# A What Hilbert space for a spinless particle?

1. Jul 18, 2017

### David Olivier

I'm looking for a rigorous mathematical description of the quantum mechanical space state of, for instance, a particle with no internal states.

At university we were told that it the Hilbert state of wave functions. They gave us no particular restrictions on these functions, such as continuity, apart from the fact that they should be quadratically integrable, so that we can define inner products. But then we are given a basis of kets of the form $| \vec r>$, representing "Dirac functions", that precisely are not quadratically integrable. We were told to shut up and calculate.

I'd like to clear this up a bit. I know the Dirac "functions" can be defined rigorously as distributions. But how does that fit into a Hilbert space?

Does someone know of a mathematically rigorous, but elementary, introduction to this issue?

2. Jul 18, 2017

### DrDu

Dirac functions are indeed not elements of the Hilbert space. However, the Hilbert space may be enlarged to encompass these functions. The new space is called "rigged Hilbert space".
https://en.wikipedia.org/wiki/Rigged_Hilbert_space
Whether you want to do QM with rigged or ordinary Hilbert spaces is rather a matter of taste, i.e. whether you like the formal simplicity of Diracs formalism or whether you rather consider distributions as unphysical idealizations.

3. Jul 18, 2017

### David Olivier

Thanks. I'll look it up.

If you want to stick to plain Hilbert spaces, what bases do you have? You don't have the "particle in a precise position" kets, nor the "particle with a precise momentum" ones; so what do you use?

4. Jul 18, 2017

### DrDu

For example harmonic oscillator eigenstates.

5. Jul 18, 2017

### David Olivier

OK, but that's not for a free particle.

Though I suppose that you could use them for a free particle (no potential), even though they would no longer be eigenstates of the Hamiltonian. They would still form a basis. But that looks like an awkward choice.

6. Jul 18, 2017

### David Olivier

7. Jul 18, 2017

### Staff: Mentor

As the OP says, those won't work for a free particle because they would just be plane waves, which are not normalizable and so are not in the (non-rigged) Hilbert space.

Not if you want your basis states to be in the (non-rigged) Hilbert space.

8. Jul 18, 2017

### David Olivier

If I remember well, the wave functions of a harmonic oscillator are something like $\psi_0(x) = a_0 e^{-b x^2}$, $\psi_1(x) = a_1 x e^{-b x^2}$...

These are not plane waves, and are normalizable.

Do they not form a basis of the Hilbert space? It seems to me that the Hilbert space is independent from the Hamiltonian. Of course, the above wave functions will evolve differently in the absence of the quadratic potential, not being eigenstates anymore of the Hamiltonian, but as elements of the Hilbert space, if they form a basis in the presence of that potential, they form one in the absence of it too.

9. Jul 18, 2017

### DrDu

Yes, that's completely correct.
For operators like p or r, it is not necessary to have a set of eigenvectors in Hilbert space. It is sufficient that they can be expressed as an integral over a projection valued measure. Von Neumanns book is still a classic on this theory, and relatively easy to read.

10. Jul 18, 2017

### Staff: Mentor

These aren't eigenstates of the harmonic oscillator Hamiltonian (or the free particle Hamiltonian, for that matter; they're not the same). They are coherent states, which are eigenstates of the harmonic oscillator annihilation operator $\hat{a}$. They are normalizable, yes, but I don't know if they can be used as a basis.

Eigenstates of the free particle Hamiltonian, which are the ones I was referring to, are plane waves, i.e., wave functions of the form $\psi = a e^{i k x}$. These are not normalizable.

Eigenstates of the harmonic oscillator Hamiltonian are eigenstates of the number operator, which is $\hat{a}^\dagger \hat{a}$. These are normalizable, and can be used as a basis. (I think I misspoke in my previous post, since these states might have been the basis @DrDu had in mind.)

