# Completeness relations in a tensor product Hilbert space

• I
Hello,

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?

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vanhees71
Gold Member
You also need to include the spin in the position-spin basis. Then everything is consistent (I suppose you deal with one particle with spin moving in an external em. field). The position spin-basis is ##|\vec{x},\sigma \rangle## and the momentum-spin basis is ##|\vec{k},\sigma \rangle##. In nonrelativistic physics these are simple Kronecker products, ##|\vec{x},\sigma \rangle=|\vec{x} \rangle \otimes |\sigma \rangle## and ##|\vec{p},\sigma \rangle =|\vec{p} \rangle \otimes \sigma \rangle##. The completeness relations read
$$\int_{\mathbb{R}^3} \sum_{\sigma} |\vec{x},\sigma \rangle \langle \vec{x},\sigma| = \mathbb{I} \equiv \mathbb{I}_{\text{space}} \otimes \mathbb{I}_{\text{spin}}.$$
$$\langle \vec{x},\sigma|\vec{x},\sigma' \rangle=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma \sigma'}$$
and
$$\langle \vec{p},\sigma|\vec{x},\sigma' \rangle=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\sigma \sigma'}$$
And the momentum-spin eigenfunctions in the position representation are given by
$$u_{\vec{p},\sigma}(\vec{x},\sigma')=\langle \vec{x},\sigma'|\vec{p},\sigma \rangle = \frac{1}{(2 \pi \hbar)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p}) \delta_{\sigma,\sigma'}.$$
All matrix elements have both a spatial and a spin label, e.g.,
$$G_0(z; \vec{x},\sigma;\vec{x}',\sigma')=\langle \vec{x},\sigma|\hat{G}_0(z)|\vec{x}',\sigma' \rangle.$$

You also need to include the spin in the position-spin basis. Then everything is consistent,

Yes this is definitely a way to solve my issue, but it does not really help me understand what actually happens in these derivations I am reading. Perhaps it would help to give a more specific example, consider the Lippmann-Schwinger equation, $$| \psi, \nu^{(+)}\rangle = |\mathbf{k}, \nu\rangle + G^0V | \psi, \nu^{(+)}\rangle.$$ The free particle boundary condition has been expressed in the momentum spin basis. Now something that I keep seeing is a derivation of the following form, for example for the position representation $$\langle \mathbf{r}| \psi, \nu^{(+)}\rangle = \langle \mathbf{r}|\mathbf{k}, \nu\rangle + \int d^3r' \int d^3r'' \ d^3 \langle \mathbf{r}|G^0| \mathbf{r}'\rangle \langle \mathbf{r}' |V | \mathbf{r}''\rangle \langle \mathbf{r}''| \psi, \nu^{(+)}\rangle.$$ Once this form has been written one may exploit the locality of the potential to simplify the integral. However this only works because I inserted a set of position states only, not a set of spin states. If I do that then I get the matrix element ##\langle \mathbf{r', \nu}|V| \mathbf{r''}, \nu'\rangle##, which is usually not simply diagonal. So it seems like in this case it is very convenient to just insert position states and leave the spin states as the identity operator on that space. Obviously however, this is only allowed if the completeness relation in my original post is valid, which is the point I am not sure about.

vanhees71
Gold Member
I don't understand your notation then. As I understand it, it's not making sense. You need always complete bases defined by a complete set of compatible observables. You can have the position-spin eigenbasis ##|\vec{r},\sigma \rangle## (where ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}## where ##s \in \{0,1/2,1,\ldots \}## is the spin of the particle) or the momentum-spin eigenbasis ##|\vec{p},\sigma \rangle##. You cannot have the scalar product of ##|\vec{r} \rangle## with ##|\vec{k},\sigma \rangle##, because you cannot have a scalar product between vectors of two different spaces.

Perhaps it helps, if you specify the concrete scattering problem you are looking at or cite the textbook/paper you are reading!

I have been thinking about this more as well. I do not mean a scalar product of ##| \mathbf{r}\rangle## with ##|\mathbf{k, \sigma}\rangle##, which as you said does not make sense. Rather what I do mean is if I view ##\langle\mathbf{r}|## as a projection operator, then I can surely form an operator product ##\langle\mathbf{r}| \otimes \mathbb{I}_{\mathrm{spin}}## and apply that to ##|\mathbf{k, \sigma}\rangle##. I guess I will then get some sort of partial projection of the momentum-spin vector, which is still an abstract vector in spin space. It reminds me somewhat of taking a partial trace in quantum optics.

As for the book I am reading, here is a small excerpt. I refrained from directly citing it before since it does not use Dirac notation explicitly. But perhaps the following passage will help,

Book: Roger G. Newton: Scattering Theory of Waves and Particles (Great book by the way, has been of tremendous help to me).

