phys208_trygve_slides.pdf

7/29/2019 phys208_Trygve_SLIDES.pdf

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1 Harmonic Crystals

To describe a crystal, we start with balls placed in a regular way with

springs holding them in place. A ball moved form its equilibrium posi-tion will experience a force. Since this force is the result of a deviation froman equilibrium position, the Taylor expansion of the potential cannot have alinear term. Also ignoring the non-interesting constant term the potential is

V(qi) =1

2

ij(qj

qj0)

2 +

O (q q0)3 , (1)

where qj0 is the equilibrium position. Thus for q q0 sufficiently small we canapproximate the potential to be the harmonic potential. If we also assumethat all springs are the same, we can write the potential as

V(q) =

2(q q0)2. (2)

1.1 Mono-atomic chain

For simplicity, consider the one-dimensional case. Also we assume that ourchain is infinitely long, so that we wont have to care about boundary effects.


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We take all balls and all springs to be the same. The balls have mass m andthe equilibrium distance between two balls are a. We let ui denote ball isdeviation from its equilibrium position. The Lagrangian of the chain is,

L = T V =j

m2

u2j 2 (uj+1 uj)2

, (3)

which gives the equations of motion:

0 =d

dt

L

ui L

ui = mui + (ui ui1) (ui+1 ui)

= mui + (2ui ui+1 ui1)To solve this set of coupled equations, we try to find a solution on the formun = ue

i(kant). We find that this is a solution provided that and k are

related by

2 =

m

2 eika eika

=2

m(1 cos ka) = 4

msin2

ka

2.

(4)


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The solution un = uei(kant) is a wave with wavenumber k and angular fre-

quency .

Equation (4) is called the dispersion relation of the wave. [...]

1.2 Density of states

To find the density of states, it is easiest to start with a finite chain. But westill do not want to be bothered by boundary effects. To avoid this, we impose

periodic boundary conditions. If the chain has N balls the periodic boundaryconditions tell us that eikan and eika(n+N) describe the same mode. Thus theseN balls can sustain no more than N different modes. eikan = eika(n+N) implieskaN = 2, thus the k-values for these N modes is evenly distributed over ainterval of 2/a. We take this interval to be /a < k < /a. From this

argument we see that independent of then size ofN the density of k-states is(k) =

N a

2. (5)

To find the density of modes , we use

() = 2(k)k. (6)


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The factor 2 is because there are two branches of k. Rearranging (6) andtaking N we find

() = 2(k)

k1 N= 2(k)k

1

(7)

On the branch 0 < k < /a the dispersion relation is = 0 sinka2 , where we

have defined =

4/m. By differentiating it we find

k = 0a2 cos ka2 = a2

20 20 sin2 ka2 = a2

20 2. (8)

Putting everything together, we find that the mode density is

() =2N

1

20 2

. (9)

For 0 can expand the mode density in /0,

() =2N

0

1

1 0

2=

2N

0

1 + O (/0)2 (10)


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which shows that for low frequencies the mode density is essentially constant.When 0 the density diverges.

1.3 Di-atomic chain

We now make a small variation of the case described above. We keep allsprings the same, but we replace every other ball with a ball with a differentmass.Coordinates ui and vi

1 are associated with the balls with mass m1 and

m2, respectively. The Lagrangian for this chain isL = T V =

j

m12

u2j +m22

v2j

2

(uj vj)2 + (uj vj1)2

. (11)

Note that the potential terms count each spring exactly once, as we of courseshould do. The equations of motion are

0 = ddt

Lui

Lui

= m1ui + (2ui vi vi1),

0 =d

dt

L

vi L

vi= m1vi + (2vi ui ui+1).

1We take ui and vi to be adjacent balls. For concreteness let ui be to the left ofvi


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In analogy with the mono-atomic string we look for solutions on the formun = ue

i(kant), vn = vei(kant). Inserting this into the equations of motion

we find:

m12 =

u(2u veika veika) = 21 v

ucos ka

,

m22 =

v(2v ueika ueika) = 2

1 u

vcos ka

.

Eliminating u/v, we find that the dispersion relation for the di-atomic chain

is 1 m1

2

2

1 m2

2

2

= cos2 ka, (12)

or

4 21

m1+

1

m22 +

42

m1m2sin2 ka = 0. (13)

Solving for 2 we find

2 =

1

m1+

1

m2

1

m1+

1

m2

2 4

m1m2sin2 ka

(14)


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Before we analyse this dispersion relation any further, lets see what happensif we put m1 = m2 = m.

