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The harmonic chain is a periodic array of equal masses coupled by quadratic nearest-neighbor springs, exactly reducible to independent phonon modes plus a free center-of-mass coordinate whose zero frequency must not be quantized as an oscillator.

The four-site example below makes the oscillator count explicit. A uniform displacement has no restoring force; the three relative displacements oscillate.

Helpful background. Coupled oscillators explains normal coordinates; phonons connects their quantization to collective excitations.

The calculation uses the following physical choices.

Field Baseline choice
sites N≥3N\geq3 equal masses labeled n=0,…,N−1n=0,\ldots,N-1
geometry one-dimensional periodic chain of reference spacing aa
coordinate one scalar displacement unu_n per site
momentum canonical momentum pnp_n
mass M>0M>0
coupling nearest-neighbor spring constant K>0K>0
boundary condition un+N=unu_{n+N}=u_n and pn+N=pnp_{n+N}=p_n
onsite pinning absent
anharmonicity absent
equilibrium ensemble fixed N,a,M,KN,a,M,K and temperature TT
thermodynamic limit N→∞N\to\infty at fixed a,M,Ka,M,K
zero-mode convention center of mass retained unless explicitly fixed

The site label is periodic, but the scalar displacement itself is not silently compactified. Uniformly shifting every unu_n costs no spring energy and produces a free center-of-mass degree of freedom. A pinned chain, an open chain, a diatomic chain, and an anharmonic chain are different physical models.

The microscopic canonical algebra is

[un,pm]=iℏδnm,[un,um]=0,[pn,pm]=0.\begin{aligned} [u_n,p_m] &= i\hbar\delta_{nm}, \\ [u_n,u_m] &= 0, \\ [p_n,p_m] &= 0. \end{aligned}

Before separating the uniform translation, the coordinate representation is

H=L2(RN).\mathcal H = L^2(\mathbb R^N).

Define the center-of-mass displacement and total momentum by

X=1N∑n=0N−1un,P=∑n=0N−1pn.X = \frac1N \sum_{n=0}^{N-1}u_n, \qquad P = \sum_{n=0}^{N-1}p_n.

They satisfy

[X,P]=iℏ.[X,P] = i\hbar.

After an orthogonal normal-coordinate transformation, the Hilbert space factors as

H≃L2(RX)⊗⨂λ=1N−1L2(RQλ).\mathcal H \simeq L^2(\mathbb R_X) \otimes \bigotimes_{\lambda=1}^{N-1} L^2(\mathbb R_{Q_\lambda}).

The first factor is a free coordinate. The remaining N−1N-1 factors are harmonic oscillators. The relative-coordinate theory omits the first factor. This is a reduction of the degrees of freedom, not a normalizable state with both X=0X=0 and P=0P=0: the commutator [X,P]=iℏ[X,P]=i\hbar forbids simultaneous sharp values. With

unrel=un−X,pnrel=pn−P/N,u_n^{\mathrm{rel}}=u_n-X,\qquad p_n^{\mathrm{rel}}=p_n-P/N,

the reduced variables obey

[unrel,pmrel]=iℏ(δnm−1/N).[u_n^{\mathrm{rel}},p_m^{\mathrm{rel}}] =i\hbar(\delta_{nm}-1/N).

There are only N−1N-1 independent canonical pairs. Below, “fixed center of mass” refers to this reduced theory; pinning instead adds a physical restoring force.

The baseline Hamiltonian is

H=∑n=0N−1[pn22M+K2(un+1−un)2].H = \sum_{n=0}^{N-1} \left[ \frac{p_n^2}{2M} + \frac{K}{2} (u_{n+1}-u_n)^2 \right].

Writing u=(u0,…,uN−1)T\mathbf u=(u_0,\ldots,u_{N-1})^{\mathsf T} and similarly for p\mathbf p gives

H=12MpTp+12uTCu,H = \frac{1}{2M} \mathbf p^{\mathsf T}\mathbf p + \frac12 \mathbf u^{\mathsf T}C\mathbf u,

with the periodic stiffness matrix

Cnm=K(2δnm−δn,m+1−δn,m−1).C_{nm} = K \left( 2\delta_{nm} - \delta_{n,m+1} - \delta_{n,m-1} \right).

