VI. Electrons in a Periodic Potential A. Energy Bands and Energy Gaps in a Periodic Potential B. Metals, Insulators, and Semiconductors C. Energy Bands and Fermi Surfaces in 2-D and 3-D Systems D. Blochs Theorem E. Using Blochs Theorem: The Kronig-Penney Model F. Empty Lattice Bands and Simple Metals G. Density of States for a Periodic Potential H. E(k) and N(E) for d-electron Metals I. Dynamics of Bloch Electrons in a Periodic Potential J. Effective Mass of Electrons K. Electrons and Holes A. Energy Bands and Energy Gaps in a Periodic Potential Recall the electrostatic potential energy in a crystalline solid along a line passing through a line of atoms: bare ions solid Along a line parallel to this but running between atoms, the divergences of the periodic potential energy are softened: U x A simple 1D mathematical model that 2x U ( x) U 0 U1 cos captures the periodicity of such a potential is:

a U 0 U1 0 Electron Wavefunctions in a Periodic Potential Consider the following cases: U1 0 Wavefunctions are plane waves and energy bands are parabolic: U1 0 k a U1 0 k a U1 0 k a Ae 2k 2 E 2m i ( kx t ) Electrons wavelengths much larger than a, so wavefunctions and energy bands are nearly the same as above Electrons wavelengths approach a, so waves begin to be strongly back-scattered : Aei ( kx t ) Be i ( kx t ) BA Electrons waves are strongly back-scattered (Bragg scattering) so standing waves are formed: C e i ( kx t ) e i ( kx t ) 12 A e ikx e ikx e it

Electron Energies in a Periodic Potential There are two such standing waves possible: A e e 12 A eikx e ikx e it 12 2 A cos(kx)e it 12 ikx e ikx i t 1 2 2iA sin(kx)e it * 2 A2 cos 2 ( ax ) * 2 A2 sin 2 ( ax ) These two approximate solutions to the S. E. at k a have very different potential energies. * has its peaks at x = a, 2a, 3a, at the positions of

the atoms, where U is at its minimum (low energy wavefunction). The other solution, * has its peaks at x = a/2, 3a/2, 5a/2, at positions in between atoms, where U is at its maximum (high energy wavefunction). We can do an approximate calculation of the energy difference between these two states as follows. Letting U0 = 0 for simplicity, and remembering U1 < 0: a a E E U ( x) * * dx 2 A2 U1 cos( 2ax ) cos2 ( ax ) sin 2 ( ax ) dx x 0 x 0 A 1 a cos 2ax Origin of the Energy Gap a E E 2

a 2 2x U cos 1 ( a )dx x 0 Now use the identity: cos 2 x 12 1 cos 2 x a E E U1 a U 4x a 4x a 1 cos( ) dx

x sin( a a 4 a ) 0 1 U1 E g energy gap x 0 E In between the two energies there are no allowed energies; i.e., an energy gap exists. We can sketch these 1-D results schematically: The periodic potential U(x) splits the free-electron E(k) into energy bands separated by gaps at each BZ boundary. E- Eg E+ a

kx a Different Representations of E(k) If we apply periodic boundary conditions to the 1-D crystal, the energy bands are invariant under a reciprocal lattice translation vector: E ( k G ) E ( k ) 2n G a i The bands can be graphically displayed in either the (i) extended zone scheme; (ii) periodic zone scheme; or (iii) reduced zone scheme. (i) G1 extended zone scheme: plot E(k) from k = 0 through all possible BZs (bold curve) (ii) periodic zone scheme: redraw E(k) in each zone and superimpose (iii) reduced zone scheme: all states with |k| > /a are translated back into 1st BZ B. Metals, Insulators, and Semiconductors It is easy to show that the number of k

values in each BZ is just N, the number of primitive unit cells in the sample. Thus, each band can be occupied by 2N electrons due to their spin degeneracy. E A monovalent element with one atom per primitive cell has only 1 valence electron per primitive cell and thus N electrons in the lowest energy band. This band will only be half-filled. The Fermi energy is the energy dividing the occupied and unoccupied states, as shown for a monovalent element. EEg E+ kx EF a a Metals, Insulators, and Semiconductors For reasons that will be explained more fully later: Metals are solids with incompletely filled energy bands Semiconductors and insulators have a completely filled or empty bands and an energy gap separating the highest filled and lowest unfilled band. Semiconductors have a small energy gap (Eg < 2.0 eV). Quick quiz: Does this mean a divalent

