Does quantization always work?

[dextercioby]

Of course it doesn't. But let me see if I can bring an argument on how some (perhaps elementary) things are so tricky, that their textbook treatment is almost absent.

The axiom of QM which (for some unknown reasons) is least emphasized is this one:

<For a quantum A system made up of non-identical subsystems a_i each described by a (complex, inf-dim, separable) Hilbert space Λ_i, the Hilbert space describing A is the tensor product of all Λ_i>

What is the most commonly described quantum system whose description falls under the above mentioned axiom?

That's right: the Hydrogen atom in non-relativistic QM.

My questions to you are:
1. What's the Hilbert space of the Hydrogen atom?
2. What's the quantum Hamilton operator of the Hydrogen atom equal to, operator whose action in the Hilbert space at 1. is well-defined?

I have some ideas about these 2 answers, but I'll state them after I (hopefully) see some replies. :)

 

[strangerep]

What is the most commonly described quantum system whose description falls under the above mentioned axiom?

That's right: the Hydrogen atom in non-relativistic QM.

My questions to you are:
1. What's the Hilbert space of the Hydrogen atom?
2. What's the quantum Hamilton operator of the Hydrogen atom equal to, operator whose action in the Hilbert space at 1. is well-defined?

I have some ideas about these 2 answers, but I'll state them after I (hopefully) see some replies. :)

Ha! A deceptively nontrivial question indeed!

Starting from the classical system and performing the usual separation into CoM and relative coordinates, we get essentially the free motion for the CoM part (hence needing a rigged Hilbert space).

For the relative part (having started from the classical system, with Coulomb interaction:
$$V_{Coulomb} ~=~ \frac{k_c \, e_1 e_2}{|q_1 - q_2|}
~\equiv~ \frac{e_1 e_2}{4\pi\epsilon_0 |q_1 - q_2|} ~,$$
assume we are only interested in the relative motion of the 2 bodies with charges $-Ze$ and $e$, where $Z$ is the dimensionless atomic number} and $e<0$ is the electron charge (i.e., $-Ze > 0$ is the (positive) charge of the nucleus). The Hamiltonian or the relative motion problem as
$$H ~=~ \frac{p^2}{2\mu} - \frac{Y}{q} ~,~~~~~~~~~~~
\Big[ Y := k_c Z e^2 > 0 \Big], $$
where $q:=\sqrt{q_k q_k}$ and $\mu$ is the reduced mass. $Y$ has dimensions of (energy $\times$ length).

To cut a long story short, one can find a basic dynamical group for this system, consisting of $H$, the angular momentum components $L_i$, and the Runge-Lenz (RL) vector $A_i$. The first 2 are (at least superficially) self-adjoint on the usual (3D) Hilbert space of square integrable functions (possibly with origin removed), while the 3rd must be symmetrized wrt $p_i, q_i$.

The superficial Lie (actually Poisson) algebra formed by these guys is (iirc! - I might have some factors wrong),
$$
[L_j , L_k] = i\hbar \, \varepsilon_{jk\ell} L_\ell ~,~~~~~
[L_j , A_k] = i\hbar \, A_\ell~,~~~~~
[A_j , A_k] = -2i\hbar \mu H \varepsilon_{jk\ell} L_\ell ~.
$$This is not really a Lie algebra because of the final rhs above. Nevertheless, we can proceed by decomposing the (rigged) Hilbert space in terms of energy eigenspaces. One finds 3 disjoint subspaces: $E>0$ (needs a rigged Hilbert space), $E=0$ (a bit boring -- haven't looked closely at this), and $E<0$ (bound states).

For the subspaces with $E\ne 0$, we may introduce an energy--scaled (but still self-adjoint) version of $A$, as follows:
$$A' ~:=~ \frac{A}{\sqrt{2\mu|H|}} ~,~~~~~~ (E \ne 0).$$After (quite a bit) more work, one finds a Lie algebra with 2 Casimirs (but they happen to coincide because $L$ and $A$ are orthogonal. For the $E<0$ case, we find ourselves dealing with 2 algebras essentially isomorphic to the usual su(2) algebra, whose half-integral quantum spectrum is well-known. From this, one finds that the bound state energies are discrete, and the ordinary 3D Hilbert space of square integrable functions is sufficient if we restrict ourselves to this dynamical group and don't try to play silly games with so(4,2) nonsense.

(Strictly speaking, this argument is a bit circular, since the usual su(2) spectrum is derived under the assumption of normalizable eigenstates.)

IMHO, there are numerous serious errors in Barut & Raczka on many of these points -- so many that I wonder whether anybody besides me has actually tried to verify all their claims.

And,... I think I'd better pause my rant here, in case the above is not what you were looking for... (?)

 

[dextercioby]

Hi Strangerep, thanks for your informative post, and most on the (rather saddening) comments on the B&R treatment of this issue, especially since the late A.O. Barut was one of the pioneers of the "dynamical groups" approach to quantum physics.

