# Exchange Operators & Spin Statistics - I don't the conclusions

1. Mar 15, 2010

### LukeD

Today in class, by the existence of an operator that exchanges the states of two indistinguishable particles, we attempted to derive the existence of fermions and bosons & how this relates to the symmetries of multiparticle wave functions.

The argument given in my textbook is: define an exchange operator P. "Clearly" P^2 = I. Therefore, the eigenvalues of P are +1 and -1. Systems of identical particles are eigenvectors of an exchange operator, so they are therefore either symmetric or antisymmetric under exchange of particles.

On the other hand, I don't see why we need P^2 = I. We can have P^2 = e^(i theta) for any value of theta because the only requirement we have is that exchanging twice leaves us with a state that is physically indistinguishable.

So what gives? Is the argument given in my book completely bogus? Can someone direct me to another book that has an explanation of bosons & fermions? (I'm looking at Townsend by the way)

2. Mar 15, 2010

### Ben Niehoff

Here is a perhaps clearer explanation, that is mentioned perhaps a third of the way down this article on path integrals: http://arxiv.org/PS_cache/quant-ph/pdf/0004/0004090v1.pdf

Basically, in order to see what happens, we must consider particles propagating as a function of time. In a system of two particles, simply transform to the center-of-mass frame. We don't care how the center of mass is moving; we're going to look at the relative separation vector between the particles, q. Suppose after some time T, the new relative separation vector is q'. (In the interest of brevity, I"m not going to put arrows on the vectors...but in this post, lowercase q is always a vector).

Define $D(q', q)$ to be the propagator for two distinguishable particles at some separation q to end up at some separation q' at some later time. Similarly, $D(-q', q)$ is the propagator for two distinguishable particles to end up at the same physical locations, but exchanged between themselves.

Now, in the quantum case, since the particles are indistinguishable, we cannot say whether they exchanged or not. The end result is simply two particles with separation q'; we can't say which one is which. Therefore, the total propagator $K(q',q)$ is a linear combination of both possibilities, weighted by a phase factor:

$$K(q', q) = D(q', q) + e^{i\phi} D(-q', q)$$

However, as you correctly point out, since the particles are indistinguishable, acting with the exchange operator on $K(q',q)$ must leave it invariant up to some (independent) phase:

$$P K(q', q) = K(-q', q) = e^{i\alpha} K(q', q)$$

Now, merely plug in the definition of $K(q', q)$:

$$D(-q', q) + e^{i\phi} D(q', q) = e^{i\alpha} \left( D(q', q) + e^{i\phi} D(-q', q) \right)$$

Now, merely collecting like terms on each side ought to convince you that

$$\phi = \alpha = n\pi$$

for any integer n. Therefore the phase picked up by exchanging particles must be $\pm 1$.

Note: In two dimensions, this argument breaks down, due to topological considerations. Specifically, the configuration space is no longer simply-connected, and we must include (an infinite number of) path exchanges with various winding number. This leads to anyons. The linked paper also explains this.

3. Mar 20, 2010

### councilmage

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