

$\displaystyle \displaystyle\sum_{mn}\langle\psi|\phi_{n}\rangle^ {*}A_{mn}\langle\phi_{m}|\varphi\rangle^{*}=\sum_{ mn}\langle\psi|\phi_{n}\rangle^{*}A_{nm}^{*}\langl e\phi_{m}|\varphi\rangle^{*}=\left[\sum_{mn}\langle\psi|\phi_{n}\rangle A_{nm}\langle\phi_{m}|\varphi\rangle\right]^{*}.$

Now the proof would work if $\displaystyle |\phi_{n}\rangle A_{nm}\langle\phi_{m}|=|\phi_{m}\rangle A_{mn}\langle\phi_{n}|,$ but that's not right is it? Can anyone help me to prove this? I know it's simple, but I'm still finding it a little hard.

Is this what you meant?

3. Yes, that's exactly what i meant. Sorry it wasn't so readable, but my computer failed to use the symbols you used for some reason.
Now, I've thought about it overnight and still not come to any conclusion. Although i thought, it might be an idea to just dot with the 2 arbitrary vectors, and say that since the result is just whole numbers time the matrix-element amn, then it is proven. But I still would like to do it my way? Can you help me?

4. I think what you really need to do is express the arbitrary vectors $\displaystyle |a\rangle$ and $\displaystyle |b\rangle$ as linear combinations of vectors in your basis. That is, you have

$\displaystyle \displaystyle|a\rangle=\sum_{k}a_{k}|\phi_{k}\rang le$ and

$\displaystyle \displaystyle|b\rangle=\sum_{\ell}b_{\ell}|\phi_{\ ell}\rangle.$

You can do this if the set of kets $\displaystyle |\phi_{n}\rangle$ truly are a basis. Then show that the two definitions are in an if-and-only-if relationship with each other. (One direction, namely that the arbitrary inner product implies the basis inner product, is quite straight-forward.)

Does that help?

Incidentally, in order to get nice-looking math, you need to click the Advanced button, and use the TeX button to enclose LaTeX code with math delimiters. There's a tutorial sticky for LaTeX in the LaTeX help section if you need to know more about LaTeX on the Math Help Forum.

5. Now, I'm not quite sure i get your idea. How I understand it is:
If you dot with arbitrary vectors on an operator written out as a combination of the basis vectors (as the outer products of your basis set form a basis for expanding operators), then you just end up with scalarmultiples of the matrix-elements of the operator, and hence the definition must also be true for arbitrary vectors.

6. Your proof actually does work, and here's why. Your question at the end shouldn't have been, "Is $\displaystyle |\phi_{n}\rangle A_{nm}\langle\phi_{m}|=|\phi_{m}\rangle A_{mn}\langle\phi_{n}|?$"; instead, it should have been, "Is

$\displaystyle \displaystyle\sum_{mn}|\phi_{n}\rangle A_{nm}\langle\phi_{m}|=\sum_{mn}|\phi_{m}\rangle A_{mn}\langle\phi_{n}|?$"

And now perhaps you can see that the equality holds because $\displaystyle n$ and $\displaystyle m$ are just dummy variables ranging over the same set of values; because of the nice properties of Hilbert spaces (in which I assume you're working), you can interchange the order of summation nicely.

Does that make sense?

7. Yes! Thanks alot!

8. You're welcome. Incidentally, it occurred to me that your proof is more of a Heisenberg approach (thinking about how the operators change), and my approach was more of a Schrödinger approach (thinking about how the vectors change). They are equivalent approaches.