Lecture 6: Creating Elementary Excitations

Once a populated reference state is known, the next question asks which physical operation changes it and what quantum numbers that change carries. This lecture introduces excitation as that relational object, joining band occupation to the addition, removal, spin-flip, and collective channels used by experiments. Fock-space operators, response functions, spectral measures, magnons, and Bogoliubov modes will later formalize different versions of the same idea. An excitation is a change defined relative to a reference state and selected by a preparation-and-probe operation.

Begin with three experimental acts. A photon absorbed by a semiconductor can promote an electron from an occupied valence state to an available conduction state, leaving an electron–hole pair relative to the initial solid. A Raman or neutron event can add one quantum to a silicon lattice vibration. A neutron that transfers spin as well as energy to an ordered magnet can create a spin wave. Each detector may report a dispersing peak, but the peaks are not three names for the same object. The preparation that existed before the probe, the operator through which the probe acts, and the conserved quantities changed by the event are different.

We will describe each probe event as an operation on a declared reference state and track the energy, momentum, charge, spin, and statistics changed in the electron, lattice-mode, and spin-wave examples.

Optical absorption is sensitive to an electronic dipole matrix element and creates a neutral pair even though its two constituents carry opposite charge. Lattice scattering couples to ionic displacement. A magnetic neutron channel couples to spin density. The measured energy and momentum are necessary evidence in all three cases, but they are insufficient until the coupling channel and reference state are declared.

Relative to a declared many-particle reference, what operation was performed, what energy and quantum numbers changed, and what excitation name is thereby justified?

This ledger is what lets photoemission be interpreted as electron removal, optical absorption as an electron–hole pair creation process, and lattice spectroscopy as the creation or destruction of collective-mode quanta. It also prevents two common category errors: calling every unoccupied state a hole and reading zero phonon chemical potential as zero phonon energy.

From Lecture 5, recover the meaning of the chemical potential in the relative-weight rule, the difference between an available state and an occupied state, and the group velocity of a band. From quantum mechanics, recover the energy ladder and normalized number states of one harmonic oscillator. Every thermal formula below is rebuilt from relative probabilities and normalization.

A hole inherits quantum numbers from the occupied reference

Removing an electron from an occupied mode, adding one to an empty mode, and occupying a collective normal mode are different constructions relative to a prepared reference: each assigns a change of sector and of conserved quantum numbers, even when all three are drawn as dispersing branches. Whether a given probe can create the change is a further question, decided by its matrix element rather than by the existence of the branch, so the three constructions can share one energy–momentum ledger while keeping their distinct microscopic origins. The electron bookkeeping below is exact in the noninteracting band model; the lattice-mode construction that follows is exact for the harmonic Hamiltonian and remains a controlled description of the crystal while anharmonic corrections stay small.

The band and the reference state it needs

Retain the spatial dimension ${d\in\{1,2,3\}}$ and Brillouin torus ${\mathrm{BZ}=\mathbb R^d/\Lambda^*}$ from Lecture 5. For the electronic band under discussion, the energy is a function ${\varepsilon:\mathrm{BZ}\to\mathbb R}$ , and ${\mathbf k\in\mathrm{BZ}}$ is a classical wave vector. The reference state below specifies which values of that function are occupied; the function alone does not. The terms Fermi occupation function and band filling retain their exact Lecture 5 meanings.

Hole charge, momentum, and energy

An excitation energy is an energy difference between a prepared reference and a changed state. This sentence prevents an immediate ambiguity in hole language. Let a zero-temperature electron reference have many-particle energy ${E_0}$ , electron number ${N_0}$ , and occupied one-particle states below ${\mu}$ . Removing an electron from an occupied state of energy ${\varepsilon(\mathbf k)<\mu}$ changes

$E'=E_0-\varepsilon(\mathbf k), \qquad N'=N_0-1.$

The raw energy difference depends on the arbitrary one-particle energy zero. The constrained comparison does not:

(6.1)

$\varepsilon_{\mathrm h}(\mathbf k) =\Delta(E-\mu N) =\mu-\varepsilon(\mathbf k)>0, \qquad \Delta Q=+e.$

An electronic state above ${\mu}$ is empty in the reference; what are its hole energy and hole charge? The expected answer is that neither is defined, because no removal has occurred: a hole is the change produced by removing an occupied electron. The plausible wrong answer — it is a positive-charge hole because the state contains no electron — mistakes a blank place in a diagram for an excitation. The changed state carries one less electron charge, so relative to the reference the excitation has charge ${+e}$ . Calling it a hole packages the missing contribution into an excitation variable. By contrast, a state above ${\mu}$ that was already empty has not been changed, so it carries no excitation energy, no charge change, and no hole.

Before assigning momentum, complete two ledgers, and reject any name until the change columns are filled. Leaving an already empty state above ${\mu}$ unchanged gives ${\Delta(E-\mu N)=0}$ and ${\Delta Q=0}$ : no excitation has been created. Removing an occupied state instead gives the changes calculated above. If “empty” is used as the only reason for the hole label, the reference-state step has been skipped.

Before reading Figure 6.1, compare a one-hole configuration with the filled-band reference vacuum. Which electron occupation differs, and what excitation energy and charge describe that difference?

Figure 6.1. Reference subtraction: the one-hole configuration minus the filled-band vacuum is described as a hole with positive excitation energy and charge +e. Vacuum means no excitations relative to the filled band, not an empty electron band. Dots represent sample states of one resolved band.

The first two panels compare electron occupations on the same band: their difference is one missing electron, not a subtraction of quantum state vectors. The third panel redescribes that difference as a hole. With ${\mathbf k_{\mathrm h}=-\mathbf k_{\mathrm e}}$ , differentiation of the hole excitation energy gives the same physical velocity while the charge assigned to the missing contribution is ${+e}$ .

Let a zero-temperature electronic reference have an occupied resolved one-particle state ${|\psi_{\mathbf k}\rangle}$ with ${\varepsilon(\mathbf k)<\mu}$ . A hole excitation is the change produced by removing that occupied electron from the reference. With electron charge ${-e}$ , ${e>0}$ , and the convention ${\mathbf k_{\mathrm h}=-\mathbf k\pmod{\Lambda^*}}$ , its relative data are $\varepsilon_{\mathrm h}(\mathbf k_{\mathrm h}) =\mu-\varepsilon(-\mathbf k_{\mathrm h})>0, \qquad \Delta Q_{\mathrm h}=+e, \qquad \mathbf v_{\mathrm h} =\hbar^{-1}\nabla_{\mathbf k_{\mathrm h}} \varepsilon_{\mathrm h}.$ The hole is a missing-fermion excitation relative to the filled reference, not an additional one-electron state. An orbital that was already empty and undergoes no removal operation is not a hole. A different momentum convention is admissible only if all momentum and velocity formulas are transformed with it.

Hole current, hole mass, and probe operations

With the two ledgers fixed, the missing electron’s current and effective mass follow from the filled reference by subtraction; the calculation below also tests the momentum convention that the hole definition adopted.

Take a completely filled, inversion-symmetric valence band in volume ${\mathcal V}$ , with electron charge ${-e}$ , and remove its occupied state at ${\mathbf k=\mathbf k_0+\mathbf q}$ . Calculate its grand-energy cost, crystal momentum, charge, current and mass at the hole minimum relative to the filled ground reference, taking chemical potential ${\mu>\varepsilon_{\max}}$ . Near a band maximum at ${\mathbf k_0}$ , take $\varepsilon(\mathbf k_0+\mathbf q) =\varepsilon_{\max} -\frac{\hbar^2}{2}q_i{(M^{-1})}_{ij}q_j+O({|\mathbf q|}^3),$ with ${M^{-1}}$ positive definite. Use the hole-momentum convention ${\mathbf k_{\mathrm h}=-\mathbf k}$ .

Subtracting the removed electron from the reference fixes the signs before any carrier interpretation: ${\Delta N=-1}$ , ${\Delta Q=+e}$ , ${\Delta\mathbf P_{\mathrm{crystal}}=-\hbar\mathbf k}$ modulo a reciprocal-lattice momentum, and ${\Delta(E-\mu N)=\mu-\varepsilon(\mathbf k)}$ . Choose the common convention ${\mathbf k_{\mathrm h}=-\mathbf k}$ . Then

$\varepsilon_{\mathrm h}(\mathbf k_{\mathrm h}) =\mu-\varepsilon(-\mathbf k_{\mathrm h}), \qquad \mathbf v_{\mathrm h} =\frac1\hbar\nabla_{\mathbf k_{\mathrm h}} \varepsilon_{\mathrm h} =\mathbf v_{\mathrm e}(\mathbf k).$

The velocity equality follows from the declared momentum convention. The charge current reverses because the excitation charge is ${+e}$ . A different hole-momentum convention is possible but must be stated before momenta are added or compared.