element will always be an insulator? Answer: In 1-D, yes, but not necessarily in 2-D or 3-D! Bands along different directions in k-space can overlap, so that electrons can partially occupy both of the overlapping bands and thus form a metal. But it is true that only crystals with an even number of valence electrons in a primitive cell can be insulators. C. Energy Bands and Fermi Surfaces in 2-D and 3-D Lattices Lets analyze a simple square lattice of atoms with interatomic distance a. Its reciprocal lattice will also be square, with reciprocal lattice base vector of length 2/a. The Brillouin zones of the reciprocal lattice can be identified with a simple construction: Now consider what happens if we have a weak periodic potential and have atoms with 1, 2, and 3 valence electrons per atom. A truly free electron system would have a Fermi circle to define the locus of states at the Fermi energy. Lets see how the shape of the Fermi circle is distorted by the periodic petential. 3 3

2 3 3 2 2 1 3 3 2 3 3 2 /a 2 /a 2-D Energy Bands and Fermi Circles This figure shows the Fermi circles corresponding to 2-D crystals with one, two and 3 valence electrons per atom. Note how the divalent crystal has a Fermi circle that cuts across the 1st BZ boundary (in red).

The different Fermi circles correspond to systems with successively higher EF values, as shown in this 2D band diagram. EF3 EF2 EF1 Shape of 2-D Energy Bands and Fermi Circles This 2D band diagram shows how a weak periodic potential introduces gaps in the free-electron bands at the BZ boundary. Here, bands are shown along the [10] and [11] directions in k-space. EF2 Thus, there is a similar discontinuity introduced into a free-electron Fermi circle any time it crosses a BZ boundary. This is illustrated for a 2-D divalent crystal: Note that the Fermi circle does not completely fill the 1st BZ, so some electrons are in the 2nd BZ. Schematic Shape of a 3-D Fermi Surface In 3D crystals the periodic potential distorts the shape of a Fermi sphere in the vicinity of the BZ boundary. A schematic example for a simple

cubic lattice and a crude model E(k) function is shown here: Note that the Fermi circle does not completely fill the 1st BZ but makes contact with the 1st BZ boundary along the [100] directions. Shape of 3-D Energy Bands in a Real Metal In 3D the energy bands are plotted along the major symmetry directions in the 1st BZ. Many of the high symmetry points on the 1st BZ boundary are labeled by letters. Free-electron bands in an fcc crystal The gamma point ( ) is always the zone center, where k = 0. Electron bands in Al Shape of 3-D Fermi Surface in a Real Metal Even if the free-electron Fermi sphere does not intersect a BZ boundary, its shape can still be affected at points close to the boundary where the energy bands begin to deviate from the free-electron parabolic shape. This is the case with Cu. Just a slightly perturbed free-electron sphere! D. Blochs Theorem and Bloch Wavefunctions This theorem is one of the most important formal results in all of solid state physics because it tells us the mathematical form of an electron wavefunction in the presence of a periodic potential energy. We will prove the 1-D version, which is known as Floquets theorem.

What exactly did Felix Bloch prove in 1928? In the independent-electron approximation, the timeindependent Schrodinger equation for an electron in a periodic potential is: where the potential energy is invariant under alattice translation vector T : Bloch showed that the solutions to the SE are the product of a plane wave and a function with the periodicity of the lattice: U (r T ) U (r ) ik r k (r ) uk (r )e Bloch functions 2 2 U ( r ) E 2m

and where T ua vb wc uk (r T ) uk(r ) Heisenberg and Bloch Who says physicists dont have any fun? Proof of Blochs Theorem in 1-D 1. First notice that Blochs theorem implies: ik r ik T ik r ik T ik T k (r T ) uk (r T )e e u k (r )e e k (r )e

