Completeness relations in a tensor product Hilbert space

Question

Throughout my undergrad I have gotten maybe too comfortable with using Dirac notation without much second thought, and I am feeling that now in grad school I am seeing some holes in my knowledge. The specific context where I am encountering this issue currently is in scattering theory, so I will use that as the example here.

In scattering theory we usually work with a basis of plane wave states $\left|\mathbf{k}\right>$, which are eigenstates of the free Hamiltonian. Let's say however that we also want to include particles with arbitrary spin. Then we compose the tensor product of the spatial and spin Hilbert spaces $\mathcal{H} = \mathcal{H}_ {\mathrm{spatial}} \otimes \mathcal{H}_{\mathrm{spin}}$ and arrive at new basis vectors $\left|\mathbf{k}, \nu \right> = \left| \mathbf{k}\right> \otimes \left| \nu \right>$. I denote all relevant spin indices with $\nu$, the specifics are not so important here. Now let us take for example the free Greens operator $G^0(z) = 1/(z-H^0)$, with $H^0$ the free Hamiltonian that solely acts on the spatial Hilbert space. For the sake of argument lets say that the matrixelement $\left<\mathbf{r}' \right|G^0(z)\left|\mathbf{r} \right>$ is known. In that case it is common practice in quantum mechanics to exploit completeness relations and write expressions such as $$\langle\mathbf{r}'|G^0(z)\left|\mathbf{r} \right> = \int d^3k \sum_{\nu}\langle\mathbf{r}' | G^0(z) \left|\mathbf{k} , \nu\right>\left< \mathbf{k}, \nu|\mathbf{r} \right>.$$

Here one exploits the fact that the basis $\left|\mathbf{k}, \nu \right> $ is complete, such that the complete set inserted in the operator is really just a tensor product of two identity operators, i.e., $$\int d^3k \sum_{\nu} |\mathbf{k}, \nu \rangle \langle \mathbf{k}, \nu| = \mathbb{I}_ {\mathrm{spatial}} \otimes \mathbb{I}_{\mathrm{spin}}.$$ In the book on scattering theory I am currently reading procedures like the one above are commonplace, and I myself have also been using completeness relation such as this one frequently (and perhaps with too little thought). My first point of confusion is this; clearly in this case the insertion of the spin basis states $|\nu \rangle$ may actually just be factored out since the Greens operators does not act in that space. It would thus actually be more accurate to define a new operator $\hat{G^0}(z) = G^0(z) \otimes \mathbb{I}_{\mathrm{spin}}$ that does act on the entire space. Knowing this, would it be correct to not include the spin states is this expansion at all? So instead of the completeness relation above you insert something like this, $$\int d^3k |\mathbf{k}\rangle \langle \mathbf{k}| \otimes \mathbb{I}_{\mathrm{spin}}= \mathbb{I}_ {\mathrm{spatial}} \otimes \mathbb{I}_{\mathrm{spin}}.$$ I suppose with this expression I assume that the separate basis states are still complete within their respective Hilbert spaces. This question came to me specifically because in the book I am using they will often elect to insert complete sets of position states without spin, $| \mathbf{r}\rangle$, in equations containing the momentum-spin vectors I defined at the start. Thus it would seem like they insert a set of states that is only complete in one part of the factored Hilbert space, and neglect the spin part. In contrast when they insert sets of momentum states they are very careful to always include the spin part as well.

So I have two possible explanations before me. Either you may indeed just insert a "partial" completeness relations as I have defined above, and in my book they just make the choice to do it for the momentum states and not the position states, or I am missing something about the Hilbert space formalism in general. Either way, I think my confusion probably comes down to a lack of rigour in dealing with these things. It would be greatly appreciated can help me out with my thinking here. Is my conclusion correct, or am I completely missing the ball?

Answer 1

Both expressions are equivalent. Let us write things down a bit more explicitely. You're right that $G_0(z)$ should actually be $G_0(z)\otimes \mathbb I$, but these things are often left out. We have (I'll shorten the notation)

$$ \int d^3 k \sum_\nu \langle r|G|k,\nu\rangle\langle k,\nu|r\rangle=\int d^3 k \sum_\nu (\langle r|\otimes \mathbb I) (G\otimes \mathbb I)(|k\rangle\otimes |\nu\rangle)(\langle k|\otimes\langle\nu|)(|r\rangle\otimes \mathbb I)=\\=\int d^3 k \langle r|G|k\rangle\langle k|r\rangle\otimes\sum_{\nu}|\nu\rangle\langle \nu|$$

since $\{|k\rangle\}$ forms an orthonormal basis of the spatial Hilbert space and $\{|\nu\rangle\}$ one for the spin Hilbert space, you have

$$ \int d^3 k|k\rangle \langle k|=\mathbb I_{\textrm{spatial}}\qquad\sum_{\nu }|\nu\rangle\langle \nu|=\mathbb I_{\textrm{spin}}.$$

It is then also true that

$$ \int d^3k |k\rangle\langle k|\otimes \mathbb I= \mathbb I\otimes\mathbb I= \mathbb I \otimes \sum_{\nu }|\nu\rangle\langle \nu|$$.

TL;DR the completeness relation must be independently valid for the two bases, and you could have ignored spin if $G$ only acts on the spatial part.

Answer 2

I think that everything you say is quite correct. If $H_0$ does not act on spin, you can either just ignore spin , or you could extend $H_0\to H_0\otimes {\mathbb I}_{\rm spin}$. Both have the same effect.