Does quantization always work?

[kith]

Fine, but I'm asking for product states examples in practice.

Whenever you use a state vector to describe a certain degree of freedom of a system separately from the others, you implicitly assume a product state. So the easiest and most practical example is the usual description of the spin of a particle in the absence of a magnetic field. Of course, nobody bothers to write down the spatial part of the state vector when it isn't needed to analyze the spin properties. A common example where product states are written down explicitly are the outcomes of entanglement experiments like Bell tests.

As far as composite particles are concerned, you can think of all kinds of situations where you bring isolated particles into contact (like molecule formation, reactive scattering or a system in a decohering environment) or separate a composite (induced molecule dissociation, ionization).

If you want to treat the constituents separately, you have to either do a preparation in the beginning (in the case that you isolate two systems and bring them together to form a composite) or in the end (in the case that you separate the constituents and perform further independent experiments). So you either have a product state before the experiment or afterwards.

In the end, it boils down to interpretational questions again. From the Copenhagen perspective, you can get rid of entanglement by preparation, so this works fine. From the MWI perspective you can't, so the situation seems to be more complicated. But the MWI has a problem here which Schwindt called the "factorization problem" (http://arxiv.org/abs/1210.8447). It seems to be pretty empty if you don't introduce a tensor product decomposition somehow.

 

[TrickyDicky]

You said that what I wrote in #19 implies that there can't be entanglement between the particles. But only a potential of the form I wrote down in post #23 prevents entanglement. Since the potential isn't of this form, your statement is incorrect.

We are actually saying the same thing, you already mentioned that potential in 19 and I was referring to that, since there is not much practical use for a Hilbert space product of electrón and protón without their interaction when talking about the hydrogen atom.

 

[kith]

We are actually saying the same thing, [...]

I can't really figure out what this part of the discussion is about. Do you still think that there's a problem with my post #19 like you did in #21? If yes, we are not saying the same thing.

you already mentioned that potential in 19 [...]

I didn't. The potential in product form I mentioned in post #19 describes an interaction between the particles, although clearly not the Coulomb interaction.

and I was referring to that, since there is not much practical use for a Hilbert space product of electrón and protón without their interaction when talking about the hydrogen atom.

I find your usage of language concerning the tensor product a bit confusing. There is no "Hilbert space product of electron and proton with or without interaction". Whether there's an interaction is determined by the Hamiltonian, not by the Hilbert space.

 

[TrickyDicky]

I can't really figure out what this part of the discussion is about.

See below.

I didn't. The potential in product form I mentioned in post #19 describes an interaction between the particles, although clearly not the Coulomb interaction.

You did. You mentioned two forms of the potential, I was referring to the second. Here it is:

the form U(\vec R_e) \otimes W(\vec R_p)

You mentioned it precisely to agree with me that using this form would imply entanglement and therefore can't be written like that, you said the same thing in post 23, and included a third not coulombic common potential for electron and proton.

I find your usage of language concerning the tensor product a bit confusing. There is no "Hilbert space product of electron and proton with or without interaction". Whether there's an interaction is determined by the Hamiltonian, not by the Hilbert space.

You are completely right about this and it is probably the reason why you are disagreeing with me in the first place. Of course your usage is the correct one in the context of the standard form of QM.
I unconscioussly tend to not to separate as clearly as the orthodox postulates demand states from operators and therefore from the interactions they determine. Maybe because I am much more fond of Heisenberg and Dirac(interaction) pictures(and even of the matrix mechanics 1925 view of QM from Heisenberg, Jordan and Born that only used operators) than the Schrodinger picture.

 

[atyy]

If the Hilbert spaces are not tensor products, then it seems that entanglement is not so easy to define.

 

[TrickyDicky]

To summarize, we all agree that the potential between electron and proton in the hydrogen atom cannot be written as a tensor product, and we agree that the postulate only refers strictly to the states, not the hamiltonian operator, but the implicit issue is, what does one do with $\mathcal{H}_{\text{e}} \otimes \mathcal{H}_{\text{p}}$ without coulomb interaction in a hydrogen atom? Just stare at the states rejoicing with the fact one can indeed write it in that way? What does oe gain in practice? I guess that was the OP's drift. Otherwise I don't understand the issue either.