You can see in equation 10.10 a very similar expansion as I wrote down in my previous comment. The ##\psi_0(\mathbf{k}s\nu ; \mathbf{r})## is in my notation the projection ##\langle \mathbf{r}|\mathbf{k},s, \nu\rangle## with ##s, \nu## spin quantum numbers. Here (and in several other derivations) it seems to me as if the author injects just a complete set of position states, without also injecting spin states with it. In 10.11 then you can see that the resulting wave function is still a spin vector, since the symbol ##\chi_{\nu}^{s}## is simply the ket ##\ket{\nu,s}## in my notation.

vanhees71
Gold Member
Sure, that's a great book and here the notation is pretty clear.

Here Newton works with wave fucntions, and ##\psi^{(+)}## is a spinor-valued wave function. The correct translation between this and the Dirac bra-ket notation is
$$\psi(s \nu;\vec{r})=\langle s \nu \vec{r}|\psi \rangle.$$
where ##|s\nu \vec{r} \rangle## is the position-spin eigenvector (I left out the ##s##, because it's a fixed number for a given particle species).

The additional label ##\vec{k}## is a label for indicating that the incoming asymptotic state is a plane wave with momentum ##\vec{k}##. Here you consider energy eigenstates describing the scattering of a particle in a potential (maybe spin dependent). The energy is given by the incoming free particle, i.e., ##E=\hbar^2 \vec{k}^2/(2m)##. (10.11) describes the asymptotic free scattered state, i.e., the wave function at large distance from the scattering center. It consists of the superposition of the incoming plane wave and a spherical wave which comes from the scattering on the potential.

Alright, I see how that works for the scattering state yes, but he specifically states that ##\psi_0(\mathbf{k}s\nu;\mathbf{r})## is a plane wave state. For example see the following passage,

It could very well be (in fact I am starting to think its likely) that my confusion is just simply because I am not interpreting this notation correctly. But then I do wonder how I can make it consistent. What is the dirac notation equivalent of ##\psi_0(\mathbf{k}s\nu;\mathbf{r})##? And of ##\chi_{\nu}^s##?

Thanks a lot for your help by the way.

vanhees71
Gold Member
Ok, the ##\psi_0## is the plane wave of the incoming particle. Newton writes it with all the parameters. It's a momentum eigenstate with momentum ##\vec{k}##. Since it's a free particle that's at the same time an energy eigenstate of energy ##E=\vec{k}^2/(2m)##. Further it's a particle with a specific spin-##z## component ##\nu## of particle with spin ##s##. ##\chi_{\nu}^s## is the simultaneous eigenvector of the ##\hat{\vec{s}}^2## to the eigenvalue ##s(s+1)## and ##\hat{s}_z## to the eigenvalue ##\nu## (##\nu \in \{-s,-s+1,\ldots,s-1,s \}##).

Further he normalizes the state in a specific way, convenienf for the scattering problem. It's clear that you have
$$\int_{\mathbb{R}^3} \mathrm{d}^3 r \psi_0^*(\vec{k}s\nu,\vec{r}) \psi_0(\vec{k}'s' \nu',\vec{r})=C \delta^{(3)}(\vec{k}-\vec{k}') \delta_{ss'} \delta_{\nu \nu}'.$$
Now consider ##\vec{k}## and ##\vec{k}'## written in spherical coordinates ##(k,\vartheta,\varphi)## and ##(k',\vartheta',\varphi')##. It's easy to show that the ##\delta## distribution transforms into
$$\delta^{(3)}(\vec{k}-\vec{k}')=\frac{1}{k^2 \sin \vartheta} \delta(k-k') \delta(\vartheta-\vartheta') \delta(\varphi-\varphi')=\frac{1}{k^2} \delta(k-k') \delta^{(2)}_{\Omega}(\hat{k},\hat{k}').$$
Now you can express ##k## and ##k'## by the corresponding energies
$$E=k^2/(2m) \; \Rightarrow \; k^2=2m E.$$
Since ##k>0## you can uniquely write
$$\delta(k-k')=2k \delta(k^2-k^{\prime 2})=2k \delta[2m(E-E')]=\frac{k}{m}\delta(E-E').$$
So you can choose ##C## such that the normalization condition (10.3) is fulfilled.

