Finding 2x2 Hamiltonian Matrix for Second-Quantized Hamiltonian

[RugbyRyan]

Homework Statement

I need to find the 2x2 Hamiltonian matrix for the Hamiltonian, which is written in second-quantized form as below for a system consisting of the electrons and photons.

H = h/ωb†b + E1a†1a1 + E2a†2a2 + Ca†1a2b† + Ca†2a1b,

a's are creation and annihilation operator for electrons, and b's are for photons.

Homework Equations

Need to be written in the basis of the following states

|φ1> = 1 √(n−1)! a2† (b†)^(n−1)|0>
|φ2>= 1 √(n)! a1† (b†)^(n)|0>


I'm not sure where to begin. I'm guessing I have to find the eigenvalues and vectors of the hamiltonian but not sure how. Could someone help me start this problem? Thanks.

 

[gabbagabbahey]

No need to compute eigenvalues/vectors here...Just use the fact that any operator [itex]A[/itex] can be expressed in a basis [itex]\{|v_1\rangle,|v_2\rangle,\ldots|v_n\rangle\}[/itex] as a matrix with entries given by [itex]A_{ij}=\langle v_i|A|v_j\rangle[/itex]. So, for example [itex]H_{12}=\langle \psi_1|H|\psi_2\rangle[/itex]...just calculate the 4 inner products to get your 4 components.

 

[Cyosis]

Recall [itex]\hat{H} |e_j \rangle= H_{ij} |e_i \rangle[/itex]. Summation over i implied. How can you find the matrix elements from this?

 

[RugbyRyan]

Thanks to both of you. That helped a lot. I managed to do inner product for one of the terms of the Hamiltonian for just the first component of the matrix. Seems like this is going to be a long process. Is there a trick that I maybe missing that will reduce computations?

 

[gabbagabbahey]

Just begin by calculating the effect of [itex]H[/itex] on each of your two basis states. After that, all 4 inner products should be straightforward.