Or just: ik T k (r T ) k (r )e It is easy to show that this equation formally implies Blochs theorem, so if we can prove it we will have proven Blochs theorem. 2. Prove the statement shown above in 1-D: Consider N identical lattice points around a circular ring, each separated by a distance a. Our task is to prove: 2 1 N ( x a) ( x)eika Built into the ring model is the periodic boundary condition: ( x Na ) ( x) The symmetry of the ring implies that we can find a solution to the wave equation: ( x a) C ( x) 3

Proof of Blochs Theorem in 1-D: Conclusion If we apply this translation N times we will return to the initial atom position: And has the most This requires C N 1 general solution: Or: C e 2ni / N e ika ( x Na ) C N ( x) ( x) C N e 2ni n 0, 1, 2,... Where we define the Bloch wavevector: k Now that we know C we can rewrite It is not hard to generalize this to 3-D: ( x a) C ( x) eika ( x) Q.E.D. ik T k (r T ) k (r )e

ik r k (r ) uk (r )e But what do these functions look like? 2n Na Bloch Wavefunctions This result gives evidence to support the nearly-free electron approximation, in which the periodic potential is assumed to have a very small effect on the plane-wave character of a free electron wavefunction. It also explains why the free-electron gas model is so successful for the simple metals! A Remarkable Consequence of Blochs Theorem: In the case of lattice vibrations in a crystal, we can calculate the propagation speed (group velocity) of a wave pulse from a knowledge of the dispersion relation: group velocity: (1-D) d vg dk (3-D) v g (k ) k (k ) Similarly, it can be shown using Blochs theorem that the propagation speed of an electron wavepacket in a periodic crystal can be calculated from a knowledge of the energy band along that direction in reciprocal space:

d 1 dE electron velocity: (1-D) v g dk dk 1 (3-D) v g (k ) k E (k ) This means that an electron (with a specified wavevector) moves through a perfect periodic lattice with a constant velocity; i.e., it moves without being scattered or in any way having its velocity affected! But waitdoes Blochs theorem prove too much? If this result is true, then what is the origin of electrical resistivity in a crystal? E. Using Blochs Theorem: The Krnig-Penney Model Blochs theorem, along with the use of periodic boundary conditions, allows us to calculate (in principle) the energy bands of electrons in a crystal if we know the potential energy function experienced by the electron. This was first done for a simple finite square well potential model by Krnig and Penney in 1931: U U0 x -b 0 a a+b 2a+b 2(a+b) We can solve the SE in each region of space: 0

Each atom is represented by a finite square well of width a and depth U0. The atomic spacing is a+b. 2 d 2 U ( x ) E 2 2m dx ix ix 2 2 E 2m Qx Qx 2Q 2 U0 E 2m I ( x) Ae Be

II ( x) Ce De Now do you remember how to proceed? Boundary Conditions and Blochs Theorem The solutions of the SE require that the wavefunction and its derivative be continuous across the potential boundaries. Thus, at the two boundaries (which are infinitely repeated): x=0 x=a I ( x) Ae ix Be ix II ( x) Ce Qx De Qx i ( A B) Q(C D) (2) A B C D (1) Ae ia Be ia II (a) Now using Blochs theorem for a periodic potential with period a+b: II (a ) II ( b)eik ( a b ) k = Bloch wavevector Now we can write the boundary conditions at x = a: Aeia Be ia (Ce Qb De Qb )eik ( a b )

(3) i ( Aeia Be ia ) Q(Ce Qb DeQb )eik ( a b ) (4) The four simultaneous equations (1-4) can be written compactly in matrix form Results of the Krnig-Penney Model 1 i eia ia ie 1 i e ia ie ia 1 Q e Qbe ik ( a b ) Qe Qb eik ( a b ) 1 Q eQbe ik ( a b ) Qb ik ( a b )

Qe e A B 0 C D Since the values of a and b are inputs to the model, and Q depends on U 0 and the energy E, we can solve this system of equations to find the energy E at any specified value of the Bloch wavevector k. What is the easiest way to do this? Taking the determinant and setting it equal to zero gives: Q2 2 sin a sinh Qb cosa cosh Qb cos k (a b) 2Q Numerical Solution of the Krnig-Penney Model Using the values of a, b, and U0 we can construct the energy bands as follows: 1. Choose a value of k 2. Solve for the allowed energies E1(k), E2(k), E3(k) numerically 3. Choose a new value of k 4. Repeat step 2 2 1 2m Using atomic units where , length is measured in Bohr radii (a 0 = 0.529) and energy has units of Rydbergs (1 Ry = 13.6 eV), I chose arbitrary values of: a = 5.0 a0 b = 0.5 a0

