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Proving a matrix is unitary

  1. Nov 20, 2008 #1
    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]
    [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]
    [tex]c_q = \sum_{p} U_{pq} b_p[/tex]
    [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?
  2. jcsd
  3. Nov 20, 2008 #2
    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.
  4. Nov 20, 2008 #3
    Isn't that precisely what I'm supposed to be proving?
  5. Nov 21, 2008 #4
    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
  6. Nov 23, 2008 #5
    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.
  7. Nov 23, 2008 #6
    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.
  8. Nov 23, 2008 #7
    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..
  9. Nov 23, 2008 #8
    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.
  10. Nov 23, 2008 #9
    Ok, but what about finding an equation for U?
  11. Nov 24, 2008 #10
    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]?
  12. Nov 25, 2008 #11
    That was never stated in the question, but maybe it was implied somehow. It would make sense. How would I go about doing that?
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