The Sommerfeld model of Chapter is a genuine achievement. It
gives the electronic heat capacity, the Pauli susceptibility, the Wiedemann–Franz
ratio, the thermionic emission current and the plasma frequency, all from one
assumption: that the conduction electrons form a degenerate Fermi gas in a box.
Its failures are equally systematic, and they fall into a single class.
It cannot distinguish a metal from an insulator. Every solid in
the model has a partly filled Fermi sphere, so every solid conducts. Yet
room-temperature resistivities run from
1.6×10⁻⁸ Ωm for silver to about
10²⁰ Ωm for fused quartz — a factor of \(10^{28}\).
It gives the wrong sign of the Hall coefficient for beryllium,
zinc, cadmium and aluminium. In the free-electron model
\(R_{H} = -1/ne\) is negative for any density; there is no parameter
that could flip it (the relation in Chapter 6).
It cannot explain the mean free path. At low temperature the
electrons in a pure metal travel \(10^{3}\)–\(10^{5}\) lattice spacings between
collisions. If the ions were scattering centres this would be impossible.
It says nothing about optical absorption edges, about why
silicon is opaque to visible light and transparent in the infrared, or about
the sharp threshold in the transmission spectrum of any insulator.
Every item on that list is a property of the lattice, and the model has
thrown the lattice away. The repair is to put it back.
The potential inside a crystal
The conduction electron sees the ion cores, each of which attracts it, and the
other electrons, which screen and repel. Bundle everything into an effective
one-electron potential \(V(\vb{r})\). Whatever its detailed shape, the arrangement
of the ions guarantees one property exactly:
In one dimension this reads \(V(x+a) = V(x)\), with \(a\) the lattice constant.
The effective potential seen by a
conduction electron along a line of ion cores. It is deepest at each lattice
point and shallowest midway between neighbours, and it repeats with period \(a\).
The free-electron model replaces it by the dashed constant. Everything in this
chapter follows from taking the periodicity seriously, and only the
periodicity is needed — no detail of the well shape enters Bloch's
theorem.
It cannot be solved in closed form for a realistic \(V\). What can be
extracted, exactly and with no approximation whatever, is the form of the
solution. That is Bloch's theorem.
Bloch's theorem
Theorem 7.1. (Bloch, 1928)
The eigenstates of a one-electron Hamiltonian with a periodic potential
\(V(\vb{r}+\Rvec) = V(\vb{r})\) can be chosen in the form
The two statements are equivalent: substituting the relation above into the
left-hand side of the relation above gives
\(u(\vb{r}+\Rvec)e^{i\kvec\cdot\vb{r}}e^{i\kvec\cdot\Rvec}
= u(\vb{r})e^{i\kvec\cdot\vb{r}}e^{i\kvec\cdot\Rvec}\), which is
\(e^{i\kvec\cdot\Rvec}\psi(\vb{r})\).
Proof by translation operators.
Define \(T_{\Rvec}\) by \(T_{\Rvec}f(\vb{r}) = f(\vb{r}+\Rvec)\). Because \(V\) is
periodic and \(\nabla^{2}\) is translation invariant,
\(T_{\Rvec}H\psi = HT_{\Rvec}\psi\), so
Commuting operators share eigenstates, so we may label the eigenstates of \(H\) by
the eigenvalues of \(T_{\Rvec}\). Write \(T_{\Rvec}\psi = c(\Rvec)\psi\). The
composition rule forces \(c(\Rvec)c(\Rvec') = c(\Rvec+\Rvec')\), whose only
solution is an exponential; and normalisation of \(\psi\) over an infinite (or
periodically repeated) crystal forces \(\left|c\right| = 1\). Hence
\(c(\Rvec) = e^{i\kvec\cdot\Rvec}\) for some real vector \(\kvec\), which is exactly
the relation above.
Two immediate consequences deserve to be stated separately, because both are
regularly examined.