All indices in this expression are understood modulo NN. Every row sums to zero:

∑mCnm=0.\sum_m C_{nm} = 0.

Consequently, the uniform vector

e0=1N(1,1,…,1)T\mathbf e_0 = \frac{1}{\sqrt N} (1,1,\ldots,1)^{\mathsf T}

is an exact zero eigenvector. For K>0K>0, CC is positive semidefinite and this is its only zero direction. For K<0K<0, nonuniform directions are unstable, so the usual oscillator quantization does not define a lower-bounded Hamiltonian.

Allowed wave numbers are

qm=2πmNa,q_m = \frac{2\pi m}{Na},

chosen once in a first Brillouin zone. Define

uq=1N∑ne−iqnaun,pq=1N∑ne−iqnapn.\begin{aligned} u_q &= \frac{1}{\sqrt N} \sum_n e^{-iqna}u_n, \\ p_q &= \frac{1}{\sqrt N} \sum_n e^{-iqna}p_n. \end{aligned}

Hermiticity gives

uq†=u−q,pq†=p−q,u_q^{\dagger} = u_{-q}, \qquad p_q^{\dagger} = p_{-q},

and the transformed commutator is

[uq,pq′]=iℏδq+q′,0.[u_q,p_{q'}] = i\hbar \delta_{q+q',0}.

The Hamiltonian becomes

H=12∑q[pqp−qM+Mωq2uqu−q],H = \frac12 \sum_q \left[ \frac{p_qp_{-q}}{M} + M\omega_q^2u_qu_{-q} \right],

where

ωq2=4KMsin⁡2(qa2).\omega_q^2 = \frac{4K}{M} \sin^2\left( \frac{qa}{2} \right).

Thus

ωq=2KM∣sin⁡qa2∣.\omega_q = 2\sqrt{\frac KM} \left| \sin\frac{qa}{2} \right|.

For q≠0q\ne0, introduce bosonic mode operators. One full-zone convention is

un=X+1N∑q≠0ℏ2Mωq×(bqeiqna+bq†e−iqna),u_n ={} X + \frac{1}{\sqrt N} \sum_{q\ne0} \sqrt{ \frac{\hbar}{2M\omega_q} } \times \left( b_qe^{iqna} + b_q^{\dagger}e^{-iqna} \right),

and

pn=PN−iN∑q≠0ℏMωq2×(bqeiqna−bq†e−iqna).p_n ={} \frac PN - \frac{i}{\sqrt N} \sum_{q\ne0} \sqrt{ \frac{\hbar M\omega_q}{2} } \times \left( b_qe^{iqna} - b_q^{\dagger}e^{-iqna} \right).

The mode algebra is

[bq,bq′†]=δqq′.[b_q,b_{q'}^{\dagger}] = \delta_{qq'}.

With the zero coordinate separated,

H=P22NM+∑q≠0ℏωq(bq†bq+12).H = \frac{P^2}{2NM} + \sum_{q\ne0} \hbar\omega_q \left( b_q^{\dagger}b_q + \frac12 \right).

There are N−1N-1 positive-frequency oscillator modes, not 2(N−1)2(N-1). Summing qq over the full discrete Brillouin zone gives one traveling-wave oscillator for each allowed nonzero qq. A real standing-wave convention instead combines each generic pair qq and −q-q into cosine and sine oscillators. Mixing the two conventions causes double counting.

The baseline model has:

  • continuous uniform-shift symmetry un↦un+cu_n\mapsto u_n+c, generated by PP;
  • cyclic lattice translation n↦n+1n\mapsto n+1, giving crystal momentum modulo 2πℏ/a2\pi\hbar/a;
  • reflection symmetry n↦−nn\mapsto-n;
  • time-reversal symmetry with pn↦−pnp_n\mapsto-p_n;
  • conservation of the energy in every nonzero normal mode;
  • conservation of total momentum PP.

