I have a Hamiltonian of the form $H = \Sigma_{ij} H_{ij}a^\dagger_ia_j$, and I want to diagonlize it:

Let $ H_{ij} = \Sigma_{\alpha}U_{i\alpha}\epsilon_\alpha U^*_{j\alpha} $, where U is a unitary matrix. Then I proceed by inserting this in the first equation:

$ H = \Sigma_{ij\alpha}U_{i\alpha}\epsilon_\alpha U^*_{j\alpha}a^\dagger_ia_j = \Sigma_\alpha \epsilon_\alpha \Big(\Sigma_i U_{i\alpha}a^\dagger_i\Big)\Big(\Sigma_jU^\dagger_{j\alpha}a_j\Big) $

Defining $ b^\dagger_\alpha = \Sigma_i U_{i\alpha}a^\dagger_i $ my Hamiltonian can be writen as:

$ H = \Sigma_\alpha \epsilon_\alpha b^\dagger_\alpha b_\alpha $ which is diagonal. My first question is: why is this a diagonal hamiltonian? How can I be so sure?

The second question is: how to effectvely use this diagonalization procedure?

I have now a Hamiltonian of the form: $ H = \epsilon_\alpha a^\dagger a + \epsilon_b b^\dagger b - J(a^\dagger b + b^\dagger a) $ where a and b are two modes, that can be either bosonic or fermionic.

To diagonalize this I have to use the procedure described only in the part that is multiplied by -J?


(1) I guess the best answer of this is “because it meets the definition of a diagonal Hamiltonian,” but I don’t actually have a handy definition of a diagonal Hamiltonian.

Basically what we mean is that we can write our Hilbert space as a product of $n$ smaller spaces such that our Hamiltonian has the form of $$\hat H = \hat h_1 \otimes I \otimes \dots \otimes I ~+~ I \otimes h_2 \otimes \dots \otimes I ~+~ \dots ~+~ I \otimes I \otimes \dots \otimes \hat h_n,$$ where $I$ is the identity operator.

In your case, to “be sure” you will want to confirm that your $b_i$ are commuting or anticommuting annihilators, so that they satisfy either $$ b_m b_n^\dagger \mp b_n^\dagger b_m = \delta_{mn}$$with the $-$ sign for the bosonic case and the $+$ sign for the fermionic case. If you have this, then you have a Fock space in the form of the occupation numbers for the states that the $b_n$ annihilate, and your Hamiltonian is diagonal with regard to that Fock space.

(2) Yes, you essentially want to imagine that you have a matrix looking something like $\begin{bmatrix}\epsilon_a & -J\\-J & \epsilon_b\end{bmatrix}.$

You diagonalize that and then you try to transfer that basic result to a result on the annihilators, to find the perspective from which the Hamiltonian is diagonal.

  • $\begingroup$ very well, thank you; I guess I have to read a bit more on that Fock space and second quantization thing. This is new to me and I have some doubts yet. But what you said was helpful. I will try to diagonalize the hamiltonian using this, it seems to be the right way. Thank you $\endgroup$
    – Dimitri
    Nov 19 '19 at 2:05

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