The current statement can be checked without words. For one completely filled nondegenerate band in physical volume ${\mathcal V}$ , the paired momentum grid gives $\mathbf j_{\mathrm{full}} =-\frac{e}{\mathcal V}\sum_{\mathbf k\in\mathrm{BZ}} \mathbf v_{\mathrm e}(\mathbf k)=\mathbf 0,$ because inversion pairs velocities at ${\mathbf k}$ and ${-\mathbf k}$ on the symmetric momentum grid. In the continuum Brillouin-zone integral, periodicity alone also makes the integral of the band gradient vanish. Removing the electron at ${\mathbf k}$ deletes its contribution, so $\Delta\mathbf j =+\frac{e}{\mathcal V}\mathbf v_{\mathrm e}(\mathbf k) =\frac{q_{\mathrm h}}{\mathcal V} \mathbf v_{\mathrm h}(\mathbf k_{\mathrm h}).$ Thus the hole current is not an additional postulate; it is the exact change in the electron current relative to the filled reference.

The positive hole charge, the chosen hole momentum, and the hole velocity are three linked conventions and deductions, not three independent definitions. Reverse the momentum convention and the formulas must be rewritten consistently, while the physical current change remains fixed.

In the supplied band-maximum expansion, the electron curvature is negative, while the hole energy has positive curvature. At its minimum ${\mathbf k_{\mathrm h}=-\mathbf k_0}$ , its Hessian gives ${\left(M_{\mathrm h}^{-1}\right)}_{ij} =\frac{1}{\hbar^2} \frac{\partial^2\varepsilon_{\mathrm h}} {\partial k_{\mathrm h,i}\partial k_{\mathrm h,j}} ={\left(M^{-1}\right)}_{ij}.$ This is why a nearly full simple band can often be described by positively charged, positive-mass carriers. It is a reference-state transformation, not the discovery of a new fundamental particle in vacuum.

For example, along a principal direction with mass ${m_*>0}$ , removal at ${k_0+q}$ gives ${v_{\mathrm e}=-\hbar q/m_*}$ , hole displacement from its minimum ${q_{\mathrm h}=-q}$ , and ${v_{\mathrm h}=\hbar q_{\mathrm h}/m_*=-\hbar q/m_*}$ . Thus both ledgers give ${\Delta j=-e\hbar q/(\mathcal V m_*)}$ . This direct check tests the momentum sign; the Hessian tests the positive mass. The excitation energy is unchanged if both ${\varepsilon}$ and ${\mu}$ shift by the same constant. Current density has units charge per area per time in three dimensions. The local mass description requires ${\mathbf q}$ small enough that the omitted cubic terms are negligible; the removal ledger itself does not require that expansion. It is this ledger, rather than an empty spot on a band plot, that a removal experiment probes.

The distinction affects experiments. Photoemission removes an electron and measures removal energy and momentum. Inverse photoemission adds an electron to an available state. A band plot alone cannot supply either intensity; matrix elements, occupations, and lifetimes are still required.

For optical absorption across a semiconductor gap, the bookkeeping contains both operations at once: an electron is added to a conduction state and removed from a valence state. The total charge change of the isolated solid is zero, while the excitation contains a negatively charged electron and a positively charged hole. In the independent-particle approximation the pair energy is the conduction addition energy plus the valence removal energy. Writing ${\epsilon_c}$ and ${\epsilon_v}$ for the chosen conduction and valence one-electron energies makes the cancellation explicit:

$\Delta(E-\mu N) =(\epsilon_c-\mu)+(\mu-\epsilon_v) =\epsilon_c-\epsilon_v=\Delta E, \qquad \Delta N=0,\qquad \Delta Q=-e+e=0.$

The two terms use the same reference chemical potential. This cancellation does not determine which transition is visible: the photon momentum and the optical matrix element still constrain the transition. Coulomb attraction can bind the two constituents into an exciton and shift the optical threshold below the independent-particle continuum. That possibility is outside the present calculation, but it is precisely why an absorption onset must not be identified uncritically with a bare band gap.

Excitation as a change relative to a declared reference

The removal ledger is one instance of a pattern that the lattice will instantiate a second time. Before that second construction, fix the pattern itself: what must be declared for a change of state to count as an excitation, and what a probe adds to that declaration.

Fix a model Hilbert space ${\mathcal H}$ and a self-adjoint constrained-energy generator ${\hat K}$ . Use ${\hat K=\hat H}$ in a closed fixed-number sector and ${\hat K=\hat H-\mu\hat N}$ when the preparation exchanges particles at chemical potential ${\mu}$ . Let the normalized reference eigenstate ${|\Omega\rangle}$ minimize ${\hat K}$ in the declared prepared sector.

The model must also declare a one-excitation index set ${\mathcal A}$ and a family ${\mathcal X_1=\{[|X_\alpha\rangle]:\alpha\in\mathcal A\}}$ of normalized eigenstate rays of ${\hat K}$ , supplied by a stated single-change construction such as one occupied-state removal or one harmonic-mode quantum. The brackets mean that normalized vectors differing only by an overall phase represent the same ray. The excitation data are changes relative to the reference, for example $\varepsilon_\alpha =K_{X_\alpha}-K_\Omega\geq0, \qquad \Delta Q_\alpha=Q_{X_\alpha}-Q_\Omega, \qquad \Delta\mathbf k_\alpha =\mathbf k_{X_\alpha}-\mathbf k_\Omega\pmod{\Lambda^*}.$ The nonnegativity follows from the declared stability of ${|\Omega\rangle}$ ; it would not follow for an arbitrary excited reference. An elementary excitation relative to this reference and model is a pair ${(\alpha,[|X_\alpha\rangle])}$ in the declared one-excitation family, with its construction, relative data, and model boundary stated before dilute multi-excitation combinations are formed. A curve, spectral line, or empty one-particle orbital alone is not this definition.

This block is exact for the stated model. When a later lecture uses an eigenmode of a linearized or effective Hamiltonian, it must say so and apply the same relative-data ledger within that approximation. A finite-lifetime resonance is not an exact eigenstate ray and is deferred to the spectral definitions of Lectures 13–14.

Landau’s 1956 formulation is an original source for the narrower Fermi-liquid claim that low-lying interacting states can be organized by quasiparticle occupation changes and an energy functional [Lan56, pp. 920–922, Eqs. (1), (5), and (6)]. It is historical context rather than authority for the broader course definition above, which also includes holes, phonons, dark modes, and explicitly model-relative excitation families.

Let an already defined excitation ${[|X_\alpha\rangle]}$ be tested by a weak probe perturbation proportional to a specified operator ${\hat O}$ . In first-order transition theory, a necessary coupling condition is $\langle X_\alpha|\hat O|\Omega\rangle\neq0,$ together with compatible energy, momentum, and other quantum-number transfer. A vanishing matrix element means that this probe operator does not connect the reference to the excitation in first-order transition theory. Here “first order” refers to the transition amplitude’s expansion in the probe strength, not the number of excitations. This does not remove the excitation from ${\mathcal X_1}$ . A nonzero matrix element alone does not establish its identity; the measured transfers and model prediction must also match.

The hole calculation above supplied the single-change construction required by the definition: a filled reference, one removal operation, and the resulting relative energy, charge, and momentum data. Photoemission accessibility is a separate claim about its removal operator and matrix element.

Before reading Figure 6.2, predict which two differences are reconstructed from the analyzer record and whether every eigenmode of the sample must appear as a bright ridge.

Figure 6.2. A triple-axis scattering chain reconstructs energy and momentum transfer, while a computed mode includes a dark point caused by the probe matrix element.

The analyzer yields final energy and direction; calibration and conservation then construct ${\hbar\omega=E_{\mathrm i}-E_{\mathrm f}}$ and ${\mathbf q=\mathbf k_{\mathrm i}-\mathbf k_{\mathrm f}}$ . Panel (b) enlarges the sample geometry; panel (c) is an illustrative dispersion and scan construction, not a dataset. Its open point makes the logical separation operational: the mode may exist at that ${(\mathbf q,\omega)}$ , yet the measured record can be dark when ${\langle X_\alpha|\hat O|\Omega\rangle=0}$ . A dispersion ridge is therefore evidence only after the transfer and probe channel have been declared. Van Hove’s original space–time correlation treatment makes the corresponding Born-approximation neutron cross section a function of both momentum and energy transfer [Hov54, printed pp. 249–251, Eqs. (1)–(7)].

