Does quantization always work?

[vanhees71]

I don't understand this. What other Hilbert space, do you think is needed to describe the hydrogen atom?

 

[TrickyDicky]

I don't understand this. What other Hilbert space, do you think is needed to describe the hydrogen atom?

As you know, because you have explained it here, the Hilbert spaces used depending on how realistic or how good you want the approximation are: $L^2(R^3)$ , or the $cm \times rel$ exploiting the symmetry of the potential (or even Pauli's $L^2(R^3)\otimes C^2$) but you will never see on any textbook describing the H atom with $H_e \otimes H_p$ and getting any reasonable result, the reason is that it is not a separable state which is the condition to treat H as a pure(elementary) tensor hilbert space. But for some strange reason that you might try and explain, you and other sources keep saying that it can be described by $H_e \otimes H_p$ as a matter of fact, is it any more than a rhetorical claim without any base in the practice of QM?

 

[vanhees71]

I still don't understand this statement. The Hilbert space is $L^2(\mathbb{R}^6,\mathbb{C})$, and the energy eigenfunctions are
$$u_{\vec{P},nlm}(\vec{R},r,\vartheta,\varphi)=N \exp(\mathrm{i} \vec{P} \cdot \vec{R}) R_n(r) Y_{lm}(\varphi,\vartheta),$$
with $\vec{R}=(m_e \vec{r}_e+m_p \vec{r}_p)/(m_e+m_p)$, $\vec{P}=\vec{p}_e+\vec{p}_p$, $\vec{r}=r(\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta)=\vec{r}_e-\vec{r}_p)$. As you see, the energy eigenfunctions are indeed in $\mathcal{H}_e \otimes \mathcal{H}_p$. Why do you think, it's not? Of course the energy eigenstates are not a product of electron and proton wave functions, but it's still in $\mathcal{H}_e \otimes \mathcal{H}_p$. Again, what other Hilbert space should it be, if you have a system consisting of an electron and a proton?

 

[TrickyDicky]

I still don't understand this statement. The Hilbert space is $L^2(\mathbb{R}^6,\mathbb{C})$, and the energy eigenfunctions are
$$u_{\vec{P},nlm}(\vec{R},r,\vartheta,\varphi)=N \exp(\mathrm{i} \vec{P} \cdot \vec{R}) R_n(r) Y_{lm}(\varphi,\vartheta),$$
with $\vec{R}=(m_e \vec{r}_e+m_p \vec{r}_p)/(m_e+m_p)$, $\vec{P}=\vec{p}_e+\vec{p}_p$, $\vec{r}=r(\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta)=\vec{r}_e-\vec{r}_p)$. As you see, the energy eigenfunctions are indeed in $\mathcal{H}_e \otimes \mathcal{H}_p$. Why do you think, it's not? Of course the energy eigenstates are not a product of electron and proton wave functions, but it's still in $\mathcal{H}_e \otimes \mathcal{H}_p$. Again, what other Hilbert space should it be, if you have a system consisting of an electron and a proton?

But why do you restrict the discussion to the time-independent situation, I'm referring to the general situation that includes dynamics.

 

[TrickyDicky]

When you add a time-dependent interaction between electron and proton there is no longer a completed tensor product Hilbert space for the hamiltonian of the hydrogen atom, and it can not be described as a pure tensor. I guess one can abstractly say that in the absence of interaction between proton and electron the dynamical hydrogen atom Hilbert space state could be described as the completed tensor product of the states of the electron and proton but then one needs to wonder if the object one is describing can still be called an H atom(wich is defined by this interaction).

 

[atyy]

I still don't understand this statement. The Hilbert space is $L^2(\mathbb{R}^6,\mathbb{C})$, and the energy eigenfunctions are
$$u_{\vec{P},nlm}(\vec{R},r,\vartheta,\varphi)=N \exp(\mathrm{i} \vec{P} \cdot \vec{R}) R_n(r) Y_{lm}(\varphi,\vartheta),$$
with $\vec{R}=(m_e \vec{r}_e+m_p \vec{r}_p)/(m_e+m_p)$, $\vec{P}=\vec{p}_e+\vec{p}_p$, $\vec{r}=r(\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta)=\vec{r}_e-\vec{r}_p)$. As you see, the energy eigenfunctions are indeed in $\mathcal{H}_e \otimes \mathcal{H}_p$. Why do you think, it's not? Of course the energy eigenstates are not a product of electron and proton wave functions, but it's still in $\mathcal{H}_e \otimes \mathcal{H}_p$. Again, what other Hilbert space should it be, if you have a system consisting of an electron and a proton?