For the exactly harmonic Hamiltonian,

nq=bq†bqn_q = b_q^{\dagger}b_q

commutes with HH for every q≠0q\ne0. The total phonon occupation also commutes with this quadratic Hamiltonian, but it is not a microscopic conserved charge protected by a fundamental U(1)U(1) symmetry. Anharmonic terms can create and destroy phonons while preserving the actual energy and crystal-momentum constraints.

The degeneracy ωq=ω−q\omega_q=\omega_{-q} follows from inversion and time-reversal symmetry. A perturbation that breaks those symmetries can alter that relation without changing the number of microscopic coordinates.

The spring timescale and band-edge frequency are

t∗=MK,ωmax⁡=2KM.t_* = \sqrt{\frac MK}, \qquad \omega_{\max} = 2\sqrt{\frac KM}.

At long wavelength,

ωq=c∣q∣[1−(qa)224+O((qa)4)],\omega_q = c|q| \left[ 1- \frac{(qa)^2}{24} +O((qa)^4) \right],

with sound speed

c=aKM.c = a\sqrt{\frac KM}.

The smallest nonzero frequency is

ωmin⁡=2KMsin⁡(πN)≃2πcNa.\omega_{\min} = 2\sqrt{\frac KM} \sin\left( \frac{\pi}{N} \right) \simeq \frac{2\pi c}{Na}.

Useful dimensionless quantities are

qa,kBTℏωmax⁡,N,qa, \qquad \frac{k_{\mathrm B}T}{\hbar\omega_{\max}}, \qquad N,

together with a strain criterion that tests the harmonic approximation. For example, the model should not be extrapolated to a microscopic lattice unless

⟨(un+1−un)2⟩≪a.\sqrt{ \left\langle (u_{n+1}-u_n)^2 \right\rangle } \ll a.

This inequality is a validity test, not a consequence of writing a quadratic Hamiltonian. The harmonic model itself has no parameter that describes the neglected cubic and quartic force constants.

The harmonic chain is exactly solvable after its stiffness matrix is diagonalized, but the claim has boundaries.

Quantity Status Caveat
finite-chain normal frequencies and eigenvectors exact discrete Fourier solution requires the stated periodic monatomic spring model
center-of-mass motion exact free-particle solution it is not a harmonic oscillator and has no normalizable ground state on R\mathbb R
relative-coordinate spectrum exact product of N−1N-1 oscillator spectra zero mode must be fixed, pinned, or retained separately
relative ground-state and thermal Gaussian correlations exact within the harmonic Hamiltonian a general non-Gaussian initial state does not become Gaussian under quadratic evolution
equilibrium oscillator occupations exact Bose factors for positive-frequency modes equilibrium phonon chemical potential is zero
real-time harmonic dynamics exact coherent mode propagation no damping, mode mixing, or thermalization occurs
low-wave-number continuum theory controlled asymptotic limit lattice dispersion and Brillouin-zone periodicity are discarded
material phonon spectrum not predicted by the toy model alone force constants, basis atoms, dimensions, and electronic effects are missing
transport coefficients not generically finite in the ideal chain integrability and absent scattering make ordinary relaxation assumptions fail

For a fixed center of mass, an exact eigenstate is labeled by occupations nq=0,1,2,…n_q=0,1,2,\ldots:

E{nq}=∑q≠0ℏωq(nq+12).E_{\{n_q\}} = \sum_{q\ne0} \hbar\omega_q \left( n_q+\frac{1}{2} \right).

The free dynamics is

bq(t)=e−iωqtbq(0).b_q(t) = e^{-i\omega_qt} b_q(0).

Exact propagation does not imply irreversible equilibration. It instead preserves every nqn_q indefinitely.

With the center of mass fixed, the relative partition function is

Zrel=∏q≠012sinh⁡(βℏωq/2).Z_{\mathrm{rel}} = \prod_{q\ne0} \frac{1} {2\sinh(\beta\hbar\omega_q/2)}.

The Bose occupation of a positive-frequency mode is

n‾q=1eβℏωq−1.\overline n_q = \frac{1} {e^{\beta\hbar\omega_q}-1}.