Modes with unconstrained occupation

Unconstrained mode number and Bose weights

Historical lineage

[Bos24, pp. 178–181] — Bose’s 1924 paper derives the photon-counting distribution that yields Planck’s radiation law.
[Ein24, pp. 261–267] — Einstein’s 1924 Academy paper extends Bose counting to a monatomic ideal gas.

The lattice supplies a different kind of disturbance. A normal mode is a collective displacement pattern of the coupled ions, with a definite angular frequency. In the harmonic model, the force constants and masses determine the pattern and its frequency; occupying the mode does not introduce a new pattern. Quantization assigns an integer number of quanta to each such mode. Thus the mode label specifies which pattern, whereas its occupation specifies how many quanta. A displacement drawing represents the mode’s polarization, not a classical trajectory of an oscillator energy eigenstate.

Begin with one normal mode of angular frequency ${\omega>0}$ . Its quantum Hilbert space is ${\ell^2(\mathbb N_0)}$ , with orthonormal energy states ${|n\rangle}$ and

$E_n=\hbar\omega\left(n+\frac12\right), \qquad n=0,1,2,\ldots.$

The label ${n}$ counts quanta in this mode. In thermal equilibrium, ordinary anharmonic processes can create and destroy phonons while conserving total energy. There is therefore no externally fixed equilibrium phonon number and no chemical work associated with changing ${n}$ : ${\mu_{\mathrm{ph}}=0}$ . Zero chemical potential does not mean zero mode energy; ${\hbar\omega}$ remains a separate dynamical scale. The equilibrium distribution is a preparation assumption: an isolated exactly harmonic lattice does not by itself relax arbitrary initial mode occupations to it [Gri07].

For an oscillator with ${E_n=\hbar\omega(n+1/2)}$ , ${\omega>0}$ , ${n\in\mathbb N_0}$ , and ${T>0}$ , determine every occupation probability and its mean occupation from the relative-weight rule, then check the resulting mean energy in the cold and classical regimes. There is no particle-number constraint to truncate the sum. With ${\beta=1/(k_{\mathrm B}T)}$ , the supplied reservoir rule is

$\frac{p_n}{p_0}=\mathrm e^{-n\beta\hbar\omega}.$

Normalize the full geometric series before taking its first moment, using the relative-weight principle of the single-level occupation example. The changed input is the occupation set: all ${n\geq0}$ , not only zero and one.

Let ${x=\mathrm e^{-\beta\hbar\omega}\in(0,1)}$ . Normalization gives the geometric distribution

$p_n=(1-x)x^n,$

where ${n\mapsto p_n}$ is a probability mass function on the discrete set ${\mathbb N_0}$ , not a density with respect to a continuous energy variable. Its normalization follows because

$1=\sum_{n=0}^{\infty}p_n =p_0\sum_{n=0}^{\infty}x^n =\frac{p_0}{1-x}, \qquad p_0=1-x.$

The mean is not obtained by replacing ${n}$ with a typical value. Differentiate the normalized geometric series:

$\sum_{n=0}^{\infty}n x^n =x\frac{\mathrm d}{\mathrm dx} \sum_{n=0}^{\infty}x^n =x\frac{\mathrm d}{\mathrm dx}\frac{1}{1-x} =\frac{x}{{(1-x)}^2}.$

Multiplying by the normalization factor now completes the mean calculation: $\sum_{n=0}^{\infty}n p_n =(1-x)\sum_{n=0}^{\infty}n x^n =(1-x)\frac{x}{{(1-x)}^2} =\frac{x}{1-x}.$

The result gives $\langle E\rangle=\hbar\omega\left(\frac{x}{1-x}+\frac12\right).$ Normalization is the first check: summing ${(1-x)x^n}$ returns one. At ${\beta\hbar\omega\gg1}$ , the mean occupation is ${\mathrm e^{-\beta\hbar\omega}+O(\mathrm e^{-2\beta\hbar\omega})}$ , so the mean energy approaches ${\hbar\omega/2}$ . At ${y:=\beta\hbar\omega\ll1}$ , expansion gives ${\bar n=y^{-1}-1/2+y/12+O(y^3)}$ , hence ${\langle E\rangle=k_{\mathrm B}T[1+y^2/12+O(y^4)]}$ . The half quantum cancels the ${-1/2}$ occupation term in the classical limit, while every energy term retains units of energy. This distribution exists for ${x<1}$ ; an exact zero-frequency translation is outside the oscillator calculation. The following definition names the mean we have constructed, ready to be used in a many-mode energy sum.

Fix ${T>0}$ , ${\beta={(k_{\mathrm B}T)}^{-1}}$ , and zero equilibrium mode chemical potential. The Bose occupation function is the ordinary function $n_{\mathrm B}:(0,\infty)\longrightarrow(0,\infty), \qquad n_{\mathrm B}(\omega) =\sum_{n=0}^{\infty}n p_n =\frac{1}{\mathrm e^{\beta\hbar\omega}-1}.$ For a positive-frequency bosonic mode it is the mean of the normalized integer-valued distribution ${p_n=(1-\mathrm e^{-\beta\hbar\omega}) \mathrm e^{-n\beta\hbar\omega}}$ , not a probability density over energy and not the probability of one particular occupation. Its domain excludes ${\omega=0}$ , where it diverges as ${n_{\mathrm B}(\omega)\sim k_{\mathrm B}T/(\hbar\omega)}$ . A rigid translation or other exact zero mode must therefore be constrained, removed, or treated by a separate limiting argument before a discrete mode sum is formed.

Unlike ${f}$ , ${n_{\mathrm B}}$ is not the probability of one occupation outcome and may exceed one. It is the mean of the entire integer-valued distribution. The mean energy is

$\langle E\rangle =\hbar\omega\left[n_{\mathrm B}(\omega)+\frac12\right].$

The limits explain the physical regimes. If ${k_{\mathrm B}T\ll\hbar\omega}$ , then ${n_{\mathrm B}\simeq\mathrm e^{-\beta\hbar\omega}}$ : excited quanta are rare. If ${k_{\mathrm B}T\gg\hbar\omega}$ , then

$n_{\mathrm B} =\frac{k_{\mathrm B}T}{\hbar\omega}-\frac12 +O\!\left(\frac{\hbar\omega}{k_{\mathrm B}T}\right),$

so ${\langle E\rangle\simeq k_{\mathrm B}T}$ . The zero-point term and the ${-1/2}$ in the occupation expansion combine correctly; dropping one but not the other spoils the classical limit.

Close the notes for seven minutes. For a mode with ${\omega>0}$ at ${T>0}$ , the allowed occupations are ${n=0,1,2,\ldots}$ , and the equilibrium weights obey ${p_n/p_0=\mathrm e^{-n\beta\hbar\omega}}$ . Determine the normalized probabilities, verify that they sum to one, and calculate the mean occupation. Then obtain its low- and high-temperature limits. Show the normalization and first-moment steps rather than quoting the final occupation formula.

Before viewing Figure 6.3, use a different symbol from the geometric-series ratio: ${y=\beta\hbar\omega>0}$ . Predict the small- ${y}$ behavior of both the mean occupation ${n_{\mathrm B}=1/(\mathrm e^y-1)}$ and the dimensionless thermal energy ${y n_{\mathrm B}}$ . Only one of them diverges.

Figure 6.3. The exact Bose mean occupation diverges as dimensionless frequency tends to zero, while the corresponding dimensionless thermal-energy kernel approaches one; both curves are compared with their controlled small- and large-frequency asymptotes.

Panel (a) makes the occupation divergence explicit. Panel (b) multiplies by the energy quantum and shows the finite limit ${y/(\mathrm e^y-1)\to1}$ . The acoustic continuum calculation below consumes the second combination, together with its mode measure. The graph does not assign a Bose distribution to the excluded discrete rigid-translation coordinate; it only controls the ${\omega\to0^+}$ limit of positive-frequency modes.

Using the same positive dimensionless frequency ${y=\beta\hbar\omega>0}$ , the formulas already derived distinguish number, thermal energy and full energy:

(6.2)

$\bar n=\frac{1}{\mathrm e^y-1},\qquad \frac{E_{\mathrm{th}}}{k_{\mathrm B}T}=y\bar n,\qquad \frac{\langle E\rangle}{k_{\mathrm B}T}=y\left(\bar n+\frac12\right), \qquad p_0=1-\mathrm e^{-y}.$

For this comparison, use the figure above.

