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.$\newcommand{\ket}[1]{\left|#1\right>}$ $\newcommand{\bra}[1]{\left<#1\right|}$ 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!


1 Answer 1


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


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