U0 = 5.0 Ry and obtained the following set of energy bands: Numerical Results of the Krnig-Penney Model The 1st BZ has a maximum k value given by: k max You can see the same qualitative 1-D band structure we deduced from the approximate band gap calculation earlier! rad 0.57 a b 5.5a0 a0 F. Empty Lattice Bands and Simple Metals We expect that for a very weak periodic potential the energy bands of a solid should look like those of a free-electron gas system. The limit of a vanishing potential is called the empty lattice, and the empty-lattice bands are often plotted for comparison with the energy bands of real solids. But how are the empty lattice bands calculated? It is desirable to plot them in the reduced zone scheme, so we need to represent all of the translations of the fully periodic E(k) back into the 1st BZ. The symmetry of the reciprocal lattice requires: So to represent all of the periodic bands in the reduced zone scheme:

E (k ) E (k1 G ) Where k1 is a wavevector lying in the 1st BZ. 2 2 E (k ) E (k1 G ) k1 G 2m The sign is redundant, and Myers chooses to use only the minus sign. In principle all reciprocal lattice vectors G will be represented, but electrons can only propagate when the G vector satisfies the Bragg condition (i.e.,only those G that give nonzero structure factors). Empty Lattice Bands for bcc Lattice Myers illustrates this procedure for the bcc lattice by plotting the empty lattice bands along the [100] direction in reciprocal space. (For HW you will do this for the fcc lattice and the [111] direction) The general reciprocal lattice translation vector: In our analysis of the structure factor, we used a simple cubic lattice, for which the reciprocal lattice is also simple cubic: Ghkl hA kB lC 2 2 2 A x B

y C z a a a And thus the general reciprocal lattice translation vector is: 2 2 2 Ghkl h x k y l z a a a We write the reciprocal lattice vectors that lie in the 1st BZ as: 2 2 2 k1 x x y y z z a a a The maximum value(s) of x, y, and z depend on the reciprocal lattice type and the direction within the 1st BZ, as we can see.. k Values in 1st BZ for bcc Lattice

2 2 2 k1 x x y y z z a a a We write the reciprocal lattice vectors that lie in the 1st BZ as: The maximum value(s) of x, y, and z depend on the reciprocal lattice type and the direction within the 1st BZ. For example: Remember that the reciprocal lattice for a bcc direct lattice is fcc! Here is a top view, from the + kz direction: 2 a H kx [100] 0

0

2 2 2 2 E x E0 x 2m a {G} = {110} (110) (1 1 0) (101) (10 1 ) ( 1 1 0) ( 1 10) ( 1 0 1 ) ( 1 01) (011) (01 1 ) (0 1 1 ) (0 1 1) {G} = {200} (200) ( 2 00) (020) (0 20) (002) (002 ) E E x 1 1 E x 1 1 E E x 1 1 E x 2 E E x 2 E E x 2 E E x 2 E x 4 2 2 E E0 x 1 12 E0 x 1 1 2 2

2 0 0 2 2 2 2 0 0 2 0 2 0 2 0 2 2 0

Empty Lattice Bands for bcc Lattice: Results 2 E ( k ) E0 x h k 2 l 2 Thus the lowest energy empty lattice energy bands for the bcc lattice are: 6 5 4 Series1 E/E0 Series2 Series3 3 Series4 Series6 Series7 Series8 2 1 0 0 0.2

0.4 0.6 0.8 x These are identical to the bands on p. 196 of Myers, except one higher lying band is missing from my plot. 1 G. Density of States for a Periodic Potential A weak periodic potential only perturbs the freeelectron energy bands in the vicinity of the Brillouin zone boundaries, so the density of states N(E) closely resembles the ideal free-electron form: 2k 2 E (k ) 2m N ( E ) CE 1/ 2 However, it is clear that the perturbation of free-electron bands near the BZ boundaries will introduce sharp features in the N(E) where the gradient of E(k) goes to zero. Thus we can schematically plot N(E) for the simple metals: N (E) for a 3-D system