What Bloch's theorem does and does not say
\(\left|\psi(\vb{r}+\Rvec)\right| = \left|\psi(\vb{r})\right|\): the
probability density is lattice periodic, even though the
wavefunction itself is not. The electron is not localised near one ion; it
is spread over the whole crystal with the same probability in every cell.
\(\psi(\vb{r}+\Rvec) = \psi(\vb{r})\) is false in general. The
wavefunction picks up the phase \(e^{i\kvec\cdot\Rvec}\), and it is precisely
that phase which carries the label \(\kvec\).
A Bloch state is not an eigenstate of the momentum operator
\(-i\hbar\nabla\), because \(u_{n\kvec}\) contributes. Only for \(V=0\), where
\(u\) is constant, does \(\hbar\kvec\) become true momentum.
The theorem is exact. It uses periodicity and nothing else —
no weak-potential assumption, no tight-binding assumption, no particular
well shape.
The commonest misstatement of Bloch's theorem
Writing the periodicity condition as \(\psi(x+a) = \psi(x)\) instead of
\(\psi(x+a) = e^{ika}\psi(x)\) makes the theorem trivially false and rejects
perfectly good Bloch functions. Test: the state
satisfies \(\psi(x+a) = A e^{i\pi x/a}e^{i\pi}\cos(2\pi x/a + 2\pi) = -\psi(x)\).
That is not equal to \(\psi(x)\) — and it is a legitimate Bloch function
with \(k = \pi/a\), since \(e^{ika} = e^{i\pi} = -1\). The modulus is what repeats,
not the function.
Crystal momentum, and why \(\kvec\) is defined only modulo \(\Gvec\)
The label \(\kvec\) in the relation above enters only through
\(e^{i\kvec\cdot\Rvec}\). If \(\Gvec\) is a reciprocal lattice vector then
\(\Gvec\cdot\Rvec = 2\pi\times\text{integer}\) (the relation in Chapter 1), so
the band energies are periodic functions on the reciprocal lattice. All the
distinct states are contained in one primitive cell of the reciprocal lattice,
and the conventional choice is the first Brillouin zone
(the section in Chapter 1). This is the reduced-zone scheme; the index \(n\)
that then becomes necessary to distinguish the several energies at a given
\(\kvec\) is the band index.
Because \(\hbar\kvec\) is not a momentum but does obey momentum-like conservation
rules — and only modulo \(\hbar\Gvec\), exactly as for phonons
(the relation in Chapter 4) — it is called the crystal momentum.
Apply Born–von Kármán boundary conditions to a crystal of \(N_{1}\times
N_{2}\times N_{3} = N\) primitive cells, exactly as for phonons in
the section in Chapter 4. The allowed wavevectors are
one \(\kvec\) per volume \((2\pi)^{3}/V\) of reciprocal space. The first Brillouin
zone has volume \((2\pi)^{3}/V_{\text{cell}}\), so it holds exactly \(N\) distinct
wavevectors — one for every primitive cell in the crystal, the same counting
that gave \(N\) modes per phonon branch. Including spin:
The counting rule that decides metal or insulator
\begin{equation}
\boxed{\;\text{each band holds } 2N \text{ electrons,}\;}
\end{equation}
where \(N\) is the number of primitive cells. A crystal with an odd number
of electrons per primitive cell must therefore have a half-filled band, and is a
metal. An even number may fill an integral number of bands
completely — and if a gap separates the last filled band from the next empty
one, the solid is an insulator or semiconductor. It is the number of electrons
per primitive cell, not per atom, that matters.
Which of the following can be a Bloch function for a one-dimensional lattice of
period \(a\)?
\(\psi = A\sin(\pi x/a)\)
\(\psi = A\,e^{i\pi x/(2a)}\cos(\pi x/a)\)
\(\psi = A\,e^{i\pi x/a}\cos(2\pi x/a)\)
\(\psi = A\,e^{-x^{2}/a^{2}}\)
Solution. Test each against \(\left|\psi(x+a)\right| = \left|\psi(x)\right|\),
which is necessary, and then check that the ratio is a pure phase independent of
\(x\).
