# A hamiltonial question

## Main Question or Discussion Point

So im working up to some exams and have a question regarding properties of hermitians, specifically the properties of Hamiltonian operators and trying to prove that for example if..

$$\hat{O}$$ is a hamiltonian operator then...

$$\hat{O}$$ + $$\hat{O}$$$$\dagger$$

is hermitian*.

Now what I think im having a problem with is understanding exactly what im expected to know with regard to this, as what I know about hamiltonian operators (real eigenvalues and orthogonality) don't seem to help a massive amount here (unless im meant to show that $$\hat{O}$$ with $$\hat{O}$$$$\dagger$$ is orthogonal).

Any help is appreciated, I feel this is one of them subjects where if I start to understand with one example like this I will be able to nail the rest out pretty quickly :)

*In case im explaining badly due to my limited knowledge of hermitian and hamiltonian things, the exact question says...

Show for any operator $$\hat{O}$$, that $$\hat{O}$$ + $$\hat{O}$$$$\dagger$$ is Hermitian.

edit: sigh, spelt the title wrong :(

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What is a Hamiltonian operator? You mean THE hamiltonian? Or did you mean to say a Hermitian operator? Or a Hilbert operator?

$$O + O^\dagger$$ is always Hermitian. Use the fact that $${O^\dagger}^\dagger = O$$.

olgranpappy
Homework Helper
and use the fact that
$$A+B=B+A$$

Sorry, I think I meant Hermitian operators. Thank you for the replies but it doesnt help me very much but I think thats maybe because im asking hthe question wrong! :S

What im asking is how I would recognise the answer as a Hermitian in particular? Is it hermitian because...

($$\hat{O}$$$$^{dagger}$$)$$^{dagger}$$

is $$\hat{O}$$ and thus Hermitian and thus because Hermitian Operators are commutative Hermitian + Hermitian = Hermitian ?

malawi_glenn
Homework Helper
Well you basically have everything you need:

i) a hermitian operator fulfills: $$\hat{O}^{\dagger} = \hat{O}$$

ii) $$A+B=B+A$$

Then what is:

$$(\hat{O} + \hat{O}^{\dagger}) ^{\dagger}$$

?

oh i see, so...

$$(\hat{O} + \hat{O}^{\dagger}) ^{\dagger}$$ = $$\hat{O}^{\dagger} + \hat{O}^{\dagger}^{\dagger}$$ = $$\hat{O} + \hat{O}^{\dagger}$$

thus proving it is hermitian. Ok, so, in a similar vein...

$$\hat{O}\hat{O}^{\dagger}$$ = $$(\hat{O}\hat{O}^{\dagger}) ^{\dagger}$$ = $$\hat{O}^{\dagger}\hat{O}^{\dagger}^{\dagger}$$ = $$\hat{O}\hat{O}^{\dagger}$$

thus proving IT is hermitian ?

George Jones
Staff Emeritus
Gold Member
Ok, so, in a similar vein...

$$\hat{O}\hat{O}^{\dagger}$$ = $$(\hat{O}\hat{O}^{\dagger}) ^{\dagger}$$ = $$\hat{O}^{\dagger}\hat{O}^{\dagger}^{\dagger}$$ = $$\hat{O}\hat{O}^{\dagger}$$

thus proving IT is hermitian ?
The end result is correct, but the second-last equality is wrong.

malawi_glenn
Homework Helper
I dont understand, you now want to PROOVE that $$\hat{O}$$ is a hermitian operator? That is a property that is given to you as a fact, you can't proove that unless you know what $$\hat{O}$$ explicity is. Or do you want to proove that given $$\hat{O}$$ is hermitian, the product $$\hat{O}\hat{O}^{\dagger}$$ is hermitian?

By the way: $$(AB)^{\dagger} = B^{\dagger}A^{\dagger}$$ so:

$$(\hat{O}\hat{O}^{\dagger}) ^{\dagger} = (\hat{O}^{\dagger})^{\dagger}\hat{O}^{\dagger}$$

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Or do you want to proove that given $$\hat{O}$$ is hermitian, the product $$\hat{O}\hat{O}^{\dagger}$$ is hermitian?
Yes thats correct :) Sorry, I have difficulty explaining things I dont understand very well, but im getting there.

The point of these seems to be that if you can conjugate the example and get back to your origonal statement then your statement is Hermitian (or at least this is the point of the questions it would seem).

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malawi_glenn
Homework Helper
yes, that is the thing you want to do. Then you must do as I told you in post #8

yes, that is the thing you want to do. Then you must do as I told you in post #8
Aye I did thanks! :)

Great, thank you all for your help!

malawi_glenn
Homework Helper
Aye I did thanks! :)

Great, thank you all for your help!
Great, so you agree with me that

$$(\hat{O}\hat{O}^{\dagger}) ^{\dagger} \neq \hat{O}^{\dagger}\hat{O}^{\dagger}^{\dagger}$$

?

Yeah, I was basically just being rubbish at maths/not thinking about it properly.