 

[vanhees71]

Of course, the Hilbert space of a proton-electron system is a product space. You can construct it from the basis of an arbitrary complete set of compatible observables. You can use, e.g., the 6 Cartesian position-vector coordinates for the proton and the electron, which all commute. This is a product basis, leading to the position representation of (pure) states, i.e., Schrödinger's wave function $\psi(t,\vec{x}_e,\vec{x}_p)$, which of course is not necessarily a product state but a superposition. I think the thread becomes more and more confusing because of making up problems which aren't really existent.

For a pedagogical treatment of the entanglement aspect, see

http://fma.if.usp.br/~piza/artigos/prit.pdf

 

[TrickyDicky]

If the Hilbert spaces are not tensor products, then it seems that entanglement is not so easy to define.

Of course you need the concept of tensor product as operation to define entanglement, which is not the same thing as saying that you need Hilbert spaces that are simple(not sums of products) tensors products to have entanglement, you need sum of products.

 

[TrickyDicky]

Of course, the Hilbert space of a proton-electron system is a product space. You can construct it from the basis of an arbitrary complete set of compatible observables. You can use, e.g., the 6 Cartesian position-vector coordinates for the proton and the electron, which all commute. This is a product basis, leading to the position representation of (pure) states, i.e., Schrödinger's wave function $\psi(t,\vec{x}_e,\vec{x}_p)$, which of course is not necessarily a product state but a superposition.

You should maybe clarify to avoid further confusion that here you are not describing a tensor product: $\otimes$ but a cartesian product(direct sum): $\times$ , giving 3N=6 coordinates.

 

[atyy]

Of course you need the concept of tensor product as operation to define entanglement, which is not the same thing as saying that you need Hilbert spaces that are simple(not sums of products) tensors products to have entanglement, you need sum of products.

When it is said that the Hilbert space is a tensor product, it means an arbitrary state is a sum of products, ie. the product states are basis states.

 

[TrickyDicky]

When it is said that the Hilbert space is a tensor product, it means an arbitrary state is a sum of products, ie. the product states are basis states.

We're on the same page, then. Any thought on #36?

 

[vanhees71]

What I meant is a Cartesian product, constructed via the generalized position eigenstates
$$|\vec{x}_e \rangle \otimes |\vec{x}_p \rangle.$$
A general state is given as a sum/integral of product states
$$|\Psi \rangle=\sum_{i,j} c_{ij} |\phi_i \rangle \otimes |\psi_j \rangle.$$
In the position representation with the above product basis, this reads
$$\Psi(\vec{x}_e,\vec{x}_p)=(\langle \vec{x}_e | \otimes \vec{x}_p|)\Psi \rangle=\sum_{i,j} C_{ij} \phi_i(\vec{x}_e) \psi_j(\vec{x}_p).$$
I still don't see, where the problem is :-(.

 

[atyy]

Of course, the Hilbert space of a proton-electron system is a product space.

That is the textbook treatment, and is my understanding too. But dextercioby seems to be saying there is something missing if one treats the Hilbert space of a proton-electron system as a tensor product of the proton Hilbert space and the electron Hilbert space.

 

[vanhees71]

Ok, then I really didn't understand the question right. So we need a clarification (perhaps a reference?) from dextercioby to answer the question better.

 

[atyy]

Ok, then I really didn't understand the question right. So we need a clarification (perhaps a reference?) from dextercioby to answer the question better.

Since it seems that vanhees71, kith, TrickyDicky and me are all equally mystified, here are some examples where it is stated that the Hilbert space of a multiple particle system is a tensor product. Let's keep everything in non-relativistic quantum mechanics. These are in the context of identical particles, but it is usually understood to be the same for non-identical particles except that there is no (anti)symmetrization requirement. I do agree the tensor product axiom is not as fundamental as the others, but I've always seen the discussion in the context of QFT.

http://www.scholarpedia.org/article/Second_quantization
"One of the basic principles of Quantum Mechanics (Dirac PAM, 1958) relates compositeness to the tensor product of Hilbert spaces. The state space of an assembly of systems is identified with the tensor product of the state spaces of each system."

http://yclept.ucdavis.edu/course/242/2Q_Fradkin.pdf
Eq 1.112, 1.113

http://www.phys.ens.fr/~mora/lecture-second-quanti.pdf
Eq 1

 

[kith]

You mentioned it precisely to agree with me that using this form would imply entanglement and therefore can't be written like that [...]