(a) \(\sin(\pi(x+a)/a) = -\sin(\pi x/a)\), so the ratio is \(-1 = e^{i\pi}\),
constant. This is a Bloch function, with \(k = \pi/a\). (It is the
standing wave at the zone boundary; see the section on the Nearly Free Electron Model.)
(b) The prefactor gives \(e^{i\pi/2}\) but the cosine gives \(-1\), so the ratio is
\(-i\) — still a constant phase, so this too is a Bloch function, now with
\(e^{ika} = -i\), i.e. \(k = -\pi/2a\) (equivalently \(3\pi/2a\)).
(c) Ratio \(= e^{i\pi}\times 1 = -1\): a Bloch function with \(k = \pi/a\).
(d) A Gaussian: \(\left|\psi(x+a)\right|\neq\left|\psi(x)\right|\). Not a Bloch
function — it is localised, and a Bloch state never is.
Worked example 7.2.
For a body-centred cubic lattice of cube side \(a\), which of these can serve as
the periodic amplitude \(u_{\kvec}(\vb{r})\) of a Bloch function?
Solution. The requirement is \(f(\vb{r}+\Rvec) = f(\vb{r})\) for every
bcc lattice vector, and it is enough to check the three primitive translations
\(\avec{1} = \tfrac{a}{2}(\uv{x}+\uv{y}-\uv{z})\),
\(\avec{2} = \tfrac{a}{2}(-\uv{x}+\uv{y}+\uv{z})\),
\(\avec{3} = \tfrac{a}{2}(\uv{x}-\uv{y}+\uv{z})\) (the relation in Chapter 1).
Take (a) and translate by \(\avec{1}\). The combination \(x+y\) increases by
\(\tfrac{a}{2}+\tfrac{a}{2} = a\), so its cosine argument increases by \(2\pi\) —
unchanged. The combination \(y+z\) increases by \(\tfrac{a}{2}-\tfrac{a}{2} = 0\),
and so does \(z+x\). Hence \(f(\vb{r}+\avec{1}) = f(\vb{r})\), and by the same
bookkeeping for \(\avec{2}\) and \(\avec{3}\) — in each case exactly one argument
shifts by \(2\pi\) and the other two by zero — (a) is periodic. It is
the answer.
(b) differs only in the factor of two, and that is fatal: under \(\avec{1}\) the
argument \(\pi(x+y)/a\) shifts by \(\pi\), so that cosine changes sign.
(c) fails as well: under \(\avec{1}\), \(x+y+z\) shifts by \(\tfrac{a}{2}\) and its
argument by \(\pi\).
The structural reason is worth more than the arithmetic. A lattice-periodic
function is a Fourier sum over the reciprocal lattice,
\(f = \sum_{\Gvec}f_{\Gvec}e^{i\Gvec\cdot\vb{r}}\), so the allowed cosines are
exactly \(\cos(\Gvec\cdot\vb{r})\). The reciprocal of bcc is fcc, whose shortest
vectors are \(\tfrac{2\pi}{a}(1,1,0)\) and permutations — giving arguments
\(2\pi(x+y)/a\), as in (a). The vectors \(\tfrac{2\pi}{a}(1,1,1)\) used in (c)
belong to the reciprocal lattice of fcc, not bcc. Swapping the two is
the intended trap.
bcc and fcc, the wrong way round
Because the reciprocal of bcc is fcc and the reciprocal of fcc is bcc, the
cosine arguments that work for one structure fail for the other. For a bcc
direct lattice use \((x\pm y)\), \((y\pm z)\), \((z\pm x)\) with prefactor \(2\pi/a\);
for an fcc direct lattice use \((\pm x\pm y\pm z)\) with the same prefactor.
Check by translating through one primitive vector: every argument must shift by
a multiple of \(2\pi\).
Worked example 7.3.
Show that a Bloch state is not an eigenstate of momentum, and identify what it
is an eigenstate of.
