The reason this general problem is so useful in a wide range of areas of physics is in physics we love to deal with harmonic approximations of systems.

The point of solving the problem of N harmonic oscillators in this way is that they approximate (actually, correspond to) the behaviour of the particles in an ideal gas.

Here, we work out the number of states (microstates) available to the system, the entropy of the system, and derive the energy of the corresponding ideal gas as a function of temperature.

### Question

The energy levels of a harmonic oscillator with frequency are given by \begin{equation} E_n=(n+\frac{1}{2})\hbar\omega, n=0,1,2,… \label{eq:Elevel} \end{equation} A system of `N` uncoupled and distinguishable oscillators has the total energy \begin{equation} E=\frac{N}{2}\hbar\omega + M\hbar\omega

\label{eq:totalE}

\end{equation} where `M` is a non-negative integer.

- Calculate the number of states for a given E.
- Calculate the entropy of the system using the Stirling formula for `M`>>`1` and `N`>>`1`.
- The temperature is defined as . Express the total energy as a function of the temperature and discuss the function `E(T)`.

### Solution

#### 1. Number of Microstates

At first approach, quantifying the number of states seems to be a bit tricky. In fact, the solution is quite simple if you think about it in the right way, but it’s hardly jump-off-the-page-and-slap-you-round-the-face obvious.

Consider, if an oscillator is in the energy state , we say it has ‘quanta’. The problem becomes how many ways can we distribute \(M\) identical quanta amongst \(N\) oscillators.

If we further imagine the oscillators to be boxes placed side-by-side, and the quanta as indistinguishable balls, O, within the boxes, there would be `(N-1)` identical partitions (we could call them ‘sticks’), |, between the boxes. Visually, we lay the quanta out in a row with the sticks between the (not shown) boxes: OO|O|OOO|O||OO etc.

Each arrangement of balls and sticks corresponds to one accessible microstate of the system.

Therefore, the total number of microstates is given by the Binomial distribution as

\begin{equation}

\binom{M+N-1}{M} = \frac{(M+N-1)!}{(N-1)!M!} = \Omega_M

\end{equation}

#### 2. Entropy

Entropy is given by .

Therefore, $$S=k_B [{\ln(M+N-1)! -\ln(N-1)! -\ln M!}]$$Using the Stirling approximation and neglecting the terms of the order 1, $$ \approx k_B [{(N+M)\ln{(N+M)} - (N+M) - N\ln N + N - M\ln M + M}]$$ \begin{equation} S=k_B [N \ln{ \frac{N+M}{N}} + M \ln {\frac{N+M}{M}}]

\label{eq:entropy}

\end{equation}

#### 3. Total Energy as a Function of Temperature

We are told temperature is given by the equation . Clearly at this point we aren’t simply discussing an abstract system of N harmonic oscillators, but are approximating the behaviour of an ideal gas, with each point-like molecule one of our harmonic oscillators.

Re-arranging the equation for total energy \eqref{eq:totalE}, $$M=\frac{E}{\hbar\omega}-\frac{N}{2}$$ Plugging this back into the entropy formula \eqref{eq:entropy} we obtain $$S=k_B\left[\left(\frac{E}{\hbar \omega} + \frac{N}{2}\right)\ln{\left(\frac{E}{\hbar \omega} + \frac{N}{2}\right)}-\left(\frac{E}{\hbar \omega}-\frac{N}{2}\right)\ln{\left(\frac{E}{\hbar \omega}-\frac{N}{2}\right)} - N\ln N\right]$$

\begin{equation}

\frac{\partial S}{\partial E} = \frac{k_B}{\hbar \omega} \ln{\left( \frac{\frac{E}{\hbar \omega} + \frac{N}{2}}{\frac{E}{\hbar \omega} – \frac{N}{2}} \right)}=\frac{1}{T}

\label{eq:dsde}

\end{equation}

We were asked to express the total energy as a function of the temperature, `E(T)`, so re-arrange \eqref{eq:dsde}:

\begin{equation}

E(T)=N \hbar \omega \left(\frac{1}{2} + \frac{1}{e^{\frac{\hbar \omega}{k_B T}}-1} \right)

\end{equation}

We note the first term, , is nothing more than the contribution of the zero-point energy. The second term is the excitation energy contribution, the familiar Bose-Einstein distribution.

Going a little farther, expanding `E(T)` around `T`>>`0` to understand high-temperature behaviour:

\begin{equation}

E(T) = \frac{N \hbar \omega}{2} + N k_B T

\end{equation}

We can find the heat capacity at constant volume from the formula $$C_V=\left(\frac{\partial E}{\partial T}\right)_V=Nk_B=\text{const.}$$

For 3-D systems, this would be $$C_V=3Nk_B$$

Thus, at high temperatures, quantum effects such as the zero-point energy become negligible and we recover classical behaviour (see equipartition theorem).