Here's what I really had in mind with my OP:

So I stated that axiom because it should apply to the H-atom, being a 2 particle system. So let's see how it looks like. First we notice that we expect a quantization of a classically conceivable model based on a Hamilton function. So what's the classical Hamiltonian? It's simply:

$$H_{cl} = \frac{1}{2m_{e}}\left(p_{x, e}^2 +p_{y, e}^2 +p_{z, e}^2\right) + \frac{1}{2m_{p}}\left(p_{x, p}^2 +p_{y, p}^2 +p_{z, p}^2\right) -\frac{e^2}{4\pi\epsilon_{0}} \left(|\vec{r}_{el} - \vec{r}_{p}|\right)^{-1}$$

So one should take the Hilbert space as simply $\mathcal{H}_{h} = \mathcal{H}_{e} \otimes \mathcal{H}_{p}$. But is there any way we can convert the classical Hamiltonian into a quantum version, so that the Hilbert space action is well-defined? Well, we can nicely fit the kinetic part, because we can write

$$\hat{K}_{e} = \frac{1}{2m_{e}}\left(\hat{p}_{x, e}^2 +\hat{p}_{y, e}^2 +\hat{p}_{z, e}^2\right) \otimes \hat{1}_{p}$$ and similarly:

$$\hat{K}_{p} = \hat{1}_{e} \otimes \frac{1}{2m_{p}}\left(\hat{p}_{x, p}^2 +\hat{p}_{y, p}^2 +\hat{p}_{z, p}^2\right)$$

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.
So what's the way to save it all? Even though the Hydrogen atom was first solved using matrix mechanics in 1926 by W. Pauli (http://link.springer.com/article/10.1007/BF01450175), it's only the wave mechanics by Schrödinger that helps here:
It's apparently a result of von Neumann that tells that any 2 separable Hilbert spaces are isomorphic, so we can use this here as follows.
If the wave mechanics formalism is used, then we can safely take $\mathcal{H}_{p,e} = L^2 \left(\mathbb{R}^3, d\vec{r}_{e,p}\right)$, so that the Hilbert space is then the tensor product: $L^2 \left(\mathbb{R}^3, d\vec{r}_{e}\right) \otimes L^2 \left(\mathbb{R}^3, d\vec{r}_{p}\right)$. Now we can use the result of von Neumann and re-write the Hilbert space as $L^2 \left(\mathbb{R}^6, d\vec{r}_{e} \times d\vec{r}_{p}\right)$ which not only allows us to get rid of the $\otimes$ being carried along every time, but also helps us with the potential problem, cause through its Schrödinger 'quantization', it becomes a well-defined function of 6 variables. The momenta become differential operators, of course (laplaceians) which are also well-defined on a dense subset of the L^2 (R^6).

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

 

[TrickyDicky]

Hi Strangerep, thanks for your informative post, and most on the (rather saddening) comments on the B&R treatment of this issue, especially since the late A.O. Barut was one of the pioneers of the "dynamical groups" approach to quantum physics.

Here's what I really had in mind with my OP:

So I stated that axiom because it should apply to the H-atom, being a 2 particle system. So let's see how it looks like. First we notice that we expect a quantization of a classically conceivable model based on a Hamilton function. So what's the classical Hamiltonian? It's simply:

$$H_{cl} = \frac{1}{2m_{e}}\left(p_{x, e}^2 +p_{y, e}^2 +p_{z, e}^2\right) + \frac{1}{2m_{p}}\left(p_{x, p}^2 +p_{y, p}^2 +p_{z, p}^2\right) -\frac{e^2}{4\pi\epsilon_{0}} \left(|\vec{r}_{el} - \vec{r}_{p}|\right)^{-1}$$

So one should take the Hilbert space as simply $\mathcal{H}_{h} = \mathcal{H}_{e} \otimes \mathcal{H}_{p}$. But is there any way we can convert the classical Hamiltonian into a quantum version, so that the Hilbert space action is well-defined? Well, we can nicely fit the kinetic part, because we can write

$$\hat{K}_{e} = \frac{1}{2m_{e}}\left(\hat{p}_{x, e}^2 +\hat{p}_{y, e}^2 +\hat{p}_{z, e}^2\right) \otimes \hat{1}_{p}$$ and similarly:

$$\hat{K}_{p} = \hat{1}_{e} \otimes \frac{1}{2m_{p}}\left(\hat{p}_{x, p}^2 +\hat{p}_{y, p}^2 +\hat{p}_{z, p}^2\right)$$

