# Understanding the Bose-Fermi dichotomy

I'm an amateur on a quest to understand QM. In various places (such as early in chapter III.4 of the Feynman Lectures) I have seen an argument that looks like it's trying to convince me that any class of indistinguishable particles must necessarily behave either like bosons or like fermions -- i.e., that this is not only an empirical fact, but a deductive consequence of the basic structure of QM.

I have trouble understanding these arguments. They seem to pull an assumption out of nowhere, that interchanging two of the identical particles must necessarily correspond to multiplying the wavefunction by a constant. Once this is given, I can see that this constant must be either $1$ or $-1$, but where does that assumption come from? If I write down a wavefunction without knowing in advance that the particles are identical, there are oodles of ways to write one that doesn't react to an interchange merely by a constant factor. Is there something that a priori excludes all those states from consideration?

I've been able to reconstruct an argument that works for a world containing only $2$ instances of the indistinguishable particles, to wit:

What it formally means for the particles to be indistinguishable must be that the operator $P$ which interchanges the two particle commutes with every observable, in particular with the Hamiltonian.

If we have any random state of the system, we can write it as a sum of eigenstates of $P$, and because $P^2=1$, the eigenvalues must be $\pm 1$. Now whenever time passes (we apply $H$) or we make an observation (and apply some other observable operator), the operator we use commutes with $P$, and therefore we can diagonalize $P$ and the operator simultaneously and see that the $+1$ eigenstates never contribute any amplitude to the $-1$ eigenstates and vice versa.

Thus if at some point in history we have made an experiment to determine out whether our two particles are bosons or fermions, right after the experiment the state of the world is going to be a pure $+1$ or $-1$ eigenstate -- and because of the above reasoning it is going to stay in that eigenspace for all eternity henceforth.

So in this case it's of no use to consider mixed states.

The problem with this argument is that it doesn't seem to scale beyond two particles. With $n$ particles there are a lot of different permutations of them one could consider, and these permutations don't commute internally, and so I cannot be sure that they will all diagonalize simultaneously with the operator I'm observing.

This is probably not just a problem with my proof skills, as dimension-counting shows. Let's consider a discrete system where each of $3$ particles can be in one of $3$ states. The state space has dimension $27$ over $\mathbb C$ -- but the "fermion" subspace (consisting of those states that go to their negative under an interchange of any pair of particles) has dimension $1$, and the "boson" subspace (those states that stay unchanged under any transposition) has dimension $10$ if I count it correctly. There are $16$ dimensions left unaccounted for! Is there a principled reason why all of those states are necessarily unphysical or uninteresting?

In Zee's Quantum Field Theory in a Nutshell (which I bought by mistake but don't have anywhere near the prerequisites to understand) there's some mumbling about "anyons", which as far as I can decipher it seems to claim that a strict Bose-Fermi dichotomy is only necessary if one has a continuous set of base states of dimension at least $3$ -- by some topological considerations that mostly went over my head.

Am I misunderstanding it when I think Feynman and others claim that the dichotomy is a derived result?

• – Qmechanic Aug 18 '12 at 11:39
• @Qmechanic: I probably shouldn't have mentioned the anyon thing (which I don't understand anyway) -- I only meant that as circumstantial evidence that perhaps the argument Feynman had in mind doesn't quite work anyway. What I'm trying to understand is just what's the deal in ordinary undergraduate QM. – Henning Makholm Aug 18 '12 at 12:22

## 1 Answer

There is one assumption and one empirical fact which enter the argument:

• the assumption is that the particles are truly indistinguishable
• the empirical fact is that there is no exchange degeneracy

What does this mean. You write down the wavefunction for a many particle problem. Then you want to implement the fact that the particles are really indistinguishable. That means that the Hamiltonian commutes with all the elements of the permutation group (the permutation group is a symmetry) As you probably know from quantum mechanics, symmetries generically lead to degeneracies. However in this case, there is empirical evidence that multiparticle states are degenerate not degenerate. This leave us with the fact that the wavefunctions which diagonalize the Hamiltonian have to be in a one dimensional representation of the permutation group. There are two of those: the symmetric (boson) and the antisymmetric (fermion). What is left to show is that the fact whether it is symmetric or antisymmetric depends on the spin of the particle. This can be done in the context of relativistic field theory.

Edit:

For a simple example let us discuss the case of three particles (I do not know how do discuss three states without first introducing second quantization, so let us have three particles in a box).