11. Jul 18, 2017

### David Olivier

(In reply to DrDu) In other words, the need for a basis is overdone! :)

12. Jul 18, 2017

### dextercioby

Some more comments at "A" level.

Regardless of the used interpretation, theoretically, a quantum system will always be specified by the mathematically conceivable states and observables. And here the word "mathematically" is crucial. If I had written "physically", you would have been in the position to ask me: how does one experimentally prepare and measure these states and observables? But since I've chosen to tell you about the mathematically meaningful states and observables, then let me do so.

A quantum system is mathematically defined by the Hamiltonian and the set of irreducible observables and the algebraic (addition, composition = multiplication) relations which help us construct from them the whole set of all possible observables. Depending on the system, the Hamiltonian may or may not be in the irreducible set. This set of all observables can be given the structure of an associative algebra with unity with respect to observables' addition. In this scenario, the set of irreducible observables of a quantum system is the center of this algebra, if we further endow it with the Lie product structure (commutator). For a free spinless particle in 1D, the irreducible observables are the coordinate x and momentum p, because the system is defined by H=p^2/2m. These 2 irreducible observables obey the so-called Born-Jordan commutator relation [x,p]=i.

If one then asks which are the possible PHYSICAL states for the free spinless particle in 1D, the answer is simple: it is the complex, separable, infinite dimensional Hilbert space (or, more pedantic, the set of points in the projective Hilbert space built from it is the set of pure states, as opposed to mixed states which can't be given the structure of a projective Hilbert space) in which the commutation relation can be mapped/realized. By the known Stone-von Neumann uniqueness theorem, this space is only L^2(R), if the particle is conceived to be unrestricted in motion. The possible MATHEMATICAL states are the ones which require the mapping of the B-J commutation relation into a rigged Hilbert space with L^(R) as the Hilbert space from it.

Anything else is derived from here. The set of states of "well-defined position" (and here we venture into the realms of interpretations of QM) for a free spinless particles is not made up of physical states, but of mathematical ones. The set of states of "well-defined energy" for a free spinless particles is not made up of physical states, but of mathematical ones etc.

Last edited: Jul 18, 2017
13. Jul 18, 2017

### Physics Footnotes

In principle, yes. Here's a rough outline of how it works von Neumann style. If it sounds like the approach appeals to you, I agree with @DrDu that von Neumann's original work "Mathematical Foundations of Quantum Mechanics" remains a wonderful introduction to the subject for the mathematically paranoid ;-)

Consider a finite dimensional Hilbert Space $\mathcal H$ spanned by a set of orthonormal vectors $|\lambda_i\rangle$ labelled by the eigenvalues of the corresponding Hermitian operator$$A=\sum \lambda_i|\lambda_i \rangle\langle \lambda_i|$$Now, for any element $|\psi\rangle \in \mathcal H$ we have:$$|\psi\rangle=\sum \langle \lambda_i|\psi\rangle|\lambda_i\rangle=\sum|\lambda_i \rangle\langle \lambda_i|\psi\rangle$$which, in terms of projection operators, gives $$I=\sum |\lambda_i \rangle\langle \lambda_i|$$All the work we need to do, it turns out, (as far as the statistical and spectral analysis of $A$ are concerned) can now be done with the family of projection operators $|\lambda_i \rangle\langle \lambda_i|$ (called, by virtue of that last relation, the Resolution of the Identity for $A$), which, together with the spectrum, is used to define the Projection Valued Measure (PVM) of $A$.

What's the point of all this? Well, it turns out that the notion of a PVM can be rigorously extended to infinite dimensional Hilbert spaces, and in particular copes perfectly with continuous spectra without any need of eigenvectors. This approach, invented by von Neumann in the late 1920s, allows us to do all quantum mechanical calculations rigorously within the Hilbert Space framework, by eliminating Dirac's delta function and all the Rigged Hilbert Space machinery that it requires for a rigorous treatment.

14. Jul 18, 2017

### Staff: Mentor

Its called a Rigged Hilbert Space.