2 =2

m1 1 sin

2 ka = m

(1

cos ka) . (15)

Picking the minus-sign this is exactly the dispersion relation for the mono-atomic chain, (4). But what about the plus-sign? ch

Now it is time to take a closer look at equation (14) in the case where themasses are different. To simplify the notation we write the dispersion relation

in terms of 1/ = 1/m1 + 1/m2 and M = m1 + m2,

2 =

1

1 4

Msin2 ka

. (16)

Since < 4M the discriminant is always non-negative, and 1. This isreassuring since then will be real for all combinations of m1 and m2.

If 0 < k /a equation (16) can be expanded to give2 =

1

1 2

Msin2 ka + O sin4 ka

=

1 1 2

M(ka)2 + O (ka)

4

.(17)


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2 Heat capacity in a solid

2.1 The Einstein model

The Einstein model approximates the heat capacity in a solid with the assump-tion that each oscillator is independent and has a certain frequency . Sincewe work in three dimensions a solid consisting of N atoms will have 3N suchoscillators. Since all oscillators are identical, the ratio of number of oscillatorsin the (n + 1)th quantum state to number of oscillators in the nth quantumstate is ch

Nn+1Nn

= eh. (18)

Thus Nn enh. Then the average excitation number of an oscillator is

n = sesh esh . (19)The summation in the denominator is simple since

xs =

1

1

x

. (20)


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Using a simple trick, we can do the summation in the nominator almost justas simply,

sxs = xd

dx xs =x

(1 x

)

2. (21)

Thus the average occupation number is

n = eh

1 eh =1

eh 1 . (22)

The average thermal energy of an oscillator is then

n

h, thus for the whole

sample, containing 3N such oscillators, the thermal energy is

U = 3Nnh = 3N heh 1 (23)

To find the heat capacity we simply differentiate the thermal energy withrespect to temperature,

CV =

U

T

V

= 3NkB

h

kBT

2 eh(eh 1)2

= 3NkB ET

2eE/T

(eE/T

1)2

,

(24)


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where we have defined the Einstein temperature E = h/kB. At low temper-atures, T E we find

CV 3N kB ET 2

e

E/T (25)

which shows that the heat capacity is exponentially decaying when T 0. Athigh temperatures, T E we expand the exponentials to find

CV 3NkB ET 2

1

(1 + E/T 1)2 = 3N kB (26)

which is the result of Dulong and Petit.

2.2 The Debye model

The Debye model is a simple extension of the Einstein model where the as-sumption that all oscillators have the same frequency is removed. Instead weassume that the oscillator frequencies can take any value up to a yet unknowncut-off, D, and that the modes have a quadratic distribution g() = F .


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For each mode the calculation of average occupation number we did for theEinstein model still applies. Thus

n =1

eh 1 .With these assumptions the thermal energy is

U =

D

0

h

eh

1

F 2d. (27)

To fix the value of F we integrate the density of states up to the cut-off andrequire this to be equal to the total number of oscillators,

3N =

D

0

F 2d =1

3F 3D. (28)

Thus the density of states is

g() =9N 2

3D. (29)


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To calculate the heat capacity we differentiate the thermal energy with respectto temperature

CV

= UT

V

=

T

9Nh

3DD0

3

eh 1d

=9Nh2

kBT23D

D0

4eh

(eh 1)2 d

= 9NkB T

D3

D/T

0

x4ex

(ex

1)2

dx,

(30)

where we have defined the Debye temperature, D = hD/kB. At low temper-atures, T D we integrate all the way to infinity to. Then the integral istemperature independent and

CV T3 (31)as T 0 in agreement with experiments. At high energies T D,

CV 9N kB

T

D

3 D/T0

x2dx = 3N kB (32)

just as for the Einstein model.


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3 Electrons in a periodic potential

3.1 Blochs theorem

Blochs theorem states that the eigenstates of the one-electron Hamiltonian

H = h2 2

2m+ U(r), U(r + R) = U(r). (33)

can be written as

nk(r) = eikrunk(r), unk(r + R) = unk(r). (34)

Consider a one-dimensional infinite atomic lattice. The potential set up bythis lattice is naturally periodic, and can be written as

V(x) =

(x xi) =

n=

U(x na), (35)

where a is the lattice constant. U are the independent wells localised aroundeach atom. We now consider the solution on a subset of size L = Na. We


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start by expanding the wave function (x) and the potential U(x) in Fourierseries,

(x) = (kn)eiknx, (36)U(x) =

U(Kn)e

iKnx. (37)

Note that the allowed wave numbers in the two expansion are not the samesince the periodicity of requires (x) = (x + Na) giving

kn =

2n

aN , (38)whereas U has period equal to the lattice spacing, U(x) = U(x + a) giving

Kn =2n

a. (39)

Inserting the expansions into the one-particle Schrodinger equation we find

0 =

h22m

x + U(x)

(x)

=n

h2

2mk2n

(kn)e

iknx +m,n

U(Km)(kn)ei(kn+Km)x.


