Does quantization always work?

[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...