Spin, Statistics, CPT and All That Jazz

John Baez

March 4, 2012

This is a little grab-bag of proofs of the spin-statistics theorem. Quantum mechanics says that if you turn a particle around 360°, its wavefunction changes by a phase of either +1 (that is, not at all) or -1. It also says that if you interchange two particles of the same type, their joint wavefunction changes by a phase of +1 or -1.

The spin-statistics theorem says that these are not independent choices: you get the same phase in both cases! The phase you get by rotating a particle is related to its spin, while the phase you get by switching two goes by the funny name of "statistics". The spin-statistics theorem says how these are related.

The theorem lays out two possibilities. Some particles change phase by +1 when you rotate one by 360° or switch two of them. These are called bosons. They include photons, the W and Z boson, gluons and the Higgs boson (not found yet at this time). Others change phase by -1 when you rotate one by 360° or switch two of them. These are called fermions. They include electrons, neutrinos and other leptons, as well as a bunch of particles called quarks.

The usual proof of the spin-statistics theorem is based on axioms for quantum field theory. This book is the classic reference:

This proof, which goes back originally to Fermi, is rather intimidating and mysterious. However, there is also a more intuitive approach based on topology. You can see hints of it in Feynman's lecture here: For more on this approach, see the references in here: So, let's start by sketching two intuitive topological proofs of the spin-statistics theorem, and then outline one based on quantum field theory:
From: John Baez
Newsgroups: sci.physics,sci.math
Subject: Re: Spin/Statistics Heuristic Argument
Date: 31 Jan 1995 18:58:22 -0800

Jeff Steele wrote:

In the American Journal of Physics, Feb 1995, Roy Gould gives a nice heuristic argument for the spin statistics theorem — why spin 1/2 particles are fermions, and integer spin particles are bosons.

Thanks for the reference. Jim Dolan and I have been thinking about this kind of thing lately. The standard proof involves lots of assumptions, but there should be a topological — actually category-theoretic — essence to it, and this heuristic argument is probably pretty close to the theorem we're looking for. Actually I already know another heuristic argument, but this one might be fundamentally a bit different.

The "proof" is topological, in that if you rotate a thing (particle) with strings all over it, the strings are all twisted after 360 degrees of rotation, but after 720 degrees the strings can be untangled without moving the object. Similarly, if two things (particles) connected by lots of strings are interchanged, the strings are left twisted up exactly as if one particle had been rotated by 360 degrees. So the conclusion is that interchanging two particles is topologically indistinguishable from a rotation of one particle by 360 degrees — a particle which changes sign after a rotation will be antisymmetric with respect to pairwise interchange.

Another argument goes like this. Take a piece of ribbon and make it look like this:

|
|
 \
  \  /\
   \   \
    \  /
  /  \/
 /
|
|

Make sure the surface of the ribbon lies essentially flat against the plane. In practice one can't quite do this, but this is topology, so what I really mean is: just do your best and make sure there are no extra twists in the ribbon.

Okay, now what does this represent? Well, the vertical axis is time, the horizontal is space, and this is the worldline of a fermion (say a spin-1/2 particle like an electron). But wait! It doubles back in time! Well, when it's going backwards in time it represents a positron. So what we have here is an electron minding its own business when all of a sudden an electron-positron pair is formed. The two electrons cross paths, and the original electron is annihilated by the positron. Then the new electron continues along. It's a perfect substitute for the original electron — since all electrons are identical. It seems like perfect murder, no? The impostor looks just like the victim and his accomplice (the positron) has completely disappeared along with the corpse!

However, in this case of The Telltale Phase, there is one clue left behind!! Since the murderer and the victim switched places, and they were both identical fermions, the wavefunction has changed sign!! Of course, your average detective is unable to spot when something's wavefunction has changed sign. However, Holmes had cleverly planned a double-slit experiment before the crime, and was able to detect the resulting destructive interference.

Hmm, I'm getting a little carried away here. And indeed the whole story is somewhat unrealistic, as those Holmes cases always are: in reality, when an electron/positron pair forms or is annihilated, some photons are involved. But I want to keep the number of suspects down.