The relative internal energy and heat capacity are

Urel=∑q≠0ℏωq(n‾q+12),U_{\mathrm{rel}} = \sum_{q\ne0} \hbar\omega_q \left( \overline n_q+\frac{1}{2} \right),

and

CV,rel=kB∑q≠0(βℏωq)2×eβℏωq(eβℏωq−1)2.C_{V,\mathrm{rel}} ={} k_{\mathrm B} \sum_{q\ne0} (\beta\hbar\omega_q)^2 \times \frac{e^{\beta\hbar\omega_q}} {(e^{\beta\hbar\omega_q}-1)^2}.

In the thermodynamic limit, the vibrational density of states per chain length L=NaL=Na is

g(ω)L=2πaωmax⁡11−(ω/ωmax⁡)2,\frac{g(\omega)}{L} = \frac{2} {\pi a\omega_{\max}} \frac{1} {\sqrt{1-(\omega/\omega_{\max})^2}},

for 0<ω<ωmax⁡0<\omega<\omega_{\max}. It approaches

g(ω)L⟶1πc\frac{g(\omega)}{L} \longrightarrow \frac{1}{\pi c}

at low frequency and has a one-dimensional van Hove divergence at the band edge.

The thermodynamic-limit zero-point energy per site is

EzpN=ℏωmax⁡π=2ℏπKM.\frac{E_{\mathrm{zp}}}{N} = \frac{\hbar\omega_{\max}}{\pi} = \frac{2\hbar}{\pi} \sqrt{\frac KM}.

In the acoustic quantum window

ℏωmin⁡≪kBT≪ℏωmax⁡,\hbar\omega_{\min} \ll k_{\mathrm B}T \ll \hbar\omega_{\max},

the thermal energy density and heat capacity are

U(T)−U(0)L=π6(kBT)2ℏc,\frac{U(T)-U(0)}{L} = \frac{\pi}{6} \frac{(k_{\mathrm B}T)^2} {\hbar c},

and

CVL=πkB2T3ℏc.\frac{C_V}{L} = \frac{\pi k_{\mathrm B}^2T} {3\hbar c}.

At high temperature, every fixed-center oscillator approaches classical equipartition:

CV,rel⟶(N−1)kB.C_{V,\mathrm{rel}} \longrightarrow (N-1)k_{\mathrm B}.

For a finite chain at temperatures below ℏωmin⁡/kB\hbar\omega_{\min}/k_{\mathrm B}, the nonzero modes freeze out exponentially. The linear-TT law therefore requires a thermodynamic or sufficiently large-system window.

For a fixed center of mass, the absolute displacement variance is

⟨(un−X)2⟩T=ℏ2MN∑q≠01ωq×coth⁡(βℏωq2).\left\langle (u_n-X)^2 \right\rangle_T ={} \frac{\hbar}{2MN} \sum_{q\ne0} \frac{1}{\omega_q} \times \coth\left( \frac{\beta\hbar\omega_q}{2} \right).

Because ωq∼c∣q∣\omega_q\sim c|q|, this quantity grows as log⁡N\log N at T=0T=0 and as NN at any fixed T>0T>0. The unpinned one-dimensional chain therefore does not support size-independent absolute localization around the chosen reference sites in the thermodynamic limit.

Relative displacements behave better. The variance of

Bn=un+1−unB_n = u_{n+1}-u_n

contains an additional factor 4sin⁡2(qa/2)4\sin^2(qa/2) that removes the acoustic infrared divergence. Local bond strain can remain finite even when absolute displacements wander. In the classical thermal limit the finite-ring sums can be evaluated exactly:

⟨(un−X)2⟩=kBT12KN2−1N,⟨Bn2⟩=kBTK(1−1N).\left\langle(u_n-X)^2\right\rangle =\frac{k_{\mathrm B}T}{12K}\frac{N^2-1}{N}, \qquad \langle B_n^2\rangle=\frac{k_{\mathrm B}T}{K}\left(1-\frac1N\right).

The first grows with NN; the second tends to a constant. The factor 1−1/N1-1/N in the bond variance records the periodic constraint ∑nBn=0\sum_n B_n=0. These classical expressions require temperature above the phonon band; at lower temperature use the full quantum covariance before taking the large-NN limit.

