Physics Friday 76

Part 1 of ?: Distinguishable and Indistinguishable particles

Suppose one has a pair of objects which interact with each other, such as two colliding billiard balls. For classical objects, we can distinguish between them. That is to say, if we name them ‘object 1’ and ‘object 2’, then following whatever set of interactions we have, we can look at one of the objects and, in theory at least, know whether it is object 1 or 2.
There are two ways we can distinguish the objects. The first is differences in the objects themselves; differences in mass, physical marks, etc. Two macroscopic objects are almost certain to be at least slightly different. The second way is to watch the objects as they interact and track their paths; you can follow 1 and 2 from their initial to final positions.

Now, let us compare to this the problem of a pair of microscopic particles of the same kind, such as two electrons, or two protons. Here, the physical properties of the two particles are identical; all electrons have the exact same mass, charge, etc. So the first method fails. For the second method, the principles of quantum physics tell us that between measurements, the particles do not have definite positions, instead being described probabilistically by the wavefunctions. As there are no definite trajectories, we cannot track the particles.

Together, this means that identical particles are indistinguishable; we cannot, even in principle, tell particles of the same species apart. This has important consequences for the joint wavefunctions of multiple particles, and for statistical mechanics.
In particular, suppose we have a pair of identical particles. We describe them with a joint wavefunction. Both particles’ state vectors occupy identical Hilbert spaces H, so that the joint wavefunction is in the Hilbert space formed by the tensor product HH. If one of the particles is in a state a, and the other in state b, one might assume the resulting state is , as this is the standard for such a tensor product space. However, this implies that we can identify the particle in state a as ‘particle 1’, and the other as ‘particle 2’; would have them swapped. This violates what we said earlier about indistinguishability.

Let us consider a linear operator P, which swaps the values of the two state vectors; this is equivalent to swapping the two arbitrary labels we’ve attached to the particles (and their single-particle spaces):
.
This operator is its own inverse, and is Hermitian and unitary. Indistinguishability means that we should expect no physical observable result from the application of P to our state Ψ; at most, P can cause a change in the complex phase:
, with |λ|=1.
This means that valid states under indistinguishability are eigenvectors of P. But the fact that P is its own inverse means that its eigenvalues obey λ2=1, so that λ=±1. Thus, the joint wavefunction must be either symmetric or antisymmetic under particle exchange; we have

for a symmetric state, or

for an antisymmetric state (the Cs are normalization constants). For three or more particles, this still holds, with the wavefunction either symmetric or antisymmetric under the exchange of any two of the particles.

The question remaining is when are the wavefunctions symmetric, and when are they antisymmetric? The answer is that it depends on the particle species; there are two kinds of particles: one type, called bosons, which are always symmetric under exchange; and one type, called fermions, which are always antisymmetric. Which of these two groups a particle species belongs to is determined by an intrinsic particle property called “spin,” which we will discuss in our next part.

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2 Responses to “Physics Friday 76”

  1. Physics Friday 78 « Twisted One 151’s Weblog Says:

    […] and other half-integer n values. What, though, does this have to do with our previous parts on indistinguishable particles and particle spin? The answer is that the above provides an alternate picture of spin1. We identify […]

  2. Physics Friday 79 « Twisted One 151’s Weblog Says:

    […] while fermions, with half-integer spin, have wavefunctions antisymmetric under exchange. In our first part, we noted that in terms of the tensor product Hilbert space H⊗H of the joint wavefunction, […]

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