# Bose Enhancement Factor

How may one explain the fact that the probability of a boson transferring to a state with an occupation number n is 'enhanced' by a factor of (1+n), compared to the classical case? (In the classical case, the probability is supposed to be independent of the occupation of the final state.)

• What do you mean by "explain"? If you've seen that fact, you've also seen its derivation, haven't you? – ACuriousMind Nov 8 '14 at 17:54
• I came across this issue while trying to understand the collision integral in the Boltzmann equation for particles obeying different statistics. I can see why the factor is 1 for classical particles and 1-n for fermions, but I'm not so sure about the bosonic case. I did see one 'proof' somewhere, but was not convinced by it. I hope you may shed some light on it. Thanks. – H L Nov 8 '14 at 20:21

A simple way to see this is by considering the fact that the probability to transition from a particular initial state to a particular final state is the same as for the inverse process where one considers the transition from the final to the initial state. This is because the square of the modulus of the matrix element is the same for both cases.

This means that instead of considering the transition to a state where there are already n bosons present and one will be added, we can just as well consider the transition rate from the final state containing n+1 bosons to the initial state where one of these bosons will go somewhere else. Obviously, this will have to be proportional to the number of bosons in the final state, so the rate will be proportional to n+1.

Note that in a classical setting you can also get to this result, but then "classical" refers to treating the field of the bosons in a classical way (take e.g. the electromagnetic field, in that case you can derive the formula for stimulated emission, in a full quantum mechanical treatment you get the classical result by ignoring certain commutators which then yields a factor of n instead of n+1).

Instead, if the classical limit is taken where the volume of the system is effectively sent to infinity so that the density of states goes to zero, then the occupation numbers go to zero as well and then n+1 becomes 1.

I will start from rather far away, but will get to the point eventually. Of course, it is possible to rigorously derive kinetic equations, or consider some statistical field theory (as suggested in some of the answers to this question), but I want to take a simpler approach. I will discuss (effectively) non-interacting particles. My answer is partly inspired by the stuff that I saw there.

In the classical case, the probability to find $n$ particles in a state does not depend on the number of particles $n$ in that state. Hence, the probability is simply a product of (independent!) $n$ single-particle transition probabilities $P_1$, such that $$P_n^\text{Classic}=(P_1)^n.$$

Obviously, this is not the case for fermions, since if a single fermion is already occupying a state, a transition to that state is forbidden. In a fancy manner this fact can also be seen from an antisymmetrization requirement for two fermions. If there are two fermions, and they can access two states $\alpha$ and $\beta$, the wavefunction is $$\psi_A=\frac{1}{\sqrt{2}}\Big(\psi_\alpha(1)\psi_\beta(2)-\psi_\alpha(2)\psi_\beta(1)\Big).$$ However, if we want to put both of them into a single state, we have to set $\alpha = \beta$ in the preceding equation, and the wavefunction becomes zero. In conclusion, two fermions cannot occupy the same state. It turns out that this fact can be put into a kinetic equation by saying that $$P_n^\text{Fermi} \sim (1-n),$$ which looks trivial for integer $n$, but turns to also be correct for the situations where $n$ is the average occupation (noninteger between 0 and 1).

Finally, consider two bosons. They are also indistinguishable (just like fermions were), but their exchange gives a plus instead of a minus, so the wavefunction is symmetric: $$\psi_S=\frac{1}{\sqrt{2}}\Big(\psi_\alpha(1)\psi_\beta(2)+\psi_\alpha(2)\psi_\beta(1)\Big).$$ Continuing the same game, we put both of them to the same state by setting $\alpha = \beta$. Instead of zero, we have $$\psi_S={\sqrt{2}}\psi_\alpha(1)\psi_\alpha(2).$$ We now want to know the probability distribution of finding both bosons in this single state. Thus, we calculate the absolute value squared of this many-body wavefunction: $$|\psi_S|^2=2|\psi_\alpha(1)|^2|\psi_\alpha(2)|^2.$$ Had we started with a (quasiclassical) situation of distinguishable particles, the total wavefunction would have been $\psi_\alpha(1)\psi_\beta(2)$ or equivalently $\psi_\alpha(2)\psi_\beta(1)$. In that case, after identifying the two states, the probability distribution would have been merely $|\psi_\alpha(1)|^2|\psi_\alpha(2)|^2$, hence, there is a twofold enhancement. Playing around with symmetrizing more particles, it turns out that the factor (which is $2$ here) generalizes to $n!$. This means that $$P_{n+1}^\text{Bose} = (n+1)! P^\text{Classic}_{n+1} = (1+n) n! P_1 P^\text{Classic}_{n} = (1+n) P_1 P_n^\text{Bose}.$$ This $1+n$ is the Bose enhancement factor that we have been looking for.

• Please correct me if I'm wrong. But isn't the factor $1/\sqrt{2}$ just a normalization factor for the wave function when the two bosons are in two different states? If the two were in the same state, the properly symmetrized wave function would read $\Psi_{\alpha}(1) \Psi_{\alpha}(2)$. In general, a symmetrized wave function should be proportional to the sum of all distinct permutations of products of individual single particle wave functions, with a normalization constant equal to $\sqrt{\frac{N_1!N_2!...N_s!}{N!}}$. $N_i$'s are the number of bosons in the i-th state and they sum up to $N$. – H L Jan 23 '16 at 19:29