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Multiplying through by eikix and using orthogonality this simplifies to

h2

2mk2i (ki) +

m

U(Km)(ki Km) = 0. (40)

Note that for a fixed ki equation (40) couples only the Fourier modes with wavenumber ki, ki + K, ki + 2K , . . ., where K is the (uniform) spacing of the modesin the expansion ofU(x). This means that the original eigenvalue problem hasbeen split up into N independent problem, one for each ki in the first Brillouin

zone. For each of these problems ki(x) can be written on the form

ki(x) =m

(ki Km)ei(kiKm)x = eikixm

(ki Km)eiKmx. (41)

Thus we have proved Blochs theorem (in one dimension).

3.2 Tight binding

To study the energy bands arising from the periodic structure in a crystal weknow introduce the approximation of Linear Combination of Atomic Orbits


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(LCAO). The basic assumption is that the solutions (r) to the one atomproblems has small overlap and that the extra potential energy due to thecrystal is small compared to the atomic potential energy. If the Hamiltonianfor one single atom is

Ha(r) = T + U(r), (42)

the Hamiltonian for the crystal is

Hc(r) = T + V(r) = T +

RU(r R). (43)

The solution can now be written as

(r) =R

c(R)(r R), (44)

where (r) is the solution to the single-atom problem. The coefficients c(R)can found by diagonalising the Hamiltonian, but we will rather exploit the factthat the solutions must follow Blochs theorem. Assume therefore that c(R)can be written on the form

c(R) = eikR. (45)


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The energy of a state is

E =

k,E|H|k,E

k,E

|k,E

. (46)

The only problem now is that if there are N possible values for R the numera-tor will contain N3 terms and the denominator N terms. But here comes thetight binding approximation to our rescuesince we assume that the single-atom wave functions have small overlap only the interaction with the nearestneighbours needs to be considered. Before continuing we do yet a simplifica-

tion: we assume that the solutions (r) are spherically symmetric (s-states),such that only the distance to the neighbours come in. The nearest neighbourterms then give

= d3r(r)U(r d)(r) + d

3r(r d)U(r)(r), (47)

where |d| is the distance to the nearest neighbour. Using this the energysimplifies to

E(k) =N E0 N

d

Nd

eikd

N E0

d

deikd, (48)


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where E0 is the single-atom energy and N ( 3N 1)3. As a simple appli-

cation, consider a cubic lattice where there are 6 nearest neighbours. Then

E(k) = E062(cos kxa +cos kya +cos kza) E066+ a2k2, (49)where the last approximation is good only for small kx, ky, kz.

3.3 Weak binding

To learn more about the band structure it is also useful to consider the weakbinding approximation. If there is no potential at all the electrons can bedescribed by plane waves, eikr, with energy E = h2k2/2m. If we contain theelectrons to move in a box of linear size L each of the the momentum will be

quantised,E =

h2

2m

2

L

2(n2x + n

2y + n

2z). (50)

Having periodic boundary conditions we should only consider wave vectorsover an interval 2/a, where a is the lattice spacing. We will take this interval


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to be /a < k/a. Wave vectors outside this interval should be moved in.In a weak potential the picture will be very similar, but close to the atomsin the lattice the potential cannot be completely neglected. The effect is thatfor k close to n/a there will be deviations from the no potential solutionproducing gaps in the energy spectrum.

