# Proving a matrix is unitary

1. Nov 20, 2008

### Threepwood

1. The problem statement, all variables and given/known data
I have been given the Hamiltonian
$$H = \sum_{k}\left(\epsilon_k - \mu\right) c_k^{\dag} c_k + \gamma \sum_{kp}c_k^{\dag} c_p$$
and also that
$$c_p = \sum_{q} U_{pq} b_q$$
I have to prove that this matrix $$U_{pq}$$ is unitary, and find an equation for $$U_{pq}$$.

2. Relevant equations
This is equivalent to proving that
$$\{b_q, b_p\} = 0$$
and
$$\{b_q , b_p^{\dag}\} = \delta_{pq}$$
where $$b$$ and $$c$$ are creation and annihiliation operators.

3. The attempt at a solution
Knowing that
$$c_p = \sum_{q} U_{pq} b_q$$
then
$$c_q = \sum_{p} U_{pq} b_p$$
and
$$\{b_q , b_p\} = b_q b_p + b_p b_q$$
$$c_p b_p = \left(\sum_{q} U_{pq} b_q\right) b_p$$
$$b_q c_q = b_q \left(\sum_{p} U_{pq} b_p\right)$$
So that
$$c_p b_p + b_q c_q = \left(\sum_{q} U_{pq} b_q\right) b_p + b_q \left(\sum_{p} U_{pq} b_p\right)$$

Hmm, now what?

2. Nov 20, 2008

### borgwal

You should use that the c operators satisfy the same anticommutation relations that the b's also satisfy. On the other hand, c_p and b_q do not, in general, satisfy such relations.

3. Nov 20, 2008

### Threepwood

Isn't that precisely what I'm supposed to be proving?

4. Nov 21, 2008

### borgwal

No, you have to prove U is unitary.

Edit: you already seem to know that U being unitary is equivalent to the b's satisfying the same anticommutation relations as the c's. But that's all there is to it....

Last edited: Nov 21, 2008
5. Nov 23, 2008

### Threepwood

I need to prove those relations. How do I prove that
$$\{b_q , b_p\} = 0$$ and $$\{b_q , b_p^{\dag} \} = \delta_{pq}$$?

And also, beyond that, how do I find an equation for U? I don't need to solve the equation for U, just find it.

6. Nov 23, 2008

### borgwal

You need more information to prove any of those relations. You must have been given some info about what the b's are supposed to be, for instance. I assumed that you had been told that the b's are fermionic annihilation operators.

7. Nov 23, 2008

### Threepwood

Yes, they are. At the moment I'm more interested in finding this equation for U, but I have no idea where to even start. I've just been playing around with the relations, like taking
$$c_p c_q^{\dag} + c_q^{\dag} c_p = \delta_{pq}$$
applying $$c_q$$ to the left
$$c_q c_p c_q^{\dag} + c_q c_q^{\dag} c_p = c_q \delta_{pq}$$
because $$c_p c_q = - c_q c_p$$, then
$$-c_p c_q c_q^{\dag} + c_q c_q^{\dag} c_p = c_q \delta_{pq}$$
and $$c_q c_q^{\dag} = 0$$, so
$$c_q \delta_{pq} = 0$$
Hmm! Is this useful relation? Probably not..

8. Nov 23, 2008

### borgwal

If the b's are fermionic annihilation operators, then that *means* they satisfy the anticommutation relations that, as you figured out, are equivalent to U being unitary. Done.

9. Nov 23, 2008

### Threepwood

Ok, but what about finding an equation for U?

10. Nov 24, 2008

### borgwal

You clearly did not state the full problem so I have to keep guessing: were you supposed to diagonalize the Hamiltonian and find U such that [tex] H=\sum_k E(k) b^+_kb_k [/itex]?

11. Nov 25, 2008

### Threepwood

That was never stated in the question, but maybe it was implied somehow. It would make sense. How would I go about doing that?