Solution. Apply \(\vb{p} = -i\hbar\nabla\) to \(\psi = ue^{i\kvec\cdot\vb{r}}\):
The second term spoils the eigenvalue equation unless \(\nabla u = 0\), i.e.\
unless the potential is constant. So \(\hbar\kvec\) is not the momentum, and a
measurement of momentum on a Bloch state does not return \(\hbar\kvec\) with
certainty: it returns \(\hbar(\kvec+\Gvec)\) with probability
\(\left|u_{\Gvec}\right|^{2}\), where \(u_{\Gvec}\) are the Fourier coefficients of
\(u\).
What the state is an eigenstate of is the translation operator
\(T_{\Rvec}\), with eigenvalue \(e^{i\kvec\cdot\Rvec}\) — which is the content of
the relation above, and the reason \(\kvec\) is a good quantum number at all.
The Kronig–Penney Model
Bloch's theorem fixes the form of the solution but not the spectrum. To
see bands and gaps we need an actual potential, and the realistic one —
overlapping Coulomb wells — cannot be solved analytically. Kronig and Penney
(1931) replaced it by a row of rectangular barriers, which keeps the periodicity
and the essential physics of alternating deep and shallow regions, and which can
be solved exactly.
The model
The Kronig–Penney potential: wells of
width \(a\) and depth zero alternating with barriers of width \(b\) and height
\(V_{0}\), period \(a+b\). The wells stand where the ion cores are, the barriers
between them. Real potentials are not square, but nothing in the band picture
depends on the shape — only on the fact that the potential repeats.
Take
\begin{equation}
V(x) = \begin{cases}
0, & 0 < x < a \quad\text{(region I)},\\
V_{0}, & -b < x < 0 \quad\text{(region II)},
\end{cases}
\qquad V(x + a + b) = V(x),
\end{equation}
and consider an electron with \(0 < E < V_{0}\). In the two regions
the relation above becomes
\(\alpha = \sqrt{2mE}/\hbar\) has dimensions of inverse length. Statements such as
“\(\alpha = \hbar^{2}k^{2}/2m\)” — which appear in more than one set of
circulating notes — are dimensionally impossible. The correct free-electron
statement is \(\alpha = k\), hence \(E = \hbar^{2}k^{2}/2m\). Likewise \(\alpha a\) is
a pure number, which is why every plot in this section uses it as the abscissa.
Matching, and the transcendental equation
Continuity of \(\psi\) and \(\mathrm{d}\psi/\mathrm{d}x\) at \(x=0\) gives two
equations. Two more come from the Bloch condition
the relation above, which relates the solution at \(x=a\) to the solution at
\(x=-b\) one period away:
(Equivalently one may work with the periodic amplitude \(u_{k}\) and demand
\(u_{k}(a) = u_{k}(-b)\) together with \(u_{k}'(a) = u_{k}'(-b)\); the algebra is
longer but the result identical.)
Four homogeneous linear equations in \(A,B,C,D\) have a non-trivial solution only
if the determinant of coefficients vanishes:
Expanding and simplifying gives the central result of the model:
\begin{equation}
\boxed{\;\frac{\beta^{2}-\alpha^{2}}{2\alpha\beta}\,
\sinh(\beta b)\sin(\alpha a) + \cosh(\beta b)\cos(\alpha a)
= \cos k(a+b) . \;}
\end{equation}
The sign in the prefactor
The coefficient is \((\beta^{2}-\alpha^{2})\), not \((\alpha^{2}+\beta^{2})\).
The two agree in the delta-function limit taken below — which is why the error
survives in several sets of notes — but they differ for any finite barrier
width. A three-line check settles it: build the transfer matrix
\(M = M_{\text{II}}M_{\text{I}}\) for one period and use \(\cos k(a+b)
= \tfrac{1}{2}\operatorname{Tr}M\). With \(a=1\), \(b=0.4\), \(V_{0}=8\) and \(E=3\) (in
units \(\hbar = m = 1\)) the trace gives \(-1.2038\), which is what
the relation above returns; the \((\alpha^{2}+\beta^{2})\) version returns
\(-0.3980\).