3.4 Dynamics of electrons in a lattice

We start out with Schrodinger equation, and assume a Bloch-type solution

2

,,k(x) = eikxu,k,

h2

2m

x+ ik

2+ U(r)

u,k(x) = (k)u,k(x) (51)

2For simplicity we work in one dimension only


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Differentiating with respect to k, multiplying from the left by u,k and inte-grating over x we find

h

mu,kih

x

+ hku,kdx + u,kHu,kk

dx

=

u,k

ku,k + u

,ku,k

k

dx,

(52)

where H is the operator on the left-hand side of equation (51). Since is aneigenvalue of H, u,k(H

) = 0 and the equation reduces to

h

m

u,k

ih

x+ hk

u,kdx =

u,k

ku,kdx =

k, (53)

or in terms of ,k(x)

1

m,kih

x,kdx = 1

h

k

. (54)

But the left-hand side is just p/m, thus we have found that the expectationvalue of the velocity of an electron in the lattice is

v = 1h

(k)

k. (55)


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Now we go on to consider the time development of the velocity (for brevitywell denote the velocity v rather then v),

dvi

dt=

d

dt1

h

ki = 1

h

ki

d

dt=

1

h

ki

kj

dkj

dt = 1

h2

2

kikj

d

dt(hkj) (56)

Since hk is something like the momentum of the particle, its time derivativemust be like a force. But then equation (56) looks like Newtons law, but witha somewhat strange mass

ai

= 1m

ij

Fj

,1

m=

2

kikj. (57)

The tensor 1/m is the inverse effective mass. Note that the effective masscan be more or less anything, including a negative number!

4 The Fermi gas model

To describe an electron gas we have to take into account that electrons arefermions. Such particles respect the Pauli principle, i.e. no more than oneparticle can occupy a certain state.


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But there is also another aspect in which the electrons are substantially dif-ferent from classical particlesall electrons look exactly the same. Indeed, iftwo electrons are interchanged this must be considered the same configuration.This means that ifP(r1, r2) is the probability for finding electron 1 in positionr1 and electron 2 in position r2 (and all other quantum numbers identical) wemust have

P(r1, r2) = P(r2, r1). (58)

To write down a wave function that conforms with these two facts the wavefunction have to be antisymmetric under interchange of any two electrons.If we assume the electrons to be non-interaction (independent electron ap-proximation) such a wave function can be built from the single particle wavefunctions,

(x1, x2, . . . , xn) = det

1(x1) 1(x2) 1(xn)2(x1) 2(x2) 2(xn)

... . . .

n(x1) n(x2) n(xn)

. (59)This construction is called a Slater determinant. Interchanging two electronscorresponds to interchanging two rows (or equivalently, two columns), which


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produces the minus-sign we need. Also, if two electrons are in the same state,two rows will be identical and the determinant vanishes.

4.1 Fermi gas at T = 0

Considering a cube with side lengths L and imposing periodic boundary con-ditions the wave function of an electron must satisfy

(x,y,z ) = (x + L,y,z )

(x,y,z ) = (x, y + L, z)

(x,y,z ) = (x,y,z + L)

(60)

Since we use the independent electron approximation, the wave functionsshould simply be plane waves. This means, for instance, that eikxx = eikx(x+L)

which implies that the momentum is quantised, kxL = nx2. Since the mo-mentum is quantised, so is also the energy,

Enxnynz =h

2m

2

L

n2x + n

2y + n

2z

. (61)


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At T = 0 all the lowest states will be occupied by one electron. Since electronscan have spin up or down, there may be two electrons in each momentumstate. Thus if there are N electrons in a solid of volume V, we can find themaximum momentum kF as

N = 2V

d3k

(2)3(kF k) = V

2

kF0

k2dk =V

32k3F. (62)

We could also do the same in terms of energy. Since = h2k2/2m, d =

(h

2

/m)kdk. Let g() be the density of states. Then,

N =

F0

g()d =h2

m

F0

g()kdk. (63)

Comparing with equation (62) we find that

g() =V

22

2m

h2

3/21/2. (64)


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4.2 Fermi gas at T > 0

To the Fermi energy we can associate a temperature, the Fermi temperature

TF = F

kB= h

2k2F2mkB

= (32n)2/3h2mkB

(65)

where n is the electron density. Putting in typical numbers we find that TF isof the order of 103 104 K. Only at this high temperatures will a substantialamount of the electrons be in excited states. At more moderate temperatures

only the electrons which are already close to the Fermi energy can be excited.Electrons with energy far below the Fermi energy cannot be excited since thePauli principle denies them to go to a state which is already filled. This explainswhy the electron contribution to the heat capacity is so smallonly those fewelectrons which are within kBT of the Fermi energy can contribute.To proceed, replace (63) with

N =

0

g()f()d (66)


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where f() is the Fermi distribution function,

f() =1

e() + 1. (67)

Note that at T = 0 the exponential will be 0 or + depending on the sign of . Thus

f(; T = 0) =

1 < 0 >

(68)

To be consistent with equation (63) we must have = F at T = 0.

