= Solution
A periodic <harmonic chain> with $N$ masses $m_0$, spacing $a$ and nearest-neighbour spring constant $K$ has
$$
H=\sum_{j=0}^{N-1}\left[\frac{p_j^2}{2m_0}+\frac K2(u_{j+1}-u_j)^2\right],\qquad u_N=u_0.
$$
Hamilton's equations give $m_0\ddot u_j=K(u_{j+1}+u_{j-1}-2u_j)$. The <discrete Fourier transform> $u_j=N^{-1/2}\sum_k Q_ke^{ikja}$, $p_j=N^{-1/2}\sum_k P_ke^{ikja}$, with $k=2\pi l/(Na)$ in one <Brillouin zone>, diagonalizes the quadratic form:
$$
\boxed{\omega_k=2\sqrt{K/m_0}\,|\sin(ka/2)|,\qquad H=\sum_k\left[\frac{P_kP_{-k}}{2m_0}+\frac{m_0\omega_k^2}{2}Q_kQ_{-k}\right].}
$$
The <harmonic-chain normal-mode reality condition> means $Q_k^\dagger=Q_{-k}$, not that $Q_k$ and $Q_{-k}$ are independent real coordinates. For <canonical quantization>, impose $[u_j,p_l]=i\hbar\delta_{jl}$, equivalently $[Q_k,P_{k'}]=i\hbar\delta_{k,-k'}$. For $k\ne0$ set
$$
Q_k=\sqrt{\frac{\hbar}{2m_0\omega_k}}(a_k+a^\dagger_{-k}),\quad
P_k=-i\sqrt{\frac{m_0\hbar\omega_k}{2}}(a_k-a^\dagger_{-k}),\quad
[a_k,a^\dagger_{k'}]=\delta_{kk'}.
$$
Then the <quantum harmonic chain> is a set of independent <phonon> oscillators:
$$
\boxed{H=\frac{P_{\rm tot}^2}{2Nm_0}+\sum_{k\ne0}\hbar\omega_k(a_k^\dagger a_k+1/2).}
$$
The <centre-of-mass zero mode of a harmonic chain> is the free coordinate $X=N^{-1}\sum_ju_j$, with $[X,P_{\rm tot}]=i\hbar$; it cannot be normalized by dividing by $\sqrt{\omega_0}$. Retaining it gives
$$
u_j=X+\frac1{\sqrt N}\sum_{k\ne0}\sqrt{\frac{\hbar}{2m_0\omega_k}}(a_ke^{ikja}+a_k^\dagger e^{-ikja}),
$$
with the corresponding momentum expansion and uniform term $P_{\rm tot}/N$. The nonzero modes alone give the site <commutator> $i\hbar(\delta_{jl}-1/N)$; the uniform term supplies the missing $1/N$. Each oscillator occupation is a nonnegative integer, and adding one <phonon> costs $\hbar\omega_k$ and crystal <momentum> $\hbar k$ modulo a reciprocal-lattice vector. <Applications of Quantum Mechanics by David Tong>[§4.1.1, printed pp.103–105, monatomic-chain normal modes and quantization]{ref}
In the <continuum limit of a harmonic chain>, $\rho=m_0/a$, $Y=Ka$ and
$$
L=\frac12\int dx\,[\rho\dot u^2-Y(\partial_xu)^2],\qquad
\Pi=\rho\dot u,\quad [u(x),\Pi(x')]=i\hbar\delta(x-x'),\qquad
\boxed{c=\sqrt{Y/\rho}=a\sqrt{K/m_0}.}
$$
The <wave equation> and linear low-$k$ <dispersion relation> follow. The lattice gives an ultraviolet cutoff; the acoustic approximation does not reproduce the zone-edge flattening.
With the centre of mass fixed, the oscillator <partition function> and <Bose-Einstein distribution> are
$$
Z_{\rm osc}=\prod_{k\ne0}\frac1{2\sinh(\beta\hbar\omega_k/2)},\qquad
\langle a_k^\dagger a_k\rangle=\frac1{e^{\beta\hbar\omega_k}-1}.
$$
For a long chain, above its finite-size level spacing and below the zone-edge scale, integrating the two acoustic propagation directions gives $C/(Na)=\pi k_{\rm B}^2T/(3\hbar c)$. At high <temperature> each retained oscillator instead contributes $k_{\rm B}$ to the <heat capacity>.
\Image[/past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2013/paper-1-qcm2013-harmonic-chain.png]
{title=Original monatomic-chain mode sketch and its acoustic lattice dispersion}
{source=paper-1-qcm2013-harmonic-chain.py}
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