As I understand it, dextercioby thinks the Coulomb potential is not expressible as a sum of terms in which each term is a product of factors acting on the single particle Hilbert spaces. In order to say the Hilbert space is a tensor product, we usually also mean that the operators are expressible as a sum of products of operators defined on the single particle Hilbert spaces.

I don't know if there is a proof that the Coulomb potential does not have an expansion as a sum of separable terms, but I also don't find any known separable expansion of the Coulomb potential.

 

[vanhees71]

This is also a misunderstanding. Of course, the Coulomb potential is well-defined to act on the product space, which is simply $\mathcal{H}_e \otimes \mathcal{H}_l \sim L^2(\mathbb{R}^6,\mathbb{C})$. You can use the simultaneous generalized position eigenvectors of electron and positron as a product basis, and then the states are represented by the square integrable complex functions (modulo a factor as usual). The operator, representing the Coulomb potential then acts on these functions in the straight-forward way
$$\hat{V} \Psi(\vec{x}_e,\vec{x}_p)=-\frac{e^2}{4 \pi |\vec{x}_e-\vec{x}_p|} \Psi(\vec{x}_e,\vec{x}_p).$$
There's no need to expand it in a sum of products of functions of $\vec{x}_e$ and $\vec{x}_p$.

I also don't understand the principle problem for general states. The time evolution of the wave function is governed by the two-body Schrödinger equation with the Hamiltonian
$$\hat{H}=-\frac{1}{2m_e} \Delta_e - \frac{1}{2m_p} \Delta_p-\frac{e^2}{4 \pi |\vec{x}_e-\vec{x}_p|}.$$

 

[atyy]

The operator, representing the Coulomb potential then acts on these functions in the straight-forward way
$$\hat{V} \Psi(\vec{x}_e,\vec{x}_p)=-\frac{e^2}{4 \pi |\vec{x}_e-\vec{x}_p|} \Psi(\vec{x}_e,\vec{x}_p).$$
There's no need to expand it in a sum of products of functions of $\vec{x}_e$ and $\vec{x}_p$.

Everything you say is standard textbook. If I understand dextercioby correctly, he is saying that in order to say the Hilbert space is a tensor product, one must also meet the requirement to expand the potential in products of functions of $\vec{x}_e$ and $\vec{x}_p$.

 

[vanhees71]

Of course, it's all standard textbook. I thus still don't understand the problem and why should one need to expand the interaction potential in this kind of products. The point is, that it is a two-body operator, making the system interacting. For more complicated forces like the nuclear force you have a whole cluster expansion with two-body, three-body,... N-body pieces.

 

[dextercioby]

From my point of view, everything that needed to be said was said. This is not going into a positive direction. I can sum up my posts here as follows:

The H-atom in its simplest description (a non-specially relativistic Hamiltonian without spin) is a 2-particle system made up of an electron interacting with a proton through the Coulomb potential. You need an axiom (or at least a statement) of QM to describe how the Hilbert space of the whole system is built, if you already know the Hilbert space of the subsystems. I claimed that natural proposal of the tensor product is useless for the H atom (and as reinforced by the posters here also for other models), because the potential operator is ill-defined in such a tensor product space. For the potential operator to be well-defined, it needed to be described as a tensor product. There's a mathematical statement which cures this issued mentioned in the OP which tells us that the the correct Hilbert space is L²(R⁶) which solves the problem raised by the potential operator.
Since most textbooks of QM don't speak about the Hilbert spaces of the various models being described (and the H-atom appears in almost every textbook), I think that this argument I made is useful. Entanglement and means to solve the spectral problem of the Hamiltonian are only secondary.

 

[vanhees71]

I disagree. It's all well-defined, and the Hilbert space of a hydrogen atom in this approximation is the tensor-product space of the electron and proton Hilbert spaces. I've given the reasons for this in the previous postings. It's standard textbook knowledge. At least you find it already in Sommerfeld's "Atombau und Spektrallinien".

 

[ShayanJ]

It seems to me that this is relevant here and may help. The author proves that the tensor product of infinite dimensional Hilbert spaces do not exist.

 

[vanhees71]

Well, then why does the standard formalism work so well for the hydrogen atom? We are working in rigged Hilbert space, by the way!