2V 8 3 dS k E H. Energy Bands and N(E) for Transition Metals One new feature arises in the transition metalsthe appearance of a series of partially filled d-electron bands. The sbands overlap the d-bands in the first row transition metals, leading to band hybridization, as we can see in the energy bands of the transition metals: For elements with the same type of crystal lattice, an increasing Z is accompanied by an increase in the Fermi energy, populating more and more of the d-electron states in the lattice. (Note the series V Cr Fe and also the series Co Ni Cu) N(E) for Transition Metals The d-bands are much narrower than s- and p-bands due to the greater localization of the d-electrons. As a result, the transition metals with partially filled d-bands have a much higher N(EF) than the simple metals. This means that these transition metals: are often very chemically reactive (especially with O) make good catalysts (supply electrons to molecules that stick to metal surface) have very large heat capacities and magnetic susceptibilities often display superconductivity I. Dynamics of Bloch Electrons in an E Field

Earlier we noted that metals have partially-filled upper bands, while semiconductors and insulators have completely filled upper bands. What is the qualitative difference between filled and partially-filled bands? For a single electron: j nev For a collection of electrons: 1 For each electron: v k E (k ) But the symmetry of the energy bands requires: J ne v (k ) k E ( k ) E ( k ) Thus we conclude: v ( k ) v(k ) So for a filled band, which has J ne v (k ) 0 an equal number of electrons 1 stBZ

with k positive and negative, Filled energy bands carry no current! We will see that this is true even when an electric field is applied. Note: the electrons in filled bands are not stationarythere are just the same number moving in each direction, so the net current is zero. Dynamics of Bloch Electrons in an E Field The same argument that we made for filled bands also applies to a partially filled band in the absence of an electric field. However, when an external electric field is applied to a periodic solid, the electrons all experience a force F that causes a change in their k values: The rate at which each electron absorbs dW F dr dE (k ) F v g energy from the field is the absorbed power: dt dt dt

1 Substituting for the group velocity and dE (k ) dk F k E (k ) k E (k ) using the chain rule, we have: dt dt dk By comparing both sides, we have: The acceleration theorem F dt dp c F We could go on to write: where pc k crystal momentum dt But rememberpc is not the electrons momentum, because F is only the external force and does not include the force of the lattice on the moving electron. The Bottom Line Now we see that the external electric field causes a change in the k vectors of all electrons: dk F

eE dt E a v dk eE dt If the electrons are in a partially filled band, this will break the symmetry of electron states in the 1st BZ and produce a net current. But if they are in a filled band, even though all electrons change k vectors, the symmetry remains, so J = 0. kx When an electron reaches the 1st BZ edge (at k = / a) it immediately reappears at the opposite edge (k = -/a) and continues to increase its k value. kx As an electrons k value increases, its velocity increases, then decreases to zero and then becomes negative when it re-emerges at k = -/a!!

a Thus, an AC current is predicted to result from a DC field! (Bloch oscillations) Do We Ever Observe This? Not until fairly recently, due to the effect of collisions on electrons in a periodic but vibrating lattice. However In both experiments the periodic potential was fabricated in an artificial way to minimize the effect of collisions and make it possible to observe the Bloch oscillations of electrons (or atoms!). But in Reality Bloch oscillations are not routinely observed because the electrons in a periodic system undergo collisions with ions in the lattice much too frequently. In practice this occurs on the time scale of the collision time (10-14 s). Lets analyze the effect of an external electric field in this more realistic case: dk eE k dt

eE x k x ky The steady-state situation can be represented (for a nearly free electron metal) by a very small shift in the Fermi sphere: This leads to a net current, since there is no longer perfect cancellation of terms corresponding to k values, as when E = 0. kx J x ne v x (k ) 1 stBZ We can also write this as: kx Ex Jx = density of e uncompensated e- J x N ( EF )E e vFx

velocity of uncompensated eE energy shift of Fermi sphere Nearly-free Electron Conductivity We can simplify this expression through the chain rule: Now using the approximation of a free-electron band E(k): dE k x J x N ( EF )E e vFx evFx N ( EF ) dk x EF eEx J x evFx N ( EF ) vFx k x evFx2 N ( EF ) F J x e 2 vFx2 F N ( EF ) E x For a spherical Fermi surface, we can write Which then yields: vF2 vFx2 vFy2 vFz2 3vFx2 1 J x e 2 vF2 F N ( EF ) E x E x 3 1 vFx2 vF2 3 1