Limit or modification Result Main caution
q→0q\to0 ωq≃c∣q∣\omega_q\simeq c\lvert q\rvert exact q=0q=0 remains a free coordinate
N→∞N\to\infty continuous acoustic band absolute displacement fluctuations diverge in one dimension
K→0+K\to0^+ all frequencies collapse the system approaches NN free masses, not NN soft oscillators with fixed Fock normalization
M→∞M\to\infty vibrational frequencies vanish thermal and quantum limits must be reordered carefully
weak onsite pinning acoustic branch gains a gap uniform-shift symmetry is explicitly broken
continuum limit free elastic scalar field lattice cutoff and Brillouin-zone structure are lost
high temperature (N−1)kB(N-1)k_{\mathrm B} relative heat capacity the free center of mass requires its own ensemble
weak anharmonicity finite shifts, scattering, and lifetimes mode occupations cease to be exact constants

For the continuum limit, define mass density and elastic coefficient

ρ=Ma,κ=Ka.\rho = \frac Ma, \qquad \kappa = Ka.

Then c2=κ/ρc^2=\kappa/\rho, and the relative Hamiltonian approaches

H=∫dx[π(x)22ρ+κ2(∂xu)2].H = \int dx \left[ \frac{\pi(x)^2}{2\rho} + \frac{\kappa}{2} (\partial_xu)^2 \right].

Harmonic Oscillator to Fields explains the continuum-field correspondence.

The q=0q=0 sector is

H0=P22NM.H_0 = \frac{P^2}{2NM}.

On X∈RX\in\mathbb R, it has continuous spectrum and no normalizable ground state. Its unconstrained canonical position integral is proportional to the available translation length and diverges if no finite translation range or reduction to relative variables is specified. The uniform translation is a physical symmetry, not automatically a gauge redundancy.

Three common choices define different finite systems:

  1. Retain the center of mass. Quantize H0H_0 as a free particle and specify the allowed range or wave-packet state of XX.
  2. Use relative coordinates only. Remove the free factor and retain N−1N-1 independent oscillator pairs with the projected commutator above. This does not impose simultaneous eigenvalue constraints on the original XX and PP.
  3. Pin the chain. Add an onsite potential
Hpin=MΩ22∑nun2,H_{\mathrm{pin}} = \frac{M\Omega^2}{2} \sum_n u_n^2,

which changes the dispersion to

ωq2=Ω2+4KMsin⁡2(qa2).\omega_q^2 = \Omega^2 + \frac{4K}{M} \sin^2\left( \frac{qa}{2} \right).

The limits N→∞N\to\infty, Ω→0\Omega\to0, and T→0T\to0 need not commute for infrared-sensitive observables. An accidental small eigenvalue from rounding is a numerical error, not a physical pinning model. Classify the known translation vector as zero within a justified tolerance. If a positive regulator is deliberately added, state its operator: pinning only the center of mass differs from the onsite term above, which shifts every mode.

For even NN, the zone-boundary momentum q=π/aq=\pi/a is its own negative modulo a reciprocal vector. Generic qq and −q-q form degenerate pairs, but this self-conjugate point contributes only one oscillator. Correct mode counting always returns N−1N-1 positive-frequency modes after the center of mass is fixed.

Natural observables include:

  • displacement and momentum covariances;
  • bond strain Bn=un+1−unB_n=u_{n+1}-u_n;
  • normal-mode occupations nqn_q;
  • vibrational energy and heat capacity;
  • group velocity and wave-packet propagation;
  • dynamical structure factors and spectral functions;
  • local energy density and energy current.

Inside 0<q<π/a0<q<\pi/a, the group velocity is

vg(q)=dωqdq=ccos⁡(qa2).v_g(q) = \frac{d\omega_q}{dq} = c\cos\left( \frac{qa}{2} \right).

It approaches cc near the zone center and vanishes at the band edge. The existence of a normal mode does not guarantee that a particular probe sees it: scattering intensity also depends on polarization, form factors, and operator selection rules. Structure Factors explains those response conventions.