 

[TrickyDicky]

It seems to me that this is relevant here and may help. The author proves that the tensor product of infinite dimensional Hilbert spaces do not exist.

Shyan, thanks a lot for finding that proof!

Well, then why does the standard formalism work so well for the hydrogen atom?

But the standard formalism never uses the tensor product for systems like the hydrogen atom in practice as we have seen in this thread! As other posters said it is only used (when continuous operators and dynamic interactions are involved, I'm not obviously talking about 2-dimensional Hilbert spaces like those involving spin) as a very crude approximation for certain problems in chemistry.

We are working in rigged Hilbert space, by the way!

Obviously, and?

I also don't understand the principle problem for general states. The time evolution of the wave function is governed by the two-body Schrödinger equation with the Hamiltonian
$$\hat{H}=-\frac{1}{2m_e} \Delta_e - \frac{1}{2m_p} \Delta_p-\frac{e^2}{4 \pi |\vec{x}_e-\vec{x}_p|}.$$

This operator doesn't have the form of a pure tensor and therefore cannot be the Hamiltonian acting linearly on a completed tensor product Hilbert space state.
The two-body Schrodinger equation with center of mass and reduced mass electron doesn't use that Hamiltonian and it is not in a $H_e \otimes H_p R^6$ space but an $L^2(R^3)$ space like the usual solution just with a small correction by using the reduced mass electron(taking advantage of the $SO^4$ symmetry of the problem).

 

[my2cts]

As you see, the energy eigenfunctions are indeed in $\mathcal{H}_e \otimes \mathcal{H}_p$. Why do you think, it's not? Of course the energy eigenstates are not a product of electron and proton wave functions, but it's still in $\mathcal{H}_e \otimes \mathcal{H}_p$.

If this is true, and I believe so, then the wave function can be explicitly written as an expansion of terms of the form $\mathcal{H}_e \otimes \mathcal{H}_p$. A sum can describe electron-proton correlation even if the individual terms do not. I would like to see the result of such an exercise.

 

[vanhees71]

Yes, of course
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \langle \vec{x}_e,\vec{x}_p |\Psi \rangle.$$
That writes any state of the electron-proton system in terms of (generalized) product states, $|\vec{x}_e,\vec{x}_p \rangle$.

 

[TrickyDicky]

Yes, of course
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \langle \vec{x}_e,\vec{x}_p |\Psi \rangle.$$
That writes any state of the electron-proton system in terms of (generalized) product states, $|\vec{x}_e,\vec{x}_p \rangle$.

In the discrete case.

 

[my2cts]

Yes, of course
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \langle \vec{x}_e,\vec{x}_p |\Psi \rangle.$$
That writes any state of the electron-proton system in terms of (generalized) product states, $|\vec{x}_e,\vec{x}_p \rangle$.

And what is the result for NR H ?

 

[strangerep]

But the standard formalism never uses the tensor product for systems like the hydrogen atom in practice as we have seen in this thread!

Well, it is used actually, but not in the naive form of a tensor product of free particle Hilbert spaces $\mathcal{H}_e \otimes \mathcal{H}_p$. Indeed, the free case has strictly positive energy, hence would have a hard time describing negative-energy bound states.

The useful tensor product involves a product between spaces representing functionally independent degrees of freedom. I.e., the CoM and relative degrees of freedom. This is even intuitively "natural": we think of the H-atom as sitting motionless in front of us, hence (in our imagination) we have already separated out these functionally independent degrees of freedom.

But this tends to be done when we're still thinking in classical terms, with a classical phase space. Then we bother only about the detailed quantization of the relative degrees of freedom. But, strictly speaking, we still have some kind of (possibly generalized) tensor product of the (quantizations of) the CoM and relative degrees of freedom.

BTW, this also relates to the earlier confusion about "entanglement" in the context of this problem: if one has a non-elementary system, it is possible (in general) that decomposing it one way may result in entangled pieces, but decomposing it a different way might not. I.e., entanglement is not necessarily invariant under different decompositions of a given system. (Ref: ch 21 of Ballentine, 2015 edition.)