The delta-function limit
The relation above is exact but opaque. Kronig and Penney's simplification is
to make the barriers infinitely high and infinitesimally thin while keeping
their area fixed:
\begin{equation}
\boxed{\;F(\alpha a) \equiv P\,\frac{\sin\alpha a}{\alpha a}
+ \cos\alpha a = \cos ka . \;}
\end{equation}
\(P\) measures how strongly the lattice binds the electron: large \(P\) means deep,
opaque barriers and nearly isolated atoms; \(P=0\) means no lattice at all. Note
that \(\hbar^{2}\), not \(h^{2}\), appears in the relation above.
Reading the answer off the graph
The right-hand side of the relation above is \(\cos ka\), which for real \(k\) can
only lie between \(-1\) and \(+1\). The left-hand side is a function of energy alone
(through \(\alpha = \sqrt{2mE}/\hbar\)) and oscillates with a slowly decaying
envelope, taking values well outside that range. Energies for which
\(\left|F(\alpha a)\right| > 1\) have no real \(k\) and therefore no propagating
solution: they are forbidden. This is the entire origin of the band gap.
The
left-hand side of the relation above against \(\alpha a \propto \sqrt{E}\), for
\(P = 3\pi/2\). Since the right-hand side is \(\cos ka\), only the shaded strips
where \(\left|F\right|\le1\) correspond to real \(k\) and hence to propagating
states. Every band ends exactly at \(\alpha a = n\pi\), where \(F = \cos n\pi
= \pm1\); the bands widen as the energy rises, because the envelope \(P/\alpha a\)
decays. For this value of \(P\) the first four bands span \(\alpha a = \)
\(2.25\)–\(3.14\), \(4.71\)–\(6.28\), \(7.42\)–\(9.42\) and \(10.28\)–\(12.57\), of widths
\(0.89\), \(1.57\), \(2.01\) and \(2.28\).
Three features of the figure above carry all the physics.
What the Kronig–Penney graph shows
The spectrum is a sequence of allowed bands separated by
forbidden gaps. Nothing was put in by hand: it follows from the boundedness
of \(\cos ka\).
Band edges occur at \(\alpha a = n\pi\), since there
\(\sin\alpha a = 0\) and \(F = \cos n\pi = (-1)^{n}\), exactly at the limit of
the allowed range. Because \(\cos ka = \pm1\) there, the band edges sit at
\(ka = n\pi\), i.e. at Brillouin zone boundaries
\(k = \pm\pi/a, \pm2\pi/a,\ldots\)
Bands widen with increasing energy, because the amplitude of
the \(P\sin(\alpha a)/\alpha a\) term falls off as \(1/\alpha a\); and
bands narrow as \(P\) increases, because a larger \(P\) pushes more of
the curve outside the \(\pm1\) strip. Deeply bound electrons form narrow
bands; weakly bound ones form wide bands.
The two limits
\(P\to\infty\): isolated atoms.
For the equation to have a solution at all, \(\sin(\alpha a)/\alpha a\) must be
driven to zero, so
\begin{equation}
\sin(\alpha a) = 0 \;\Longrightarrow\; \alpha a = n\pi
\;\Longrightarrow\;
E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} = \frac{n^{2}h^{2}}{8ma^{2}} .
\end{equation}
The bands shrink to points and the spectrum becomes that of a particle in an
infinite square well of width \(a\): a line spectrum, one sharp level per
well, which is the electronic structure of isolated atoms.
\(\sin\), not \(\sinh\)
This condition is \(\sin(\alpha a) = 0\). Several circulating solutions write
\(\sinh(\alpha a) = 0\), which has the single root \(\alpha a = 0\) and destroys the
result. There is no hyperbolic function left in the relation above at all —
the \(\sinh\) was absorbed when the barrier was made infinitely thin.
\(P\to0\): free electrons.
