# I A question on Bose enhancement & Pauli blocking

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1. Aug 7, 2016

### "Don't panic!"

Say I have $n_{a}$ bosons in some state $a$, then the transition rate from some state $b$ to state $a$, $W^{boson}_{b\rightarrow a}$, is enhanced by a factor of $n_{a}+1$ compared to the corresponding transition probability for distinguishable particles, $W_{b\rightarrow a}$, i.e. $$W^{boson}_{b\rightarrow a}=(n_{a}+1)W_{b\rightarrow a}$$ and so the transition rate is from a state $b$ to a state $a$ is enhanced by the number of identical bosonic particles already in the state $a$.

Conversely, for a fermion, the transition rate is suppressed by a factor of $1-n_{a}$, i.e. $$W^{fermion}_{b\rightarrow a}=(1-n_{a})W_{b\rightarrow a}$$

My question is, how does one derive these to relations? How does one show that transition rate for bosons are enhanced due to Bose-Einstein statistics, whereas transition rates for fermions are suppressed due to Fermi-Dirac statistics (heuristically I get that in the case of fermions it is due to the Pauli exclusion principle, so called "Pauli blocking")?!

2. Aug 7, 2016

### vanhees71

You can get it in deriving the Boltzmann equation using quantum transition amplitudes. The most simple thing is to use the Born approximation for $2 \rightarrow 2$ scattering matrix elements. Then you get a collision term with the integrand
$$\propto |\mathcal{M}_{12 \leftrightarrow 34}|^2 [f_3 f_4 (1 \pm f_1)(1 \pm f_2)-f_1 f_2 (1\pm f_3)(1 \pm f_4).$$
The upper signt is for bosons the lower for fermions.

Here are some hand-written notes on this from the last winter semester ;-) (in German but with a large "formula density"):

http://th.physik.uni-frankfurt.de/~hees/neq-therm-WS15/quantum-boltzmann-eq-a-la-greiner-book.pdf

It's taken from one Greiner's quantum mechanics textbooks.

3. Aug 7, 2016

### "Don't panic!"

Thanks for the notes, my German isn't great, but hopefully I'll understand the equations nonetheless ;-)

So is this how it was originally derived? In texts that I've read on statistical mechanics the result is simply stated with little motivation :-/

4. Aug 8, 2016

### vanhees71

That's a good question. Of course, the Bose enhancement goes as far back as to Einstein 1917, where he derived the Planck spectrum by thinking about the transition rates and detailed balance. To get the correct Planck spectrum he had to assume that additionally to induced emission, i.e., emission of a photon from an excited state triggered by the already present radiation, there exists spontaneous emission, i.e., that an excited state can spontaneously deexcite by emitting a photon, which leads to the $1+n$ factors in the collision term. The quantum derivation for photons goes back to Dirac in 1927, when he introduced the annihilation and creation operators for photons:

http://rspa.royalsocietypublishing.org/content/royprsa/114/767/243.full.pdf