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) Lets provexx = yy = zz = (1)

xx = dr x(r) x(r) U(r) =

dr x2 |(r)|2 U(r)

= dr y2 |(r)|2 U(r) =

dr y(r) y(r) U(r) = yy

(2)

Now0 = xx yy =

dr (x2 y2) |(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 = xx yy = dr (x y) |(r)|2 U(r) = xy = 0 (4)

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

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

xy(k) = eikxadr xy(r)

(((x a)2 + y2 + z2) 12)U(r)

eikxadr xy(r)

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

) 12)U(r)

eikyadr x(y a)(r)

((x2 + (y a)2 + z2) 12)U(r)

eikyadr x(y + a)(r)

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

) 12)U(r)

eikzadr xy(r)

((x2 + y2 + (z a)2) 12)U(r)

eikzadr xy(r)

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

) 12)U(r) = 0

(6)

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

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

|((k) Ep)ij + ij + ij(k)| = 0 (7)

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First we will prove that

(k) Ep + + xx(k) = (k) lon0(k) + 40cos(12kya

)cos

(12kza

)(8)

, where 0(k),0,1 and 2 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 xx(k) in this approximation:

xx(k) =R

eikRxx(R) =

ei a2 (kx+ky)dr x

(x a

2

)(r)

(((x a

2

)2+(y a

2

)2+ z2

) 12)U(r)

ei a2 (kxky)dr x

(x a

2

)(r)

(((x a

2

)2+(y +

a

2

)2+ z2

) 12)U(r)

ei a2 (kx+ky)dr x

(x+

a

2

)(r)

(((x+

a

2

)2+(y a

2

)2+ z2

) 12)U(r)

ei a2 (kxky)dr x

(x+

a

2

)(r)

(((x+

a

2

)2+(y +

a

2

)2+ z2

) 12)U(r)

ei a2 (kx+kz)dr x

(x a

2

)(r)

(((x a

2

)2+ y2 +

(z a

2

)2) 12)U(r)

ei a2 (kxkz)dr x

(x a

2

)(r)

(((x a

2

)2+ y2 +

(z +

a

2

)2) 12)U(r)

ei a2 (kx+kz)dr x

(x+

a

2

)(r)

(((x+

a

2

)2+ y2 +

(z a

2

)2) 12)U(r)

ei a2 (kxkz)dr x

(x+

a

2

)(r)

(((x+

a

2

)2+ y2 +

(z +

a

2

)2) 12)U(r)

ei a2 (ky+kz)dr x2(r)

((x2 +

(y a

2

)2+(z a

2

)2) 12)U(r)

ei a2 (kykz)dr x2(r)

((x2 +

(y a

2

)2+(z +

a

2

)2) 12)U(r)

ei a2 (ky+kz)dr x2(r)

((x2 +

(y +

a

2

)2+(z a

2

)2) 12)U(r)

ei a2 (kykz)dr x2(r)

((x2 +

(y +

a

2

)2+(z +

a

2

)2) 12)U(r)

(10)

Notice that

dr x

(x a

2

)(r)

(((x a

2

)2+(y a

2

)2+ z2

) 12)U(r) = 2 (11)

dr x2(r)

((x2 +

(y +

a

2

)2+(z +

a

2

)2) 12)U(r) = 0 + 2 (12)

2

Thus we have:

xx(k) =(ei

a2 (kx+ky) + ei

a2 (kxky) + ei

a2 (kx+ky) + ei

a2 (kxky)

+ eia2 (kx+kz) + ei

a2 (kxkz) + ei

a2 (kx+kz) + ei

a2 (kxkz)

)2

+(ei

a2 (ky+kz) + ei

a2 (kykz) + ei

a2 (ky+kz) + ei

a2 (kykz)

)(2 + 0)

(13)

And

xx(k) =(2cos

(a2(kx + ky)

)+ 2cos

(a2(kx ky)

)+ 2cos

(a2(kx + kz)

)+ 2cos

(a2(kx kz)

))2

+(2cos

(a2(ky + kz)

)+ 2cos

(a2(ky kz)

))(2 + 0) =(

4cos(12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

))2

+ 4cos(12kya

)cos

(12kza

)(2 + 0)

(14)

Now lets get back to (6):

(k) Ep + + xx(k) =

(k) Ep + +(4cos

(12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

))2+

4cos(12kya

)cos

(12kza

)(2 + 0) =

(k) Ep + +(4cos

(12kxa

)cos

(12kya

)+ 4cos

(12kxa

)cos

(12kza

)+

4cos(12kya

)cos

(12kza

))2 + 4cos

(12kya

)cos

(12kza

)0 =

(k) 0(k) + 40cos(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 0 = Ep12240 Ep = 0+12 2+4 0;X - k =

(2pia , 0, 0

), where 0 1.

e1()e0

= 1 +8 2e0

(1 cos (pi )) (16)

e2()e0

= 1 +8 2 + 4 0

e0 (1 cos (pi )) (17)

the energy band (16) is double degenerate; L - k = 2pia (, , ), where 0 12e1()e0

= 1 +12 2 + 4 0 4 1

e0 (sin2 (pi )) (18)

the energy band (18) is double degenerate,

e2()e0

= 1 +12 2 + 4 0 + 8 1

e0 (sin2 (pi )) (19)

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Figure 1: Energy bands along X.

Figure 2: Energy bands along L.

Problem 2 Solution was kindly provided by Professor Vinay Ambegaokar

a) Best done in scalar potential gauge:

H =p2

2m+ e(~x) =

p2

2m e ~E ~x

~p =1i~[~p, ~H] = e ~E

(20)

b) Best done in the vector potential gauge:

H = ei~~E~xtH exp

i

~~E ~xt+ e ~E ~x = 1

2m(~p+

e

c~Et)2 (21)

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

i~

t= [

12m

(~p+e

c~Et)2 + U(x)], (22)

where U(x) is periodic potetial

a(t) d3x ei~k0~x(~x, t)

b(t) d3x ei(~k0 ~K)~x(~x, t)

(23)

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Assume that only two Fourier components of U are inportant

U(x) = Ukei~K~x + Ukei

~K~x

i~da

dt=

~2

2m( ~k1 +

e

c~Et)2a+ Ukb

i~db

dt=

~2

2m( ~k0

+e

c~Et)2b+ Uka

~k0 = ~k0 ~K

(24)

Note:in this problem inversion symmetry UK = UK is assumed. Time-reversal invariance assures UK = Uk

c) Remove the diagonal terms by the transformation

a = eih

R t0 (t

)dta

b = eih

R t0

(t)dtb(25)

to obtain

i~da

dt= e

ih

R t0 [(t

)(t)]dtUk b

i~db

dt= e

ih

R t0 [(t

)(t)]dtUk a(26)

The most important time is when the levels cross ~k = ~k = ~K2 . The integrand in the expression (20) is then~2m

e~~E ~K(t t0). Define em ~E ~K. Note that has units of (time)2. For very weak UK , a = const 1.

i~db

dt= ei

2 (tt0)2e

i2 t

20UK

i~b = UKdt ei2 (tt0)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 ei2 (tt0)2 = i

2pi

(28)

So finally one has

|b|2 = |b|2 = 2pi|Uk|2

~22(29)

which is the answer given.

d) |b2| 1 is the condition|Uk|2 ~

2eEK

m(30)

Since K 1a0 a0 is an atomic length and ~2

ma20 F a typical electron energy The condition is

eEa0 |UK |2

f(31)

The opposite of the condition for no electrical breakdown.

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