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.
So what's the way to save it all? Even though the Hydrogen atom was first solved using matrix mechanics in 1926 by W. Pauli (http://link.springer.com/article/10.1007/BF01450175), it's only the wave mechanics by Schrödinger that helps here:
It's apparently a result of von Neumann that tells that any 2 separable Hilbert spaces are isomorphic, so we can use this here as follows.
If the wave mechanics formalism is used, then we can safely take $\mathcal{H}_{p,e} = L^2 \left(\mathbb{R}^3, d\vec{r}_{e,p}\right)$, so that the Hilbert space is then the tensor product: $L^2 \left(\mathbb{R}^3, d\vec{r}_{e}\right) \otimes L^2 \left(\mathbb{R}^3, d\vec{r}_{p}\right)$. Now we can use the result of von Neumann and re-write the Hilbert space as $L^2 \left(\mathbb{R}^6, d\vec{r}_{e} \times d\vec{r}_{p}\right)$ which not only allows us to get rid of the $\otimes$ being carried along every time, but also helps us with the potential problem, cause through its Schrödinger 'quantization', it becomes a well-defined function of 6 variables. The momenta become differential operators, of course (laplaceians) which are also well-defined on a dense subset of the L^2 (R^6).

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

I must be missing something, I thought the particles in a composite were entangled by definition and therefore not separable states. So doesn't it follow naturally from this that the hydrogen atom as 2-particle system is a 3N dim Hilbert space as you conclude?

 

[atyy]

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

But if you use the fact that the Hilbert spaces are isomorphic, then isn't it good enough to use the single particle Hilbert space?

I don't think the tensor product rule is as fundamental as the other axioms. It's more like a rule of thumb to non-rigourously guess things almost correctly, before the mathematical physicists tidy it up.

 

[dextercioby]

@TrickyDicky : This is a nice point, I wasn't thinking about entanglement, but rather of an argument which is missing from a textbook discussion of the H-atom, which should start with the axioms.

@atyy : Well, I think the tensor product of spaces is missing in textbooks altogether, if the particles/sub-systems are not identical, so I believe it should be states somewhere as a necessary assuption, for multiparticle systems, such as atoms and molecules are the cornerstones of QM.

 

[atyy]

@atyy : Well, I think the tensor product of spaces is missing in textbooks altogether, if the particles/sub-systems are not identical, so I believe it should be states somewhere as a necessary assuption, for multiparticle systems, such as atoms and molecules are the cornerstones of QM.

If I understand you correctly, the single particle and the two particle Hilbert spaces are isomorphic. If that is the case, doesn't that mean we can use the Hilbert space for a single particle, even though we are describing a two particle system?

Edit: It's a little bit like Marolf's comment in http://arxiv.org/abs/1409.2509: "Indeed, any (Hamiltonian) quantum theory of gravity defined on a separable Hilbert space is completely equivalent to some local field theory - and in fact to a quantum mechanical theory describing a single particle in one dimension - via a sufficiently non-local map. One simply uses the fact that all separable Hilbert spaces are isomorphic to transcribe the Hamiltonian to the Hilbert space of a single non-relativistic particle."

 

[dextercioby]

We can, but only because we're forced to use this general isomorphism, since the Coulomb potential won't allow for a separation. Books usually don't treat this issue, I've yet to find an argument which begins with the principles, and the futility of using a tensor product in the presence of a Coulomb interaction. Take for example Baym's <Lectures on QM> (Westview, 1969). On page 170 he begins with $\psi (\vec{r}_{1},\vec{r}_{2},t)$, thus he starts with L²(R⁶) (in disguise, of course) already.

 

[atyy]

We can, but only because we're forced to use this general isomorphism, since the Coulomb potential won't allow for a separation. Books usually don't treat this issue, I've yet to find an argument which begins with the principles, and the futility of using a tensor product in the presence of a Coulomb interaction. Take for example Baym's <Lectures on QM> (Westview, 1969). On page 170 he begins with $\psi (\vec{r}_{1},\vec{r}_{2},t)$, thus he starts with L²(R⁶) (in disguise, of course) already.

Is the issue a bit like whether the Hilbert space in an interacting QFT is a Fock space? In non-rigourous QFT, it is always taken as a Fock space together with the wrong derivation of the interaction picture formulas. But one does hear warnings that there are problems with non-rigourous derivation and that in rigourous QFT the Hilbert space is in some sense "not a Fock space". However, the sense in which the Hilbert space is "not a Fock space" cannot be too naive, because of the general isomorphism. So it is more something like the transformation is not a unitary transformation between the operators and the Hilbert spaces.

 

[dextercioby]

To my knowledge, "rigorous QFT" exists only for non-interacting systems, where the Fock space is alive and well and a true Hilbert space, per se. Our colleague @DarMM can tell more. He's the expert here.

 

[atyy]

To my knowledge, "rigorous QFT" exists only for non-interacting systems, where the Fock space is alive and well and a true Hilbert space, per se. Our colleague @DarMM can tell more. He's the expert here.

I didn't mean "non-interacting" so literally, I also meant nonlinear field theories like a quartic self-interaction term, something like what DarMM says in post #51 of https://www.physicsforums.com/threads/bosons-and-fermions-in-a-rigorous-qft.575529/.