The equation becomes \(\cos\alpha a = \cos ka\), so \(\alpha = k\) and
\begin{equation}
E = \frac{\hbar^{2}\alpha^{2}}{2m} = \frac{\hbar^{2}k^{2}}{2m} :
\end{equation}
the free-electron parabola, with no gaps. Between the two limits lies every real
solid. The lattice does not so much create the electronic structure as
interpolate between the atomic limit and the free-electron limit, and the
band width measures where a given material sits on that scale.
Extended, reduced and repeated zone schemes
Solving the relation above for \(k\) at each allowed energy gives \(E(k)\)
directly. Because \(\cos ka\) is unchanged by \(k \to k + 2\pi n/a\) and by
\(k\to-k\), the same energy can be labelled by many wavevectors, and there are
three conventional ways to draw the result.
The exact
Kronig–Penney bands for \(P = \pi/2\), obtained by solving
the relation above for \(k\) at each allowed energy. (a) Extended-zone
scheme: each band occupies its own Brillouin zone, and the curve follows the
free-electron parabola except near the zone boundaries \(k = n\pi/a\), where it
flattens and a gap opens. (b) Reduced-zone scheme: every band is
translated by a reciprocal lattice vector into the first zone, where it
acquires a band index \(n\). The two panels contain identical information —
the relation above guarantees it. Note that successive gaps appear
alternately at the zone boundary and at the zone centre in (b); in (a) they
all sit at zone boundaries.
In the extended-zone scheme the \(n\)th band is drawn in the \(n\)th
Brillouin zone. In the reduced-zone scheme every band is translated by
a suitable \(\Gvec\) into the first zone; this is the standard presentation, and
it is why real band diagrams for silicon or copper show many curves over one
small range of \(k\). In the repeated-zone scheme the whole reduced-zone
picture is copied into every zone, which makes the periodicity
the relation above explicit and is convenient for following an electron
that is accelerated past a zone boundary (the section on semiclassical Dynamics: Velocity, Effective Mass and Holes).
Worked example 7.4.
An electron moves in a one-dimensional lattice of period
\(a = \text{3 Å}\) described by the Kronig–Penney model with
\(P = 3\pi/2\). Find the energy of the bottom and top of the lowest band, and its
width.
Solution. From the figure above the first band runs from
\(\alpha a = 2.2505\) (where \(F = +1\)) to \(\alpha a = \pi\) (where \(F = -1\)). With
\(E = \hbar^{2}\alpha^{2}/2m\),
a band width of 2.03 eV. The band bottom lies above
zero: an electron in a periodic array of barriers has a zero-point energy, just
as in a single well. Note also that the top of the band, \(\alpha a = \pi\), is
exactly the infinite-\(P\) level \(n=1\) of the relation above — the isolated
atomic level always sits at the upper edge of the band it broadens into,
for this potential.
Worked example 7.5.
Show that the number of distinct states in one Kronig–Penney band equals twice
the number of unit cells, and hence that a one-dimensional crystal of monovalent
atoms is always a metal.
Solution. A band spans \(ka\) from \(0\) to \(\pi\) on the positive side and
\(0\) to \(-\pi\) on the negative side, i.e. the full first zone
\(-\pi/a < k \le \pi/a\), of length \(2\pi/a\). With Born–von Kármán conditions
on a chain of \(N\) cells of total length \(L = Na\), the allowed values are
\(k = 2\pi p/L\), spaced \(2\pi/L\) apart (the relation in Chapter 4). The number of them
in one zone is
\begin{equation*}
\frac{2\pi/a}{2\pi/L} = \frac{L}{a} = N .
\end{equation*}
Each accommodates two electrons of opposite spin, so the band holds \(2N\)
electrons, confirming the relation above. With one valence electron per
atom the band is exactly half filled, the Fermi level lies inside a band,
and arbitrarily small fields can excite electrons into nearby empty states: a
metal. Two electrons per atom would fill the band and — provided the gap to
the next band exceeds \(k_{B}T\) — give an insulator.
Worked example 7.6.
In a Kronig–Penney lattice the barrier strength is doubled. What happens to
(i) the band widths, (ii) the gaps, (iii) the effective mass at the bottom of
the lowest band?