This highlights the lesson of this thread, (according to me, anyway), that for quantization of a complex system to work, an essential technique is that one must think in terms of functionally independent degrees of freedom (and associated symmetries) of the total system, rather than arbitrarily restricting oneself to one particular (fictitious) decomposition of that system

 

[atyy]

From my point of view, everything that needed to be said was said. This is not going into a positive direction. I can sum up my posts here as follows:

The H-atom in its simplest description (a non-specially relativistic Hamiltonian without spin) is a 2-particle system made up of an electron interacting with a proton through the Coulomb potential. You need an axiom (or at least a statement) of QM to describe how the Hilbert space of the whole system is built, if you already know the Hilbert space of the subsystems. I claimed that natural proposal of the tensor product is useless for the H atom (and as reinforced by the posters here also for other models), because the potential operator is ill-defined in such a tensor product space. For the potential operator to be well-defined, it needed to be described as a tensor product. There's a mathematical statement which cures this issued mentioned in the OP which tells us that the the correct Hilbert space is L²(R⁶) which solves the problem raised by the potential operator.
Since most textbooks of QM don't speak about the Hilbert spaces of the various models being described (and the H-atom appears in almost every textbook), I think that this argument I made is useful. Entanglement and means to solve the spectral problem of the Hamiltonian are only secondary.

I disagree. It's all well-defined, and the Hilbert space of a hydrogen atom in this approximation is the tensor-product space of the electron and proton Hilbert spaces. I've given the reasons for this in the previous postings. It's standard textbook knowledge. At least you find it already in Sommerfeld's "Atombau und Spektrallinien".

Two questions.

1) Is what I said in post #68 a correct representation of dextercioby's view?

2) Since the question comes down to whether the operator is well-defined depending on the Hilbert space, maybe some references as to the definitions being used should be given?

 

[TrickyDicky]

Well, it is used actually, but not in the naive form of a tensor product of free particle Hilbert spaces $\mathcal{H}_e \otimes \mathcal{H}_p$. Indeed, the free case has strictly positive energy, hence would have a hard time describing negative-energy bound states.

The useful tensor product involves a product between spaces representing functionally independent degrees of freedom. I.e., the CoM and relative degrees of freedom. This is even intuitively "natural": we think of the H-atom as sitting motionless in front of us, hence (in our imagination) we have already separated out these functionally independent degrees of freedom.

But this tends to be done when we're still thinking in classical terms, with a classical phase space. Then we bother only about the detailed quantization of the relative degrees of freedom. But, strictly speaking, we still have some kind of (possibly generalized) tensor product of the (quantizations of) the CoM and relative degrees of freedom.

Yes, I know. I went a bit overboard in #74 trying to emphasize my point and on rereading it I think some explanation is in order. What you (and vanhees) say about the two-body functionally independent treatment of the hydrogen atom is of course true in the context of E<0. But my point was that when applying the naive tensor product one obtains 6 coordinates, 3 for the free center of mass and three for the relative degrees of freedom, and we obviously concentrate on the latter to solve the bound state problem using the spherical symmetry, this was what I meant when I referred to $L^2(R^3)$,that in practice we are using just those three degrees of freedom to solve the problem, i.e. the translational invariance of the potential allows us to treat it as a one particle with reduced mass problem. So as you say we are not using the naive $\mathcal{H}_e \otimes \mathcal{H}_p$ tensor product but a bit less naive $\mathcal{H}_{CoM} \otimes \mathcal{H}_{rel}$(BTW this was already pointed out by vanhees in a previous post) that in actuality gets us back to $L^2(R^3)$ FAPP.
What I'm saying is that even after addressing the problem regarding separability of electron and proton by using the functional degrees of freedom this tensor product might still be naive for instance if one takes seriously the proof about the impossibility of infinite dimensional Hilbert spaces tensor products linked above by Shyan. Vanhees71 seemed to suggest that the use of rigged Hilbert spaces would be important here but I would like to know how.

BTW, this also relates to the earlier confusion about "entanglement" in the context of this problem: if one has a non-elementary system, it is possible (in general) that decomposing it one way may result in entangled pieces, but decomposing it a different way might not. I.e., entanglement is not necessarily invariant under different decompositions of a given system. (Ref: ch 21 of Ballentine, 2015 edition.)

Yes, entanglement is basis-dependent, therefore context dependent. But I still think that in the dynamic case(so I'm not referring here to the time-independent Schrodinger equation that we have been discussing above but the time-dependent situation in which the time evolution operator needs to include an interaction Hamiltonian that is not in the form of a pure tensor), there is no way to avoid entanglement and therefore there is no separability, so the tensor product postulate in the OP loses generality if this is not added. This obviously even before entering on the validity of the above-mentioned proof about impossibility of tensor products for infinite-dimensional HS.