The Hamiltonian is $$H = -\frac{d^2}{dx_1^2} -\frac{d^2}{dx_2^2} -\frac{d^2}{dx_3^2} + V(x_1) + V(x_2) + V(x_3)$$ with $$V(x) = \begin{cases} 0, & 0< x < 1 \\ \infty, &\text{else.}\end{cases}$$ it clearly commutes with any possible permutation of the coordinates of the three particles $x_1, x_2, x_3$ (as it should to have them indistinguisable, in fact all allowed operators have to commuted with the permutation group). The Hilbert space (naively) is given by $\Psi(x_1,x_2,x_3) \in L^2(\mathbb{R}^3)$.

One can now solve the single particle problem $$\left[- \frac{d^2}{dx} + V(x) \right] \psi_n(x) = \epsilon_n \psi_n(x)$$ with $\psi_n(x) = \sin(\pi n)$ and $\epsilon_n = \pi^2n^2$. Then one can try to put three particles into the states $\psi_1$, $\psi_2$, $\psi_3$.

The naive approach would be to say $\Psi(x_1,x_2,x_3) = \psi_1(x_1) \psi_2(x_2) \psi_3(x_3)$. The fact that any permutation commutes with the Hamiltonian then immediately tells you that the state $\Psi$ has the same energy as for example the state with the wavefunction $\psi_2(x_1) \psi_1(x_1) \psi_3(x_3)$, and all the other wavefunctions which you get by permuting the coordinates $x_1, x_2, x_3$. In the end you have a degeneracy of each level of the order of $N!$ (where $N$ is the number of particles). This degeneracy is not observed.

So the question is: can you find eigenstates such that applying a permutations does not produce a different state but gives you back the same state. In representation theory, you are searching for a one-dimensionaly represenation of the permutation group. For $N>2$ there are only two of those: the symmetric and the antisymmetric. In our case these are \begin{align}\Psi_{S/A} &= \frac{1}{\sqrt{3!}} [\psi_1(x_1) \psi_1(x_2) \psi_1(x_3) + \psi_1(x_2) \psi_1(x_3) \psi_1(x_1) + \psi_1(x_1) \psi_1(x_2) \psi_1(x_3)\\ &\quad \pm[ \psi_1(x_1) \psi_1(x_3) \psi_1(x_2) + \psi_1(x_3) \psi_1(x_2) \psi_1(x_1) +\psi_1(x_2) \psi_1(x_1) \psi_1(x_3)]\end{align}. Both of these states map onto themselves if a permutation is applied, so there is no degeneracy of the eigenstates. In the end, what we have to do to describe bosons or fermions, we have to restrict the Hilberspace to allow only for symmetric/antisymmetric wavefunctions.

• No, I don't "probably know from quantum mechanics, symmetries generically lead to degeneracies". As I said, I'm an amateur and probably reading the wrong books in the wrong order, so I don't have a clear understanding of what "degeneracies" and "degenerate" means in this context. Your "This leaves us with" seems to jump over the very point that confuses me. Sure, eventually I'd like to understand spin-statistics, but for the nonce I'm just trying to to figure out the "statistics" part. – Henning Makholm Aug 18 '12 at 11:40
• "the fact that the wavefunctions which diagonalize the Hamiltonian have to be in a one dimensional representation of the permutation group." I must be misunderstanding this somehow, because it seems to say that the Hamiltonian in my 3-particles-in-3-states example can only have 1+10 dimensions of eigenspace in total -- but unless I'm really misremembering linear algebra, the sum of the eigenspace dimensions for a diagonalizable matrix must equal the total dimension of the space it operates on. What's going on here? Am I counting the eigenspace dimensions wrong? – Henning Makholm Aug 18 '12 at 11:45
• @HenningMakholm: this is exactly what I am saying. If you have three states and three particles and you put one particle in each of the states you have (up to a phasefactor) exactly one possible state - this is an empirical fact (found already in 19$^{th}$ century and called Gibbs paradox and using representation theory (which as far as I understand you are not yet familiar with) one can see that the missing exchange symmetry leave just bosons and fermions as options. – Fabian Aug 18 '12 at 11:56
• I know some basic definitions of representation theory, but I fail to see how you get it to enter here. As far as in can see in the 3×3 example there is no possible Hamiltonian that one can write down such that every eigenvector of the Hamiltonian is either symmetric or antisymmetric, because there would be 16 dimensions missing. What is wrong in that argument? – Henning Makholm Aug 18 '12 at 12:08
• I don't understand your argument, but I added some more information to my answer. – Fabian Aug 18 '12 at 12:32