Take N=4N=4. The allowed momenta may be represented by

q∈{0,π2a,−π2a,πa}.q \in \left\{ 0, \frac{\pi}{2a}, -\frac{\pi}{2a}, \frac{\pi}{a} \right\}.

The frequencies are

ω0=0,ω±π/(2a)=2KM,ωπ/a=2KM.\begin{aligned} \omega_0 &= 0, \\ \omega_{\pm\pi/(2a)} &= \sqrt{\frac{2K}{M}}, \\ \omega_{\pi/a} &= 2\sqrt{\frac KM}. \end{aligned}

The q=0q=0 eigenvector is uniform. The zone-boundary eigenvector is proportional to (1,−1,1,−1)(1,-1,1,-1). The q=±π/(2a)q=\pm\pi/(2a) traveling waves are degenerate and can be replaced by real sine and cosine standing waves.

After fixing the center of mass, the ground energy is

E0,rel=ℏKM(1+2).E_{0,\mathrm{rel}} = \hbar\sqrt{\frac KM} \left( 1+\sqrt2 \right).

The exact relative partition function is

Zrel=[2sinh⁡(βℏ22KM)]−2×[2sinh⁡(βℏKM)]−1.Z_{\mathrm{rel}} ={} \left[ 2\sinh\left( \frac{\beta\hbar}{2} \sqrt{\frac{2K}{M}} \right) \right]^{-2} \times \left[ 2\sinh\left( \beta\hbar \sqrt{\frac KM} \right) \right]^{-1}.

The real normal vectors can be chosen proportional to (1,1,1,1)(1,1,1,1), (1,0,−1,0)(1,0,-1,0), (0,1,0,−1)(0,1,0,-1), and (1,−1,1,−1)(1,-1,1,-1). The middle two span the same degenerate space as the traveling waves at q=±π/(2a)q=\pm\pi/(2a).

Four real displacement patterns on a four-site periodic chain: uniform translation, two degenerate relative patterns, and the alternating pattern.

Each arrow gives a site’s scalar displacement; a dot without an arrow has zero displacement. Vector amplitudes are schematic, not a common normalization. The uniform pattern is a free coordinate. The other three are oscillator directions; the two middle patterns are independent despite sharing a frequency.

For a compact matrix calculation, use

N=8,M=K=a=ℏ=1.N=8, \qquad M=K=a=\hbar=1.

Diagonalize the mass-weighted stiffness matrix

Dnm=2δnm−δn,m+1−δn,m−1.D_{nm} = 2\delta_{nm} - \delta_{n,m+1} - \delta_{n,m-1}.

The analytic eigenvalues and frequencies are

Mode labels mm Degeneracy ωm2\omega_m^2 ωm\omega_m
00 11 00 00
1,71,7 22 2−22-\sqrt2 2−2\sqrt{2-\sqrt2}
2,62,6 22 22 2\sqrt2
3,53,5 22 2+22+\sqrt2 2+2\sqrt{2+\sqrt2}
44 11 44 22

Useful decimal frequencies are

2−2=0.7653668647301795…,2=1.4142135623730951…,2+2=1.8477590650225735….\begin{aligned} \sqrt{2-\sqrt2} &= 0.7653668647301795\ldots, \\ \sqrt2 &= 1.4142135623730951\ldots, \\ \sqrt{2+\sqrt2} &= 1.8477590650225735\ldots. \end{aligned}

Independent matrix invariants are

D1=0,tr⁡D=16,D\mathbf1 = 0, \qquad \operatorname{tr}D = 16,

and

pdet⁡D=∏λj>0λj=64.\operatorname{pdet}D = \prod_{\lambda_j>0}\lambda_j = 64.

The fixed-center zero-point energy is

Ezp=1+2+2−2+2+2=5.027339492125848….\begin{aligned} E_{\mathrm{zp}} ={}& 1+\sqrt2 +\sqrt{2-\sqrt2} +\sqrt{2+\sqrt2} \\ =& 5.027339492125848\ldots. \end{aligned}

For the lowest nonzero mode in a sequence of chain lengths,

ω1c∣q1∣=sin⁡(π/N)π/N=1−π26N2+O(N−4).\frac{\omega_1}{c|q_1|} = \frac{\sin(\pi/N)}{\pi/N} = 1- \frac{\pi^2}{6N^2} +O(N^{-4}).