 

[dextercioby]

I see now. Indeed, I'm not familiar with Glimm and Jaffe's work, so I allow myself to step back from the QFT discussion. AFAIK $\phi_{4}^{4}$ is mathematically ill-defined (?).

 

[atyy]

I see now. Indeed, I'm not familiar with Glimm and Jaffe's work, so I allow myself to step back from the QFT discussion. AFAIK $\phi_{4}^{4}$ is mathematically ill-defined (?).

Yes, so far there doesn't seem to be a rigourous $\phi_{4}^{4}$. Anyway, the analogy is only that IFIC in rigourous QFT if we write a canonical commutation relation, the operators in the relation cannot act on the Fock space in the same way that the operators in the canonical commutation relation do. In your example, IFIC, the Hilbert space is a tensor product of the single particle spaces (by the general isomorphism), but it seems that the potential cannot be built from a tensor product of single particle operators?

 

[atyy]

As an aside, here is another problem where the result is known for operators that can be written as tensor products, but apparently since one can conceive of operators that are not built from tensor products there is an open problem http://arxiv.org/abs/0812.4305.

 

[dextercioby]

Yes, so far there doesn't seem to be a rigourous $\phi_{4}^{4}$. Anyway, the analogy is only that IFIC in rigourous QFT if we write a canonical commutation relation, the operators in the relation cannot act on the Fock space in the same way that the operators in the canonical commutation relation do. In your example, IFIC, the Hilbert space is a tensor product of the single particle spaces (by the general isomorphism), but it seems that the potential cannot be built from a tensor product of single particle operators?

Well, the potential mixes the coordinates in a very non-trivial way which prevents a natural separation. You'd expect that $\hat{V} = V \left(\hat{\vec{r}}_e \right) \otimes V \left(\hat{\vec{r}}_p \right)$, but for a Coulomb potential this is not possible. On the other hand, as I stated above, "the Fock space" for me makes sense iff the quantum system is (self-)interaction free.

 

[strangerep]

$$H_{cl} = \frac{1}{2m_{e}}\left(p_{x, e}^2 +p_{y, e}^2 +p_{z, e}^2\right) + \frac{1}{2m_{p}}\left(p_{x, p}^2 +p_{y, p}^2 +p_{z, p}^2\right) -\frac{e^2}{4\pi\epsilon_{0}} \left(|\vec{r}_{el} - \vec{r}_{p}|\right)^{-1}$$

So one should take the Hilbert space as simply $\mathcal{H}_{h} = \mathcal{H}_{e} \otimes \mathcal{H}_{p}$. But is there any way we can convert the classical Hamiltonian into a quantum version, so that the Hilbert space action is well-defined? Well, we can nicely fit the kinetic part, because we can write

$$\hat{K}_{e} = \frac{1}{2m_{e}}\left(\hat{p}_{x, e}^2 +\hat{p}_{y, e}^2 +\hat{p}_{z, e}^2\right) \otimes \hat{1}_{p}$$ and similarly:

$$\hat{K}_{p} = \hat{1}_{e} \otimes \frac{1}{2m_{p}}\left(\hat{p}_{x, p}^2 +\hat{p}_{y, p}^2 +\hat{p}_{z, p}^2\right)$$

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.

Let's recall some classical mechanics: a system is not necessarily separable in any (phase space) coordinate system, but in many interesting cases we can find such a coordinate system, and things become nicer.

So what's the way to save it all? Even though the Hydrogen atom was first solved using matrix mechanics in 1926 by W. Pauli (http://link.springer.com/article/10.1007/BF01450175), it's only the wave mechanics by Schrödinger that helps here:
It's apparently a result of von Neumann that tells that any 2 separable Hilbert spaces are isomorphic, so we can use this here as follows.
If the wave mechanics formalism is used, then we can safely take $\mathcal{H}_{p,e} = L^2 \left(\mathbb{R}^3, d\vec{r}_{e,p}\right)$, so that the Hilbert space is then the tensor product: $L^2 \left(\mathbb{R}^3, d\vec{r}_{e}\right) \otimes L^2 \left(\mathbb{R}^3, d\vec{r}_{p}\right)$. Now we can use the result of von Neumann and re-write the Hilbert space as $L^2 \left(\mathbb{R}^6, d\vec{r}_{e} \times d\vec{r}_{p}\right)$ which not only allows us to get rid of the $\otimes$ being carried along every time, but also helps us with the potential problem, cause through its Schrödinger 'quantization', it becomes a well-defined function of 6 variables. The momenta become differential operators, of course (laplaceians) which are also well-defined on a dense subset of the L^2 (R^6).

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

In the classical case, one starts with a Cartesian product of phase spaces, and the equations of motion define a solution subspace therein -- which may have fewer symmetries than the large ambient phase space. The ambient phase space is just a mathematical aid -- the physics lies in the solution subspace.

