Question

A quantum state of energy \(E\) can be occupied by any number \(n\) of bosons, including \(n=0 .\) At the absolute temperature \(T,\) the probability of finding \(n\) particles in the state is given by \(P_{n}=N \exp \left(-n F / k_{\mathrm{B}} T\right),\) where \(k_{\text {is }}\) is Boltzmann's constant and \(N\) is the normalization factor determined by the requirement that all the probabilities sum to unity, Calculate the mean value of \(n\), that is, the average occupation, of this state, given this probability distribution

Short answer

Answer: The average occupation of the quantum state is given by \(\langle n \rangle = \frac{1}{\exp(F / k_B T) - 1}\).

Step-by-step solution

Find the normalization factor, N

First, we need to find the normalization factor, N, by the requirement that all the probabilities sum to unity. This means summing over all possible values of n: \(\sum_{n=0}^{\infty} P_n = 1\) Replacing \(P_n\) with the given expression, we get: \(\sum_{n=0}^{\infty} N \exp(-nF / k_B T) = 1\) N can be factored out: \(N \sum_{n=0}^{\infty} \exp(-nF / k_B T) = 1\) Let's denote \(\exp(-F / k_B T)\) as \(x\), so: \(N \sum_{n=0}^{\infty} x^n = 1\) The sum is a geometric series, and its sum is given by: \(\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x}\) Therefore, we can now find the normalization factor N: \(N = 1 - x = 1 - \exp(-F / k_B T)\)

Calculate the mean value of n, i.e., the average occupation

Now that we have the normalization factor, we can calculate the mean value of \(n\), given the probability distribution. We will use the following formula: \(\langle n \rangle = \sum_{n=0}^{\infty} n P_n\) Substituting \(P_n\) and the normalization factor \(N\), we have: \(\langle n \rangle = \sum_{n=0}^{\infty} n (1 - \exp(-F / k_B T))\exp(-nF / k_B T)\) To get the sum, we can differentiate the sum of the infinite geometric series with respect to \(x\): \(\frac{d}{dx}\left[\sum_{n=0}^{\infty} x^n\right] = \frac{d}{dx}\left[\frac{1}{1 - x}\right]\) Using the chain rule, we get: \(\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1 - x)^2}\) Now, substitute \(x = \exp(-F / k_B T)\) back into the sum: \(\sum_{n=1}^{\infty} n \exp(-nF / k_B T) = \frac{1}{(1 - \exp(-F / k_B T))^2}\) Ultimately, the mean value of \(n\) becomes: \(\langle n \rangle = (1 - \exp(-F / k_B T))\cdot\frac{1}{(1 - \exp(-F / k_B T))^2}\) Simplifying, we get: \(\langle n \rangle = \frac{1}{\exp(F / k_B T) - 1}\) So, the average occupation of the quantum state is given by: \(\langle n \rangle = \frac{1}{\exp(F / k_B T) - 1}\)