A valid implementation should:

  1. construct periodic neighbors without duplicating or omitting the wraparound spring;
  2. verify row sums and the exact null vector before diagonalization;
  3. compare eigenvalues, degeneracies, trace, and pseudodeterminant;
  4. form frequencies only after classifying negative, zero, and positive eigenvalues;
  5. compare eigenvector subspaces rather than individual vectors inside degenerate pairs;
  6. state whether the zero mode is retained, fixed, or pinned;
  7. test the N−2N^{-2} acoustic convergence rather than only one chain length.

Open, free-end, and fixed-end chains have standing-wave spectra and different zero-mode structure. Reusing periodic wave numbers for them gives the wrong mode count and endpoint behavior.

The onsite term proportional to Ω2\Omega^2 gaps the acoustic branch and renders absolute displacement fluctuations infrared finite. It also explicitly breaks uniform-shift symmetry, so it is not a harmless numerical regulator unless the Ω→0\Omega\to0 limit is demonstrated for the observable being calculated.

A diatomic or multi-atom unit cell produces several branches. Acoustic and optical labels refer to long-wavelength displacement patterns, while exact frequencies require a matrix-valued dynamical problem. The general treatment belongs to Phonons as Many-Body Excitations.

Translation-invariant longer-range springs change ωq\omega_q but preserve Fourier diagonalization. Random masses or spring constants destroy momentum as an exact label and can localize vibrational modes.

Cubic and quartic displacement terms couple normal modes. They generate thermal expansion, frequency shifts, finite lifetimes, and scattering. The harmonic occupations nqn_q then cease to be exact constants of motion.

Several displacement polarizations and lattice directions produce a dynamical matrix at every crystal momentum. The long-wavelength continuum becomes an elastic field theory rather than a single scalar field.

Insert a plane wave un(t)=uqei(qna−ωt)u_n(t)=u_qe^{i(qna-\omega t)} into the classical equations of motion and derive the exact lattice dispersion.

Solution

Hamilton’s equations give

Mu¨n=K(un+1+un−1−2un).M\ddot u_n = K \left( u_{n+1}+u_{n-1}-2u_n \right).

For the plane wave,

−Mω2=K(eiqa+e−iqa−2).-M\omega^2 = K \left( e^{iqa}+e^{-iqa}-2 \right).

Using 2−2cos⁡qa=4sin⁡2(qa/2)2-2\cos qa=4\sin^2(qa/2) yields

ωq2=4KMsin⁡2(qa2).\omega_q^2 = \frac{4K}{M} \sin^2\left( \frac{qa}{2} \right).

The nonnegative frequency is therefore

ωq=2KM∣sin⁡qa2∣.\omega_q = 2\sqrt{\frac KM} \left| \sin\frac{qa}{2} \right|.

Show that the uniform Fourier coordinate gives H0=P2/(2NM)H_0=P^2/(2NM) and explain why its oscillator ground-state formula is invalid.

Solution

The normalized zero-mode variables are

u0=1N∑nun=NX,u_0 = \frac1{\sqrt N} \sum_nu_n = \sqrt N X,

and

p0=1N∑npn=PN.p_0 = \frac1{\sqrt N} \sum_np_n = \frac P{\sqrt N}.

The spring term vanishes for a uniform displacement. The kinetic contribution is

H0=p0p02M=P22NM.H_0 = \frac{p_0p_0}{2M} = \frac{P^2}{2NM}.

This is a free-particle Hamiltonian in XX. Its spectrum is continuous on R\mathbb R, and it has no normalizable state with the oscillator energy ℏω0/2=0\hbar\omega_0/2=0. The factors 1/ω01/\sqrt{\omega_0} in oscillator coordinates also diverge. One must retain, fix, or pin this coordinate explicitly.

For N=4N=4, find the allowed momenta, frequencies, and fixed-center ground energy. Explain why the two modes at q=±π/(2a)q=\pm\pi/(2a) are not double counting.