Okay yes I understand all of this, but this is also all consistent with my notation so it would seem there is no problem there. But then I am just back at my original question again. Why is the author allowed to insert a set of position states only, instead of inserting position-spin states. It still appears to me as if in expression 10.10 the author inserts $$\int d^3r \ \mathbb{I}_{\mathrm{spin}}\otimes |\mathbf{r}\rangle \langle \mathbf{r}| \otimes \mathbb{I}_{\mathrm{spin}},$$ instead of inserting,$$\sum_{s \nu} \int d^3r \ |\mathbf{r}, s,\nu\rangle \langle \mathbf{r}, s, \nu|.$$ If he actually inserted the second relation I would expect to see spin vectors in the integral, which I don't see. So then I get back to my original question, are these two expressions actually equivalent?

vanhees71
Gold Member
That's ok, because you can write
$$|\vec{r},s,\nu \rangle=|\vec{r} \rangle \otimes |s,\nu \rangle$$
and thus
$$|\vec{r},s,\nu \rangle \langle \vec{r},s,\nu| = |\vec{r} \rangle \langle \vec{r}| \otimes |s,\nu \rangle \langle s,\nu \rangle|.$$
Then summing over ##s## and ##\nu## only gives ##|\vec{r} \rangle \langle \vec{r}| \otimes \mathbb{I}##. The completeness relation in the latter form is however what you need for matrix elements of arbitrary one-particle operators, because for a general operator ##\hat{A}## you need the matrix elements
$$A(\vec{r},s,\nu;\vec{r}',s',\nu') = \langle \vec{r},s,\nu|\hat{A}|\vec{r}',s',\nu' \rangle.$$
Maybe Newton has operators in mind which trivially act on the spin, i.e., ##\hat{A}=\hat{A}(\vec{x},\vec{p}) \otimes \mathbb{I}_{\text{spin}}##, because then
$$A(\vec{r},s,\nu;\vec{r}',s',n')=\tilde{A}(\vec{r},\vec{r}') \delta_{ss'} \delta_{\nu \nu'}?$$

Decimal
Great! This is exactly what I was hoping for. Indeed for arbitrary particle operators if you want the full matrix element as just a c-number you will need ##\langle\mathbf{r}, s, \nu|\hat{A}| \mathbf{r}', s',\nu'\rangle##. But perhaps what Newton means here is that he actually computes the 'partial' element $$\langle\mathbf{r}|V|\mathbf{r}'\rangle = \mathbb{I}_{\mathrm{spin}} \otimes \langle\mathbf{r}|V|\mathbf{r}'\rangle \otimes \mathbb{I}_{\mathrm{spin}}.$$ This is then still an operator in spin space, but not in position space. I would assume that as long as you keep that in mind you are not yet doing anything illegal. That would also make all my derivations consistent with each other I think.

Thanks again!

kith
$$\langle\mathbf{r}| \otimes \mathbb{I}_{\mathrm{spin}}$$
$$\int d^3r \ \mathbb{I}_{\mathrm{spin}}\otimes |\mathbf{r}\rangle \langle \mathbf{r}| \otimes \mathbb{I}_{\mathrm{spin}},$$
$$\langle\mathbf{r}|V|\mathbf{r}'\rangle = \mathbb{I}_{\mathrm{spin}} \otimes \langle\mathbf{r}|V|\mathbf{r}'\rangle \otimes \mathbb{I}_{\mathrm{spin}}.$$
These expressions don't look correct. In the first you have a tensor product of a vector and a matrix. In the second, you have a tensor product of three operators which would correspond to a particle with two different spin-like degrees of freedom. It also isn't consistent with the last equation in your original post. Check again @vanhees71's post #10.

Oh yeah I see, I got a little carried away inputting the identity matrices. Indeed it should be, $$\int d^3 r \ |\mathbf{r}\rangle \langle \mathbf{r}| \otimes \mathbb{I}_{\mathrm{spin}},$$ and, $$V = \int d^3 r \int d^3 r' \ |\mathbf{r}\rangle \langle \mathbf{r}| V |\mathbf{r}'\rangle \langle \mathbf{r}'|\otimes \mathbb{I}_{\mathrm{spin}}.$$ But I dont think that changes my conclusion, these are now still consistent with Newton I think. In this case I would still interpret the second expression as an operator in spin space.

I am however not so sure what ##\langle\mathbf{r}|## would be formally then. It does appear to me as if Newton only projects onto position states, since for example in equation 10.11 in my post #5 the spin vectors are still present in the equation. If he had projected onto ##\langle \mathbf{r}, s, \nu|## I would have just expected some delta functions there.

vanhees71
Great! This is exactly what I was hoping for. Indeed for arbitrary particle operators if you want the full matrix element as just a c-number you will need ##\langle\mathbf{r}, s, \nu|\hat{A}| \mathbf{r}', s',\nu'\rangle##. But perhaps what Newton means here is that he actually computes the 'partial' element $$\langle\mathbf{r}|V|\mathbf{r}'\rangle = \mathbb{I}_{\mathrm{spin}} \otimes \langle\mathbf{r}|V|\mathbf{r}'\rangle \otimes \mathbb{I}_{\mathrm{spin}}.$$ This is then still an operator in spin space, but not in position space. I would assume that as long as you keep that in mind you are not yet doing anything illegal. That would also make all my derivations consistent with each other I think.