Proving a matrix is unitary

Threepwood

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

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

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

Hmm, now what?

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.

Threepwood

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

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....

Threepwood

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

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.

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.

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
[tex]c_p c_q^{\dag} + c_q^{\dag} c_p = \delta_{pq}[/tex]
applying [tex]c_q[/tex] to the left
[tex]c_q c_p c_q^{\dag} + c_q c_q^{\dag} c_p = c_q \delta_{pq}[/tex]
because [tex]c_p c_q = - c_q c_p[/tex], then
[tex]-c_p c_q c_q^{\dag} + c_q c_q^{\dag} c_p = c_q \delta_{pq}[/tex]
and [tex]c_q c_q^{\dag} = 0[/tex], so
[tex]c_q \delta_{pq} = 0[/tex]
Hmm! Is this useful relation? Probably not..

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.

Threepwood

Ok, but what about finding an equation for U?

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]?

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?