This highlights the lesson of this thread, (according to me, anyway), that for quantization of a complex system to work, an essential technique is that one must think in terms of functionally independent degrees of freedom (and associated symmetries) of the total system, rather than arbitrarily restricting oneself to one particular (fictitious) decomposition of that system

Agreed.

 

[my2cts]

But what about the potential? Well, I claim that it's not possible to write it in a tensor product way, due to the complicated form: square root, power of -1, sum.

Yes we can :
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \frac{-e^2}{4 \pi |\vec{x}_e-\vec{x}_p|}\langle \vec{x}_e,\vec{x}_p |.$$
This puts the potential on a basis of product states, $|\vec{x}_e,\vec{x}_p \rangle$.
This type of integrals is routinely calculated in Quantum Chemistry codes for the description of e-e repulsion on a basis set.

 

[TrickyDicky]

Yes we can :
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \frac{-e^2}{4 \pi |\vec{x}_e-\vec{x}_p|}\langle \vec{x}_e,\vec{x}_p |.$$
This puts the potential on a basis of product states, $|\vec{x}_e,\vec{x}_p \rangle$.
This type of integrals is routinely calculated in Quantum Chemistry codes for the description of e-e repulsion on a basis set.

We can(in the stationary case) basically because of what dextercioby explained, the isomorphism between separable infinite dimensional Hilbert spaces. The tensor product salvaged by this isomorphism is basically a cartesian product that dodges the impossibility proof by Garrett mentioned previously.
Usually the QM postulates include the condition of separability in the first postulate, and maybe this is the reason why it is not always specified in the tensor product postulate.
The separability condition should have consequences when one cannot find a separable decomposition, specifically in the dynamic and continuous operator case with interactions involving nonseparability, i.e. entangled cases where one cannot avoid the entanglement by any decomposition, or preparation(superselection sectors...). Of course in such cases the usual postulates(not only the tensor product one but the state vector, collapse and time evolution postulates) don't seem adequate.

 

[Physics Monkey]

I gather this has already been said by various people, but I wanted to add my voice to those pointing out that the Coulomb potential can be defined in the tensor product Hilbert space.

To make things conceptually simple, let us agree to regulate everything, say by putting the system in a box of some arbitrary large size and introducing a lattice cutoff. The details are not crucial and can be negotiated but it renders most objections about infinite dimensional Hilbert spaces and so forth moot. Furthermore, I claim no physics can depend on the presence of a cutoff provided it is made suitably small (or large) as appropriate.

Then we could write the potential as
\hat{V} = \sum_{x_e, x_p} V(x_e - x_p) |x_e \rangle \langle x_e| \otimes |x_p \rangle\langle x_p |.
There are also other versions of this formula, e.g. in many particle physics, where we replace the projectors onto electron and proton positions with densities.

 

[my2cts]

Many-electron systems such as molecules are certainly entangled. Quantum chemists have treated these for many decades, using an isomorphism that they were likely not aware of. At least I was not and I never met or read a quantum chemist that discussed this isomorphism. How is this possible? It is nice to have a mathematical proof but it is obvious that with sufficiently large basis sets and using a sufficiently large number of product functions, in casu slater determinants, any many-particle function can be described.

 

[dextercioby]

Yes we can :
$$|\Psi \rangle = \int \mathrm{d}^3 \vec{x}_e \int \mathrm{d}^3 \vec{x}_p |\vec{x}_e,\vec{x}_p \rangle \frac{-e^2}{4 \pi |\vec{x}_e-\vec{x}_p|}\langle \vec{x}_e,\vec{x}_p |.$$
This puts the potential on a basis of product states, $|\vec{x}_e,\vec{x}_p \rangle$.
This type of integrals is routinely calculated in Quantum Chemistry codes for the description of e-e repulsion on a basis set.

I gather this has already been said by various people, but I wanted to add my voice to those pointing out that the Coulomb potential can be defined in the tensor product Hilbert space.

To make things conceptually simple, let us agree to regulate everything, say by putting the system in a box of some arbitrary large size and introducing a lattice cutoff. The details are not crucial and can be negotiated but it renders most objections about infinite dimensional Hilbert spaces and so forth moot. Furthermore, I claim no physics can depend on the presence of a cutoff provided it is made suitably small (or large) as appropriate.

