How to write the matrix of the Hamiltonian in case of 2 subspaces?

Question

Let's imagine that our Hamiltonian acts as follows

\(\mathcal{H}: \mathcal{H}_1 \otimes \mathcal{H}_2 \to \mathcal{H}_1 \otimes \mathcal{H}_2\)

For simplicity, let's suppose that 

\(\mathcal{H} = (a^\dag b + b^\dag a), \)

where \(a, \,\, a^\dag\) \(\)are annihilation and creation operators acting in \(\mathcal{H_1}\) (similarly for b in \(\mathcal{H_2}\)).

We can calculate matrix element

\(<l,p|\mathcal{H}|n, k> = \sqrt{n+1} \sqrt{k} \delta_{l, n+1} \delta_{p,k-1} + \sqrt{n} \sqrt{k+1} \delta_{l,n-1} \delta_{p,k+1}\).

It (matrix of the operator) gives us, formally, 4-dimensional array. Is there a convenient way, how to rewrite it in form of square matrix? Because I'm interested in eigenvectors and in this form it's very uncomfortable to work with the 4-dimensional array. 

Answer 1

Order your states \(|n,k\rangle = |n\rangle\otimes|k\rangle\)into a sequence of your choice. E.g.

\(|0\rangle=|0\rangle\otimes|0\rangle\), \(|1\rangle=|0\rangle\otimes|1\rangle\),...,\(|N\rangle=|0\rangle\otimes|N\rangle\),\(|N+1\rangle=|1\rangle\otimes|0\rangle\), \(|N+2\rangle=|1\rangle\otimes|1\rangle\),...

I think you get the idea.

Comments

@Flamma Thank you for the answer. I know this idea, but I was thinking about a bit different. Previously I saw that people write the Hamiltonian in terms of tensor product of matrices of operators and unit matrices. And I'd like to see how it looks like. 

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Suppose you have Hilbert spaces \(A\) and \(B\), with Hamiltonians \(H_A\) and \(H_B\), respectively. Let there further be an interaction \(W\) between the systems described by \(A\) and \(B\). \(W\) is defined on the composite system \(A\otimes B\), and the Hamiltonian of the composite system is\[H_{A\otimes B}=H_A\otimes Id_{B}+Id_{A}\otimes H_B+W\] with \(Id_{A}\) the identity on \(A\) and likewise for \(B\). The Hamiltonian you have specified is interaction only, it corresponds to \(W\). If you want to express \(H_{A\otimes B}\) in matrix form, then you have to choose an ordered basis on \(A\otimes B\). Which brings you back to my answer above.

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@Flamma Thank you for the answer.