# Separable Hilbert space's postulate

1. Jul 22, 2014

### lfqm

The first postulate of quantum mechanics says that every physical system is associated with a separable complex Hilbert space, however this does not hold for a free particle, where the basis is uncountable (all the momentum kets). I think it also does not hold for a free falling particle (V=mgz), where the eigenfunctions are the Airy functions.

how is that?

Thanks

2. Jul 22, 2014

### bhobba

No. Or rather that is Von Neumanns formulation - which is not the modern one based on Rigged Hilbert Spaces.

Here is the single axiom from, for simplicity, we will assume is a finite dimensional complex vector space:

An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

Then we apply Gleason's Theroem to show that a positive operator of unit trace exists the probability of outcome i is Trace (PEi). By definition we call P the system state.

Now to get to your question the physically realizable states (to be precise the pure states) are assumed to be finite dimensional which makes the math a lot simpler. Then for mathematical convenience we consider the linear functionals defined on that space. In the weak topology of that space these functionals are the limit of a sequence of the space these are defined on (usually its in the strong topology as well, but no need to go into details like that initially). So to make life easier you extend it to include such limits.

It is this space that includes things like the free particle state.

You can find the technical detail here:
http://arxiv.org/pdf/quant-ph/0502053.pdf

Thanks
Bill

Last edited by a moderator: May 6, 2017
3. Jul 22, 2014

### dextercioby

Hi Bhoba, I understand how you start, but the assumption you start with is that the set of outcomes is countable, which can't be said for the free (Galilean) particle.

@the OP: indeed, it's quite amazing, the Hilbert space is actually the starting point. You soon discover that it's too small to account for the full mathematical description of even the simplest system, the Galilean free particle in 1D.

Last edited: Jul 22, 2014
4. Jul 22, 2014

### rubi

It actually does hold. The Hilbert space for a free particle is just $\mathcal H = L^2(\mathbb R^3)$, which is separable. The "momentum kets" don't form a basis. They aren't even elements of $\mathcal H$. Instead, they are generalized eigenvectors of the momentum operator and belong to a different space.

5. Jul 22, 2014

### bhobba

Well actually that has a countable basis as well - but that's not the point I am trying to make.

I start with the assumption the physically realizable states have a basis that is not only countable but in fact are from a finite dimensional space - its isomorphic to the space of sequences of finite length.

We consider the space of all such vectors - while each vector is finite the resultant space has an infinitely countable basis. Now for the trick - we consider all the linear operators defined on such a space. That space is also a countably infinite vector space but is much larger than the space its defined on - in fact its isomorphic to the space of all infinite sequences.

The free particle function you are talking about belongs to that space. To see this consider the space of continuously differentiable functions of finite support - the so called test functions from distribution theory. They are dense in the Hilbert space of square integrable functions hence one can find a basis of those functions. Simply consider all the functions that can be formed from a finite sum of this basis. The free particle state is a linear functional defined on such a space by simply taking its integral. BTW this shows that function has a countable basis as well.

In QM, for mathematical convenience we extend the space to include such functions - it includes all sorts of other weird stuff as well like the dreaded Dirac Delta function.

The dual contains the usual Hilbert space being the sequences whose squares are summable.

This is the Gelfland Triple in Rigged Hilbert space language with the Hilbert space stuck in the middle. One takes various subsets of the Hilbert space with nice properties such as say fairly good functions and looks at its dual (its the space of well tempered distributions) which contains much more such as the free particle function since if I remember correctly it is a well tempered distribution.

Terry Tao wrote a nice little article on it:
http://www.math.ucla.edu/~tao/preprints/distribution.pdf

He explains some of the jargon like weak convergence etc I used.

Basically Rigged Hilbert spaces are simply Hilbert spaces with distribution theory stitched on - hence the name rigged - like rigging on a ship.

I am however pretty sure you know that already - I probably wasn't clear initially - so hopefully its clear now.

Thanks
Bill

Last edited: Jul 22, 2014
6. Jul 23, 2014

### stevendaryl

Staff Emeritus
Someone who is more familiar with the mathematics than I am can correct me if I am wrong, but the way I understand it, non-separable Hilbert spaces can crop up whenever you have a composite system with an infinite number of components.

Quantum field theory of course deals with an unbounded number of particles, but not an actually infinite number. That is, the Hilbert space for QFT is Fock space, which is a sum of zero-particle states, one-particle states, etc. There is no state with infinitely many particles.

A sort-of physically meaningful model with an infinite number of particles might be an infinite lattice of particles, representing a solid. The quantum-mechanical description of such a system would require a non-separable Hilbert space unless you made the assumption that at any time, all except finitely many particles are in the ground state. With this assumption, you could model such a lattice using Fock space.

One thing I don't understand, though, is how it is possible to model zero-mass particles, such as photons, with a separable Hilbert space. It seems to me that there is no physical reason for there being only finitely many "soft" photons. It seems that you could have an infinite collection of very-low-energy photons whose energies add up to a finite number. Physically, this seems possible, but I don't see how the situation could be modeled using a separable basis.

7. Jul 23, 2014

### vanhees71

The problem with the photons is that the plane-wave states we use to describe free photons (i.e., the theory of a free massless vector field) are not the proper asymptotically free states of the interacting theory. The reason precisely is that the photon field is massless. A good introduction into this issue is

Kulish, P.P., Faddeev, L.D.: Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys. 4, 745, 1970
http://dx.doi.org/10.1007/BF01066485

For a complete elaboration, see

Kibble, T. W. B.: Coherent Soft‐Photon States and Infrared Divergences. I. Classical Currents, Jour. Math. Phys. 9, 315, 1968
http://dx.doi.org/10.1063/1.1664582

Kibble, T. W. B.: Coherent Soft-Photon States and Infrared Divergences. II. Mass-Shell Singularities of Green's Functions, Phys. Rev. 173, 1527–1535, 1968
http://dx.doi.org/10.1103/PhysRev.173.1527

Kibble, T. W. B.: Coherent Soft-Photon States and Infrared Divergences. III. Asymptotic States and Reduction Formulas, Phys. Rev. 174, 1882–1901, 1968
http://dx.doi.org/10.1103/PhysRev.174.1882

Kibble, T. W. B.: Coherent Soft-Photon States and Infrared Divergences. IV. The Scattering Operator, Phys. Rev. 175, 1624, 1968
http://dx.doi.org/10.1103/PhysRev.175.1624