Similarly with the H-atom: the full tensor product of (rigged!) Hilbert spaces is a mathematical aid, a background framework. The physics is contained in the rather complicated subspace thereof which I described in post #2.

If we are dealing with a system which has $N$ (finite) degrees of freedom, then an $L^2(R^?)$ Hilbert space (where $? \ge N$) is probably a reasonable starting point. But if we can analyze the dynamical symmetries of the system we might be able to find a more precise lower bound on the necessary dimensions.

Another wrinkle on this whole issue is the difference in the energy spectra. The free case has only $E\ge 0$, but in the interacting case we can have negative energies (unless we shift by the ground energy -- but to find the latter we must solve the whole problem).

 

[TrickyDicky]

@TrickyDicky : This is a nice point, I wasn't thinking about entanglement, but rather of an argument which is missing from a textbook discussion of the H-atom, which should start with the axioms.

My point was that there is arbitrariness in whether one considers the "hydrogen atom" system as a composite particle in which the subsystems electron and proton are by definition entangled and therefore the tensor product cannot be used and one has the Hilbert space of one particle, as opposed to the situation where one tries to describe the system as the tensor product of the two particles in R6, I don't think this last option is possible for the hydorgen atom or for a composite particle in general, precisely due to the entanglement, in this case it is the potential that doesn't allow the subparticles to be separable as you say, but that is just the particular cause of their entanglement.
This would answer in part why the tensor product axiom has this "absent treatment" for this kind of poblems. Simply put the tensor product postulate doesn't apply.
The tensor product postulate is usually referred to "composite quantum systems", so I guess it refers to many-particle systems that cannot themselves be considered as single particles in the sense hadrons, atoms, molecules like buckyballs... are.
I can't think off-hand of an example of a system of distinct particles where one can apply properly the tensor product postulate. Can anyone give some example?
It also makes one wonder about the concept of quasiparticle-colective excitation in condensed matter physics.

 

[ShayanJ]

I'm a bit confused here. I always thought its really natural to use e.g. $L^2(\mathbb R^6,\vec{dr}_e \times \vec{dr}_p)$ and I didn't feel the need to get this from some axiom. You may say its because of the way textbooks are written and I'm not as expert as you guys to be free enough from textbooks to think about it myself, but I think I have a point here. If the statement A is an axiom for a theory, then it should cover every case that the theory is going to be applied to. Then maybe in some situations, we get a special case of A so things get apparently different. But here the statement

<For a quantum A system made up of non-identical subsystems ai each described by a (complex, inf-dim, separable) Hilbert space Λi, the Hilbert space describing A is the tensor product of all Λi>

clearly doesn't cover all the situations that the theory is going to be used in, you guys gave a counterexample. So it seems obvious to me that this can't be an axiom of QM but in fact the use of something like $L^2(\mathbb R^6,\vec{dr}_e \times \vec{dr}_p)$ should be an axiom because its more general. We use this from the beginning and in some special cases, it happens that we can decompose the Hilbert space! Am I missing something here?

 

[kith]

But is there any way we can convert the classical Hamiltonian into a quantum version, so that the Hilbert space action is well-defined?

I don't see a fundamental problem related to the tensor product postulate here. In the composite Hilbert space $\mathcal{H}_{e} \otimes \mathcal{H}_{p}$, the position operators of the electron and the proton are $\vec R_e \otimes 1$ and $1 \otimes \vec R_p$. It's true that the potential $V(\vec R_e \otimes 1, 1 \otimes \vec R_p )$ can't be written in the form $U(\vec R_e) \otimes W(\vec R_p)$ but why is this a problem? The tensor product postulate doesn't require the operators to be of product form.

 

[TrickyDicky]

I'm a bit confused here. I always thought its really natural to use e.g. $L^2(\mathbb R^6,\vec{dr}_e \times \vec{dr}_p)$ and I didn't feel the need to get this from some axiom. You may say its because of the way textbooks are written and I'm not as expert as you guys to be free enough from textbooks to think about it myself, but I think I have a point here. If the statement A is an axiom for a theory, then it should cover every case that the theory is going to be applied to. Then maybe in some situations, we get a special case of A so things get apparently different. But here the statement clearly doesn't cover all the situations that the theory is going to be used in, you guys gave a counterexample. So it seems obvious to me that this can't be an axiom of QM but in fact the use of something like $L^2(\mathbb R^6,\vec{dr}_e \times \vec{dr}_p)$ should be an axiom because its more general. We use this from the beginning and in some special cases, it happens that we can decompose the Hilbert space! Am I missing something here?

But something like $L^2(\mathbb R^6,\vec{dr}_e \times \vec{dr}_p)$ uses the cartesian product instead of the tensor product and assumes identical non-interacting particles, and it is not general either, it would only valid for fermions, so no axiom here.