Now for a twist in the plot. Holmes takes the ribbon and stretches it tight in the vertical direction! Go ahead, do it — perhaps with a thin strip of paper if you haven't a ribbon at hand. Lo! It is just a ribbon with a 360 degree twist in it! In other words, the process just described is topologically equivalent to one in which there was only a single electron all along, which made a 360 degree rotation!

And of course, if an electron turns all the way around like that, it also gets a minus sign in its phase. "So," concludes Holmes, "but for the spacetime metric which made these two processes geometrically distinct, it would be impossible to tell, even using the telltale phase, whether a fiendish murder had been committed, with the murderer switching places with the victim, or whether the chap had merely turned around 360 degrees during the course of the day!"

This is a kind of topological "proof" that interchange of two identical particles and a 360 degree rotation of one must have the same effect on the phase. One should be able to make a real theorem out of it using ideas from category theory, in particular the C*-quantum categories discussed in Froehlich and Kerler's book.

What isn't clear to me is why spin 1/2 particles undergo a sign change after a single rotation whereas integer spin particles do not. In the context of the above discussion, it would seem that fermions have some kind of "hairiness" or connection to distant points in space that bosons lack. Is it possible to explain this difference at an elementary level, or must one resort to field theory ?

Even field theory does not "explain" the difference (apart from one possible scenario mentioned below). Basically, all we can say at present is that there are two possibilities in nature, the bosonic and the fermionic, and nature chooses to make use of them both. I should add that if spacetime were 3 dimensional there would be myriads of possibilities, and to read a bit more about these "anyons", as they're called, try reading my webpage on braids.

In the so-called Skyrme model for hadrons, where hadrons are phenomenologically treated as topological solitons — "twists" in some field which can dissipate by virtue of topology — you can get the baryons to act like fermions the way they should by including an appropriate "topological term" in the Lagrangian for the theory. I think this was discovered by Witten — he wrote about it, in any event, and you can find that paper and much similar stuff in a book edited by Shapere and Wilczek.

So in other words, the Lagrangian of a seemingly bosonic theory can make some particles appearing in it act like fermions! This possibility is important in some condensed matter systems, I believe, but it's not part of the standard model of fundamental particles. Just a trick in the particle theorist's bag so far!

Now for something a bit more technical:

Here's a brief sketch of some ideas that appear in the proof of the spin-statistics theorem. It is crucial to realize that this proof relies on the formalism of relativistic quantum field theory, not just quantum mechanics. The key axioms for quantum fields that one must assume in order to deduce this theorem are that the fields are invariant under the Poincare group (the special relativistic symmetry group), that there is a vacuum state that is invariant under this group, that all states can be built up from the vacuum by applying field operators, that the Hamiltonian is bounded below, and locality, in the sense that the fields either commute or anticommute at spacelike separations. The theorem then says that integer-spin fields commute at spacelike separations, while half-integer-spin fields anticommute. To really make the theorem precise one must make ones axioms precise. One can use, for example, the Garding-Wightman axioms (which is what my sketch implicitly does), or else one can work with the Haag-Ruelle axioms, which use C*-algebras and are arguably more fundamental, though further from the language of "practical" quantum field theory.

The proof I am outlining comes from this wonderful book, which is out of print:

All the mistakes are mine; some are deliberate attempts to make the proof seem simpler than it is, but there may well be some accidental mistakes too. I will skip all sorts of steps and try to present the logical outline.

Let $\Phi(x)$ be a field of spin $s$ (here $s$ either integer or half-integer; I'm not worrying about left-handed and right-handed spinors). Let $v$ be the vacuum state and let $$ f(x - y) = \langle v, \Phi(x) \Phi^*(y) v\rangle $$ $$ g(x - y) = \langle v, \Phi^*(x) \Phi(y) v\rangle $$

Here $\Phi^*$ is the adjoint of $\Phi$, and of course the expectation values really do depend only on $x - y$ by translation invariance. Now suppose the wrong commutation relations hold, i.e., for $x$ and $y$ spacelike separated we have $$\Phi(x) \Phi^*(y) = -(-1)^{2s} \Phi^*(y) \Phi(x)$$

instead of what we "should have" (no extra minus sign). This implies $$ f(x) + (-1)^{2s} g(-x) = 0 . \qquad \qquad 1) $$

Now we will express $g(-x)$ in terms of $g(x)$. $g(x)$ is called a "Wightman function" and these are known to transform in a nice way under the Poincare group, because of the Poincare symmetry of the field theory. In fact, one can use the fact that the Hamiltonian is bounded below to analytically continue the Wightman functions to part of the complexification of Minkowski space ($\mathbb{C}^4$ instead of $\mathbb{R}^4$), and the Wightman functions then transform in a nice way under the complexification of the Poincare group. This is rather technical, and it requires care to do correctly, but heuristically one can just let all your coordinates $(t,x,y,z)$ become complex and hope that your formulas extend in the "obvious way".

