# Why Hilbert space?

1. Sep 7, 2007

### cesiumfrog

It's fairly well known that "the possible states of a quantum mechanical system are represented by unit vectors (called "state vectors") residing in a complex separable Hilbert space", and that a Hilbert space is basically just the generalisation "from the two-dimensional plane and three-dimensional space to infinite-dimensional spaces".

However, the only examples I know of seem to be very low-dimensional: for example, in the DCQE, the state can be described using a basis consisting of just two vectors (eg. "slit A" and "slit B", or else some rotation thereof such as "in phase" and "anti phase", i.e. A+B and A-B). Looking at such examples, I can't see why the description needs to be a Hilbert space.

Can someone explain a simple QM example for which an infinite-dimensional space is necessary?

2. Sep 7, 2007

### meopemuk

A single free particle is described by a Hilbert space of infinite (actually, uncountable) dimension. The number of dimensions in this Hilbert space is the "number of distinct points" in the position (x-y-z) space, or the "number of points" in the momentum space.

Eugene.

3. Sep 7, 2007

### Anonym

You should read carefully what is written. The Hilbert spaces are the functional spaces and have nothing to do with two, three or four dim geometrical spaces and variables that you use to use in the classical mechanics (for that reason we call them the dynamical variables) which are represented by the special class of the linear operators in Hilbert space called self-adjoint operators.
The mathematical realization (theory) is defined in physics by the experimental results and not by the assumptions made by the mathematician even that great as D. Hilbert. The complex Hilbert space is suitable for the non-relativistic version of QM with the proper adjustment in both directions: the space is not necessary separable and it may be finite dim starting from n=1.

The double slit experiment is described by non-separable infinite-dimensional (continuous) Hilbert space. Your second ref in Wiki is written by the mathematician that doesn’t know and doesn’t understand anything. From his POV that Hilbert space even doesn’t exist. If you will read R.P.Feynman, you will see that the result of the double slit experiment is obtained exactly by one line of the mathematical calculations. If that do not convince you why the description needs to be a Hilbert space, I have no additional arguments.

Regards, Dany.

4. Sep 7, 2007

### cesiumfrog

Eugene, you're saying that any field (for example, an electric field in classical electrostatics) is a Hilbert space element?

And consequently that states in QM belong to a Hilbert space, not due to the quantisation per se, but purely by virtue of being "probability fields"? Why then is it that descriptions (e.g., on Wikipedia) of only QM, and not classical theories, always tend to explicitly mention Hilbert spaces?

Dany, I'm not sure which publication you're referring to but, since it's only one line, could you reproduce it here for me?

5. Sep 7, 2007

### meopemuk

I was talking only about quantum-mechanical description of particles. More specifically, I was talking about a single particle.

The correspondence between Hilbert spaces of quantum mechanics and electromagnetic fields of Maxwell's theory is a trickier subject. I have my own ideas about how this correspondence works. However, I should warn you that these ideas are far from "mainstream". They are mentioned briefly in post #5 of

A more detailed discussion can be found in chapter 11 of

http://www.arxiv.org/abs/physics/0504062

Eugene.

6. Sep 7, 2007

### Hurkyl

Staff Emeritus
Every finite-dimensional complex (or real) inner product space is a Hilbert space. There simply aren't any non-Hilbert spaces available to use in finite dimension!

7. Sep 7, 2007

### Hurkyl

Staff Emeritus
It has uncountable dimension as a vector space, but it has countable dimension as a Hilbert space.

In a Hilbert space, the existence of limits of Cauchy sequences are part of the structure, so the Hilbert subspace spanned by a set of vectors consists not only of finite linear combinations, but also of all convergent infinite linear combinations.

Last edited: Sep 7, 2007
8. Sep 7, 2007

### meopemuk

I am not sure how you can differentiate "vector space" and "Hilbert space" dimensions. Presumably, we want to have well-localized states as members of our state (Hilbert) space. This means that delta-functions $\delta(\mathbf{r}- \mathbf{r}_0)$ should correspond to valid state vectors for any $\mathbf{r}_0$. Apparently, such wavefunctions with different values $\mathbf{r}_0$ are orthogonal, and one can built a basis from these wavefunctions. Then the number of basis vectors is equal to the number of points in the position space, i.e. uncountable.

This could be inconsistent with the strict mathematical definition of Hilbert space, but it is more consistent with physics, in my opinion.

Another interesting point is that delta-functions $\delta(\mathbf{r}- \mathbf{r}_0)$ cannot serve as representatives of normalizable state vectors, because they are not square-integrable. I think, it makes more sense to use square roots $\sqrt{\delta(\mathbf{r}- \mathbf{r}_0)}$ as position-space basis vectors.

Eugene.

9. Sep 8, 2007

### Hurkyl

Staff Emeritus
It's just as I said; the Hilbert space spanned by a set of vectors is the set of infinite linear combinations of your vectors. The vector space spanned by those vectors is the set of finite linear combinations of your vectors. These different notions can lead to different values for the dimension of a Hilbert space.