At T = 0 f() will no longer be a step function, and will no longer be exactlyequal to F. But at T TF will still be close to F and f() will look muchlike a step function, only with a bit smoothening at the step.

To find the value of at finite temperature we must calculate the integral (66).To calculate this integral, lets define () such that

g() =

().

Then

N =

0

g()f() =

0

f() = ()f()|0

0

()f

(69)


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Since () vanishes at = 0 and f() vanishes when the boundaryterm does not contribute. Changing the integration variable to = ( )we find

N = ()

f

d =

()

(e/2 + e/2)2 d. (70)

To proceed, let T TF /kB. Then . We also expand around = 0,

() = (0) +

=0

+2

2

2

2 =0

+ . . .

= (0) +2

2

g

=0

+ . . .

=V

32

2m

h2

3/21 +

3

8

(kBT)2

22 + . . .

Inserting this into our integral, we find

N V32

2m

h2

3/2

d

1

(e/2 + e/2)2 +

3

8

(kBT)2

22

(e/2 + e/2)2

(71)


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Now that we have determined we can go on to compute other quantities.For instance, the energy of the Fermi gas is given by

E =0 g()f()d, (72)

and then the heat capacity can be found simply by differentiating with respectto temperature.

5 p-n junction

If a semiconductor is n-doped an extra energy level per doping atom i addedjust below the conduction band. Correspondingly for a p-doped semiconductoran extra energy level per doping atom is added just above the valence band.

To make a p-n junction we p-dope part of the semiconductor and n-dope theother part. But since thermodynamics require the chemical potential to be thesame in all the sample at equilibrium something must happen close to wherethe p-doping and the n-doping meet. What happens is that in this regionthe electrons reorganise such that the potential on the n-side is lowered. The


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reason this happens is that on the p-side there is surplus of holes, and on then-side there is surplus of electrons. The electrons and holes annihilates. Butsince both the n-side and p-side originally were neutral this means that closeto the interface there will be a positively charged zone on the n-side and anegatively charged zone on the p-side.

But as the positive charge builds up on the n-side and the negative chargebuilds up on the p-side there is created a electric field pushing the electronsback to the n-side and the holes back to the p-side. At some stage this fieldwill get strong enough to stop the electron and holes meeting to annihilate.

To find the electrical potential at which the equilibrium is reached, we mustequate the different currents involved. The process bringing electrons to-wardsthe p-side and holes to-wards the n-side is diffusion. The diffusion current isgiven by Ficks law,

J = Dn (73)where D is the diffusion constant and n is the number density of carriers. Toavoid double-counting we should consider either the electrons or the holes.Choosing the electrons we find that the charge current caused by diffusion is

jdiff = eJ = eDn (74)


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The current caused by the electric field which is build up is

jdrift = E = enE, (75)

where again n is the number density and is the mobility. At equilibrium

0 = jdrift +jdiff = en(r)E + eDn(r) (76)

Assuming that there only is change in the x-direction, the equation becomes

n(x)Ex(x) = D dn(x)dx

, (77)

which gives us the electrical field in terms of the electron distribution:

E(x) =

D

1

n

dn

dx

. (78)

But we are really more interested in the potential,

(x) = () x

Ex(x)dx. (79)


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Since the interface between the n-doped and the p-doped semiconductor isvery depleted of charge carriers, the resistance there will be much larger thanin the rest of the semi conductor. Thus practically all of the potential fall willbe in this region. Thus the potential difference between the p-side and the

n-side is

Vc = () () =

Exdx =D

n()n()

dn

n=

D

log

n()n() . (80)

Far from the interface the number densities are that of p-doped and n-dopedsemiconductors, respectively. Since most of the electrons introduced by theimpurities are free at normal temperatures, n() = ND. In the p-region weuse the semiconductor equation to find n() NA = n2i (T). Here NA andND are the densities of acceptor and donor atoms, respectively. Thus

n(

)

n() =ND

NA

n2i (T).

But to complete this we need to know what D/ is in terms of known quan-tities. To establish the relationship between D and consider a constant Ex


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giving the equation

D dndx

= Exn(x),

which has solution

n(x) eExx/D. (81)But from Boltzmann statistics we have

n(x) eeExx/kBT. (82)

Since these to exponentials must be the same to ensure consistency, we findthat

D

=

kBT

e, (83)

which gives

Vc =

kBT

e log

NDNAn2i (T) . (84)

Now we go on to study the charge density distribution, (x). We assumethat (x) has a constant value in a region of thickness dp on the p-side of the

junction and another constant value in a region of thickness dn on the n-side


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of the junction. Everywhere else we assume (x) = 0. We also assume that allelectrons from the region dn fill all the holes in the region dp. Together withcharge conservation this implies

NAdp = NDdn.From Poissons equation we know that dEx/dx = 4(x). Thus for dp < x dn

(87)

Continuity at x = 0 gives

2eNAd2p = Vc 2eNDd2n.

fi d d d d d N d N


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Defining d = dp + dn, and using dnND + dpNA we can write

dn =NA

NA + NDd, dp =

NDNA + ND

d.

