PHYS 635 Solid State Physics

Take home exam 1

Gregory Eremeev

Fall 2004Submitted: November 8, 2004

Problem 1:Ashcroft & Mermin, Ch.10, p.189, prob.2

a) Let’s proveβxx = βyy = βzz = β (1)

βxx = −∫dr ·Ψ∗

x(r) ·Ψx(r) ·∆U(r) = −∫dr · x2 · |φ(r)|2 ·∆U(r)

= −∫dr · y2 · |φ(r)|2 ·∆U(r) = −

∫dr ·Ψ∗

y(r) ·Ψy(r) ·∆U(r) = βyy

(2)

Now0 = βxx − βyy = −

∫dr · (x2 − y2) · |φ(r)|2 ·∆U(r) (3)

If we here make the change of variables: x′ = x− y; y′ = x+ y, which is basically rotation in space, then wewill get

0 = βxx − βyy = −∫dr′ · (x′ · y′) · |φ(r′)|2 ·∆U(r′) = βxy = 0 (4)

b) In the case of a simple cubic Bravais lattice with γij(R) negligible for all but the nearest-neighborR let’s calculate γxy(k). We should take into account 6 neighbors:

R = a(±1, 0, 0); a(0, 0,±1); a(0,±1, 0) (5)

γxy(k) =− eikxa

∫dr · xyφ(r)φ

(((x− a)2 + y2 + z2

) 12)

∆U(r)

− e−ikxa

∫dr · xyφ(r)φ

(((x+ a)2 + y2 + z2

) 12)

∆U(r)

− eikya

∫dr · x(y − a)φ(r)φ

((x2 + (y − a)2 + z2

) 12)

∆U(r)

− e−ikya

∫dr · x(y + a)φ(r)φ

((x2 + (y + a)2 + z2

) 12)

∆U(r)

− eikza

∫dr · xyφ(r)φ

((x2 + y2 + (z − a)2

) 12)

∆U(r)

− e−ikza

∫dr · xyφ(r)φ

((x2 + y2 + (z + a)2

) 12)

∆U(r) = 0

(6)

, because under rotation transformation x→y, y→-x each integral changes sign.

c) The energies are found by setting to zero the determinant:

|(ε(k)− Ep)δij + βij + γij(k)| = 0 (7)

1

First we will prove that

ε(k)− Ep + β + γxx(k) = ε(k)− lon0(k) + 4γ0cos

(12kya

)cos

(12kza

)(8)

, where ε0(k),γ0,γ1 and γ2 are those from A & M. In nearest-neighbor approximation we will have the 12nearest neighbors of the origin at

R =a

2(±1,±1, 0);

a

2(±1, 0,±1);

a

2(0,±1,±1) (9)

In order to prove (6) we should calculate γxx(k) in this approximation:

γxx(k) =∑R

eikRγxx(R) =

− ei a2 (kx+ky)

∫dr · x

(x− a

2

)φ(r)φ

(((x− a

2

)2

+(y − a

2

)2

+ z2

) 12)

∆U(r)

− ei a2 (kx−ky)

∫dr · x

(x− a

2

)φ(r)φ

(((x− a

2

)2

+(y +

a

2

)2

+ z2

) 12)

∆U(r)

− ei a2 (−kx+ky)

∫dr · x

(x+

a

2

)φ(r)φ

(((x+

a

2

)2

+(y − a

2

)2

+ z2

) 12)

∆U(r)

− ei a2 (−kx−ky)

∫dr · x

(x+

a

2

)φ(r)φ

(((x+

a

2

)2

+(y +

a

2

)2

+ z2

) 12)

∆U(r)

− ei a2 (kx+kz)

∫dr · x

(x− a

2

)φ(r)φ

(((x− a

2

)2

+ y2 +(z − a

2

)2) 1

2)

∆U(r)

− ei a2 (kx−kz)

∫dr · x

(x− a

2

)φ(r)φ

(((x− a

2

)2

+ y2 +(z +

a

2

)2) 1

2)

∆U(r)

− ei a2 (−kx+kz)

∫dr · x

(x+

a

2

)φ(r)φ

(((x+

a

2

)2

+ y2 +(z − a

2

)2) 1

2)

∆U(r)

− ei a2 (−kx−kz)

∫dr · x

(x+

a

2

)φ(r)φ

(((x+

a

2

)2

+ y2 +(z +

a

2

)2) 1

2)

∆U(r)

− ei a2 (ky+kz)

∫dr · x2φ(r)φ

((x2 +

(y − a

2

)2

+(z − a

2

)2) 1

2)

∆U(r)

− ei a2 (ky−kz)

∫dr · x2φ(r)φ

((x2 +

(y − a

2

)2

+(z +

a

2

)2) 1

2)

∆U(r)

− ei a2 (−ky+kz)

∫dr · x2φ(r)φ

((x2 +

(y +

a

2

)2

+(z − a

2

)2) 1

2)

∆U(r)

− ei a2 (−ky−kz)

∫dr · x2φ(r)φ

((x2 +

(y +

a

2

)2

+(z +

a

2

)2) 1

2)

∆U(r)

(10)

Notice that

−∫dr · x

(x− a

2

)φ(r)φ

(((x− a

2

)2

+(y − a

2

)2

+ z2

) 12)

∆U(r) = γ2 (11)

−∫dr · x2φ(r)φ

((x2 +

(y +

a

2

)2

+(z +

a

2

)2) 1

2)

∆U(r) = γ0 + γ2 (12)

2

Thus we have:

