Proving Unitarity of Matrix U_{pq}

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In summary, the student is trying to find an equation for U such that [tex] H=\sum_k E(k) b^+_kb_k [/itex]
  • #1
Threepwood
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Homework Statement


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].

Homework 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.

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?
 
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  • #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.
 
  • #3
Isn't that precisely what I'm supposed to be proving?
 
  • #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:
  • #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.
 
  • #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.
 
  • #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..
 
  • #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.
 
  • #9
Ok, but what about finding an equation for U?
 
  • #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]?
 
  • #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?
 

What is a unitary matrix?

A unitary matrix is a square matrix that has an inverse which is equal to its conjugate transpose. This means that when a unitary matrix is multiplied by its conjugate transpose, the result is the identity matrix.

How can I prove that a matrix is unitary?

To prove that a matrix is unitary, you can use the definition and check if the matrix satisfies the condition of having an inverse equal to its conjugate transpose. Another way is to perform matrix multiplication and see if the result is the identity matrix.

What is the importance of proving a matrix is unitary?

Unitary matrices are important in various fields of science, including quantum mechanics, signal processing, and linear algebra. They have properties that make them useful in solving equations and performing transformations.

Can a matrix be both unitary and non-unitary?

No, a matrix cannot be both unitary and non-unitary. A matrix is either unitary or non-unitary based on the definition and properties of unitary matrices.

Are all matrices with real entries unitary?

No, not all matrices with real entries are unitary. A matrix must satisfy the condition of having an inverse equal to its conjugate transpose to be considered unitary, regardless of its entries being real or complex numbers.

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