Therefore? No. Of course, the Coulomb interaction leads to entanglement as well. Both the correct Coulomb interaction and the wrong tensor product form I mentioned in post #19 lead to entanglement. The potential I wrote down in post #23 doesn't.

Your usage of terminology still seems strange to me. I am wondering whether this is purely semantic but I can't pinpoint the issue.

 

[dextercioby]

That is the textbook treatment, and is my understanding too. But dextercioby seems to be saying there is something missing if one treats the Hilbert space of a proton-electron system as a tensor product of the proton Hilbert space and the electron Hilbert space.

I didn't say <something is missing>, but stated that what's missing is a treatment of the H atom starting with the initial (that is non quasiparticle separated) Hilbert space as H_el ⊗ H_p derived from the axiom of states description for multiparticle systems. To this original tensor product one has to say that it's of no use, because of the non-separability of the potential. That's why I invoked the isomorphism which allows for a use of f(x_e,x_p,t) as a wafunction, that is L^2(R^6).

 

[atyy]

I didn't say <something is missing>, but stated that what's missing is a treatment of the H atom starting with the initial (that is non quasiparticle separated) Hilbert space as H_el ⊗ H_p derived from the axiom of states description for multiparticle systems. To this original tensor product one has to say that it's of no use, because of the non-separability of the potential. That's why I invoked the isomorphism which allows for a use of f(x_e,x_p,t) as a wafunction, that is L^2(R^6).

So you're saying that in writing an expression like Eq 20 in http://www.phys.ens.fr/~mora/lecture-second-quanti.pdf, the Hilbert space is not the tensor product, and one has to use the theorem that all infinite dimensional separable Hilbert spaces are isomorphic?

 

[dextercioby]

Yes, because the Coulomb potential makes impossible to define V on the tensor product of 2 copies of the single particle H-space (just like in the case of the H-atom), so that V as defined by that expression makes sense on a Fock space built on L^(R^6).

 

[atyy]

Yes, because the Coulomb potential makes impossible to define V on the tensor product of 2 copies of the single particle H-space (just like in the case of the H-atom), so that V as defined by that expression makes sense on a Fock space built on L^(R^6).

Is it right to say that the wave functions can still be written as sums of products, but the potential itself cannot be written as sums of products?

 

[dextercioby]

Well, there's no product constriction anymore in the wavefunction, because you're not forced to take f(x_1,x2) = g(x_1)h(x_2). Why would separation of variables be a requirement?

 

[atyy]

Well, there's no product constriction anymore in the wavefunction, because you're not forced to take f(x_1,x2) = g(x_1)h(x_2). Why would separation of variables be a requirement?

Well, we usually still take the product states as basis states, so is it simply that this is a good enough approximation?

 

[dextercioby]

Yes, it's the argument by JanoL in post#27 in which he discusses atoms/molecules.

 

[atyy]

Is it possible to prove that there is no separable expansion for the Coulomb potential?

 

[kith]

I didn't say <something is missing>, but stated that what's missing is a treatment of the H atom starting with the initial (that is non quasiparticle separated) Hilbert space as H_el ⊗ H_p derived from the axiom of states description for multiparticle systems. To this original tensor product one has to say that it's of no use, because of the non-separability of the potential.

Just to get you right: you think that the problem is that the potential can't be written in product form, not the specific \frac{1}{|\vec r_e - \vec r_p|} form of the Coulomb potential?

If yes, could you sketch how the solution would look like for an interaction in product form and where the difference between product and sum of products becomes important?

 

[TrickyDicky]

I didn't say <something is missing>, but stated that what's missing is a treatment of the H atom starting with the initial (that is non quasiparticle separated) Hilbert space as H_el ⊗ H_p derived from the axiom of states description for multiparticle systems. To this original tensor product one has to say that it's of no use, because of the non-separability of the potential. That's why I invoked the isomorphism which allows for a use of f(x_e,x_p,t) as a wafunction, that is L^2(R^6).