It's the former notion that's most important (AFAIK). For example, consider the quantum harmonic oscillator; an important fact is that every state can be written as an infinite linear combination of eigenstates of the Hamiltonian; there are only countably many of those.

No we wouldn't. The most important thing about a quantum state is that observables have an expectation on it. Mathematically, the expectation formula breaks down. Physically, most observables (anything involving momentum) are nonsense for such a thing. So, we should be disinclined to call such a thing a quantum state.

Incidentally, we don't always want one master Hilbert space in which all of our states are representable: for example, in a one-charged-particle situation, we might prefer to have one Hilbert space for the positively charged states and one Hilbert space for the negatively charged states.

I'm not sure how one could make a square root of delta makes any sense at all.

If I wave my hands and consider the only heuristic meaning that comes to mind, I observe that, for a wavefunction $\phi$, if
$$\int_{-\infty}^{+\infty} \delta(x) \phi(x) \, dx = \phi(0),$$
then shouldn't
$$\int_{-\infty}^{+\infty} \sqrt{\delta(x)} \phi(x) \, dx = 0?$$
And so these square root-deltas would be rather useless for any practical purpose.

10. Sep 8, 2007

### George Jones

Staff Emeritus
A (Hamel) basis for a vector space is a subset that consists of linearly independent vectors that span the vector space. Even in infinite-dimensional spaces, the definitions of linear independence and span involve sums of vectors that have a finite number of terms.

An orthonormal basis for a Hilbert space H is a set B of orthonormal vectors such that any vector in H can be expressed as a possibly countably infinite linear combination (which involves a limiting process) of vectors from B. If B is countable, then H is separable. A Hamel basis for a Hilbert space H that has a countable orthonormal basis is uncountable. When H is finite-dimensional, the two types of basis have the same cardinality and every orthonormal basis is also a Hamel basis.

As Hurkyl has said, orthonormal bases are usually used in quantum theory and functional analysis.

11. Sep 8, 2007

### Haelfix

Everytime I see a Delta function in quantum mechanics, and rigorous mathematical talk about Hilbert spaces. I immediately urge the mathematically inclined reader to forget the standard construction and instead move to a Gelfland triple, or else vast amounts of unending confusion arises.

You can also choose to think of the delta function as shorthand for say a limiting sequence of certain square integrable functions (and thus clutter up notation without actually buying you anything)

12. Sep 8, 2007

### meopemuk

By writing $\delta(\mathbf{r} - \mathbf{r}_0)$ I presumed that physically it is possible to prepare a state of the particle localized at the point $\mathbf{r}_0$. Practically, it may be impossible to do that, because of limited precision of our instruments. However, it is important that by spending some extra efforts we can always improve particle localization, and there is no any natural limit for such an improvement. So, theoretically, the assumption of the possibility of perfect localization seems to be reasonable.

The same can be said about particle localization with wavefunctions $\delta(\mathbf{p} - \mathbf{p}_0)$ in the momentum space.

If we agree that such perfectly localized states should be present in our formalism, then we should conclude that the number of mutually orthogonal vectors in the (Hilbert) space of states of our particle is uncountable.

Your formula (1) expresses the amplitude of finding the particle exactly at point x. This amplitude is, of course, vanishing. However, it wouldn't be correct to say that this amplitude is exactly zero. If we sum up squares of all these (seemingly vanishing) amplitudes over all points in space (there is an infinite uncountable number of them) we should get exacly "1" - the probability of finding the particle somewhere. So, basically, we have here an uncertainty of the type "zero $\times$ infinity". I understand that usual quantum mechanics is not formulated in this language, however, there exists a mathematical formalism which, in my opinion, can help to make sense of such undefined expressions. This is called "non-standard analysis". I think that proper formulation of quantum mechanics should be done in "non-standard" Hilbert spaces. There is some activity along these lines in the literature, but it hasn't reached the "mainstream" yet.

Eugene.

13. Sep 8, 2007

### Hurkyl

Staff Emeritus
That doesn't follow. Lower bounds are not always reachable; you could make all the improvements you want, but you need to do something fundamentally different if you actually want to arrive at the conclusion that perfect localization makes sense!

If we agree that such perfectly localized states should be present in a continuum formulation of QM, then the current mathematical tools are inadequate

But if we don't insist that they are actual quantum states, then there is no technical problem with using deltas or other generalized functions.