Solution. (i) The envelope of \(F\) scales as \(P/\alpha a\), so a larger
\(P\) pushes more of the curve outside \([-1,1]\): every band narrows. (ii)
Since the band edges \(\alpha a = n\pi\) are fixed, narrower bands mean
wider gaps. (iii) A narrower band means a flatter \(E(k)\), hence a smaller
curvature \(\mathrm{d}^{2}E/\mathrm{d}k^{2}\) and, by
the relation above below, a larger effective mass. All three
statements are the same statement: strong binding localises the electron, and a
localised electron is heavy, slow and unable to conduct.
The Nearly Free Electron Model
The Kronig–Penney model is exactly soluble but artificial: no real crystal has
square barriers. What survives when the potential is realistic? The nearly free
electron (NFE) model answers this for the opposite extreme from
the section on the Kronig–Penney Model. Instead of strong barriers we take the periodic
potential to be weak — a perturbation on the free-electron gas of
Chapter — and ask what the smallest amount of periodicity
does. The answer is startling in its economy: a weak potential leaves the
free-electron parabola almost untouched everywhere except at the
Brillouin zone boundaries, where it opens a gap whose size is fixed by a single
Fourier coefficient.
That such a feeble perturbation can have so violent an effect is the signature
of degenerate perturbation theory, and it is worth being clear at the outset
about why. Ordinary perturbation theory shifts a level by
\(|\langle m|U|n\rangle|^{2}/(E_{n}-E_{m})\), which is small when the matrix
element is small. But when two unperturbed states are degenerate, the
denominator vanishes and the argument collapses. In that case the splitting is
first order in \(U\), not second, and no amount of weakening the potential removes
it. Zone boundaries are exactly where the free-electron states become
degenerate.
The empty lattice: bands before there is a potential
Begin with \(U = 0\). The eigenstates are plane waves \(e^{i\vb{q}\cdot\vb{r}}\) with
\(E = \hbar^{2}q^{2}/2m\), and the spectrum is a single parabola — no bands, no
gaps. But we now know from the relation above that the Bloch label \(\kvec\) is
defined only modulo a reciprocal lattice vector. If we insist on labelling every
state by a \(\kvec\) inside the first Brillouin zone, we must write
\(\vb{q} = \kvec + \Gvec\) and record the energy as
one branch for each \(\Gvec_{n}\). This is the empty lattice model: the
free-electron parabola cut at the zone boundaries and folded back into the first
zone. It produces a set of bands with no gaps between them — the branches
simply cross. Nothing physical has happened; we have only relabelled.
Folding is bookkeeping, gaps are physics
The empty-lattice construction shows that multiple branches \(E_{n}(\kvec)\)
are a consequence of periodicity alone, present even when \(U = 0\). What the
potential does is not create the branches — it lifts the degeneracies
where they cross. Bands are geometry; gaps are dynamics.
The crossings occur wherever two folded branches have equal energy,
Setting \(\Gvec = 0\) and writing \(\Gvec' = -\Gvec\) for brevity, this reduces to
\(2\kvec\cdot\Gvec = G^{2}\), i.e. \(\kvec\cdot\uv{G} = G/2\): the plane that
perpendicularly bisects \(\Gvec\). That is precisely the Bragg condition of
the section in Chapter 1 and precisely the Brillouin zone boundary of
the section in Chapter 1. The three statements
Bragg reflection \(\Longleftrightarrow\)
zone boundary \(\Longleftrightarrow\)
degenerate free-electron states
are one statement in three languages. An electron whose wavevector satisfies the
Bragg condition is diffracted by the lattice; a diffracted wave cannot propagate;
a wave that cannot propagate is a standing wave; and standing waves come in pairs
with different energies. The gap follows.