e 2 vF2 F N ( EF ) 3 This result reduces to the free-electron gas expression (see right) when the FEG value of N(EF) is substituted, but it is more general and highlights the importance of the N(EF) and F in determining the conductivity of a metal. ne 2 m J. Band Effective Mass of an Electron We can use our previous results to write the equation of motion of a Bloch electron in 1-D: dv dv dk ax x x x dt dk x dt 2 This gives: a x And we can write: dvx dvg d 1 dE 1 d 2 E

dk x dk x dk x dk x dk x2 Also, from the acceleration theorem: dk x Fx dt 1 d E Fx 2 dk x Or, in the form of Newtons 2nd law: This allows us to define a band effective mass 2 Fx 2 a x d E dk x2 2 m* 2 d E dk x2 or 1 1 d 2E m * 2 dk x2

By contrast to the free-electron mass, the band effective mass varies depending on the electrons energy and thus its location in the band! Physical Meaning of the Band Effective Mass 2 k x2 Of course, for a free electron, E 2m In a 3-D solid we would find that m* is a second-order tensor with 9 components: 2 and m* 2 m m 1 1 2 E 2 m * ki k j i , j x, y , z The effective mass concept if useful because it allows us to retain the notion of a freeelectron even when we have a periodic potential, as long as we use m* to account for the effect of the lattice on the acceleration of the electron. But what does it mean to have a varying effective mass for different materials? Physical Meaning of the Band Effective Mass The effective mass is inversely proportional to the curvature of the energy band. Near the bottom of a nearly-free electron band m* is approximately constant, but it increases dramatically near the inflection point and even becomes negative (!) near the zone edge.

K. Electrons and Holes Remember that the Hall Effect gives us a way to measure the sign and concentration of the charge carriers in a metal. If electrons are the charge carriers, we expect the Hall coefficient RH = 1/ne to be negative. However, for some elements (Zn, Cd, Bi) R H is positive! To understand this behavior, first consider the excitation of an electron from a filled band into an unfilled band: By the band symmetry, k-j = -kj This is what happens in a semiconductor. The electron in the upper band can now conduct electric current under a field. But how does the lower band now behave? When the lower band was filled, we had a net electron wavevector of zero: N / 2 i 1 ki 0 Electrons and Holes N / 2 This can also be written:

ki k j 0 i j N / 2 ki k j k j i j So when the state +j is empty in a band, the (incomplete) band has effective wavevector k -j. Now we analyze the current flow in the incomplete band under the influence of a field E: N / 2 i 1 N / 2 evi evi evj 0 i j ev j

This shows that an incomplete band (with state +j empty) behaves just like a positive charge moving with the same velocity an electron would have in that state. Thus the properties of all of the remaining electrons in the incomplete band are equivalent to those of the vacant state j if the vacant state has: a. A k-vector k-j b. A velocity v+j c. A positive charge +e We call this vacant state a positive hole (h +) Dynamics of Electrons and Holes If this hole is accelerated in an applied electric field: The corresponding equation for the electron is: But earlier we deduced that the hole velocity is the same as that of the corresponding missing electron: So by equating the derivatives we find: eE eE mh me

dvh mh eE dt dve me eE dt vh ve mh me However, note that near the top of a band the band curvature is negative, so the effective electron mass is also negative. The corresponding hole mass is then positive! So the equation of motion of a hole in an electromagnetic field is: This explains why dk h F e E vh B some metals have dt positive R ! H Dynamics of Electrons and Holes in K-Space Remember that in a metal with the Fermi energy near the top of the band, we can EITHER describe the properties of the electrons OR the properties of the hole(s). The hole has no tangible existence apart from the electrons in the band. The motion of electrons and holes in k-space is tricky to understand, because a hole has the opposite wavevector as

the missing electron. So we need to use the following notation to describe hole motion in the presence of an electric field: position of electron in k-space position of hole at ke (site of hole) position of hole at kh So we see that electrons and holes in the same band move rigidly in the same direction in k-space without altering their relative position--like beads on a string.