Or, we could treat the situation the way we normally would:
$$\int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty} \delta(x - y) f(x) \, dx \right|^2 \, dy = 1.$$

Now, if you were going to invoke nonstandard analysis, and define things on a hyperfinite lattice (rather than on a continuum) whose points are separated by 1/H (for some transfinite H), and range over the interval [-H, H], then, we have
$$\langle \varphi(x) \mid \psi(x) \rangle = \int_{-\infty}^{+\infty} \varphi(x)^* \psi(x) \, dx \approx \sum_{n = -H^2}^{H^2} \frac{1}{H} \varphi(n/H)^* \psi(n/H),$$
where the approximation symbol denotes the fact they are infinitessimally close. In particular, we have
$$\sum_{n = -H^2}^{H^2} \frac{1}{H} |\varphi(n/H)|^2 \approx 1.$$

Now, we could define a function d(x) by:
$$d(x) = \begin{cases} H & x = 0 \\ 0 & x \neq 0 \end{cases}$$
If you have a standard point a, then you can find a point b of the lattice that is infinitessimally close to a, and we have that

$$\langle d(x - b) \mid \varphi(x) \rangle \approx \sum_{n = -H^2}^{H^2} \frac{1}{H} d(n/H - b)^* \varphi(n/H) = \varphi(b) \approx \varphi(a)$$

You could take it's square root to get a unit vector, if you so desired. Then, it is true that
$$\begin{multiline} \begin{split} \sum_{n = -H^2}^{H^2} \left| \langle \sqrt{d(x - n/H)} \mid \varphi(x) \rangle \right|^2 &= \sum_{n = -H^2}^{H^2} \left| \frac{1}{\sqrt{H}} \langle d(x - n/H) \mid \varphi(x) \rangle \right|^2 \\&= \sum_{n = -H^2}^{H^2} \frac{1}{H} \left|\varphi(n/H) \right|^2 \\&\approx 1. \end{split} \end{multiline}$$

However, you are wrong on a different point: this sum must be taken over all the points of the lattice. In particular, each ordinary point a has lots of nonstandard points near it, and we must also sum over transfinite points, which are not near any standard point.

In other words, you cannot use nonstandard analysis to say that you have one basis vector per standard point.

Last edited: Sep 8, 2007
14. Sep 8, 2007

### meopemuk

I accept as a physical fact that position observable exists and that the Hermitian operator of position has a continuous spectrum and an uncountable number of distinct eigenvectors. The same can be said about the observable of momentum. If our current mathematical tools (e.g., separable Hilbert spaces) cannot describe this situation, I see nothing wrong in trying to find other, more adequate, tools. The rules of the game and appropriate mathematical models should be derived from physical observations, not the other way around.

Eugene.

15. Sep 8, 2007

### quetzalcoatl9

yes, trace the canonical commutation relation:

$$Tr([x,p]) = Tr(i\hbar1)$$

this equation is invalid unless the operators are of infinite dimension.

16. Sep 9, 2007

### George Jones

Staff Emeritus
Many of the standard examples treated in quantuims mechanics courses have infinite-dimensional state spaces. For example, the state spaces of a particle in a box, a harmonic oscillator, and a hydrogen atom all are infinite-dimensional Hilbert spaces.

This is related to this thread, in which I used the canonical commutation relation to "prove" 0 = 1.

Another way to look at this is the following. If a pair of observables satisfies the canonical commutation relation, then at least one of the observables must be unbounded. Since all operators on finite-dimensional vector spaces are bounded, any quantum system which has a pair of observables that satisfy the canonical commutation relation must have a state space that is infinite-dimensional.

Last edited: Sep 9, 2007
17. Sep 9, 2007

### quetzalcoatl9

i believe that it was von Neumann who first pointed this out. it wouldn't be 0=1 it would be 0=ihN where N is the dimensionality.

i think that the easiest way to show this is:

$$Tr(xp - px) = \sum_i^N (xp)_{ii} - \sum_i^N (px)_{ii} = i\hbar \sum_i^N 1$$

but in the limit that $$N \rightarrow \infty$$ the RHS is divergent. by equality, the LHS must then also diverge but this can occur in the limit of infinite N since the matrix product (xp) is:

$$xp = \sum_i^N \sum_j^N x_{ik} p_{kj}$$

this is equivalent to saying that it is sufficient for the matrices to be of finite dimension for the cyclic property of the trace to hold - if not, then the LHS is not 0.

18. Sep 9, 2007

### Anonym

The Feynman Lectures on Physics. Since I use the Russian translation, it is ch. “Quantum behavior”, par.3 “Experiment with waves” my numbers (37.2, 37.3, 37.4) and (37.7) in par.7 “Initial principles of QM”.

Sorry, I don’t like to type math. I feel pleasure doing that manually.

Regards, Dany.

19. Sep 9, 2007

### Anonym

Otherwise, it is not metric space and therefore useless in the physical applications.

Reilly, you are right and QM is perfectly O.K. This is wrong question and therefore it has the wrong answer. Tr(i*I)not = i*Tr(I). Tr(i*I)=Tr(i)=0; i is 2x2 traceless antisymmetric matrix. The complex numbers are 2-dim quadratic normal division algebra. And it is also connected with our Classical friend E. Schrödinger Cat from Alice Wonderland.

Regards, Dany.

Last edited: Sep 9, 2007
20. Sep 9, 2007

### Hurkyl

Staff Emeritus
And therein lies our difference. This is not known to be a physical fact; while it may ultimately prove true, currently every indication says otherwise.