Solution

The allowed momenta are 00, π/(2a)\pi/(2a), −π/(2a)-\pi/(2a), and π/a\pi/a. Substitution into the dispersion gives

ω={0,2KM,2KM,2KM}.\omega = \left\{ 0, \sqrt{\frac{2K}{M}}, \sqrt{\frac{2K}{M}}, 2\sqrt{\frac KM} \right\}.

After fixing the zero coordinate, there are three oscillators. Their zero-point energies sum to

E0,rel=ℏ2[22KM+2KM]=ℏKM(1+2).\begin{aligned} E_{0,\mathrm{rel}} &= \frac{\hbar}{2} \left[ 2\sqrt{\frac{2K}{M}} + 2\sqrt{\frac KM} \right] \\ &= \hbar\sqrt{\frac KM} (1+\sqrt2). \end{aligned}

The qq and −q-q traveling waves are independent directions of propagation. Equivalently, their two-dimensional degenerate subspace has independent cosine and sine standing waves. Either basis contains two oscillators, not four.

Exercise 4: Derive the low-temperature heat capacity

Section titled “Exercise 4: Derive the low-temperature heat capacity”

Use the acoustic dispersion and thermodynamic-limit mode density to derive the leading thermal energy and heat capacity per length.

Solution

Both signs of qq contribute. With ω=c∣q∣\omega=c|q|,

U(T)−U(0)L=1π∫0∞dq ℏcqeβℏcq−1.\frac{U(T)-U(0)}{L} = \frac1\pi \int_0^\infty dq\, \frac{\hbar cq} {e^{\beta\hbar cq}-1}.

Set x=βℏcqx=\beta\hbar cq. Then

U(T)−U(0)L=(kBT)2πℏc∫0∞dx xex−1.\frac{U(T)-U(0)}{L} = \frac{(k_{\mathrm B}T)^2} {\pi\hbar c} \int_0^\infty dx\, \frac{x}{e^x-1}.

Using

∫0∞dx xex−1=π26\int_0^\infty dx\, \frac{x}{e^x-1} = \frac{\pi^2}{6}

gives

U(T)−U(0)L=π6(kBT)2ℏc.\frac{U(T)-U(0)}{L} = \frac{\pi}{6} \frac{(k_{\mathrm B}T)^2} {\hbar c}.

Differentiating at fixed length yields

CVL=πkB2T3ℏc.\frac{C_V}{L} = \frac{\pi k_{\mathrm B}^2T} {3\hbar c}.

The derivation requires ℏωmin⁡≪kBT\hbar\omega_{\min}\ll k_{\mathrm B}T so that the momentum sum may be replaced by an integral.

Exercise 5: Compare absolute and relative fluctuations

Section titled “Exercise 5: Compare absolute and relative fluctuations”

Use the low-qq form of the covariance to determine the infrared scaling of ⟨(un−X)2⟩\langle(u_n-X)^2\rangle at zero and nonzero temperature. Explain why the bond variance does not have the same divergence.

Solution

At zero temperature, the long-wavelength contribution is proportional to

∫qmin⁡dqωq∼1c∫qmin⁡dqq∼log⁡N.\int_{q_{\min}} \frac{dq}{\omega_q} \sim \frac1c \int_{q_{\min}} \frac{dq}{q} \sim \log N.

At fixed T>0T>0 and sufficiently small qq,

coth⁡(βℏcq2)∼2kBTℏcq.\coth\left( \frac{\beta\hbar cq}{2} \right) \sim \frac{2k_{\mathrm B}T} {\hbar cq}.

The integrand is then proportional to 1/q21/q^2, so

∫qmin⁡dqq2∼1qmin⁡∝N.\int_{q_{\min}} \frac{dq}{q^2} \sim \frac1{q_{\min}} \propto N.

For a bond displacement, Fourier transformation supplies

∣eiqa−1∣2=4sin⁡2(qa2)∼q2a2.|e^{iqa}-1|^2 = 4\sin^2\left( \frac{qa}{2} \right) \sim q^2a^2.

This factor cancels enough of the small-qq singularity to make the local bond variance infrared finite. Absolute positional order and local elastic strain therefore have different thermodynamic behavior.