There is something I find confusing here, L^2(R^6) is the tensor product for 2 separable particles(it is naturally reduced to a cartesian product due to separability), so why the need to invoke the infinite dmensional Hilbert space isomorphism?

 

[dextercioby]

The isomorphism is needed to go from an abstract Hilbert space H to L²(R³) or to L²(R⁶) or to L²(R³) ⊗ L²(R³)

 

[TrickyDicky]

The isomorphism is needed to go from an abstract Hilbert space H to L²(R³) or to L²(R⁶) or to L²(R³) ⊗ L²(R³)

If by the Hilbert space H you mean the hydrogen atom wavefunction, why would you go to L²(R⁶)=L²(R³) ⊗ L²(R³) if its subsystems e and p are clearly not separable states?
Your question was about the application of the tensor product postulate to a composite system like the hydrogen atom, the answer is that, at least in the form written in the OP, the postulate only applies to separable subsystems states, and for some reason this part is not usually specified in the postulate.

 

[vanhees71]

Somehow this simple issue becomes more and more confusion. So I try again:

The Hilbert space for a hydrogen atom in the non-relativistic approximation, neglecting spin (i.e., what you deal with in QM 1 usually after a few weeks of introduction to QT) is the tensor product of the Hilbert space of a single electron and a single proton (in sense of Hilbert spaces of course, i.e., inducing the scalar product from the spaces in the product).

It consists of all superpositions of an arbitrary product basis (or generalized basis). One example for a product basis are the simultaneous generalized position eigenvectors,
$$|\vec{x}_e \rangle \otimes |\vec{x}_p \rangle=|\vec{x}_e,\vec{x}_p \rangle.$$
An arbitrary state $$|\Psi \rangle \in \mathcal{H}_e \otimes \mathcal{H}_p$$ is then represented by the wave functions
$$\Psi(\vec{x}_e,\vec{x}_p)=\langle \vec{x}_e,\vec{x}_p|\Psi \rangle.$$

Now, it's clear that due to the Coulomb-interaction term in the Hamiltonian, the energy eigenstates are not product states of an electron and a proton state. That's all. There's no reason to confusion.

The usual solution of the eigenvalue problem is also clear: You exploit the symmetries and express everything in terms of another representation of the electron-proton Hilbert space, namely the tensor product of center-mass and relative motion. This is equivalent to a Hilbert space of two types of particles, neither being electrons or protons but "quasiparticles". The one is the freely moving atom as a whole, having mass $m_e+m_p$ and the other is a particle, moving in a Coulomb potential of a point charge in the origin with a mass $\mu=m_e m_p/(m_e+m_p)$.

In this representation the energy eigenvectors are product states, because the Hamiltonian is the sum of two commuting parts involving the center-mass and the relative coordinates separately. The reason for this is the symmetry under special Galilei transformations (Galilei boosts).

 

[TrickyDicky]

Somehow this simple issue becomes more and more confusion. So I try again:

The Hilbert space for a hydrogen atom in the non-relativistic approximation, neglecting spin (i.e., what you deal with in QM 1 usually after a few weeks of introduction to QT) is the tensor product of the Hilbert space of a single electron and a single proton (in sense of Hilbert spaces of course, i.e., inducing the scalar product from the spaces in the product).

It consists of all superpositions of an arbitrary product basis (or generalized basis). One example for a product basis are the simultaneous generalized position eigenvectors,
$$|\vec{x}_e \rangle \otimes |\vec{x}_p \rangle=|\vec{x}_e,\vec{x}_p \rangle.$$
An arbitrary state $$|\Psi \rangle \in \mathcal{H}_e \otimes \mathcal{H}_p$$ is then represented by the wave functions
$$\Psi(\vec{x}_e,\vec{x}_p)=\langle \vec{x}_e,\vec{x}_p|\Psi \rangle.$$
Now, it's clear that due to the Coulomb-interaction term in the Hamiltonian, the energy eigenstates are not product states of an electron and a proton state. That's all. There's no reason to confusion.

I don't think it is so simple when dealing with wave functions and position as opposed to spin. Even before we get to the hamiltonian. You seem to be assuming that the hydrogen atom is a separable state, If the tensor product you write above can be written in general you are getting rid of the distinction separable/entangled for states, making all of them separable by definition.