Unfortunately mathematically rigorous and elementary are not found in this area. Attached is a rigorous treatment - but far from elementary. Its Dr Madrid's PhD Thesis on it and goes into much more detail than the articles by him you have probably read eg areas like Nuclrar spaces etc.

However the following, while not fully explaining what going on is approachable:
https://www.amazon.com/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

It should really be known by all physicists, and indeed applied mathematicians. Its worth it for its treatment of Fourier Transforms alone which otherwise becomes bogged down in issues of convergence,

Thanks
Bill

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15. Jul 19, 2017

### DrDu

No, these are really eigenstates of the harmonic oscillator and not coherent states. You can easily show that e.g. $\psi_1(x)$ isn't an eigenstate of $\hat{a}$.

16. Jul 19, 2017

### atyy

If you measure position, you still need the position states which are Dirac Deltas and not in the Hilbert space. You need the position Delta functions, simply because they come along as part of the definition of an exact measurement of position.

The modification of rule-of-thumb basic QM that is needed is that after an exact position measurement, the wave function does not collapse into a position eigenstate (since that doesn't belong to the Hilbert space). The collapse rule is given in Eq 3 and 4 of https://arxiv.org/abs/0706.3526, and an example of the wave function after an exact position measurement is discussed in section 2.3.2 on p7.

17. Jul 19, 2017

### DrDu

I don't think there is a modification of any QM rules needed. The point is rather that we can't measure a continuous variable arbitrarily precisely, so the limit of delta functions is unphysical.

18. Jul 19, 2017

### David Olivier

Thanks. As noted above, I already had found a (much shorter) pdf by the same author, "The role of the rigged Hilbert space in Quantum Mechanics". What I now lack is time to digest all this.

19. Jul 19, 2017

### DrDu

I would like to dwell on this a little bit more.
Let's generalize this to infinite dimensions and take the position operator as an example. Then,

$X=\int dx x \delta(x'-x) \delta(x''-x)=\int dx x \delta(x'-x) \delta(x''-x')= \delta(x''-x') \int x d \theta(x-x').$
The last step should be familiar from classical statistics, where we express the integral over a probability density function by an integral with the cumulative distribution function as integrand.

20. Jul 19, 2017

### atyy

At least in the formalism, a sharp measurement of position is possible. After a sharp measurement of position, the wave function collapses to something that is not a position delta function.

21. Jul 19, 2017

### DrDu

I fear, to convice me, you would have to provide a sound reference.

22. Jul 19, 2017

### DrDu

Isn't the center of the algebra formed by all those operators which commute with all elements of the algebra? Then how can p and x be in the center if they don't even commute among themselves?

23. Jul 19, 2017

### Staff: Mentor

Hmmm. The multiplication of two Dirac Delta functions. That's rather interesting. Its legit mind you - but is mathematicaly a bit hairy:
https://arxiv.org/abs/1308.0257

Do physicists really want this sort of mathematical sophistication?

Thanks
Bill

24. Jul 19, 2017

### DrDu

Here, the point is rather the result than the way to derive it. At the end we can express the operator X in terms of projectors, which here are simple Heaviside theta functions in position space. The defining property of a projector P is that it is idempotent :$P^2=P$ from which it follows that it has only eigenvalues 0 or 1.
These projectors appear as measure in the integral, hence projector valued measure.
In contrast to the products of delta functions from which we started, the projectors are the most well behaved operators we can think of in Hilbert space. As we see, they are sufficient to characterize the whole position operator.

25. Jul 19, 2017

### Staff: Mentor

Eigenstates of the harmonic oscillator what?

Eigenstates of the harmonic oscillator Hamiltonian are eigenstates of the number operator, $\hat{a}^\dagger \hat{a}$. In the position representation, these are fairly complicated expressions involving Hermite polynomials. They aren't just functions of the form $a e^{-bx^2}$ (although factors of that form appear in them).