After reconstructing the normalized distribution, predict what happens to mean occupation and thermal energy when the positive frequency decreases at fixed temperature. Drag either hollow marker on Figure 6.3 from ${y=10}$ down to ${0.01}$ , or click its following number to type a value. The linked markers show mean occupation and thermal energy divided by ${k_{\mathrm B}T}$ ; the readouts also retain the full mean energy and the ground-state probability. The nine displayed probabilities cover occupations 0 through 8; a separate readout gives their sum and the probability in all higher occupations. With ${x=\mathrm e^{-y}}$ and ${p_n=(1-x)x^n}$ , these readouts are $\sum_{n=0}^{8}p_n=1-x^9, \qquad \sum_{n=9}^{\infty}p_n=x^9.$ The displayed bars are not renormalized: a large omitted tail at small positive ${y}$ is part of the distribution, not a normalization failure. Both horizontal axes are logarithmic. Reset returns to ${y=1}$ without clearing a separately pinned inspection point.

The mean number grows while the thermal-energy ratio tends to one. The full mean energy includes zero-point energy and must not be mistaken for the thermal contribution. No state at exactly zero frequency is included, and a mean number larger than one is not a probability larger than one.

Normal-mode energy and the zero-point contribution

Ordinary solid-state physics supplies the normal-mode reduction, but students may have seen it only for a monatomic chain. The needed operation is to write the coupled harmonic equations in mass-weighted coordinates and diagonalize the Hermitian dynamical matrix at each crystal momentum. The next display states the resulting eigenproblem in its multi-atom form; Advanced Appendix A works it through for a diatomic chain.

A crystal has many modes. For a finite periodic harmonic lattice, let ${\mathcal U\cong\mathbb R^{dN_{\mathrm a}N_{\mathrm c}}}$ be its real mass-displacement space, with ${N_{\mathrm a}}$ atoms per cell and ${N_{\mathrm c}}$ cells. Translation symmetry reduces the real equations of motion to matrix-valued functions

$D:\mathrm{BZ}\to\operatorname{Herm}(\mathbb C^{dN_{\mathrm a}}), \qquad D(\mathbf q)\mathbf e_{\lambda}(\mathbf q) =\omega_{\lambda}^2(\mathbf q) \mathbf e_{\lambda}(\mathbf q).$

The branch label ${\lambda}$ and wave vector ${\mathbf q}$ identify a mode; ${\mathbf e_\lambda}$ is its mass-weighted polarization coefficient vector; ${\omega_\lambda}$ is its angular frequency. These are not an occupation.

Before reading Figure 6.4, predict which atoms move in each branch at the zone boundary of a diatomic chain.

Figure 6.4. A diatomic chain with mass ratio two has an acoustic and an optical branch; at the zone center the acoustic pattern is the excluded rigid translation, and at the zone boundary each branch moves one sublattice alone.

For the finite sample, define the positive-frequency index set $\mathcal M_+ =\left\{(\mathbf q,\lambda): \omega_{\mathbf q\lambda}>0\right\}.$ The ${d}$ rigid translations at ${\mathbf q=\mathbf0}$ are exact zero modes of an unpinned crystal. They describe center-of-mass coordinates, not internal harmonic oscillators with the positive-frequency distribution above. Here the center of mass is fixed, so those modes are excluded from ${\mathcal M_+}$ . In Figure 6.4 the zone-center acoustic row is exactly this pattern: both atoms move together, and the mode is excluded. Quantization attaches an integer ${n_{\mathbf q\lambda}}$ to each member of ${\mathcal M_+}$ and gives

(6.3)

$E_{\{n\}} =\sum_{(\mathbf q,\lambda)\in\mathcal M_+} \hbar\omega_{\mathbf q\lambda} \left(n_{\mathbf q\lambda}+\frac12\right),$

Thermal averaging uses the Bose occupation derived above and produces a different mathematical object—a mean over the equilibrium probability distribution:

(6.4)

$U_{\mathrm{ph}} =\sum_{(\mathbf q,\lambda)\in\mathcal M_+} \hbar\omega_{\mathbf q\lambda} \left[n_{\mathrm B}(\omega_{\mathbf q\lambda})+\frac12\right].$

The first expression is the energy of a specified occupation configuration; the second is the equilibrium mean. The formal creation and annihilation operators that implement changes of ${n_{\mathbf q\lambda}}$ are deliberately deferred to Lecture 7. Nothing in this lecture requires them to interpret the energy spacing or mean occupation [Gri07].

Although ${n_{\mathrm B}(\omega)}$ diverges at zero frequency, the thermal energy carried by a positive-frequency oscillator has the finite limit $\lim_{\omega\to0^+} \hbar\omega n_{\mathrm B}(\omega)=k_{\mathrm B}T.$ This limiting statement justifies the lower endpoint of a continuum acoustic integral; it does not reinstate the excluded discrete rigid-translation coordinate as an internal oscillator.

The same temperature, two very different occupation patterns

Use the normalized mean in Example 6.2 to decide whether a single mode is thermally active, and separate its thermal and total energy. Take ${\hbar\omega=20\,\mathrm{meV}}$ and compare ${T=20\,\mathrm K}$ with ${T=600\,\mathrm K}$ . Use the rounded thermal energies ${k_{\mathrm B}T=1.72\,\mathrm{meV}}$ and ${51.7\,\mathrm{meV}}$ , respectively, and assess the corresponding low- and high-temperature estimates. For a same-temperature comparison use the metal in the parabolic-metal estimate, with ${E_{\mathrm F}\simeq6.8\,\mathrm{eV}}$ , at ${600\,\mathrm K}$ .

Evaluate the mean occupation first; multiplying by one quantum gives thermal energy, while adding half a quantum gives the total mean. At ${20\,\mathrm K}$ , ${k_{\mathrm B}T\simeq1.72\,\mathrm{meV}}$ and

$n_{\mathrm B} =\frac1{\mathrm e^{20/1.72}-1} \simeq9.0\times10^{-6}.$

The exponential low-temperature estimate agrees. At ${600\,\mathrm K}$ , ${k_{\mathrm B}T\simeq51.7\,\mathrm{meV}}$ and

$n_{\mathrm B} =\frac1{\mathrm e^{20/51.7}-1} \simeq2.12.$

The leading classical estimate ${k_{\mathrm B}T/(\hbar\omega)\simeq2.59}$ is visibly rough, while including the next ${-1/2}$ term gives ${2.09}$ . At both temperatures the zero-point energy remains ${\hbar\omega/2=10\,\mathrm{meV}}$ .

Using the rounded thermal scales stated here, the thermal energies are ${20n_{\mathrm B}\simeq1.8\times10^{-4}\,\mathrm{meV}}$ and ${42.3\,\mathrm{meV}}$ ; the total means are consequently about ${10.00018\,\mathrm{meV}}$ and ${52.3\,\mathrm{meV}}$ . At the colder temperature the omitted next exponential is of relative order ${10^{-5}}$ . At the hotter temperature the total mean is about 1.2 percent above ${k_{\mathrm B}T}$ , consistent with the positive ${y^2/12}$ correction. These independent limit checks also show why a mean occupation above one is permitted: it averages unrestricted integer occupations.

At the same ${600\,\mathrm K}$ , the metal has ${k_{\mathrm B}T/E_{\mathrm F}\simeq51.7/6800=7.6\times10^{-3}}$ , whereas this phonon has mean occupation ${2.12}$ . Only a narrow fraction near the electronic Fermi boundary is thermally rearranged, while the oscillator is strongly populated. A shared temperature does not create a shared occupation law. To turn this single-mode result into a lattice energy, the next calculation must count the available modes.

Many acoustic modes turn one occupation law into a power law

The single-mode calculation becomes a thermodynamic prediction only after the number of modes near each frequency is counted. The following integral carries that count’s consequence: its temperature scaling and low-frequency boundary supply the physics, while its dimensionless constant is a number that can be looked up afterward. Here the acoustic DOS is supplied, not derived: it replaces the discrete positive-frequency mode sum by a continuum count. This requires temperatures above the lowest positive mode energy divided by Boltzmann’s constant, but well below the acoustic cutoff energy divided by that constant. The detailed inequalities appear in the calculation below.

Determine the low-temperature thermal-energy and heat-capacity scaling for a supplied acoustic DOS, then compare a flat optical branch of frequency ${\omega_0>0}$ and mode density ${\nu_0}$ per physical volume. Keep the oscillator weights from Example 6.2 and change only the spectrum. Consider one acoustic branch in ${d}$ spatial dimensions with linear dispersion ${\omega(\mathbf q)=v_{\mathrm s}|\mathbf q|}$ up to a cutoff ${\omega_{\mathrm D}}$ . The mode density per physical volume has the form

$g_d(\omega)=A_d\frac{\omega^{d-1}}{v_{\mathrm s}^{d}}, \qquad 0<\omega<\omega_{\mathrm D},$

where the geometric constant ${A_d}$ includes the reciprocal-space measure and any declared polarization degeneracy. Assume ${d>0}$ , ${v_{\mathrm s}>0}$ and a temperature low compared with ${\hbar\omega_{\mathrm D}/k_{\mathrm B}}$ , but high enough for the continuum acoustic DOS to apply.