The parity/time-reversal operator $PT$ lives in the connected component of the complexified Lorentz group (or more precisely, its double cover). This group, by the way, is just $SL(2,\mathbb{C}) \times SL(2,\mathbb{C}$) (direct sum of 2 copies of the double cover of the Lorentz group). $PT$ corresponds to the map $$ (t,x,y,z) \mapsto -(t,x,y,z) $$

which we may think of as being a 180 degree rotation in the $(x,y)$ plane followed by a 180 degree rotation in the complexified $(t,z)$ plane. (One needs to complexify precisely to get away with the second rotation.) Thus we should be able to derive how a spin-s field should transform under $PT$ assuming we know how it transforms under the complexified Lorentz group. And by analytic continuation, this should be calculable from knowing how the field transforms under the plain old Lorentz group—i.e., its spin!

But this is a calculation I don't want to do here, since I haven't figured out how to do it simply, even cheating a little here and there. Maybe sometime I'll explain this part and finish writing up a nice little spin-statistics FAQ... but I will offer a raincheck on this one.

Let me simply assert, then, that $PT$ transforms the field in such a way that the covariance of the Wightman functions implies $$ g(x) = (-1)^{2s} g(-x) . $$

It follows from 1), then, that $$ f(x) + g(x) = 0 , $$

or recalling what f and g are, $$ \langle \Phi^*(x)v, \Phi^*(y)v \rangle + \langle \Phi(x)v, \Phi(y)v \rangle = 0 $$

for all spacelike separated $x, y$. (Here one should really smear the fields by test functions!) Now letting $y$ approach $x$ along a suitable spacelike direction one gets in the limit $$ \|\Phi^*(x)v\|^2 + \|\Phi(x)v\|^2 = 0. $$

Thus $\Phi(x)v = 0$ for all $x$. A big fat theorem called the Wightman reconstruction theorem says one can reconstruct a quantum field from its Wightman functions, and using this we can show $\Phi(x) = 0$ for all $x$. Thus no (nonzero) field can have commutation relations that don't match its spin correctly.

Regarding the closely allied $CPT$ theorem, here's a nice sketch of how it goes:

Subject: Re: Does the arrow of time reverse when the crunch starts?
From: Ron Maimon
Newsgroups: sci.physics
Date: 28 Feb 1994 19:48:45 GMT

Vijay Fafat writes:

Why, exactly, is the $CPT$ theorem considered so holy? Specifically, what might be the effects if $CPT$ were not to hold?

Because the $CPT$ theorem is an almost-consequence of Lorentz invariance.

If you have a Lorentz invariant theory, then you can change the coordinates of space time with the following matrix and leave the theory the same

| cosh(y) 0    0    sinh(y) |
| 0       1    0    0       |
| 0       0    1    0       |
| sinh(y) 0    0    cosh(y) |

and this is true for any y.

An interesting property of quantum mechanics is that the amplitudes you calculate are analytic functions of the variables in the problem, this is an obvious fact in perturbation theory, where the amplitudes are just rational functions of the momenta coming in and going out, but its more general than that. The amplitudes are analytic in a wide range of circumstances.

And this means that the theory is invariant for any complex value of $y$, by the principle of analytic continuation. In particular, for $$ y= \pi \sqrt{-1} = \pi i $$ and if you stick in $y = \pi i$, you get that the matrix above does a $PT$ transformation. So that a theory of a bunch of scalar particles is $PT$ invariant.

A more general result, if you allow charged particles, is that a general theory is $CPT$ invariant; the argument is essentially the same.

Ron Maimon


© 2012 John Baez
baez@math.removethis.ucr.andthis.edu

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