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The mystery of boson statistics
Posted:
Jun 14, 1996 2:59 AM


I am now serving as Michael Weiss' personal posting service. First something by him, then some comments from me.
>Date: Thu, 13 Jun 1996 22:06:15 0400 >From: Michael Weiss <columbus@osf.org> >To: baez@guitar.ucr.edu >Subject: The mystery of boson statistics ReplyTo: columbus@osf.org (Michael Weiss)
Here's what I had in mind when I said, with careless abandon, that you get the creation operator by symmetrizing the operation of tensoring on the left.
Let's write ab for a tensor b. Also let's say we have a huge bucketful of particles, all in orthogonal normalized states, a,b,c,...
First let's keep tossing new distinguishable particles into the brew:
a > ab ab > abc abc > abcd etc.
(OK, so I'm tensoring on the right. But I *drive* on the left...)
Now let's do it with bosons (to the tune of "Send in the Clowns"). Symmetrize both sides:
a > ab + ba ab + ba > abc + acb + bac + bca + cab + cba abc + ... + cba > abcd + ... + dcba etc.
We don't bother with pesky little factors 'cause we're imbued with the freewheeling spirit of the 60s. Anyway, why this factor rather than that one?
A simple calculation now shows that the norm of
v_1 = a is sqrt (1!) v_2 = ab + ba is sqrt (2!) v_3 = abc + ... + cba is sqrt (3!) etc.
So let w_n be v_n/v_n, so v_n = sqrt(n!) w_n. So we have:
sqrt (n!) w_n > sqrt((n+1)!) w_{n+1}
or
w_n > sqrt(n+1) w_{n+1}
Now we blithely assume that this works also when we add particles in the *same* state, because (take your pick):
a) of continuity. b) aa demands equal treatment with ab, or it threatens to sue. c) it does. d) we're feeling really blithe today, having just overdosed on Shelley.
I claim this for the above derivation, *and no more*:
1) It's simple. 2) It gets the right answer.
==========================
Yes, I am aware of the defects in this argument. After spending a couple of hours staring at the treatments in Dirac, in Feynman, and in Weinberg's new QFT book, I conclude that there is no totally convincing argument no proof that it *has* to be that way. The most anyone can do is make it look plausible. Of course, if you *start* with commutation relations [C,C*] = 1, then you can come up with the factor sqrt(n) by the usual route.
Right now, the closest I have to an "aha" remark is this observation. The obvious symmetrizing operator is
S(ab...g) = (1/n!) Sum (all n! permutations of a,b,...,g)
since that's what gives you a projection from the nth tensor power onto the symmetrized power (which we can regard as a subspace of the tensor power).
But then the norm of S(ab...g) works out to 1/sqrt(n!). Well, no wonder: ab...g isn't in the symmetric subspace, so *of course* it gets shorter when you project it. But S(aa...a) stays the same length, since it's already symmetric. Which means (somehow) that aa...a deserves to be n! more probable than S(ab...g). I have a feeling that the next "aha" will have to wait until I truly grok quantum statistical mechanics.
Post freely.

So I guess you looked at chapter 4 of the 3rd volume of the Feynman Lectures in Physics, "Identical Particles"? It's a nice intuitive approach to seeing where the sqrt(n+1) factor comes from, though probably not the "proof" one might want.
In my earlier disquisition on bosons and fermions and their correspondence to symplectic and orthogonal geometry I came pretty close to taking the (anti)commutation relations for annihilation and creation operators as my starting point. Actually I started with the (anti)commutation relations for the field operators. This certainly has its merits. But you are right that it's also tempting to start by trying to understand what it means to annihilate or create a particle...
Let me just sketch for fermions how it all fits together very sweetly. Here we start with the singleparticle space H, which is a Hilbert space, but we focus attention more on its orthogonal structure (the real part of the inner product) than its symplectic structure (the imaginary part).
We can define the Fock space K over H as the exterior algebra over H, suitably completed. (This is the algebra of *anti*symmetrized tensors over H.) We can define the creation operator c(v) for any v in H to be left multiplication by v:
c(v) x = vx
and define the annihilation operator to be the adjoint a(v) of c(v):
<a(v)x,y> = <v,c(v)x>
Now the product in the exterior algebra is defined so that vw = wv, so we have the canonical anticommutation relations
c(v)c(w) + c(w)c(v) = 0
and taking adjoints
a(v)a(w) + a(w)a(v) = 0
A bit more thought also shows that
a(v)c(w) + c(w)a(v) = <v,w>
which is the really interesting canonical anticommutation relation. Let's see how that goes. Suppose we can show
a(v)c(v) + c(v)a(v) = Re<v,v>
for all v. Then applying this equation to a bunch of different linear combinations of v and w and taking a linear combination of all the equations we come up with, we get
a(v)c(w) + c(w)a(v) = Re<v,w>
(This juggling trick is called "polarization" and is a very useful thing to know if you are lazy like me.)
So lets figure out what
a(v)c(v) + c(v)a(v)
does to any vector x in K. Well, we can assume x is an exterior product of orthonormal basis vectors, since everything is a linear combination of those. Normalizing v we can take our basis to contain v. So x is a product which either contains v or doesn't. If it doesn't, c(v)a(v)x = 0 and c(v)a(v)x = <v,v>x. If it does, it goes the other way around. Either way we get
a(v)c(v) + c(v)a(v) = <v,v> = Re<v,v>
as desired.
So we get a representation of the "canonical anticommutation algebra" or "Clifford algebra" of H on the exterior algebra. The Clifford algebra in question has generators and relations
c(v)c(w) + c(w)c(v) = 0 a(v)a(w) + a(w)a(v) = 0 a(v)c(w) + c(w)a(v) = <v,w>
This may not look like a Clifford algebra to the untutored eye. To make it look like a textbookstyle Clifford algebra, we can define field operators
F(v) = (a(v) + c(v))/sqrt(2)
The annihilation and creation operators can be expressed in terms of these field operators so the algebra generated by the F's is the same as that generated by the a's and c's. (Exercise: solve for the a's and c's in terms of the F's. Hint: use c(iv) = ic(v), a(iv) = ia(v).) But the F's satisfy:
F(v)F(w) + F(w)F(v) = (1/2)((a(v) + c(v))(a(w) + c(w)) + (a(w) + c(w))(a(v) + c(v))) = (1/2)(<v,w> + <w,v>) = Re<v,w>
using the canonical commutation relations for the a's and c's. So our Clifford algebra is generated by the field operators, or F's, and has relations
F(v)F(w) + F(w)F(v) = Re<v,w>
depending only on the orthogonal structure of H. This is a textbookstyle Clifford algebra. Note that F(iv) is not iF(v), so we should really now think of H as a real vector space with an orthogonal structure, and think of the Clifford algebra as being the algebra generated by this real vector space with the above sort of anticommutation relation.
In quantum field theory the case where H is infinitedimensional is the most interesting, but the case of finitedimensional H is nice too. In the latter case the Clifford algebra above is isomorphic to a matrix algebra. It is curious and fascinating that infinitedimensional Clifford algebras show up as algebras of field operators in the quantum field theory of fermions, while the finitedimensional ones show up in the study of spinors.
So: all of this stuff has its analog for bosons, too. Instead of the exterior algebra we use the symmetric algebra, instead of the Clifford algebra we get the Weyl algebra, instead of the orthogonal structure we use the symplectic structure... all just a bunch of changes of sign.



