…By quantizing we the get the following Hamiltonian operator
$$\hat{H}=\sum_{\mathbf{k}}\hbar \omega(\mathbf{k})\left(\hat{n}(\mathbf{k})+\frac{1}{2} \right)$$
where $\hat{n}(\mathbf{k})=\hat{a}^{\dagger}(\mathbf{k})\hat{a}(\mathbf{k})$ is the number operator of oscillator mode $\mathbf{k}$ with eigenvalues $n_{\mathbf{k}}=0,1,2,\dots$.Using the quantum canonical ensemble show that the internal energy $E(T)$ is given by>
$$E(T)=\langle H \rangle = E_0 + \sum_{\mathbf{k}}\frac{\hbar \omega(\mathbf{k})}{e^{\beta\hbar \omega(\mathbf{k})}-1}$$
where $E_0$ is the sum of ground state energies of all the oscillators.
I started this by calculating the partition function
$$\begin{align} Z &= \sum_{\Gamma}e^{-\beta \mathcal{H}(\Gamma)} \\
&= \sum_{\Gamma}e^{-\beta (\sum_{\mathbf{k}}\hbar \omega(\hat{n}(\mathbf{k})+\frac{1}{2}))}
\end{align}$$
($\Gamma$ is a microstate of the system)
but I cannot see the thought process behind evaluating these, particularly with respect to the summations. This is a common problem I have found.
I would then go on to use $E=-\frac{\partial \ln Z}{\partial \beta}$
Best Answer
Quantum mechanically the general expression you want for the partition function is $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right),$$ where $\mathrm{Tr}$ means the trace (i.e. sum over micro-states). Now you can use the fact that the modes are independent, so that quantum Boltzmann operator $\mathrm{e}^{-\beta H}$ factorises into a product. This means that you can evaluate the trace over each oscillator mode separately: $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right) = \mathrm{Tr} \left( \prod_\mathbf{k}\mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} Z_\mathbf{k} $$ where $\mathrm{Tr}_\mathbf{k}$ means the trace over only the Hilbert space of mode $\mathbf{k}$, and $$H_\mathbf{k} = \hbar\omega(\mathbf{k})\left(\hat{n}(\mathbf{k}) + \frac{1}{2}\right).$$ Now $\mathrm{Tr}_\mathbf{k}$ means simply averaging over all the possible states in the Hilbert space, which you might as well choose to be the eigenstates of the number operator $\hat{n}(\mathbf{k})\lvert m_\mathbf{k}\rangle= m_\mathbf{k} \lvert m_\mathbf{k}\rangle$, with $m_\mathbf{k} = 0,1,2,\ldots$. So you have to evaluate $$ Z_\mathbf{k} = \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \sum_{m_\mathbf{k}=0}^\infty \langle m_\mathbf{k} \rvert \mathrm{e}^{-\beta H_\mathbf{k}} \lvert m_\mathbf{k} \rangle. $$