 

[TrickyDicky]

I don't see a fundamental problem related to the tensor product postulate here. In the composite Hilbert space $\mathcal{H}_{e} \otimes \mathcal{H}_{p}$, the position operators of the electron and the proton are $\vec R_e \otimes 1$ and $1 \otimes \vec R_p$. It's true that the potential $V(\vec R_e \otimes 1, 1 \otimes \vec R_p )$ can't be written in the form $U(\vec R_e) \otimes W(\vec R_p)$ but why is this a problem? The tensor product postulate doesn't require the operators to be of product form.

But then you are assuming the electron and the proton in the hydrogen atom are not interacting, and are not entangled, as they can be written as produc states. I think it was you who in a recent thread explained to me how this doesn't make much sense for composite particles.

 

[TrickyDicky]

I would still like to hear about any useful examples of quantum systems where the tensor product axiom can be put to work, all the examples that I'm finding use the the tensor product to explain issues related to entanglement precisely to point out that the entangled components are not product states because they can't be simply written in terms of tensor products.

 

[kith]

But then you are assuming the electron and the proton in the hydrogen atom are not interacting, and are not entangled, as they can be written as produc states.

That's not true. How do you conclude that the state remains a product state under time evolution? This would require a potential of the form $V(\vec R_e) \otimes 1 + 1 \otimes V(\vec R_p)$ which is a common external potential for both particles. The coulomb interaction between them can't be written in this form, of course.

I think it was you who in a recent thread explained to me how this doesn't make much sense for composite particles.

I think we have been talking past each other, then. We can resume the discussion in the other thread but I don't have much time right now.

 

[kith]

I would still like to hear about any useful examples of quantum systems where the tensor product axiom can be put to work, all the examples that I'm finding use the the tensor product to explain issues related to entanglement precisely to point out that the entangled components are not product states because they can't be simply written in terms of tensor products.

The tensor product postulate doesn't imply that the composite system occupies only product states. The tensor product says something about the Hilbert space of the composite system. This space includes product states as well as entangled states. So all the examples you are finding are proper examples of the postulate.

 

[TrickyDicky]

That's not true. How do you conclude that the state remains a product state under time evolution?

I'm obvioulsly concluding they don't. But there is not much use for the product states in a time-independent situation, for bound states there is not much use for the product states of electron and proton, no?

 

[TrickyDicky]

The tensor product postulate doesn't imply that the composite system occupies only product states. The tensor product says something about the Hilbert space of the composite system. This space includes product states as well as entangled states. So all the examples you are finding are proper examples of the postulate.

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

 

[Jano L.]

Well, the potential mixes the coordinates in a very non-trivial way which prevents a natural separation. You'd expect that $\hat{V} = V \left(\hat{\vec{r}}_e \right) \otimes V \left(\hat{\vec{r}}_p \right)$, but for a Coulomb potential this is not possible. On the other hand, as I stated above, "the Fock space" for me makes sense iff the quantum system is (self-)interaction free.

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

Well, I think the tensor product of spaces is missing in textbooks altogether, if the particles/sub-systems are not identical, so I believe it should be states somewhere as a necessary assuption, for multiparticle systems, such as atoms and molecules are the cornerstones of QM.

Books usually don't treat this issue, I've yet to find an argument which begins with the principles, and the futility of using a tensor product in the presence of a Coulomb interaction. Take for example Baym's <Lectures on QM> (Westview, 1969). On page 170 he begins with ψ(r⃗ 1,r⃗ 2,t), thus he starts with L2(R6) (in disguise, of course) already.

Well, the potential mixes the coordinates in a very non-trivial way which prevents a natural separation. You'd expect that V̂ =V(r⃗ ̂ e)⊗V(r⃗ ̂ p), but for a Coulomb potential this is not possible.

You seem to assume that tensor product is something important to be involved in deriving the Schroedinger equation for atoms and molecules, or generally for many-particle systems. I think people who use the equation for calculations concerning atoms and molecules would disagree or care very little. Not because they are not into math or do not like tensors, but because it seems like a very bad idea.

Schroedinger's equation for many-particle system is easy to write based on obvious generalization of the known successful applications to light atoms and simple molecules. Apart from spin, it is essentially directly based on the classical Hamiltonian function of system of charged particles and is natural to use one such equation for system with many strongly interacting particles (EM interaction) rather than many equations for each particle.

Tensor products are used in chemical physics for atoms/molecules too, even for strongly interacting subsystems, but not to formulate the above equation, but merely as a formal device to express simplified and heavily modified version of it for reasons of desired tractability.

I'll explain.

When you have two molecules (atoms...) each with $N$ particles far away from each other, blind following of the standard algorithm would lead you to write Hamiltonian acting on Hilbert space of normalized functions of $2\times3N$ variables. Computational demand for such equations gets out of hand with more molecules added and although going through with the whole procedure used for hydrogen atom could be viewed mathematically correct way to find basis of functions to use , it is just never done exactly. The expected results are believed to be unphysical anyway (eigenfunctions would represent hypothetical long-distance correlations ("entanglement") of large objects). So, what people do instead is replace the big Hilbert space by a simpler one, and the obvious way to do it is to assume that each molecule has its unperturbed Hamiltonian, use pairs of their eigenfunctions as new basis for the combined system and voila, we have tensors in theory, because the Hilbert space thus constructed is tensor product of the separate Hilbert spaces of the molecules.