Integrate the energy of one thermally occupied mode against the supplied DOS; for the flat branch, use its discrete mode count. The mode density has units ${{(\text{volume}\times\text{angular frequency})}^{-1}}$ . The thermal part of the energy density is therefore

$u_{\mathrm{th}}(T) =\int_0^{\omega_{\mathrm D}}\mathrm d\omega\, g_d(\omega)\, \frac{\hbar\omega}{\mathrm e^{\beta\hbar\omega}-1}.$

At ${k_{\mathrm B}T\ll\hbar\omega_{\mathrm D}}$ , the scale ${\omega\sim k_{\mathrm B}T/\hbar}$ sets a broad thermal frequency range, not a sharp boundary between occupied and empty modes. Frequencies far above this scale are exponentially suppressed, so extending the upper limit to infinity introduces an exponentially small cutoff correction. With ${x=\hbar\omega/(k_{\mathrm B}T)}$ ,

$\begin{aligned} u_{\mathrm{th}}(T) &\simeq \frac{A_d}{v_{\mathrm s}^{d}} \frac{{(k_{\mathrm B}T)}^{d+1}}{\hbar^{d}} \int_0^\infty\mathrm dx\, \frac{x^d}{\mathrm e^x-1},\\ C_V/V &=\frac{\partial u_{\mathrm{th}}}{\partial T} \propto T^d. \end{aligned}$

The first line is the error-localization checkpoint: the entire temperature dependence is explicit before the dimensionless integral is evaluated. In three dimensions it gives the Debye ${T^3}$ law.

The Jacobian contributes one power of ${T}$ , the DOS contributes ${d-1}$ , and each quantum contributes one: this accounts for ${T^{d+1}}$ . Near zero, the remaining integrand behaves as ${x^{d-1}}$ , so it is integrable for ${d>0}$ ; at infinity it behaves as ${x^d\mathrm e^{-x}}$ . The prefactor has units energy per ${d}$ -dimensional volume, providing a second check independent of the power count.

For the flat optical branch, the integral is replaced by its actual mode count: $u_{\mathrm{opt}}=\nu_0\frac{\hbar\omega_0}{\mathrm e^{x_0}-1}, \qquad x_0=\frac{\hbar\omega_0}{k_{\mathrm B}T}, \qquad c_{\mathrm{opt}}=\nu_0 k_{\mathrm B} \frac{x_0^2\mathrm e^{x_0}}{{(\mathrm e^{x_0}-1)}^2}.$ Differentiating uses ${\mathrm dx_0/\mathrm dT=-x_0/T}$ . For ${x_0\gg1}$ , ${u_{\mathrm{opt}}\sim\nu_0\hbar\omega_0\mathrm e^{-x_0}}$ and ${c_{\mathrm{opt}}\sim\nu_0 k_{\mathrm B}x_0^2\mathrm e^{-x_0}}$ . For ${x_0\ll1}$ , ${c_{\mathrm{opt}}\to\nu_0 k_{\mathrm B}}$ , the classical heat capacity of the counted modes. Both branches exclude temperature-independent zero-point energy from heat capacity. A dispersive gapped branch can add a power-law prefactor to its activation; a flat branch does not determine that prefactor. A finite acoustic sample also requires ${k_{\mathrm B}T}$ above its lowest internal level spacing for the continuum DOS to apply. The criterion below collects precisely these acoustic premises.

Suppose that in ${d}$ spatial dimensions each retained acoustic polarization has ${\omega_{\mathbf q\lambda}\sim c_\lambda|\mathbf q|}$ near ${\mathbf q=\mathbf0}$ , with finite positive sound speed, declared polarization count, and the exact rigid translation excluded from the discrete oscillator set. Then the low-frequency mode density satisfies ${g_d(\omega)\propto\omega^{d-1}}$ , and the convergent continuum limit gives $u_{\mathrm{th}}(T)\propto T^{d+1}, \qquad C_V/V\propto T^d.$ The isolated value ${n_{\mathrm B}(0)}$ remains undefined; convergence follows instead from ${\hbar\omega n_{\mathrm B}(\omega)\to k_{\mathrm B}T}$ together with the mode measure. A gap, quadratic dispersion, or different low-frequency mode density requires a new scaling analysis.

For a finite sample, let ${\omega_{\min}>0}$ be its lowest retained internal mode frequency. The continuum acoustic power law requires a temperature window $\hbar\omega_{\min}\ll k_{\mathrm B}T\ll\hbar\omega_{\mathrm D},$ with the thermally relevant frequencies also lying in the linear-dispersion regime. Below the lower scale, use the discrete mode sum rather than extrapolate the continuum power law to ${T=0}$ at fixed sample size. The excluded rigid translation is not a zero-frequency oscillator to add back into that sum.

A gapped optical branch with ${\omega(\mathbf q)\geq\omega_0>0}$ changes the decisive step: the lower frequency limit no longer reaches zero, and its low-temperature contribution is exponentially suppressed as ${\mathrm e^{-\hbar\omega_0/(k_{\mathrm B}T)}}$ rather than following a power law. The Bose occupation has not changed; the mode density and its lower-frequency boundary have.

Two checks prevent common mistakes. The zero-point term is temperature independent in the harmonic model and therefore contributes no heat capacity. Also, ${T^d}$ is tied to a linear acoustic dispersion in ${d}$ dimensions; a quadratic mode would change the density-of-states power and must be recalculated rather than assigned the Debye exponent.

For a controlled finite-cutoff comparison, give both models the same total mode density ${\nu}$ . Let the acoustic density be ${g_d(\omega)=\nu d\omega^{d-1}/\omega_{\mathrm D}^d}$ on ${0<\omega<\omega_{\mathrm D}}$ , and put every optical mode at ${\omega_0=\omega_{\mathrm D}}$ . The normalization is $\int_0^{\omega_{\mathrm D}}g_d(\omega)\,\mathrm d\omega=\nu.$ Here ${g_d(\omega)\,\mathrm d\omega}$ counts modes per volume in a frequency interval; it is not an occupation probability. Multiply this count by the mean thermal energy of one mode before integrating. With ${\theta=k_{\mathrm B}T/(\hbar\omega_{\mathrm D})}$ and ${s=\omega/\omega_{\mathrm D}}$ , the two factors become $g_d(\omega)\,\mathrm d\omega=\nu d s^{d-1}\,\mathrm ds, \qquad \hbar\omega n_{\mathrm B}(\omega) =\hbar\omega_{\mathrm D}\frac{s}{\mathrm e^{s/\theta}-1}.$ Their product explains the power ${s^d}$ in the acoustic integral. A flat optical branch instead contributes the same energy for every counted mode. The resulting thermal energies are

(6.5)

$\frac{u_{\mathrm{ac}}}{\nu\hbar\omega_{\mathrm D}} =d\int_0^1\frac{s^d}{\mathrm e^{s/\theta}-1}\,\mathrm ds, \qquad \frac{u_{\mathrm{opt}}}{\nu\hbar\omega_{\mathrm D}} =\frac{1}{\mathrm e^{1/\theta}-1}.$

Before changing a control, predict which spectrum loses thermal energy faster on cooling and justify the prediction from its low-frequency modes.

Choose one, two or three dimensions and lower the temperature to test it. Compare the thermal energies of the gapless acoustic spectrum and an equal number of flat optical modes. The acoustic contribution falls as a power at low temperature, while the optical contribution is activated. The curves use the finite cutoff in Equation (6.5), not a low-temperature formula extrapolated to all temperatures. Only thermal energy is plotted; temperature-independent zero-point energy is omitted. The graph and readouts use base-ten logarithms so that exponentially small optical energies remain visible rather than being rounded to zero.

The displayed range is ${0.02\leq\theta\leq1}$ . The acoustic continuum requires a sample whose lowest internal level spacing is well below ${k_{\mathrm B}T}$ , with the rigid translation excluded. These equal-count models compare energy per counted mode; changing dimension does not hold a dimensionful material density fixed across different-dimensional spaces. They isolate the effect of frequency support and are not fits to the silicon experiment or a claim that its entire dispersion is linear.

Figure 6.5 records the same comparison for all three dimensions on one graph, so that the power laws and the activated branch can be read without the controls.

Figure 6.5. With equal mode counts, the thermal energy of a gapless acoustic spectrum falls as a power of temperature in one, two and three dimensions, while a flat optical branch at the cutoff frequency is exponentially activated.

