Fall 2013 - Physics 143 bProblem Set 1 Solutions

Shubhayu Chatterjee

1. Griffiths 5.15

In the ground state of the electron gas, all the momentum eigenstates with momentum less than theFermi momentum kF will be filled. Since electrons are spin half particles, there are two electrons(of opposite spin) for each value of allowed momenta. ∗

So, the total energy in the ground state is given by

Etotal = 2Σ|~k|<kFE(~k)

For large lattice sizes, we can use the continuum approximation, where we assume that each allowed

point in ~k space occupies a volume of (2π)3

L3 , and replace the summation by an integral over momenta.More precisely, we can evaluate the total energy as

Etotal = 2Σ|~k|<kFE(~k)

= 2

∫|~k|<kF

d3k

(2π)3/L3E(~k)

= 2L3

(2π)3

∫|~k|<kF

E(~k)

Now, we note that E(~k) = h̄2k2

2m depends only on the magnitude of the vector ~k, we move to sphericalpolar coordinates in the momentum space to simplify evaluating the integral.

Etotal = 2L3

(2π)3

∫ kF

0

dk k2 h̄2k2

2m

∫ π

0

dθ sin(θ)

∫ 2π

0

dφ

= 2L3

(2π)3

h̄2

2m4π

∫ kF

0

dk k4

=L3h̄2k5

F

10π2m

The total number of electrons is simply equal to the number of occupied states. The volume of the

occupied sphere is 43πk

3F , and each allowed value of momenta in ~k space occupies a volume of (2π)3

L3 .Multipying by a factor of 2 for spin,

N = 24

3πk3

F /

((2π)3

L3

)=L3k3

F

3π2

∗The allowed momenta are determined by the boundary conditions - Griffiths uses hard boundary conditions, butI am going to use periodic boundary conditions as done in class. In the end, the static properties turn out to beindependent of the boundary conditions.

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Thus, in the ground state of electron gas, the average energy per electron is

Etotal/N =3h̄2k2

F

10m=

3

5EF

2. Griffiths 5.16

a) From the previous problem, we can find the Fermi momentum kF in terms of number density n† of electrons. Since the number of free electrons in Cu is the same as the number of atoms, this isthe same as the number density of Cu atoms.

n = N/L3 = k3F /3π

2

Hence, we have kF =(3π2n

)1/3, so

EF =h̄2

2mk2F = 1.13× 10−18J = 7.05eV

b) The corresponding velocity can be estimated by vF =√

2EF /m = 1.57× 106m/s .

This is roughly 0.005 times the speed of light, so the non-relativistic approximation should give ussensible results. We could see this in another way - the rest mass energy of the electron (mc2) isaround 511 keV, which is much larger than the maximum kinetic energy of a few electron volts. So,we can treat the problem non-relativistically and still get sensible results.

c) The Fermi temperature is TF = EF /kB ≈ 81800K .

This is much larger than the room temperature, which has very important physical implications.Thermal fluctuations can only excite electrons to states close to the Fermi surface, and the densityof states at the Fermi surface therefore determines a lot of physical properties of metals, such asspecific heat or susceptibility to magnetic fields.

d) The degeneracy pressure of Cu in the electron gas model can be evaluated by using equation 5.46in Griffiths.

P =(3π2)2/3h̄2n5/3

5m= 3.83× 1010N/m2

Note: This is the pressure at zero temperature, since we have only used the ground state energy tocalculate it. But since room temperature is very low compared to the typical temperature scale ofCu (which is TF ), the thermodynamic properties are determined mainly by the ground state, andthis is still a good approximation.

†Griffiths uses ρ for number density

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3. Griffiths 5.17

The (isothermal) bulk modulus is defined as the fractional change of volume, per unit increase ofpressure on a system, keeping the number of particles and temperature constant.

B = −V ∂P∂V|N,T

Equation 5.46 in Griffiths gives us the pressure at zero temperature in terms of density, which wecan easily convert to a function of volume and number of particles.

P =(3π2)2/3h̄2n5/3

5m=

(3π2)2/3h̄2N5/3

5mV −5/3

Hence, the bulk modulus at zero temperature is given by

B = −V(−5

3

(3π2)2/3h̄2N5/3

5mV −8/3

)=

5

3P

Plugging in the value of P from the previous problem, we find that the bulk modulus of Cu is

B = 6.38× 1010N/m2

4. Griffiths 5.18

a) Following Griffiths 5.59 and 5.60, we can use Bloch’s theorem for a particle moving in a periodicpotential to say

ψ(x) = A sin(kx) +B cos(kx), (0 < x < a)

ψ(x+ a) = eiKaψ(x) =⇒ ψ(x) = e−iKa [A sin(k(x+ a)) +B cos(k(x+ a))] , (−a < x < 0)

Using continuity of the wavefunction at x=0, we have

B = e−iKa [A sin(ka) +B cos(ka)] =⇒ A = B

(eiKa − cos(ka)

sin(ka)

)We can use this to write

ψ(x) = B

(eiKa − cos(ka)

sin(ka)

)sin(kx) +B cos(kx)

= C[

sin(kx) + e−iKa ( cos(kx) sin(ka)− sin(kx) cos(ka))]

Here C is a constant in terms of B and ka. By using trigonometric addition formulae, the abovecan be re-written as

ψ(x) = C[

sin(kx) + e−iKa sin(k(a− x))], 0 < x < a

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b) From figure 5.6 in Griffiths, we can see that at the top of a band, one of the following two holds.

1. ka = (2n+ 1)π and cos(Ka) = −1

2. ka = 2nπ and cos(Ka) = 1

If we use either of these pair of conditions in equation 5.60, we see that Bloch’s theorem is automat-ically satisfied for arbitrary coefficients A and B. In other words, equation 5.60 reduces to

ψ(x) = A sin(kx) +B cos(kx), − a < x < 0

This is automatically continuous at x = 0, so imposing continuity at x = 0 does not give us any newcondition.

But even more than that, we actually have the same functional form for both x < 0 and x > 0. Thismeans that unlike the generic case of the delta function potential, the derivative of the wavefunctioncannot have any discontinuity across x = 0. This is only possible if the wavefunction is zero at x = 0(so that it cannot feel the effect of the delta function at all). ‡ Hence, the form of the wavefunctionmust be

ψ(x) = A sin(kx)

‡I argued physically, but you can show that the same answer comes out of explicitly evaluating the discontinuityof the derivative across x = 0

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