# Is a state space with indistinguishable particles a quotient? Of what?

1. Mar 1, 2014

### andrewkirk

I’ve been trying to understand how having indistinguishable particles in a system changes the nature of the state space.

The QM texts I have gloss over this.

A typical approach is to define the symmetric and anti-symmetric kets that serve as a basis for the eigenspace containing $|ω_1,ω_2⟩$ and $|ω_2, ω_1⟩$ where $ω_1$ and $ω_2$ are eigenvalues of an operator Ω corresponding to a measurement that can be made on either of the identical particles, and then say that the system state after a measurement of Ω for both particles must be either one of the symmetric or antisymmetric kets, according to whether the particles are bosons or fermions.

But they don’t say anything about how, if at all, the state space that is subject to this adjustment differs from the raw direct product space.

My guess is that, if there are two identical particles then the state space $\mathscr{V}$ is a quotient space of the unrestricted state space $\mathscr{V}_0$ that is obtained as the direct product of the state spaces for the individual particles, but quotient over what subspace?

Again I have a guess. That is that we take the quotient of $\mathscr{V}_0$ over the subspace $\mathscr{S}$ of $\mathscr{V}_0$ generated by $\cup_{\omega_1,\omega_2\in Eig(\Omega)} \mathscr{G}_{\omega_1 +\omega_2}$ where $\mathscr{G}_{\omega_1 +\omega_2}$ is the two-dimensional eigenspace of the $\mathscr{V}_0$-operator $\Omega^{(2)}$ that measures $\Omega$ for both particles, corresponding to eigenvalue $(\omega_1 +\omega_2)$ of $\Omega^{(2)}$, and $Eig(\Omega)$ is the set of eigenvalues of $\Omega$.

However I have a feeling that this quotient may remove too much from the state space, not leaving enough distinguishable states.

I would like to work out what, if anything is the correct quotient space, and then (1) work out how the quotient changes if the indistinguishable particles change between being all bosons and being all fermions and (2) extend it to the case of more than two indistinguishable particles.

Does anybody have any suggestions? Am I completely on the wrong track? I've googled things like 'indistinguishable particles quotient space' with no luck. Is the space still a Hilbert Space once it contains indistinguishable particles?

Thanks for any suggestions.

2. Mar 1, 2014

### ChrisVer

Isn't what you ask for the Fock Space (for identical particles)?

3. Mar 1, 2014

### marmoset

I think what you are looking for is the concept of exterior and symmetric powers of a vector space. If your one particle Hilbert space is H, the Hilbert space for n fermions is the nth exterior power of H, and the Hilbert space for n bosons is the nth symmetric power of H. These terms are defined in lectures 18 and 21 here http://people.virginia.edu/~mah7cd/Math5651/. Those notes also describe how to form these spaces as quotients of the tensor powers of H, like you ask.

This is related to the fermion and boson Fock spaces which are the exterior and symmetric algebras built from H; these are sums of all the exterior/symmetric powers of H, just like the tensor algebra is a sum of all the tensor powers of H. (All these spaces include the 0th power which is taken as the scalars, they represent the 0 particle state in quantum mechanics)

If your H is infinite dimensional I think you have to use products which are slight generalizations of the ones defined in the notes above. You want the sums/products of H to be complete so they too are Hilbert spaces. I don't know much about this though, best to ask someone else (the constructions are defined here http://en.wikipedia.org/wiki/Hilbert_space#Constructions).

4. Mar 1, 2014

### andrewkirk

Thank you both. That is exactly the sort of thing I had in mind. I shall now go and refresh my understanding of wedge products. Those Uni of Virginia lecture notes look like a good place to do that.

It's a pity that QM texts don't often deal with this rigorously. I can understand they choose not to because the maths is difficult. But it puts the theory into a broader perspective that, to me at least, makes more sense of it once one grasps it. A particularly nice thing about the Fock space is that it appears to give a very natural framework for describing the creation of new particles and their subsequent disappearance.

5. Mar 1, 2014

### WannabeNewton

Well it doesn't really add to any understanding of the physics. The math is actually not difficult at all with regards to what you're talking about. The difficult parts are essentially all contained in the study of operators. Regardless, see section 19.6 of the following (brilliant) text: https://www.amazon.com/Quantum-Theory-Mathematicians-Graduate-Mathematics/dp/146147115X

Fock spaces are the natural state spaces used in QFT (but not the only way to formulate things).

6. Mar 1, 2014

### andrewkirk

It does for anyone that wants the physics to be a logical development from the postulates, rather than a collection of brute facts.

The postulates tell us that the state space is a Hilbert Space, and that is repeatedly relied on. If we start identifying elements of the Hilbert space with one another, as is done when we analyse systems containing indistinguishable particles, we can no longer be sure that it is still a Hilbert Space. If it ceases to be a Hilbert Space, the whole enterprise collapses.

The Fock Space approach ensures that the identification is done in such a way that the space remains a Hilbert Space, and hence underpins the validity of any reasoning done with systems containing indistinguishable particles.

That 'QM for mathematicians' looks like just the sort of book I would enjoy. Thanks for pointing it out.

7. Mar 1, 2014

### WannabeNewton

I've found it much easier to hold on to that mindset in GR than in QM. With the latter I find myself quickly drowned in a sea of math I can't understand

No problem! It's an extremely valuable reference, especially when you find yourself being cheated out of some proper definitions and justifications in standard QM texts.

8. Mar 1, 2014

### strangerep

That's not what is done. See below.

For composite systems, one forms the tensor product space (not merely the direct product that you mentioned earlier). Then we can indeed be sure we're still dealing with a Hilbert space.

Indeed, but that's why one uses the tensor product rather than the ordinary direct product.

BTW, since you expressed dissatisfaction other QM texts earlier, you might also get something from reading Ballentine ch17 (and maybe also sect 8.3).