Inserting this for dn and dp we find

Vc = 2e

NAN

2D + NDN

2A

(NA + ND)2

d2 = 2ed2

NANDNA + ND

,

or solving for d,d =

Vc

2e NA + ND

NAND. (88)

If an external voltage over the p-n junction, the resulting current is stronglydependent on the direction of the voltage. Since the voltage drop over the

p-n junction happens over such a small distance the electric field here is verylarge. Therefore very high external fields must be applied to change the driftcurrent appreciably. But the diffusion current can influenced by changing thedensities. That is, if we connect the positive side of a battery to the p-sidewe remove electrons from the depletion zone. Thus we make room for new

l diff h j i D i h h d dd


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electrons to diffuse across the junction. Doing the other way around we addelectrons where they are already abundant, causing less diffusion.

At V = 0 the current over the p-n junction is

j = jdiffjdrift = j0 j0 = 0.If V = 0 jdrift will not be much changed, but in the drift current there will bean extra Boltzmann factor eeV/kBT. Thus the current with an external field is

j = (eeV/kBT 1)j0. (89)At V = 0 we retain j = 0 as we should. At V > 0 there is a current growingexponentially as we increase V, and at large negative values of V we are leftwith only the drift current which is very small.

6 Magnetic properties of solids

If a solid is exposed to a static magnetic field, there will be a response to theapplied field. What this response is, however, is strongly dependent on the ma-terial of which the solid is made. The basic types of responses are diamagnetic,

ti d f ti A di ti t i l d b tti


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paramagnetic and ferromagnetic. A diamagnetic material responds by settingup a weak field opposing the applied field. A paramagnetic material sets up aweak field in the same direction as the applied field, whereas a ferromagneticmaterial sets up a strong field in the same direction as the applied field.

6.1 Diamagnetism

Diamagnetism is present in all materials, but it is so weak that if the material isparamagnetic the diamagnetism will not be seen. But if there are no competingeffects the solid will respond to an external magnetic field by setting up a weakfield in the opposite direction. Diamagnetism can be viewed as a manifestationof Lenzs lawwhen the material is exposed to an external magnetic field, theelectronic orbits adjust so as to set up a magnetic field opposing the appliedfield.

6 1 1 Cl i l l ti


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6.1.1 Classical evaluation

To find an expression for the magnetic susceptibility of a material, consider acharge revolving in a circle inside the solid. (This charge will be a classical

electron). First we have to fix the sign conventions; Assume that the motionis in the xy plane, and let the positive z-direction define the positive revolutiondirection. The angular frequency of the revolution must be treated as a signedquantity according to this definition. In the radial direction we define to-wardsthe origin as the positive direction.

If the particle revolving has charge q, the radius of the circle is a and theangular frequency is 0 we have the current

I =q02

,

which gives rise to a magnetic moment

m =IA

c=

q0a2

2c(90)

If the charged particle has mass m, there must be a radial force Fr = m20a to

keep it in its orbit. Now we apply an external magnetic field H in the positive

z direction This gives rise to an extra radial force Assuming that the particle


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z-direction. This gives rise to an extra radial force. Assuming that the particleis confined to move only in its circle of radius a the magnetic force is

Fm =

qaH

c

, (91)

where is the frequency with the external field present. Since the radial forcenow is modified, but the circle of motion remains the same there must havebeen a frequency shift. Thus = 0. To find the frequency shift we write theequation of motion for the system with the applied magnetic field

m2a = Fr + Fm = m20a qaH

c ,

or

2 20 = qH

mc. (92)

Assuming that the magnetic field is weak, such that 0 we find

= qHmc

+ 0

qH2mc

. (93)

is known as the Larmour frequency. The associated energy shift is

E= h = BH.

= eh2mc is known as the Bohr magneton


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B = eh2mc is known as the Bohr magneton.

