Regarding exercise 1 of the Einstein solid, what I’m struggling with is understanding why the BE distribution gives the expected number of excitations of a single quantum harmonic oscillator. In statistical mechanics we learned that the BE distribution gives the average number of bosons that will all share the same single-particle state with a certain energy E(s). So we’re considering multiple harmonic oscillators.
So when <n(\hbar\omega)>=\frac{1}{e^{\frac{\hbar\omega}{k_{b}T}}-1} is given, I would either think this is the number of harmonic oscillators that are all in the quantum state with energy \hbar\omega, or the probability of a single oscillator being in that state. Now if you would like to know the expected number of oscillators that are in the state with energy 2\hbar\omega, that would be <n(2\hbar\omega)>\frac{1}{e^{\frac{2\hbar\omega}{k_{b}T}}-1}, and etcetera for higher energy states. However that’s not done when calculating the total expected energy E, that’s simply \frac{1}{e^{\frac{\hbar\omega}{k_{b}T}}-1}\cdot \hbar\omega+\frac{1}{2}\hbar\omega, now I would think this is only the number of oscillators that are in the state with energy \hbar\omega. Shouldn’t we sum over all possible energies to obtain the expected energy, like so:
E=<E>=\sum_{n=0}^{\infty} \frac{1}{e^{\frac{\hbar\omega(\frac{1}{2}+n)}{k_{b}T}}-1}( n+\frac{1}{2})\hbar\omega. Now this expression would give the expected energy of a quantum harmonic oscillator, but apparently not.
Could someone please help me with resolving this confusion.
Thank you.
Greetings,
Jelle
Your description captures an important aspect of the problem, but mixes the Boltzmann distribution and the Bose-Einstein one.
The probability of the system to have a unique state with energy E is proportional to \exp(-E/k_BT). Let’s apply this to a harmonic oscillator. It has states with all possible n excitations, and probabilities P_n \propto \exp[-(n+1/2)\hbar\omega/k_BT]. To obtain the actual probabilities, we need to normalize these so that they add up to 1.
P_n = \frac{\exp[-(n+1/2)\hbar\omega/k_BT]}{\sum_{i=0}^{\infty}\exp[-(i+1/2)\hbar\omega/k_BT]}=\frac{\alpha^n}{\sum_{i=0}^{\infty}\alpha^i}=\alpha^n(1-\alpha),
with \alpha = \exp(-\hbar\omega/k_BT).
To obtain the average value of n we need to compute \langle n \rangle = \sum_{i=0}^{\infty}i \alpha^i(1-\alpha). Let’s try it out:
\langle n \rangle = \sum_{i=0}^{\infty}i \alpha^i(1-\alpha) = \sum_{i=0}^{\infty}(i \alpha^i -i\alpha^{i+1})\\=\sum_{i=0}^{\infty}[(i+1) \alpha^{i+1} -i\alpha^{i+1}]=\sum_{i=1}^{\infty}\alpha^i=\frac{\alpha}{1-\alpha},
where for the third equality we rewrote \sum_{i=0}^\infty i \alpha^i = \sum_{i=0}^\infty (i+1)\alpha^{i+1}. The idea of this change is to make the powers of \alpha match in two terms.
Substituting the value for \alpha, we obtain… 
\langle n \rangle = \frac{\exp(-\hbar\omega/k_BT)}{1- \exp(-\hbar\omega/k_BT)} = \frac{1}{\exp(\hbar\omega/k_BT)-1}\equiv n_B,
which is exactly the Bose-Einstein distribution.
We can obtain the Fermi-Dirac distribution in a very similar way.
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Thank you so much for this clear derivation, I now understand the connection between the Boltzmann factor and Bose-Einstein distribution. Awesome.
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