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Topic: The mystery of boson statistics
Replies: 1   Last Post: Jun 14, 1996 8:34 PM

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john baez

Posts: 57
Registered: 12/6/04
The mystery of boson statistics
Posted: Jun 14, 1996 2:59 AM
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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 <>
>Subject: The mystery of boson statistics

Reply-To: (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

(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

We don't bother with pesky little factors 'cause we're imbued with the
free-wheeling 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!)

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}


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

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 n-th 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 single-particle 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

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 textbook-style Clifford algebra, we can define field

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>) =

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

F(v)F(w) + F(w)F(v) = Re<v,w>

depending only on the orthogonal structure of H. This is a
textbook-style 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 infinite-dimensional is the
most interesting, but the case of finite-dimensional H is nice too.
In the latter case the Clifford algebra above is isomorphic to a matrix
algebra. It is curious and fascinating that infinite-dimensional
Clifford algebras show up as algebras of field operators in the quantum
field theory of fermions, while the finite-dimensional 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.

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