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.