8—Multivariable Calculus 231The second example said that you have several different allowed energies, typical of what happens in
quantum mechanics. The total number of particles and the total energy are given, how are the particles distributed
among the different energies?
If there areN particles and exactly two energy levels,E 1 andE 2 ,
N=n 1 +n 2 , and E=n 1 E 1 +n 2 E 2these are two equations in two unknowns and all you have to do is solve them for the numbersn 1 andn 2 of
particles in each state. If there are three or more possible energies the answer isn’t determined by just two
equations, and there can be many ways that you can put particles into different energy states and still have the
same number of particles and the same total energy.
If you’re dealing with four particles and three energies, you can perhaps count the possibilities by hand.
How many ways can you put four particles in three states? (400), (310), (301), (220), 211),etc. There’s only
one way to get the (400) configuration: All four particles go into state 1. For (310) there are four ways to do it;
any one of the four particles can be in the second state and the rest in the first. Keep going. If you have 1020
particles you have to find a better way.
If you have a total ofNparticles and you placen 1 of them in the first state, the number of ways that you
can do that isNfor the first particle,(N−1)for the second particle,etc. =N(N−1)(N−2)···(N−n 1 +1) =
N!/(N−n 1 )!. This is over-counting. You don’t care which one went into the first state first, only that it’s there.
There aren 1 !rearrangements of thesen 1 particles, so you have to divide by that to get the number of ways that
you can get this number of particles into state 1:N!/n 1 !(N−n 1 )!For example,N= 4,n 1 = 4as in the (400)
configuration in the preceding paragraph is4!/0!4! = 1.
Once you’ve gotn 1 particles into the first state you want to putn 2 into the second state (out of the
remainingN−n 1 ). Then on to state 3.
The total number of ways that you can do this is the product of all of these numbers. For three allowed
energies it is
N!
n 1 !(N−n 1 )!
. (N−n^1 )!
n 2 !(N−n 1 −n 2 )!
. (N−n^1 −n^2 )!
n 3 !(N−n 1 −n 2 −n 3 )!
=
N!
n 1 !n 2 !n 3!(23)
There’s a lot of cancellation and the final factor in the denominator is one because of the constraintn 1 +n 2 +n 3 =
N.
Lacking any other information about the particles, the most probable configuration is the one for which
Eq. ( 23 ) is a maximum. This calls for Lagrange multipliers because you want to maximize a complicated function