# Completeness in tensor product basis


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