# How to construct a 2-partite matrix

Let's assume we have an internal hamiltonian $$H_0 = \mid 1\rangle \langle 1\mid$$.

Now let's assume we have two systems with identical Hamiltonians $$H^1_0$$,$$H^2_0$$ and I want to compute the joint Hamiltonian

$$H^{(2)}_0 = H_0^1\otimes \mathbb{I}+ \mathbb{I}\otimes H_0^2$$

I have a little bit of a hard time understanding how exactly $$H^{(2)}_0$$ is constructed.

I approach this the following way

$$H^{(2)}_0 = \begin{bmatrix} 1&0\\0&0\end{bmatrix} \otimes \begin{bmatrix} 1&0\\0&1\end{bmatrix} + \begin{bmatrix} 1&0\\0&1\end{bmatrix} \otimes\begin{bmatrix} 1&0\\0&0\end{bmatrix}$$

$$=\begin{bmatrix} 1&0&0&0& \\ 0&0&0&0& \\ 0&0&1&0& \\ 0&0&0&0& \end{bmatrix} + \begin{bmatrix} 1&0&0&0& \\ 0&1&0&0& \\ 0&0&0&0& \\ 0&0&0&0& \end{bmatrix} = \begin{bmatrix} 2&0&0&0& \\ 0&1&0&0& \\ 0&0&1&0& \\ 0&0&0&0& \end{bmatrix}$$

Is this correct? I'm doing this mostly via gut feeling. I remember my professor mentioning that depending on where the Hamiltonian is, like in $$H_0^1\otimes \mathbb{I}$$ where it is on the first entry or $$\mathbb{I}\otimes H_0^2$$ where it is on the second entry I can see on what sub-system it acts.

What I'm asking is 1. if my calculations were correct and to give me some conceptual context. Also a link that could explain it to me would be greatly appreciated. (Spent some time looking for it).

Yes, this looks quite right. I'll offer how I think about constructing such an object.

First, it is usually written sloppily (cough: Griffiths) as

$$H^{(2)} = H^{(1)} + H^{(2)}$$

(please allow me to drop the subscripts). However, if we truly wish to construct the $$2$$ particle hilbert space we must write

$$H^{(2)} = H^{(1)} \otimes 1^{(2)} + 1^{(1)} \otimes H^{(2)}.$$

where the superscripts indicate on which space ($$\mathcal{H}_1$$ or $$\mathcal{H}_2$$) the operators correspond to. Of course, our goal is to explicitly construct the matrix representation of $$\mathcal{H}_1 \otimes \mathcal{H}_2$$.

We are told that the hamiltonian of a single particle is given by

$$H = | 1 \rangle \langle 1 |$$

Note what has been assumed but not stated explicitly in the problem: (1) We assume that the hamiltonian describes a two state system and (2) that we are choosing the regular representation of that system and (3) $$|1 \rangle$$ represents the eigenstate of $$H$$ corresponding to eigenvalue $$1$$. (Number (3) is actually conventional to assume, but I thought I'd mention it for completeness.)

That is, we are choosing to represent the eigenstates of the 2 state system as

$$| 1 \rangle \dot{=} {1 \choose 0}$$

This is convenient as it makes the Hamiltonian diagonal.

At any rate, what's very nice about Dirac notation is that, after choosing our representation, all we have to do is just follow the recipe and know basic computational linear algebra. That is,

$$|1 \rangle \langle 1| = {1 \choose 0}(1\ \ 0) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$

We may then use the kronecker product to compute the direct product of two matrices $$A, B$$( where $$[A] = m\times n$$ and $$[B]= p\times q$$)

$$\mathbf {A} \otimes \mathbf {B} ={\begin{bmatrix}a_{11}\mathbf {B} &\cdots &a_{1n}\mathbf {B} \\\vdots &\ddots &\vdots \\a_{m1}\mathbf {B} &\cdots &a_{mn}\mathbf {B} \end{bmatrix}}.$$

to construct the complete hamiltonian in the 2-particle hilbert space (which it looks like you did correctly).