In an atom with closed shells all electron spins are paired, so that they sum to0. Therefore there is no permanent magnetic moment. However, if a magneticfield is applied the orbital frequency will be changed by the Lamour frequency.Substituting the Larmour frequency into (90), we find that as a response tothe applied field a magnetic moment

m = q2a2H

4mc2(94)

is induced. So far we have viewed the orbital motion as confined to a plane. Inreal life, however, the electrons move in three dimensions. This means that a2

in reality should be the mean square perpendicular distance from the magneticfield axis through the nucleus to the electron. If H is taken to point in the zdirection, a2 = x2 + y2. This should be related to the mean square radialdistance r2 = x2 + y2 + z2. If the atom is spherical x2 = y2 = z2which gives r2 = 3a2/2. To compare the result with experiments we shouldcalculate the susceptibility = mN/V H, where N is the number of atoms inthe volume V. From (94 we find

= pe2N

6mc2Vr2 (95)

where p is the number of electron in the outer shell 3 Note that is negative


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where p is the number of electron in the outer shell. Note that is negative,as Lenzs law told us it should be.

6.1.2 Quantum evaluation

We apply an external magnetic field B = (0, 0, Bz). Since B = A anrealisation of this field in terms of A is A = (Bzy/2, Bzx/2, 0). If we nowintroduce the minimal coupling,

p p e

c A (96)

3Strictly speaking we should include all electrons, each with its own value for r2i. But

since usually only the outer shell electrons have a sizable contribution to r2, this is a goodapproximation.

in the Hamiltonian we get


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in the Hamiltonian we get

H =(p e

cA)2

2m+ V(r)

= p22m

e2mc

p A + A p + e2

2mc2A2 + V(r)

= H0 e2mc

L B + e2

8mc2(x2 + y2)B2z

= H0 m B + .

(97)

The energy of the magnetic field can also be written

H =1

8

B HdV = 1

8

B (B4M)dV = 1

8

B2(14)dV. (98)

Since diamagnetism is only important when all spins pair to zero, we consider

m = 0. Thus the contribution containing must come from the -term.Thus

= N p2V

e2

8mc2x2 + y2 = pe

2N

24mc2Vr2. (99)

6.2 Paramagnetism


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6.2 Paramagnetism

If an atom has angular momentum (including spin) not equal to zero, it willhave a permanent magnetic moment. An orbital angular momentum L con-

tributes L = BL to the magnetic moment, whereas an unpaired electronspin contributes S = geB/2. ge is called the the gyromagentic ratio of theelectron, and is approximately equal to 2. Coupling the orbital angular mo-mentum to the spin, the total angular momentum J is found. The magneticmoment of the atom is then

m = gJBJ = BL + 2S (100)gJ is called the Lande g-factor.

6.2.1 Classical evaluation

If a magnetic field H is applied the magnetic moment m will contribute Em =m H to the energy. Thus by aligning the magnetic moments to the appliedfield, the magnetic energy will be minimised. However, thermal excitationwill partially destroy the alignment. Neglecting any other effect which might

hinder alignment, the resulting magnetisation of the material is


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hinder alignment, the resulting magnetisation of the material is

M =mN

Vcos

=m

N

Vcos eEmd

eEmd

=mN

V

11

cos emHcos d(cos )11

emHcos d(cos )

=N

HVmH

mH

xexdxmHmH

exdx

=mN

V

emH + emH

emH emH 1

mH

=N

V 1

tanh mH 1

mH

(101)

If mH kBT the susceptibility is

=2mN

3kBT V=

C

T, (102)

where we have defined the Curie constant C = 2mN/2kBT V.


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m / B

6.2.2 Quantum evaluation

Quantum mechanically the magnetic moment of a atom cannot take an arbi-trary value. The projection of J on H can only take 2J + 1 different values.Then the magnetic contribution to the energy is Em = mH = mJgJBH,J < mJ < J, and the magnetisation is

M = NV

JmJ=JgJBmJemJgJBHJmJ=J

emJgJBH= N

VgJBJBJ(y), (103)

where

BJ(y) =

2J + 1

2J tanh

12J + 12J y

1

2J tanh

1 y2J

, (104)

with y = gJBJH. BJ(y) is called the Brillouin function. For large J theBrillouin function asymptotically approaches the classic result.