Where the sound approximation meets measured phonons

The thermal power law uses only the long-wavelength part of an acoustic branch. How far can its straight-line dispersion be extrapolated? The conservation laws in Figure 6.2 still apply, but the analyzer geometry drawn there is not the direct-geometry time-of-flight method used for the linked silicon experiment. Figure 6.6 supplies its physical system and measurement bridge.

Figure 6.6. Diamond-cubic silicon and the functional ARCS measurement path for the linked experiment. The cell and sample are at separate scales; sphere radii and the sample drawing are schematic. Colors distinguish basis positions, not elements. The instrument panel explains the readout and is not a hardware reconstruction or scale drawing. [Kim+18, Materials and Methods and Fig. 1 caption; Abe+12, instrument description]

The simulation in Figure 6.7 uses the two-atom primitive cell of diamond-cubic silicon, giving three acoustic and three optical branches. An empirical harmonic force-constant model is calibrated to silicon’s elastic stiffnesses ${C_{11}=165.7}$ , ${C_{12}=63.9}$ , and ${C_{44}=79.6}$ GPa and tabulated zone-center and zone-boundary phonon frequencies. These calibration inputs are not independent predictions. The stiffness tensor fixes long-wavelength sound speeds; atomistic force constants are needed away from that limit. The model starts from the Fan silicon potential and refines its harmonic force constants through six neighbor shells using tabulated silicon frequencies and elasticity benchmarks. At each momentum transfer, the one-phonon intensity uses phonon polarizations, interference between the two silicon sites, Bose occupation at 300 K, and Debye–Waller attenuation. Event-based neutron transport uses ${E_{\mathrm i}=97.5}$ meV and 200 rotations in ${0.5^\circ}$ increments about ${[110]}$ , as reported for the experiment, with nominal ARCS detector tubes and pixels. The simulation includes source and chopper timing, finite sample geometry, elastic scattering and a multiphonon approximation based on the phonon density of states. Detector readout supplies reconstructed energy and momentum. We average accessible reciprocal zones, divide by an independent simulated flat-scattering response, and subtract an independent multiphonon template whose scale is fitted in reconstructed-energy sidebands. This normalization is a model response correction, not measured-vanadium calibration. The map uses 0.5 meV bins and overlapping local neighborhoods of radius ${0.055\,\mathring{\mathrm A}^{-1}}$ in folded momentum space, including symmetry vertices. It shows the expected response under assumed conditions; these do not establish an equivalent experimental duration. For comparison with the reference’s processed spectrum, the positive-energy residual is divided by the phonon-creation factor ${n_{\mathrm B}(E)+1}$ , equivalently multiplied by ${1-\exp[-E/(k_{\mathrm B}T)]}$ at 300 K. This thermal correction is not elastic-background subtraction. The momentum window is a tested teaching choice, not a reported experimental resolution; narrower windows reduce averaging but leave fewer simulated events per bin.

Figure 6.7. Expected simulated silicon neutron spectrum after event reconstruction, simulated-response normalization and multiphonon subtraction. Energy is in meV; relative signal uses signed-log colors, including negative residuals. The U versus K discontinuity is a path break. Elastic events remain. Overlapping momentum cuts are correlated; finer sampling does not improve instrumental resolution. The preceding figure supplies the physical system and measurement principle. No experimental pixels or measured intensity arrays enter the calculation.

The acoustic ridges bend away from their small-wave-number slopes, while optical branches remain at finite energy near the zone center. One sound line cannot describe either the full spectrum or which modes appear strongly to neutrons. The calibrated harmonic simulation supplies the mode spectrum; the virtual instrument and reduction determine the displayed intensity. In HTML, Show harmonic phonon dispersion overlays the same model’s six mode energies: solid lines are acoustic and dashed lines are optical; degenerate branches coincide. The overlay starts off. These are frequencies at the nominal path, not fitted intensity maxima. Finite momentum integration and instrument resolution can shift apparent ridges relative to the lines. Identical finite-volume cuts act on sample, template and normalization events; no image interpolation or gap filling is used. Weak-bin Monte Carlo precision and the integrated elastic intensity remain unresolved, so use this map to identify dispersing ridges rather than to infer quantitative intensities. The model does not predict anharmonic lifetimes or temperature-dependent frequency shifts. The initial azimuth, run-specific detector masks and chopper settings remain assumptions; the sample environment and extinction are incomplete. Frequencies between the fitted symmetry points remain model predictions. The response-weighted zone average is not a periodic neutron structure factor or an exact reproduction of the experimental reduction. For the experimental comparison, open Kim et al., silicon phonon measurements, Fig. 1(C), 300 K [Kim+18, fig. 1(C), p. 1993]. The preceding silicon and ARCS schematic supplies the shared physical and measurement context.

Historically, aluminum measurements also separated longitudinal and transverse phonon groups and showed the limits of a sound-velocity guide [BS55, fig. 3 and discussion, p. 757]. The linked silicon result provides a denser experimental comparison; its intensities also contain material and instrumental effects omitted from the present harmonic material model and approximate virtual instrument.

Diagnosing an excitation from its defining data

The electron, hole, and phonon examples suggest a reusable diagnostic. Before naming a disturbance, state:

the reference state;
the physical addition, removal, or collective displacement;
the energy and momentum of the change;
its charge, spin, and statistics; and
whether its number is constrained in equilibrium and which probe can create or detect it.

Apply two stages. Model existence and classification require (i) the declared model and stable reference state or preparation, (ii) the stated single-change construction and one-excitation family, (iii) the resulting relative energy, momentum, charge, spin, statistics, and equilibrium number constraint, and (iv) the exact or effective status and first omitted interaction or phase boundary. Experimental accessibility and assignment add (v) a specified probe channel, its matrix element or selection rule, and compatible transferred quantum numbers. A spectral line, empty orbital, charge value, or dispersion by itself satisfies neither stage. A dark excitation may satisfy the first stage while failing the second for one probe.

Use Figure 6.8 to apply those five questions to electron, hole, phonon, magnon, and Bogoliubov-quasiparticle candidates. Predict which pairs remain indistinguishable if only electric charge is measured.

Figure 6.8. A comparison schematic classifies electron, hole, phonon, magnon, and Bogoliubov-quasiparticle disturbances by reference state, statistics, charge, spin, number constraint, and representative probe; magnon and Bogoliubov entries are marked as forward references.

The map is useful because several columns are required. In their simplest settings, a phonon and a magnon are both neutral bosonic collective modes; charge and statistics alone cannot distinguish them. Lattice polarization and momentum dependence support a phonon assignment, whereas spin transfer and field dependence support a magnon assignment. Likewise, an electron and a Bogoliubov quasiparticle are both fermionic, but only the latter presupposes a paired reference and mixes electron-addition and electron-removal character.

Use the map actively: choose one neutral sharp mode and one fermionic dispersing feature, cover their names, and infer only what the visible ledger fields warrant. Then name the spin, polarization, or paired-reference evidence needed to recover the hidden labels.

The two rightmost entries are forward references, not compressed derivations. In an ordered magnet, exchange can support quantized collective rotations of the order parameter. The Zeeman coupling

$\hat H_{\mathrm Z}=-\hat{\boldsymbol\mu}\mathbin{\cdot}\mathbf B$

explains why a spin-sensitive probe or field can couple, but Lecture 18 builds the actual spin-wave spectrum. In a conventional paired superconductor, fermionic Bogoliubov quasiparticles occur above an excitation gap and the condensate supports the flux quantum

$\Phi_0=\frac{h}{2e}.$

These observations identify the reference phase; Lecture 20 derives the quasiparticle mixture and gap. Neither forward reference supplies a result used earlier in this lecture.

Counterexample. A scattering experiment observes a sharp, neutral, dispersing line. That evidence does not prove the line is a phonon. A magnon can share all three features. The assignment requires the changed spin or lattice polarization, a field dependence, and the probe coupling. “Sharp dispersing mode” is an observation; “phonon” is a model-supported inference.

The three opening systems can now be compared without flattening their physics.

System and preparation	What the probe changes	What the ideal model supports	First inference that exceeds it
Semiconductor with occupied valence states	Optical absorption produces a conduction electron and a valence hole	Pair energy, charge ledger, and momentum/selection-rule bookkeeping	The absorption threshold equals the independent-particle gap; an exciton may intervene
Silicon-like harmonic lattice	Raman or neutron transfer changes a normal-mode occupation by an allowed integer	Mode frequency, Bose occupation, and lattice-polarization assignment	The line position and width remain temperature independent; anharmonicity produces shifts and decay
Ordered magnet	Spin-sensitive neutron transfer changes a collective spin mode	A magnon assignment when an ordered reference and spin-transfer channel exist	Every neutral magnetic line is a magnon; a quantum-disordered magnet may have no ordered reference for that construction

The rows are intentionally unequal. The electron–hole case requires a constrained charge and energy ledger. The phonon case requires a number law and a distinction between harmonic frequency and anharmonic lifetime. The magnon case is only a diagnostic preview here because its construction requires the ordered magnetic reference developed later. What unifies the rows is the method of inference, not their microscopic dynamics.

