Does quantization always work?

[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?

 

[TrickyDicky]

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.

Right. It was addressed in #87. Is it also not useful?

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.

Is that different from introducing quasiparticle states either in the H atom like above or in the lattice?

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?

I tried to explain that the only problem is time-dependency in the inf-dim. case(and it is not an issue just for the tensor product postulate). The rest is fine. Do you agree?

 

[dextercioby]

I'm still having a trouble to understand why those projectors can be used in a tensor product decomposition, since the $|x_e\rangle\$ from that product is not even known, not to mention that the (possibly rigged) Hilbert space in which it "lives" is also not known.

 

[my2cts]

There are perhaps 20 or so quantum chemistry codes around that spend huge amounts of cpu time on calculating zillions of two-particle matrix elements of the kind discussed here. So the problem is formal, not practical.

 

[kith]

Right. It was addressed in #87. Is it also not useful?

I tried to explain that the only problem is time-dependency in the inf-dim. case(and it is not an issue just for the tensor product postulate). The rest is fine. Do you agree?

I didn't get your point. If you are okay with the fact that the interaction Hamiltonian is not in pure tensor form for finite dimensional spaces (like two distinguishable spin-1/2 systems), which step exactly fails in the infinite dimensional case?

Is that different from introducing quasiparticle states either in the H atom like above or in the lattice?

Why should it lead to quasiparticles? You get Physics Monkey's result this way.

 

[kith]

I'm still having a trouble to understand why those projectors can be used in a tensor product decomposition, since the $|x_e\rangle\$ from that product is not even known, not to mention that the (possibly rigged) Hilbert space in which it "lives" is also not known.

$|x_e\rangle$ lives in $\mathcal{H}_e$ and we can forget about the rigged Hilbert space if we use Physics Monkey's approximation. I don't really understand what your issue is. Do you think it is also present if we have two (distinguishable) spin-1/2 systems? If not, what's the crucial difference?

 

[TrickyDicky]

I didn't get your point. If you are okay with the fact that the interaction Hamiltonian is not in pure tensor form for finite dimensional spaces (like two distinguishable spin-1/2 systems), which step exactly fails in the infinite dimensional case?

The composite 2-dimensional HS case is clearly different, any sum of products can be expressed as a pure tensor with a unitary transformation, the projection operator is orthogonal so it is not affected by time-dependent perturbation.

You get Physics Monkey's result this way.

But that approximation amounts to ignoring the interaction(the projection operators are not unitary here). Unless the interaction is so weak that allows you to do it(not the case for hydrogen), you need some simplyfying geometric or energy decomposition in the form of Com/rel or simlar, and that only for the time independent case.

 

[rubi]

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.

This is very misleading. The author just shows that in the category $Hilb$ of Hilbert spaces, there are no tensor products in the sense of tensor products of modules. However, not all tensor products are tensor products of modules. The general definition of tensor product is captured by the concept of a monoidal category. The category of ($R$-)modules is just a particular example of a monoidal category. In $Hilb$, we have a different definition of tensor product, so this is just a terminology issue. When we talk about tensor products in $Hilb$, we really mean the completion of the tensor product of modules with respect to the canonically defined inner product. With this definition of tensor product, $Hilb$ can be shown to satisfy all properties of a monoidal category and so it is legitimate to talk about tensor products of Hilbert spaces.

In practice, all of this is irrelevant, since everyone except the author of that paper has already been using the correct definition of tensor product of Hilbert spaces and it doesn't even require any knowledge of category theory. No theorems need to be fixed and no textbooks need to be rewritten.

 

[Physics Monkey]

I do not concede that there is any problem with the infinite dimensional statements, but for conceptual simplicity I think its best to first be clear about the finite dimensional case (which, again, I claim is sufficient for physics). So let us consider $\mathbb{C}^n \otimes \mathbb{C}^n$. Then

1. Not every operator O is equivalent under unitary transformation to a pure tensor product. For example, consider a density operator (positive, normalized, hermitian) on the tensor product. Because the spectrum of the operator is invariant under unitary transformations we can get a contradiction by counting eigenvalue degrees of freedom. A generic pure tensor product density matrix has $2(n-1)$ free eigenvalues ($2$ for the two product density matrices and $n-1$ because each product density matrix must be normalized). Meanwhile a generic density operator on the tensor product has $n^2-1$ free eigenvalues.