Setting up the perturbation
Expand the periodic potential in a Fourier series over reciprocal lattice
vectors — the only wavevectors allowed by periodicity (the section in Chapter 1):
We may set \(U_{0} = 0\), since a constant merely shifts the zero of energy. If
\(U\) is real then \(U_{-\Gvec} = U_{\Gvec}^{*}\), and if the crystal has a centre of
inversion the \(U_{\Gvec}\) may be taken real.
so \(U\) connects a plane wave only to those differing from it by a
reciprocal lattice vector. This selection rule is the entire content of the
model. Away from a zone boundary, all the states \(\kvec + \Gvec\) that \(U\) can
reach lie far away in energy, the denominators are large, and second-order
perturbation theory gives a tiny shift. Near a zone boundary, one of them is
degenerate with \(\kvec\), and everything changes.
\(U\) does not connect arbitrary wavevectors
It is tempting to think of a weak potential as scattering an electron into any
final state. A periodic potential cannot: the Kronecker delta in
the relation above allows only \(\kvec \to \kvec + \Gvec\). This is why
electrons in a perfect crystal have infinite mean free path — the perfectly
periodic lattice does not scatter Bloch waves at all. Resistivity requires
something that breaks the periodicity: phonons, impurities or defects.
Two levels at the zone boundary
Keep only the two degenerate plane waves. In one dimension with
\(G = 2\pi/a\), the pair is \(e^{ikx}\) and \(e^{i(k-G)x}\), degenerate at
\(k = G/2 = \pi/a\). In this two-dimensional subspace the Hamiltonian is
Exactly on the boundary the two unperturbed energies coincide,
\(E^{0}_{\pi/a} = E^{0}_{-\pi/a} \equiv \lambda\), the square root collapses to
\(|U_{G}|\), and
The gap that opens at the zone boundary associated with \(\Gvec\) equals
twice the modulus of the \(\Gvec\)th Fourier component of the crystal
potential — and depends on nothing else. It does not matter how deep the
potential wells are, only how much of the potential sits at that particular
spatial frequency. In particular, if \(U_{\Gvec} = 0\) for some \(\Gvec\), then
no gap opens at all on that zone face, and the free-electron branches
cross undisturbed.
The factor of two, and where students lose it
For \(U(x) = U_{0}\cos(2\pi x/a)\) the temptation is to say “the potential varies
between \(+U_{0}\) and \(-U_{0}\), so the gap is \(2U_{0}\)”. That is wrong twice
over. Write the cosine in exponentials,
\begin{equation*}
U_{0}\cos\!\left(\frac{2\pi x}{a}\right)
= \frac{U_{0}}{2}\,e^{i G x} + \frac{U_{0}}{2}\,e^{-i G x},
\qquad G = \frac{2\pi}{a},
\end{equation*}
so that \(U_{G} = U_{0}/2\), and the relation above gives \(E_{g} = 2|U_{G}| = U_{0}\).
The peak-to-peak amplitude of the potential and the gap happen to be equal here,
and it is a coincidence of the cosine, not a rule. For a general potential you
must extract the Fourier coefficient first.
Why the two states have different energies
The eigenvectors at \(k = \pi/a\) are the symmetric and antisymmetric combinations
which are shifted from one another by half a period. If the ion cores sit at
\(x = 0, a, 2a, \dots\), then \(|\psi_{-}|^{2}\) piles charge on the cores,
where the potential is deep, and \(|\psi_{+}|^{2}\) piles it between the
cores, where the potential is shallow. The lower state is the one that sits on
the attractive ions. The gap is the difference in electrostatic energy between
these two ways of arranging the same standing wave — a physical statement that
survives long after the perturbation theory that produced it has been forgotten.
At \(k=\pi/a\) the two
Bragg-reflected plane waves combine into standing waves. The cosine solution
concentrates charge on the ion cores and is pushed down; the sine
solution concentrates charge between them and is pushed up. The
splitting is \(2|U_{G}|\).
Near, but not at, the boundary
Write \(k = \pi/a + q\) with \(qa \ll 1\) and expand the relation above. Put
\(\lambda = \hbar^{2}(\pi/a)^{2}/2m\) and note that
\(E^{0}_{k} - E^{0}_{k-G} = 2\hbar^{2}\pi q/(ma) + \mathcal{O}(q^{2})\), so
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