# Does quantization always work?

1. Feb 14, 2015

### 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 ai each described by a (complex, inf-dim, separable) Hilbert space Λi, the Hilbert space describing A is the tensor product of all Λi>

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. :)

2. Feb 15, 2015

### strangerep

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... (?)

Last edited: Feb 15, 2015
3. Feb 15, 2015

### 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.

Last edited: Feb 15, 2015
4. Feb 15, 2015

### TrickyDicky

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?

Last edited: Feb 15, 2015
5. Feb 15, 2015

### atyy

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.

Last edited: Feb 15, 2015
6. Feb 15, 2015

### 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.

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7. Feb 15, 2015

### atyy

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."

8. Feb 15, 2015

### 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.

9. Feb 15, 2015

### atyy

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.

10. Feb 15, 2015

### 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.

11. Feb 15, 2015

### atyy

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/.

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12. Feb 15, 2015

### 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 (?).

13. Feb 15, 2015

### atyy

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?

14. Feb 15, 2015

### 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.

15. Feb 15, 2015

### dextercioby

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.

16. Feb 15, 2015

### strangerep

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.
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).

Last edited: Feb 15, 2015
17. Feb 16, 2015

### TrickyDicky

My point was that there is arbitrariness in whether one considers the "hydrogen atom" system as a composite particle in wich 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.

Last edited: Feb 16, 2015
18. Feb 17, 2015

### 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
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?

19. Feb 17, 2015

### kith

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.

20. Feb 17, 2015

### TrickyDicky

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.