Then various schemes to account for the interaction of the systems are considered, Hamiltonians are modified to correct the previous step where we messed up with the big equation. These are typically expressed in terms of the so constructed product basis. (The chain of modifications goes on for very long... to the point people deal with nasty caricature of the original equation).

But all the tensors come merely as a device to account for the error that was made in the first place by abandoning the "big" Schroedinger equation; their role is corrective and approximative and used mostly for systems that are not strongly interacting.

Using the same approximative scheme for, say, electron and proton in hydrogen atom would be possible formally but I doubt it would reproduce the results of ordinary Schroedinger equation. And really there is no point, Schroedinger's equation for atoms/molecules has well-known form based on Hamiltonian operator in configuration space of any system of charged particles and that does not need tensors to formulate.

Of course, the original non-relativistic Schroedinger equation is just a mathematical model of atoms/molecules, and has known drawbacks. I just wanted to point out that tensors never played role in its formulation.

 

[TrickyDicky]

You seem to assume that tensor product is something important to be involved in deriving the Schroedinger equation for atoms and molecules, or generally for many-particle systems. I think people who use the equation for calculations concerning atoms and molecules would disagree or care very little. Not because they are not into math or do not like tensors, but because it seems like a very bad idea.

Schroedinger's equation for many-particle system is easy to write based on obvious generalization of the known successful applications to light atoms and simple molecules. Apart from spin, it is essentially directly based on the classical Hamiltonian function of system of charged particles and is natural to use one such equation for system with many strongly interacting particles (EM interaction) rather than many equations for each particle.

Tensor products are used in chemical physics for atoms/molecules too, even for strongly interacting subsystems, but not to formulate the above equation, but merely as a formal device to express simplified and heavily modified version of it for reasons of desired tractability.

I'll explain.

When you have two molecules (atoms...) each with $N$ particles far away from each other, blind following of the standard algorithm would lead you to write Hamiltonian acting on Hilbert space of normalized functions of 2x3N variables. Computational demand for such equations gets out of hand with more molecules added and although going through with the whole procedure used for hydrogen atom could be viewed mathematically correct way to find basis of functions to use , it is just never done exactly. The expected results are believed to be unphysical anyway (eigenfunctions would represent hypothetical long-distance correlations ("entanglement") of large objects). So, what people do instead is replace the big Hilbert space by a simpler one, and the obvious way to do it is to assume that each molecule has its unperturbed Hamiltonian, use pairs of their eigenfunctions as new basis for the combined system and voila, we have tensors in theory, because the Hilbert space thus constructed is tensor product of the separate Hilbert spaces of the molecules.

Then various schemes to account for the interaction of the systems are considered, Hamiltonians are modified to correct the previous step where we messed up with the big equation. These are typically expressed in terms of the so constructed product basis. (The chain of modifications goes on for very long... to the point people deal with nasty caricature of the original equation).

But all the tensors come merely as a device to account for the error that was made in the first place by abandoning the "big" Schroedinger equation; their role is corrective and approximative and used mostly for systems that are not strongly interacting.

Using the same approximative scheme for, say, electron and proton in hydrogen atom would be possible formally but I doubt it would reproduce the results of ordinary Schroedinger equation. And really there is no point, Schroedinger's equation for atoms/molecules has well-known form based on Hamiltonian operator in configuration space of any system of charged particles and that does not need tensors to formulate.

Of course, the original non-relativistic Schroedinger equation is just a mathematical model of atoms/molecules, and has know drawbacks. I just wanted to point out that tensors never played role in its formulation.

This is the kind of example I was expecting. It is mostly used for weakly interacting systems and only as an approximation tool, not rigorously. And not for composite particles like an atom. So the tensor product postulate is basically a heuristic rule of thumb rather than an axiom.

 

[vanhees71]

Hi Strangerep, thanks for your informative post, and most on the (rather saddening) comments on the B&R treatment of this issue, especially since the late A.O. Barut was one of the pioneers of the "dynamical groups" approach to quantum physics.