To take a really simple example, consider an atom with one unpaired electron


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y p p , pand no orbital angular momentum. This atom has J = 1/2, thus mJ can onlytake the values 1/2. Then the magnetisation is

M = NV

12gJB(e

gJBH

eg

JBH

)egJBH + egJBH

=NgJB

2V tanh(gJBH)

. (105)

For gJBH

kBT, we expand the hyperbolic tangent and find the suscepti-

bility to be

=g2J

2BN

2kBT V. (106)

6.3 Ferromagnetism

Some materials can have a net magnetic moment even without an appliedmagnetic field. Furthermore, if an external field is applied, the material in-creases the field strength enormously. For these materials the independentatom approximation we did for the paramagnetic case does not hold. The

auto-magnetisation of the material must be due to interactions between the


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gmagnetic moments of the atoms. In 1928 Heisenberg modelled this by addingat term

Vm =

2J(rij)Si

Sj, (107)

where Si is the spin of atom i and J(rij) describes the interaction strengthbetween atoms i and j. If J > 0 it is favourable for the spins to align, andwe get what we expect for ferromagnetism. If J < 0 we get what is knownas antiferromagnetism. The origin of this interaction is actually not magnetic,but electrostatic. The interaction arise because the charge distribution, as con-

trolled by the electron wave function) must obey the Pauli principle. Assumingthat J is independent of temperature, we see that at some temperature themagnetisation must be destroyed by thermal agitation. This is indeed observedin experiments. The magnetisation vanishes when the temperature passes acritical temperature, Tc. At higher temperature the material is paramagnetic.

Before proceeding with the Heisenberg theory, lets have look at one the Weissmolecular field theory. This is a phenomenological theory which explains manyaspects of ferromagnetism. Assume that there is an effective field BE = EMwhere M is the magnetisation of the solid, acting on the individual magneticmoments. In the paramagnetic phase, T > Tc, the total field is the sum

of the applied field and E. Denoting the paramagnetic susceptibility 0 the


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magnetisation is

M = 0(H + BE) = 0(H + EM) (108)

Thus the normal susceptibility is

=M

H=

01 0E (109)

The temperature dependence of the paramagnetic susceptibility is given byCuries law, 0 = C/T. Thus

=C

T CE =C

T Tc . (110)

Note that at T T+

c . This is typical for phase transitions.Recall that the quantum theory of paramagnetism showed that

M =gJBJN

VBJ(gJBJH). (111)

To describe the ferromagnetic phase, assume that the same equation can be


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used to express the spontaneous magnetisation, Ms in the absence of an appliedfield if we replace H by BE = EMs. Thus

Ms =gJ

B

JN

V BJ(gJBJEMs). (112)

This equation cannot be solved analytically, but using a numeric method wecan solve for Ms.

Returning to the Heisenberg model we can now find approximate values for

the exchange constant by using the results from the Weiss theory. In terms ofthe Heisenberg model the effective field BE must be understood as set up bythe effect of all neighbouring spins. Thus

2j

J(rij)Sj Si = i BE = gbSi BE (113)

Using M gBSjN/V and Tc = EC = 2effN/3kBV we obtainj

J(rij) 3kBTc2S(S+ 1)

. (114)

6.4 The Hall effect


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At the end of this section we will also discuss a completely different aspect ofmagnetism in solids. Consider a slab of a conducting material. Let there be

a static electric field in the positive x-direction and a static magnetic field inthe positive z direction. The charge carriers is then subject to the force

F = q(E + v B). (115)

The applied electric field will give rise to a current in the x-direction. The

charges contributing to this current will be affected by a magnetic force trans-verse to their motion. Since F = qv B, and the sign of v is opposite forpositive and negative charge carriers, the carriers will be drawn in the nega-tive y direction regardless of the sign of their charge. Since we are consideringa finite slab of material the carriers will have to accumulate at the boundary

giving rise to a electric field in either the positive y-direction (positive carriers)or the negative y-direction (negative carriers).

E = Exex + Eyey,

B = Bzez,(116)

and the corresponding forces are


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FE = q(Exex + Eyey)

FB = qvxBzey(117)

To find Ey we equate the forces in the y direction. But we do not want to keepthe velocity of the carriers in our final answer since this is hard to measure.Therefore we expresses the current in terms of the velocity,

j = qnv

and substitute this into the equation. Thus

Ey = vxBz =jxBz

qn. (118)

We also define the Hall coefficient

RH =Ey

jxBz= 1

qn. (119)

Note that RH is only dependent on the sign of the charge carriers and thecarrier density. The former indicates that by measuring the sign of RH we

can decide the sign of the charge carriers. However, measurements show thatl h H ll ffi d h


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some metals have a positive Hall coefficient, indicating positive charge carriers.Since we know that the charge carriers in metals are electrons, this shows thatwe need to refine the model.

phys208_trygve_slides.pdf

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