For a weak probe and an incoherent mixture of initial states, the contribution from a transition ${i\to f}$ is weighted by ${p_i|M_{fi}|^2}$ , where ${p_i}$ is the initial-state probability and ${M_{fi}=\langle f|\hat O|i\rangle}$ . This comparison holds the compatible energy and momentum transfers and other response factors fixed. Increasing ${p_i}$ strengthens a coupled transition, but cannot compensate for ${M_{fi}=0}$ . This is not a complete intensity formula: one must also sum the allowed channels and account for the instrument response. Temperature can change the initial-state probabilities, while probe polarization can change the coupling. These effects should not be confused with the existence or absence of the excitation itself.

For each of five proposed disturbances, write the reference state and the physical change before stating charge, statistics, or probe. Reject any case in which an unchanged empty state is called a hole or zero chemical potential is called zero energy.

For each of three changes—adding a conduction electron, removing a valence electron, and raising a lattice mode from ${n}$ to ${n+1}$ —state the reference, the operation, the constrained energy change, the charge change, and whether equilibrium fixes the excitation number. Then explain why an unchanged empty conduction state supplies none of those changes. This five-entry ledger is the minimum evidence for assigning an excitation name in a later model.

From a preparation change to an excitation

Lectures 5–6 began with possible one-particle energies and asked what else a physical preparation must supply. The answer developed as two connected chains:

$\text{relative weights}+\text{Pauli alternatives} \Longrightarrow f(\varepsilon) \Longrightarrow g(\varepsilon)f(\varepsilon) \Longrightarrow\mathcal S_{\mathrm F},\mathbf v,m^{*-1},$

and

$\begin{aligned} \text{declared reference} &\Longrightarrow\text{addition/removal or collective displacement} \\ &\Longrightarrow\text{excitation energy and quantum numbers} \\ &\Longrightarrow\text{appropriate occupation and probe}. \end{aligned}$

The first chain distinguishes available states from occupied states. The second distinguishes an unoccupied state from a missing occupied electron and a mode frequency from its mean occupation. Together they establish the joint conclusion of Lectures 5–6: an elementary excitation is a change relative to a specified many-particle reference, characterized jointly by dispersion, statistics, conserved quantum numbers, and occupation constraints. Electron removal changes an occupied level from 1 to 0; creating a mode quantum raises its occupation from ${n}$ to ${n+1}$ . Lecture 7 introduces operators that implement such changes while retaining the reference and the associated energy and charge.

Several boundaries are part of that claim rather than footnotes.

The Fermi and Bose formulas were derived from explicit relative weights, not from an assumed statistical-mechanics course. Density operators and traces remain Lectures 9–10 material.
Band velocity and curvature describe collisionless one-band kinematics. Lifetimes, reservoirs, multiband Hall response, and Berry velocity require additional providers.
A band or mode plot supplies energies and eigenvectors, not occupations, spectral intensities, or a unique experimental assignment.
Hole, magnon, and Bogoliubov language depends on the reference state. It cannot be inferred from an isolated empty curve.

In the larger course, this is the point where a spectrum of stationary states becomes a theory of disturbances in matter. Screening will reorganize electron–hole fluctuations, neutron and optical spectra will be written as transition probabilities, magnetism will supply a reference state for magnons, and superconductivity will redefine the fermionic excitation itself. The recurring discipline is to ask what changed relative to which state before assigning a particle name to a plotted branch.

The forward question is now unavoidable. Electron and phonon examples have been described using occupation labels, but a solid contains many identical particles and modes. How can states, additions, removals, interactions, and observables be written without labeling every particle and repeatedly antisymmetrizing wave functions? Lecture 7 builds that Fock-space language.

To justify an excitation label, name the reference state, the sector-changing operator, the energy and conserved quantum numbers, and the spectral window in which the feature is sharp. An empty state, a missing occupied state, and a phonon number change therefore remain different even when their dispersions look similar. A detector peak supports the label only through its declared coupling channel; the peak is not the definition.

Problems

1. One mode and one missing electron

A phonon mode has ${\hbar\omega=15\,\mathrm{meV}}$ .

Calculate ${n_{\mathrm B}}$ at ${30\,\mathrm K}$ and ${300\,\mathrm K}$ , and compare with the appropriate asymptotic forms.
Include zero-point energy in the mean mode energy.
A separate semiconductor reference has a filled valence band with maximum energy ${\varepsilon_{\max}<\mu}$ and an empty conduction band. Contrast removing a valence electron of energy ${\varepsilon\leq\varepsilon_{\max}}$ with leaving a conduction state of energy ${\varepsilon'>\mu}$ empty. Give the energy and charge change in each case.
Explain why ${\mu_{\mathrm{ph}}=0}$ says nothing about whether the mode energy is zero.

2. Mode heat capacities from one occupation law

For one phonon mode of fixed angular frequency ${\omega>0}$ , define ${x=\hbar\omega/(k_{\mathrm B}T)}$ .

Starting from the mean equilibrium mode energy, differentiate with respect to ${T}$ and show that

$C_\omega=k_{\mathrm B}x^2 \frac{\mathrm e^x}{{(\mathrm e^x-1)}^2}.$

Derive its leading low- and high-temperature limits, and explain why the zero-point term contributes to the mean energy but not to this heat capacity when ${\omega}$ is temperature independent.

State why the high-temperature limit is a property of the occupation statistics rather than evidence that the mode frequency has vanished.
Now let the mode belong to a gapped branch in three dimensions with ${\omega(\mathbf q)=\omega_0+\alpha{|\mathbf q|}^2}$ , ${\alpha>0}$ , one polarization, and the momentum-space state count per volume used for the electron gas in Lecture 5. Show that the mode density per physical volume is ${g(\omega)={(\omega-\omega_0)}^{1/2}/(4\pi^2\alpha^{3/2})}$ for ${\omega>\omega_0}$ and zero otherwise.
For ${k_{\mathrm B}T\ll\hbar\omega_0}$ , replace the Bose occupation by its leading exponential and show that the thermal energy density and the leading heat capacity are $u_{\mathrm{th}}\simeq \frac{\hbar\omega_0}{8\pi^{3/2}\alpha^{3/2}} {\left(\frac{k_{\mathrm B}T}{\hbar}\right)}^{3/2} \mathrm e^{-x_0}\left[1+\frac{3}{2x_0}\right], \qquad c\simeq \frac{k_{\mathrm B}}{8\pi^{3/2}\alpha^{3/2}} {\left(\frac{k_{\mathrm B}T}{\hbar}\right)}^{3/2} x_0^2\,\mathrm e^{-x_0}, \qquad x_0=\frac{\hbar\omega_0}{k_{\mathrm B}T}.$ Identify which step of the flat-branch calculation in the acoustic worked example this dispersion changes, and explain why the activation exponential survives while a power-law prefactor appears.

3. Neutral excitation, charged constituents

A semiconductor reference state has an occupied valence state ${(\varepsilon_{\mathrm v}(\mathbf k_{\mathrm v})<\mu)}$ and an empty conduction state ${(\varepsilon_{\mathrm c}(\mathbf k_{\mathrm c})>\mu)}$ .

Give the constrained excitation energy and charge change for adding the conduction electron.
Give the corresponding quantities for removing the valence electron and describing the change as a hole.
Combine the two operations. Show explicitly that the pair is electrically neutral while its excitation energy is ${\varepsilon_{\mathrm c}(\mathbf k_{\mathrm c})- \varepsilon_{\mathrm v}(\mathbf k_{\mathrm v})}$ .
If the hole momentum is defined as ${\mathbf k_{\mathrm h}=-\mathbf k_{\mathrm v}}$ , state the pair crystal momentum modulo a reciprocal vector. Explain why an unchanged empty conduction state is not itself either member of an excitation pair.

(advanced): From a Local Oscillator to Polarization-Resolved Phonon Bands

The core lecture needed only one positive-frequency mode and a general normal- mode label. This optional appendix changes the physical condition: the mode is now one member of a periodic diatomic lattice, so frequency, crystal momentum, and polarization must be obtained together. Deleting the appendix leaves every core definition, Bose calculation, acoustic scaling result, and problem complete.