2. The finite dimensional approximation (and maybe its not even an approximation in quantum gravity ) in no way amounts to ignoring the effects of interactions. Interaction effects are present in both statics and dynamics. For example, one can calculate the low lying states of hydrogen to high accuracy using a lattice regulated system.

 

[TrickyDicky]

I do not concede that there is any problem with the infinite dimensional statements, but for conceptual simplicity I think its best to first be clear about the finite dimensional case (which, again, I claim is sufficient for physics). So let us consider $\mathbb{C}^n \otimes \mathbb{C}^n$. Then

1. Not every operator O is equivalent under unitary transformation to a pure tensor product. For example, consider a density operator (positive, normalized, hermitian) on the tensor product. Because the spectrum of the operator is invariant under unitary transformations we can get a contradiction by counting eigenvalue degrees of freedom. A generic pure tensor product density matrix has $2(n-1)$ free eigenvalues ($2$ for the two product density matrices and $n-1$ because each product density matrix must be normalized). Meanwhile a generic density operator on the tensor product has $n^2-1$ free eigenvalues.

2. The finite dimensional approximation (and maybe its not even an approximation in quantum gravity ) in no way amounts to ignoring the effects of interactions. Interaction effects are present in both statics and dynamics. For example, one can calculate the low lying states of hydrogen to high accuracy using a lattice regulated system.

It is not a finite versus infinite issue. The issue arises for time-dependent interactions of 3 or more dimensions Hilbert spaces states, it is there where one has to deal with unavoidable entanglement (non-separability) which is not compatible with the HS tensor product definition.
If the approximation you refer to in #84 is time-independent it should not have problem with tensor products regardless its being finite or infinite-dimensional. But I don't think the expression you use is correct, unless the projectors are orthogonal. The quantum chemistry approximations mentioned bu my2cts are also time-independent so there should be no problem there either(this was already addressed by Jano L).

 

[my2cts]

"Does quantization always work?"
My answer to the original question is yes, as quantum physics always works.
If quantization does not work the applied procedure is wrong, incorrectly executed, or both.
Then the discussion went on about tensor products of Hilbert spaces.
I wonder what is the consequence of this for a simple physical system such as the hydrogen atom.
If there is none, as I believe, the math must be wrong or incorrectly applied.

 

[TrickyDicky]

"Does quantization always work?"
My answer to the original question is yes, as quantum physics always works.
If quantization does not work the applied procedure is wrong, incorrectly executed, or both.
Then the discussion went on about tensor products of Hilbert spaces.
I wonder what is the consequence of this for a simple physical system such as the hydrogen atom.
If there is none, as I believe, the math must be wrong or incorrectly applied.

The specific question was about the hydrogen atom Hilbert space. That has been answered:$\mathcal{H}_{Com}\otimes \mathcal{H}_{rel}=L^2(R^6)$ that avoids the problem with the potential mentioned by dextercioby.
Where do you see wrong or incorrectly applied math here?

 

[bhobba]

If quantization does not work the applied procedure is wrong, incorrectly executed, or both.

That depends on what you mean by 'work'. In QM there is an operator order issue:
http://motls.blogspot.com.au/2012/12/ordering-ambiguities-and.html

If its a problem is debatable.

Thanks
Bill

 

[my2cts]

The specific question was about the hydrogen atom Hilbert space. That has been answered:$\mathcal{H}_{Com}\otimes \mathcal{H}_{rel}=L^2(R^6)$ that avoids the problem with the potential mentioned by dextercioby.
Where do you see wrong or incorrectly applied math here?

My comment is intended to provoke a physically more clear statement of the issue.
Concerning the Hilbert space of H, $\mathcal{H}_{p}\otimes \mathcal{H}_{e}=L^2(R^6)$ also does the job.

 

[my2cts]

That depends on what you mean by 'work'. In QM there is an operator order issue:
http://motls.blogspot.com.au/2012/12/ordering-ambiguities-and.html
Bill

It also depends on what is meant by quantization. In any case, the hydrogen problem poses no problems.
Thus, if there is a mathematical problem it must be the math not the atom ;-).

 

[strangerep]

I'm still having a trouble to understand why those projectors can be used in a tensor product decomposition, since the $|x_e\rangle\$ from that product is not even known, not to mention that the (possibly rigged) Hilbert space in which it "lives" is also not known.

Isn't it? One can take the (generalized) momentum eigenstates for a free electron, giving a rigged Hilbert space expressed in momentum basis. Then perform a Fourier transform...