Then we could write the potential as
\hat{V} = \sum_{x_e, x_p} V(x_e - x_p) |x_e \rangle \langle x_e| \otimes |x_p \rangle\langle x_p |.
There are also other versions of this formula, e.g. in many particle physics, where we replace the projectors onto electron and proton positions with densities.

I was thinking about the same things when I first wrote the OP. Namely try to express the potential operator simply as

\hat{V} = \frac{1}{4\pi\epsilon_0 |\vec{x}_e-\vec{x}_p|} \hat{1}_{e}\otimes\hat{1}_{p}

But now I wonder if this makes any sense because the fraction could belong to any of the 2 unit operators, or perhaps to both at the same time?

 

[TrickyDicky]

I gather this has already been said by various people, but I wanted to add my voice to those pointing out that the Coulomb potential can be defined in the tensor product Hilbert space.
...
Then we could write the potential as
\hat{V} = \sum_{x_e, x_p} V(x_e - x_p) |x_e \rangle \langle x_e| \otimes |x_p \rangle\langle x_p |.

Dirac formalism is often deceiving, see below.

I was thinking about the same things when I first wrote the OP. Namely try to express the potential operator simply as

\hat{V} = \frac{1}{4\pi\epsilon_0 |\vec{x}_e-\vec{x}_p|} \hat{1}_{e}\otimes\hat{1}_{p}

But now I wonder if this makes any sense because the fraction could belong to any of the 2 unit operators, or perhaps to both at the same time?

In the general case(that includes dynamics) to both, it multiplies the two particle wave function $ψ(r1, r2)$ and that is the problem.
At the risk of repeating myself and others but it seems my point is not getting across(if this is incorrect or not pertinent I would like to know why):
If one has a time evolution operator $U_e(t) = e^{− \frac{i}{ħ} tH_e}$ with Hamiltonian operator $H_e$, for the electron and $U_p(t) = e^{−\frac{i}{ħ} tH_p}$ with $H_p$,for the proton and if there is no interaction between the two systems, then the time evolution of the total system is described by the operator $U(t) = e^{−\frac{i}{ħ} tH}$ , with U(t) the tensor product operator $U_e(t) ⊗ U_p(t)$ and total Hamiltonian $H = H_e ⊗\hat{1}_p+\hat{1}_e⊗ H_p$.

More interesting is our situation in which the two systems interact, like in the hydrogen atom and in that case an interaction Hamiltonian $H_{12}$ (that involves that the potential affects both e and p simultaneously: this Hamiltonian-and therefore the unitary evolution operator that builds- has no longer the form of a pure tensor, the only possible form for tensor products of infinite-dimensional separable Hilbert spaces) is added to $H = H_1 + H_2$, this addition makes impossible the use of the tensor product because it breaks the separability condition as commented above. In the stationary case one can avoid this by using the symmetries, the translational invariance of the interaction potential allows one to obtain separable states by using CoM and rel systems, but that is not the general case. In this case of infinite dimensional separable Hilbert spaces all of them are isomorphic anyway and the tensor product is unique up to isomorphism so it would seem the domain used doesn't ultimately make a big difference mathematically (whether it is $R^3$ or any other $R^n$).

 

[kith]

Physics Monkey makes a good point. For a start, does everyone agree that there's no problem with interacting systems in the case of finite-dimensional spaces or is this controversial?

 

[kith]

I was thinking about the same things when I first wrote the OP. Namely try to express the potential operator simply as

\hat{V} = \frac{1}{4\pi\epsilon_0 |\vec{x}_e-\vec{x}_p|} \hat{1}_{e}\otimes\hat{1}_{p}

I don't think that this by itself is useful. \hat{1}_{e}\otimes\hat{1}_{p} is the identity in the big space. Either, the fraction is also an operator, then you can omit the identity. Or the fraction is a number, then \hat V is proportional to the identity and doesn't describe an interaction in the first place.

What you can do is insert another identity in front and decompose both identities in terms of simultaneous eigenstates of \hat{\vec X_e} and \hat{\vec X_p}. The existence of these eigenstates may be a problem but let's follow Physics Monkey's suggestion and have a look at the approximated case first.

 

[ShayanJ]

Not sure whether this is even meaningful or not, but have you considered the expansion w.r.t. Legendre polynomials and then spherical harmonics? Does it even make sense here?