# A hamiltonial question

1. Apr 25, 2008

### Bunting

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 :(

Last edited: Apr 25, 2008
2. Apr 25, 2008

### 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 $${O^\dagger}^\dagger = O$$.

3. Apr 25, 2008

### olgranpappy

and use the fact that
$$A+B=B+A$$

4. Apr 26, 2008

### Bunting

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 ?

5. Apr 26, 2008

### 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:

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

?

6. Apr 26, 2008

### 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 ?

7. Apr 26, 2008

### George Jones

Staff Emeritus
The end result is correct, but the second-last equality is wrong.

8. Apr 26, 2008

### malawi_glenn

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:

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

Last edited: Apr 26, 2008
9. Apr 26, 2008

### Bunting

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).

Last edited: Apr 26, 2008
10. Apr 26, 2008

### malawi_glenn

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

11. Apr 26, 2008

### Bunting

Aye I did thanks! :)

Great, thank you all for your help!

12. Apr 26, 2008

### malawi_glenn

Great, so you agree with me that

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

?

13. Apr 26, 2008

### Bunting

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