Here's what I really had in mind with my OP:

So I stated that axiom because it should apply to the H-atom, being a 2 particle system. So let's see how it looks like. First we notice that we expect a quantization of a classically conceivable model based on a Hamilton function. So what's the classical Hamiltonian? It's simply:

$$H_{cl} = \frac{1}{2m_{e}}\left(p_{x, e}^2 +p_{y, e}^2 +p_{z, e}^2\right) + \frac{1}{2m_{p}}\left(p_{x, p}^2 +p_{y, p}^2 +p_{z, p}^2\right) -\frac{e^2}{4\pi\epsilon_{0}} \left(|\vec{r}_{el} - \vec{r}_{p}|\right)^{-1}$$

So one should take the Hilbert space as simply $\mathcal{H}_{h} = \mathcal{H}_{e} \otimes \mathcal{H}_{p}$. But is there any way we can convert the classical Hamiltonian into a quantum version, so that the Hilbert space action is well-defined? Well, we can nicely fit the kinetic part, because we can write

$$\hat{K}_{e} = \frac{1}{2m_{e}}\left(\hat{p}_{x, e}^2 +\hat{p}_{y, e}^2 +\hat{p}_{z, e}^2\right) \otimes \hat{1}_{p}$$ and similarly:

$$\hat{K}_{p} = \hat{1}_{e} \otimes \frac{1}{2m_{p}}\left(\hat{p}_{x, p}^2 +\hat{p}_{y, p}^2 +\hat{p}_{z, p}^2\right)$$

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.
So what's the way to save it all? Even though the Hydrogen atom was first solved using matrix mechanics in 1926 by W. Pauli (http://link.springer.com/article/10.1007/BF01450175), it's only the wave mechanics by Schrödinger that helps here:
It's apparently a result of von Neumann that tells that any 2 separable Hilbert spaces are isomorphic, so we can use this here as follows.
If the wave mechanics formalism is used, then we can safely take $\mathcal{H}_{p,e} = L^2 \left(\mathbb{R}^3, d\vec{r}_{e,p}\right)$, so that the Hilbert space is then the tensor product: $L^2 \left(\mathbb{R}^3, d\vec{r}_{e}\right) \otimes L^2 \left(\mathbb{R}^3, d\vec{r}_{p}\right)$. Now we can use the result of von Neumann and re-write the Hilbert space as $L^2 \left(\mathbb{R}^6, d\vec{r}_{e} \times d\vec{r}_{p}\right)$ which not only allows us to get rid of the $\otimes$ being carried along every time, but also helps us with the potential problem, cause through its Schrödinger 'quantization', it becomes a well-defined function of 6 variables. The momenta become differential operators, of course (laplaceians) which are also well-defined on a dense subset of the L^2 (R^6).

So the way I see it, the answer is that, even though we are forced to use a tensor product, we can use a single Hilbert space even for multiparticle systems, and that even before attempting to separate variables into COM + virtual particle.

I don't understand your quibbles. The hydrogen atom in this most simple form, is of course described on the tensor-product space $\mathcal{H}_e \times \mathcal{H}_p$, and in this space you can have a lot of operators that are not tensor products or sums of tensor products. The kinetic energies are simply one-particle operators, while the interaction potential is a two-body operator, as it must be in order to describe an interaction.

Also when you solve, say, the energy-eigenvalue problem (as is usually done in QM1), you won't work in this tensor-product space, but use the symmetries of the Hamiltonian to make the task easier (or better said feasible at all) and use the fundamental conservation laws due to the space-time (Galilei) symmetry. The first thing you do in quantum as in classical mechanis is to separate off the center-mass motion of the two-body system as a whole. Then you realize that in fact you better work in the tensor-product space $\mathcal{H}_{\text{cm}} \otimes \mathcal{H}_{\text{rel}}$. This is the representation of the problem in terms of two quasiparticles, one of which is free (center-mass motion), having the mass $M=m_p+m_e$, and the other is moving in an external Coulomb potential and has the reduced mass as mass.

The center-mass motion is trivial: you can use the three momentum vectors of the center-mass motion as complete set of observables, and this is at the same time the eigenbasis of this part of the Hamiltonian (the center-mass kinetic energy).

The relative motion then has center symmetry around the origin of the relative position, leading to $E_{\text{rel}}$, $\vec{L}^2$, and $L_z$ as a good complete set of compatible observable.

Then you use the "accidental" additional symmetry, known from classical celestial mechanics, leading to an SO(4) symmetry group for the bound states, a Galilei group for the zero modes, and an SO(1,3) for the scattering states. This makes the hydrogen atom pretty special and solvable completely in algebraic terms as was indeed first shown by Pauli, using matrix mechanics.

Of course, at the end everything boils down to the harmonic oscillator, which is the one model exactly solvable in all branches of physics! It's an harmonic oscillator, because a harmonic oscillator in $N$ spatial dimensions leads to the representations of the Lie algebra $\mathrm{su}(n)$.

As to the remark concerning entanglement: It's clear that you have to tell the basis you are referring too. In the quasi-particle picture the two quasi particles are independent of each other, because their Hamiltonians commute, and thus the eigenstates of the total energy are product states of the eigenstates of the quasi-particle Hamiltonians.

Of course, you can write the energy-eigenstates also in terms of the electron-proton product basis, and there you'll find that the energy-eigenstates are superpositions of product states, and this means from this point of view you deal with entangled states.

 

[kith]

I'm obvioulsly concluding they don't.

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.

 

[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.