Lecture 3 established that translations organize any periodic linear eigenproblem by crystal momentum, and its own appendix (Lecture 3, Advanced Appendix A) already derives the diatomic mass-weighted eigensystem for its translation-band purpose. Per the bounded-recap clause of the appendix policy, this appendix recomputes that eigensystem locally — with an audit route, explicit zone-center polarizations, and the mass-ratio caution — because the excitation ledger built here must consume branch and polarization data in this lecture’s conventions; the earlier appendix remains the provider of the translation-band construction itself. Consider a one-dimensional chain with cell length ${a>0}$ , masses ${M_{\mathrm A},M_{\mathrm B}>0}$ , and identical springs of stiffness ${\kappa>0}$ between neighboring unlike atoms. Let ${u_n,v_n\in\mathbb R}$ be the longitudinal displacements of the A and B atoms in cell ${n}$ . The harmonic equations are

$M_{\mathrm A}\ddot u_n =\kappa(v_n+v_{n-1}-2u_n), \qquad M_{\mathrm B}\ddot v_n =\kappa(u_n+u_{n+1}-2v_n).$

Use the cell-phase ansatz

$u_n(t)=U(k)\mathrm e^{\mathrm i(kna-\omega t)}, \qquad v_n(t)=V(k)\mathrm e^{\mathrm i(kna-\omega t)}, \qquad k\in[-\pi/a,\pi/a).$

The physical amplitudes ${(U,V)}$ do not have the Euclidean inner product appropriate to a Hermitian eigenproblem because the kinetic energy is mass weighted. Define the mass-weighted polarization coefficient

$\boldsymbol\xi(k) =\begin{pmatrix}\sqrt{M_{\mathrm A}}\,U(k)\\ \sqrt{M_{\mathrm B}}\,V(k)\end{pmatrix} \in\mathbb C^2.$

Then the equations of motion become the Hermitian eigenproblem

(6.6)

$D(k)\boldsymbol\xi_\lambda(k) =\omega_\lambda^2(k)\boldsymbol\xi_\lambda(k),$

$D(k)=\kappa \begin{pmatrix} \dfrac{2}{M_{\mathrm A}} & -\dfrac{1+\mathrm e^{-\mathrm i ka}} {\sqrt{M_{\mathrm A}M_{\mathrm B}}}\\[8pt] -\dfrac{1+\mathrm e^{\mathrm i ka}} {\sqrt{M_{\mathrm A}M_{\mathrm B}}} & \dfrac{2}{M_{\mathrm B}} \end{pmatrix}.$

The matrix ${D:[-\pi/a,\pi/a)\to\operatorname{Herm}(2)}$ has units of angular frequency squared. Figure 6.4 is the worked instance of this eigensystem for ${M_{\mathrm B}=2M_{\mathrm A}}$ : both branches over the first zone and the physical displacement patterns at the zone center and the zone boundary. Its eigenvectors are normalized by

$\boldsymbol\xi_\lambda^\dagger(k) \boldsymbol\xi_{\lambda'}(k)=\delta_{\lambda\lambda'},$

which is equivalent in physical coordinates to

$M_{\mathrm A}U_\lambda^*U_{\lambda'} +M_{\mathrm B}V_\lambda^*V_{\lambda'} =\delta_{\lambda\lambda'}.$

Thus the eigenvector is a polarization section over crystal momentum; its components depend on the declared cell-phase basis, while relative physical displacements and spectral projectors are convention-covariant data.

The characteristic polynomial can be audited without guessing a branch:

$\operatorname{Tr}D =2\kappa\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right), \qquad \det D =\frac{4\kappa^2}{M_{\mathrm A}M_{\mathrm B}} \sin^2\!\left(\frac{ka}{2}\right).$

Hence

$\omega_{\pm}^2(k) =\kappa\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right) \pm\kappa \sqrt{ \left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right)^2 -\frac{4}{M_{\mathrm A}M_{\mathrm B}} \sin^2\!\left(\frac{ka}{2}\right)}.$

The minus branch is acoustic and the plus branch optical, but those names must be checked against eigenvectors rather than attached only by energy order. At the zone center,

$\omega_-(0)=0, \qquad \omega_+^2(0) =2\kappa\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right).$

Normalized mass-weighted polarizations may be chosen as

$\boldsymbol\xi_-(0) =\frac1{\sqrt{M_{\mathrm A}+M_{\mathrm B}}} \begin{pmatrix}\sqrt{M_{\mathrm A}}\\ \sqrt{M_{\mathrm B}}\end{pmatrix}, \qquad \boldsymbol\xi_+(0) =\frac1{\sqrt{M_{\mathrm A}+M_{\mathrm B}}} \begin{pmatrix}\sqrt{M_{\mathrm B}}\\ -\sqrt{M_{\mathrm A}}\end{pmatrix}.$

The acoustic vector gives ${U=V}$ , a rigid translation. Its zero frequency is required by translation invariance. Under the core convention this one discrete coordinate is fixed and excluded from ${\mathcal M_+}$ ; substituting it into ${n_{\mathrm B}(\omega)}$ would be a domain error. The optical vector has opposite displacements and satisfies ${M_{\mathrm A}U+M_{\mathrm B}V=0}$ , so the cell center of mass remains fixed.

At the zone boundary, ${1+\mathrm e^{\mathrm i\pi}=0}$ , and the dynamical matrix is diagonal. The two squared frequencies are ${2\kappa/M_{\mathrm A}}$ and ${2\kappa/M_{\mathrm B}}$ ; the corresponding motion resides on one sublattice at a time in this nearest-neighbor model. Which sublattice belongs to the upper energy-ordered branch depends on the mass ratio. This limit prevents the phrase “optical means opposite motion” from being extended beyond the zone-center statement that justified it.

Parameter changes distinguish an overall frequency scale from a change of mode shape. At fixed masses and wave number, multiplying every spring constant by ${s>0}$ multiplies both frequencies by ${\sqrt{s}}$ without changing their polarizations. Multiplying both masses by ${r>0}$ , at fixed springs, instead multiplies the frequencies by ${1/\sqrt{r}}$ and leaves the relative physical displacements unchanged. Changing the mass ratio generally changes both the frequencies and the polarizations. At equal masses the two branches meet at the zone boundary; the degenerate eigenspace does not select a unique pair of polarization vectors. Use fixed reference units when comparing these changes, so that rescaling the axes does not hide a physical frequency shift.

The excitation ledger now has executable entries. Fix the harmonic reference and choose one positive-frequency pair ${(k,\lambda)\in\mathcal M_+}$ . Raising its occupation from ${n_{k\lambda}}$ to ${n_{k\lambda}+1}$ changes the energy by ${\hbar\omega_\lambda(k)}$ , changes crystal momentum by ${\hbar k}$ modulo a reciprocal-lattice momentum, carries no electric charge, and has bosonic occupation statistics. A Raman or neutron matrix element also depends on the polarization vector; the frequency alone cannot predict intensity. Anharmonic frequency shifts and lifetimes remain outside this harmonic appendix.

Set ${M_{\mathrm B}=2M_{\mathrm A}}$ . Calculate both zone-center and zone-boundary frequencies, identify which sublattice moves in each boundary mode, and complete the excitation ledger for adding one quantum to the lower positive-frequency mode at a small nonzero ${k}$ .

References

[Abe+12] Abernathy, D. L., Stone, M. B., Loguillo, M. J., Lucas, M. S., Delaire, O., Tang, X., Lin, J. Y. Y., and Fultz, B. (2012). “Design and operation of the wide angular-range chopper spectrometer ARCS at the spallation neutron source.” Review of Scientific Instruments 83 (1), 015114. https://doi.org/10.1063/1.3680104.

[Bos24] Bose, S. N. (1924). “Plancks gesetz und lichtquantenhypothese.” Zeitschrift für Physik 26 (1), 178–181. https://doi.org/10.1007/BF01327326.

[BS55] Brockhouse, B. N., and Stewart, A. T. (1955). “Scattering of neutrons by phonons in an aluminum single crystal.” Physical Review 100 (2), 756–757. https://doi.org/10.1103/PhysRev.100.756.

[Ein24] Einstein, A. (1924). “Quantentheorie des einatomigen idealen gases.” Sitzungsberichte der Preussischen Akademie der Wissenschaften, Physikalisch-mathematische Klasse, 261–267. http://echo.mpiwg-berlin.mpg.de/MPIWG:DRQK5WYB.

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Lecture 6: Creating Elementary Excitations

6.1 A hole inherits quantum numbers from the occupied reference

6.1.1 The band and the reference state it needs

6.1.2 Hole charge, momentum, and energy

6.1.3 Hole current, hole mass, and probe operations

6.1.4 Excitation as a change relative to a declared reference

6.2 Modes with unconstrained occupation

6.2.1 Unconstrained mode number and Bose weights