Does quantization always work?

[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.

 

[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.