Finding 2x2 Hamiltonian Matrix for Second-Quantized Hamiltonian

RugbyRyan
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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.
 
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No need to compute eigenvalues/vectors here...Just use the fact that any operator A can be expressed in a basis \{|v_1\rangle,|v_2\rangle,\ldots|v_n\rangle\} as a matrix with entries given by A_{ij}=\langle v_i|A|v_j\rangle. So, for example H_{12}=\langle \psi_1|H|\psi_2\rangle...just calculate the 4 inner products to get your 4 components.
 
Recall \hat{H} |e_j \rangle= H_{ij} |e_i \rangle. Summation over i implied. How can you find the matrix elements from this?
 
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?
 
Just begin by calculating the effect of H on each of your two basis states. After that, all 4 inner products should be straightforward.
 
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The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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