γxx(k) =(ei a

2 (kx+ky) + ei a2 (kx−ky) + e−i a

2 (kx+ky) + e−i a2 (kx−ky)

+ ei a2 (kx+kz) + ei a

2 (kx−kz) + e−i a2 (kx+kz) + e−i a

2 (kx−kz))γ2

+(ei a

2 (ky+kz) + ei a2 (ky−kz) + e−i a

2 (ky+kz) + e−i a2 (ky−kz)

)(γ2 + γ0)

(13)

And

γxx(k) =(2cos

(a2(kx + ky)

)+ 2cos

(a2(kx − ky)

)+ 2cos

(a2(kx + kz)

)+ 2cos

(a2(kx − kz)

))γ2

+(2cos

(a2(ky + kz)

)+ 2cos

(a2(ky − kz)

))(γ2 + γ0) =(

4cos(

12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

))γ2

+ 4cos(

12kya

)cos

(12kza

)(γ2 + γ0)

(14)

Now let’s get back to (6):

ε(k)− Ep + β + γxx(k) =

ε(k)− Ep + β +(

4cos(

12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

))γ2+

4cos(

12kya

)cos

(12kza

)(γ2 + γ0) =

ε(k)− Ep + β +(4cos

(12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

)+

4cos(

12kya

)cos

(12kza

))γ2 + 4cos

(12kya

)cos

(12kza

)γ0 =

ε(k)− ε0(k) + 4γ0cos

(12kya

)cos

(12kza

)(15)

The same way one can verify other elements of matrix (10.33) in A & M.

d) For k=0 all three bands are degenerate and ε0 = Ep−β−12γ2−4γ0 ⇒ Ep−β = ε0 +12 ·γ2 +4 ·γ0;ΓX - k =

(2πµ

a , 0, 0), where 0 ≤ µ ≤ 1.

e1(µ)e0

= 1 +8 · γ2

e0· (1− cos (π · µ)) (16)

e2(µ)e0

= 1 +8 · γ2 + 4 · γ0

e0· (1− cos (π · µ)) (17)

the energy band (16) is double degenerate; ΓL - k = 2πa (µ, µ, µ), where 0 ≤ µ ≤ 1

2

e1(µ)e0

= 1 +12 · γ2 + 4 · γ0 − 4 · γ1

e0· (sin2 (π · µ)) (18)

the energy band (18) is double degenerate,

e2(µ)e0

= 1 +12 · γ2 + 4 · γ0 + 8 · γ1

e0· (sin2 (π · µ)) (19)

3

Figure 1: Energy bands along ΓX.

Figure 2: Energy bands along ΓL.

Problem 2 Solution was kindly provided by Professor Vinay Ambegaokar

a) Best done in scalar potential gauge:

H =p2

2m+ eφ(~x) =

p2

2m− e ~E · ~x

~p =1i~

[~p, ~H] = e ~E

(20)

b) Best done in the vector potential gauge:

H ′ = e−i~

~E·~xtH expi

~~E · ~xt+ e ~E · ~x =

12m

(~p+e

c~Et)2 (21)

Note:Eigenvalues depend on ~p(t) = ~p0 + e ~Et, ~p = e ~EShrodinger equation in this gauge

i~∂ψ′

∂t= [

12m

(~p+e

c~Et)2 + U(x)]ψ′, (22)

where U(x) is periodic potetial

a(t) ≡∫d3x · e−i~k0·~xψ′(~x, t)

b(t) ≡∫d3x · e−i(~k0− ~K)·~xψ′(~x, t)

(23)

4

Assume that only two Fourier components of U are inportant

U(x) = Ukei ~K·~x + U−ke

−i ~K·~x

i~da

dt=

~2

2m( ~k1 +

e

c~Et)2a+ Ukb

i~db

dt=

~2

2m( ~k0

′+e

c~Et)2b+ U−ka

~k′0 = ~k0 − ~K

(24)

Note:in this problem inversion symmetry U−K = UK is assumed. Time-reversal invariance assures U−K = U∗k

c) Remove the diagonal terms by the transformation

a = eih

R t0 ε(t′)dt′a

b = eih

R t0 ε′(t′)dt′b

(25)

to obtain

i~da

dt= e

ih

R t0 [ε(t′)−ε′(t′)]dt′Uk b

i~db

dt= e−

ih

R t0 [ε(t′)−ε′(t′)]dt′U∗

k a

(26)

The most important time is when the levels cross ~k = −~k′ = ~K2 . The integrand in the expression (20) is then

~2

me~~E · ~K(t− t0). Define ∆ ≡ e

m~E · ~K. Note that ∆ has units of (time)−2. For very weak UK , a = const ≈ 1.

i~db

dt= e−i ∆

2 (t−t0)2e

i∆2 t20U∗

K

i~b = U∗K

∫dt · e−i ∆

2 (t−t0)2× phase

(27)

The transition from b = 0 to finite b occures on the time scale ∆− 12 . In the small UK or strong E limit, this

time is much smaller than the mixing time for a and b, ~|UK | . One can then do the integral over all times

∫ ∞

−∞dt · e−i ∆

2 (t−t0)2

=√−i√

2π∆

(28)

So finally one has

|b|2 = |b|2 =2π|Uk|2

~2∆2(29)

which is the answer given.

d) |b2| � 1 is the condition

|Uk|2 �~2eEK

m(30)

Since K ∼ 1a0a0 is an atomic length and ~2

ma20≈ εF a typical electron energy The condition is

eEa0 �|UK |2

εf(31)

The opposite of the condition for no electrical breakdown.

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