# Homework Help: Operators and Hermiticity

1. Aug 24, 2011

### Beer-monster

1. The problem statement, all variables and given/known data

Consider two operators, A and B which satisfy:
[A, B] = B ; B†B = 1 − A

A. Determine the hermiticity properties of A and B.
B. Using the fact that | a = 0 > is an eigenstate of A, construct the other
eigenstates of A.
C. Suppose the eigenstates of A form a complete set. Determine if eigen-
states of B can be constructed, and if so, determine the spectrum of the
eigenstates of B.

2. Relevant equations

Condition for Hermiticity:

$$\int (A^{\dagger}\psi)^{*}\psi.dx = \int \psi^{*}A\psi.dx$$

3. The attempt at a solution[/b

Completely lost on this one. All I could think of was trying to work to the commutation relation from

$$\int \psi^{*} B^{\dagger}B\psi.dx$$

but that just leads to a dead end. Any help would be appreciated.

2. Aug 24, 2011

### vela

Staff Emeritus
It's easy to show A = 1-BB is Hermitian. Just take the adjoint of both sides of the equation. B, I'm pretty sure, is anti-Hermitian, but I'm not sure how to show that.

Last edited: Aug 24, 2011
3. Aug 24, 2011

### Dick

B is a ladder operator for A, isn't it? I don't think it's either Hermitian or anti-Hermitian.

4. Aug 24, 2011

### vela

Staff Emeritus
Yup, you're right.

5. Aug 24, 2011

### Dick

Beer-monster, suppose A|a>=a|a>, i.e. |a> is an eigenvector of A with real eigenvalue a. vela already told you how to show A is hermitian. Use [A,B]=B to say something about the eigenvalue of B|a>. Give us some help here.

6. Aug 25, 2011

### Beer-monster

Okay, I guess I really need to study this stuff again because I'm lost.

I've tried taking the adjoint of A but am not quite sure what to do with it. I guess I just need to hash through the maths.

How did you see that B was a ladder operator, it not really clear to me.

7. Aug 25, 2011

### vela

Staff Emeritus
What does it mean for an operator to be Hermitian?

8. Aug 25, 2011

### Beer-monster

That is satisfies the condition that I mentioned above.

$$\int (A^{\dagger}\psi)^{*}\psi.dx = \int \psi^{*}A\psi.dx$$

and also the quantities of the operators are all real.

9. Aug 25, 2011

### vela

Staff Emeritus
That's not quite it. If you have two operators $\hat{A}$ and $\hat{A}^\dagger$ that satisfy$$\int (\hat{A}^{\dagger}\psi_l)^*\psi_r\,dx = \int \psi_l^*\hat{A}\psi_r\,dx$$for all pairs $\psi_l$ and $\psi_r$ from the Hilbert space, you would say $\hat{A}^\dagger$ is the Hermitian adjoint of $\hat{A}$. Now how are the operator and its adjoint related if it's Hermitian?

10. Aug 25, 2011

### Beer-monster

The operator and its adjoint are equal if it is Hermitian?

11. Aug 25, 2011

### vela

Staff Emeritus
That's right. So you want to calculate $\hat{A}^\dagger$ and show it equals $\hat{A}$. You should find rules in your textbook on how to calculate the adjoint of products and sums, which you will find helpful.

12. Aug 25, 2011

### Beer-monster

I see it now. Here I was adding in kets and bras because I've often found it useful with commutator stuff.

The $$B^\dagger$$ becomes B and vice versa would give $$BB^\dagger$$ but because factors swap round in an adjoint you get what you started with.

I'm still not sure how you worked out B was a ladder operator. Though I had guessed something along those lines as ladder operators were the only way I could think to get an eigenspectrum with no wavefunction. i.e. the question gave it away.

13. Aug 26, 2011

### Dick

I guessed B was a ladder operator because I've seen a form like [A,B]=B before. That's the big reason. You can figure it out for yourself if you operate both sides on an eigenfunction of A, like I suggested in post #5. Or have you already figured that out?

14. Aug 27, 2011

### Beer-monster

Not yet. But I think it`s just a matter of beating my head against the maths and cracking out a textbook that goes over the mathematical manipulations of operators again.

Incidently, I noticed there was a mistake in my first post. The relation between B and A is:

$$B^{\dagger}B=1-A^{2}$$

This shouldn't effect the results too much as I would imagine if A*A is Hermitian so A would be.

15. Aug 28, 2011

### Dick

Untrue, if A=i*I where I is the identity operator then A^2 is hermitian, but A isn't. You might have a bit more work to do on this problem.