What are the properties of a Hermitian Hamiltonian operator?

[Bunting]

So I am 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 I am having a problem with is understanding exactly what I am 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 I am 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 I am 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 :(

 

[lbrits]

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]

and use the fact that
<br /> A+B=B+A<br />

 

[Bunting]

Sorry, I think I meant Hermitian operators. Thank you for the replies but it doesn't help me very much but I think that's maybe because I am asking hthe question wrong! :S

What I am 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]

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}

?

 

[Bunting]

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]

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]

I don't 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}

 

[Bunting]

Or do you want to proove that given \hat{O} is hermitian, the product \hat{O}\hat{O}^{\dagger} is hermitian?

Yes that's correct :) Sorry, I have difficulty explaining things I don't understand very well, but I am 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).

 

[malawi_glenn]

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

 

[Bunting]

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]

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}

?

 

[Bunting]

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