# Completeness in tensor product basis

I have a (probably simple) question about the completeness relation in dirac notation. Mostly I just want to check if I am understanding this correctly, because I can't actually find this mentioned anywhere.$$\newcommand{\ket}{\left|#1\right>}$$ $$\newcommand{\bra}{\left<#1\right|}$$ So for a complete set of states $$\ket{\alpha}$$ we often use the completeness relation to rewrite operator products. $$\sum_{\alpha} \ket{\alpha} \bra{\alpha} = \mathbb{I}.$$ Here $$\mathbb{I}$$ is the identity operator. In many situations we use a basis of tensor product states, i.e. a basis $$\ket{\alpha} \otimes \ket{\beta}$$, where $$\ket{\alpha}$$ should be understood to be in a different Hilbert space than $$\ket{\beta}$$ (maybe they refer to different particles). I am wondering what the completeness relation would look like in this case. Let's say I have an operator $$A$$ that acts only on the $$\ket{\alpha}$$ space. Are all the following expressions valid? $$(1) \qquad A = \sum_{\alpha} \sum_{\alpha'} \ket{\alpha'}\bra{\alpha'} A \ket{\alpha} \bra{\alpha}$$ $$(2)\qquad A = \sum_{\alpha} \sum_{\alpha'} \sum_{\beta} \ket{\alpha'}\bra{\alpha'} A \ket{\alpha} \ket{\beta} \bra{\beta} \bra{\alpha}$$ $$(2)\qquad A = \sum_{\alpha} \sum_{\beta} A \ket{\alpha} \ket{\beta} \bra{\beta} \bra{\alpha}.$$ I realise these examples are pretty arbitrary, but I hope they illustrate my question. Do the completeness relations still hold separately, also when the total Hilbert space $$\mathcal{H} = \mathcal{H}_{\alpha} + \mathcal{H}_{\beta}$$ now has the complete basis $$\ket{\alpha} \otimes \ket{\beta}$$. Am I still allowed to just insert complete outer products of either $$\ket{\alpha}$$ or $$\ket{\beta}$$ where convenient in my equations, or should I be more careful.

I think what I am doing is not wrong, since in essence I am just inserting identity operators of the two different spaces, but I also realise that what I am doing is far from rigorous. Any help would be appreciated!

Let's say I have an operator A that acts only on the $$|\alpha\rangle$$ space.

If the Hilbert space under consideration is the tensor product space $$\mathcal H = \mathcal H_\alpha \otimes \mathcal H_\beta$$, then you can't act on the elements of $$\mathcal H$$ with an operator $$A$$ which acts only on $$\mathcal H_\alpha$$ alone. You must consider the operator $$A \otimes \mathbf 1_\beta$$, with $$\mathbf 1_\beta$$ being the identity operator on $$\mathcal H_\beta$$.

The completeness relation on $$\mathcal H_\alpha\otimes \mathcal H_\beta$$ takes the form

$$\mathbf 1= \sum_{\alpha,\beta} \bigg(|\alpha\rangle\otimes |\beta\rangle\bigg)\bigg(\langle \alpha|\otimes \langle \beta|\bigg)$$

You could use this relation to yield (dropping the tensor product symbols for brevity, and to match your notation)

$$A \otimes \mathbf 1_\beta = \sum_{\alpha,\alpha',\beta,\beta'} |\alpha\rangle|\beta\rangle A_{\alpha\alpha'}\delta_{\beta\beta'} \langle \alpha'|\langle\beta'| = \sum_{\alpha,\alpha',\beta} |\alpha\rangle|\beta\rangle A_{\alpha\alpha'}\langle\alpha'|\langle \beta|$$

because $$\langle\alpha|\langle\beta| (A\otimes \mathbf 1_\beta) |\alpha'\rangle|\beta'\rangle = \langle\alpha|\langle\beta|\bigg(|A\alpha\rangle|\mathbf 1_\beta\beta'\rangle\bigg) \equiv \langle\alpha|A\alpha\rangle \cdot \langle \beta|\beta'\rangle = A_{\alpha\alpha'} \delta_{\beta \beta'}$$