Homework #9
1. Frequency dependence of the electrical conductivity.
Use the equation
m( dv / dt + v / τ ) = −eE for the electrical drift velocity v to show that the
conductivity at frequency ω is
⎛ 1 + iωτ ⎞
σ (ω ) = σ (0)⎜⎜
⎟,
2 ⎟
⎝ 1 + (ωτ ) ⎠
where σ (0) = ne 2τ / m .
Solution:
For E = E0 e −iωt , v = v0 e − iωt . Then m( dv / dt + v / τ ) = −eE becomes
eE
eτ E
m( −iωv + v / τ ) = −eE , or v = −
.
=−
m( −iω + 1 / τ )
m(1 − iωτ )
⎛ 1 + iωτ ⎞
ne 2τ
− nev
σ (0)
The conductivity is σ (ω ) =
=
=
= σ (0)⎜⎜
⎟.
2 ⎟
E
m(1 − iωτ ) 1 − iωτ
⎝ 1 + (ωτ ) ⎠
2. Dynamic magnetoconductivity tensor for free electrons.
A metal with a
concentration n of free electrons of charge –e is in a static magnetic field Bˆz . The
electric current density in the xy plane is related to the electric field by
jx = σ xx E x + σ xy E y ;
j y = σ yx E x + σ yy E y .
Assume that the frequency ω >> ω c and ω >> 1 / τ , where ω c ≡ eB / mc and τ is the
collision
time.
(a)
Solve
the
drift
velocity
equation
r r r
r
r
m(dv / dt + v / τ ) = −e( E + B × v / c ) to find the components of the magetoconductivity
tensor:
σ xx = σ yy = iω p2 / 4πω ;
σ xy = −σ yx = ω cω p2 / 4πω 2 ,
where ω p2 = 4πne 2 / m . (b) Note from the Maxwell equation that the dielectric
function tensor of the medium is related to the conductivity tensor as
r
ε = 1 + i ( 4π / ω )σ . Consider an electromagnetic wave with wavevector k = kzˆ .
Show that the dispersion relation for this wave in the medium is
c 2 k 2 = ω 2 − ω p2 ± ω cω p2 / ω .
At a given frequency there are two modes of propagation with different wavevectors
and different velocities. The two modes correspond to circularly polarized waves.
Because a linearly polarized wave can be decomposed into two circularly polarized
waves, it follows that the plane of polarization of a linearly polarized wave will be
rotated by the magnetic field.
Solution:
r r r
r
r
(a) For E = E0 e −iωt , v = v0 e − iωt . Then m(dv / dt + v / τ ) = −e( E + B × v / c ) becomes
r r r
r r
m( −iωv + v / τ ) = −e( E + B × v / c )
or
ω c ≡ eB / mc is the cyclotron frequency.
r
r
eτ r
(1 − iωτ )v + ω Cτzˆ × v = −
E
m
The equation in x, y, and z directions are:
⎧ (1 − iωτ )v x + ω cτv y = − eτE x / m
⎪
⎨− ω cτv x + (1 − iωτ )v y = − eτE y / m
⎪
(1 − iωτ )v z = − eτE z / m
⎩
Solving these three equations gives the solution:
eτ [(1 − iωτ ) E x − ω cτE y ]
⎧
⎪ v x = − m[(1 − iωτ ) 2 + (ω τ ) 2 ]
c
⎪
eτ [ω cτE x + (1 − iωτ ) E y ]
⎪
⎨v y = −
m[(1 − iωτ ) 2 + (ω cτ ) 2 ]
⎪
eτE z
⎪
vz = −
⎪
m(1 − iωτ )
⎩
r
r
With j = − nev , we have
⎧
ne 2τ [(1 − iωτ ) E x − ω cτE y ]
j
=
⎪ x
m[(1 − iωτ ) 2 + (ω cτ ) 2 ]
⎪
ne 2τ [ω cτE x + (1 − iωτ ) E y ]
⎪
⎨ jy =
m[(1 − iωτ ) 2 + (ω cτ ) 2 ]
⎪
ne 2τE z
⎪
j
=
z
⎪
m(1 − iωτ )
⎩
Then
ω p2τ (1 − iωτ )
ne 2τ (1 − iωτ )
σ xx = σ yy =
=
m[(1 − iωτ ) 2 + (ω cτ ) 2 ] 4π [(1 − iωτ ) 2 + (ω cτ ) 2 ]
ω pω cτ
ne 2ω cτ 2
=−
=−
2
2
m[(1 − iωτ ) + (ω cτ ) ]
4π [(1 − iωτ ) 2 + (ω cτ ) 2 ]
2
σ xy = −σ yx
ω pτ
ne 2τ
=
m(1 − iωτ ) 4π (1 − iωτ )
Here ω p2 = 4πne 2 / m is the plasmon frequency.
2
σ zz =
With the condition of ω >> ω c and ω >> 1 / τ ,
σ xx = σ yy ≈
− ω p2τiωτ
− 4π (ωτ ) 2
σ xy = −σ yx ≈ −
σ zz ≈
iω p2
4πω
=
iω p2
4πω
ω p2ω cτ 2
ω p2ω c
=
− 4π (ωτ ) 2 4πω 2
2
where
(b) ε = 1 + i ( 4π / ω )σ gives the dielectric tensor of
ω p2
⎞ ⎛⎜
−
1
⎟
0
ω2
ω
⎟ ⎜
2
4πiσ yy
⎟ ⎜ iω pω c
1+
0
⎟ = ⎜− ω3
ω
⎜
4πiσ zz ⎟ ⎜
0
1+
⎟⎟
ω ⎠ ⎜ 0
⎝
r
For an EM wave with wavevector k = kzˆ , E z = 0 . The
⎛
4πiσ xx
⎜1 +
ω
⎜
⎜ 4πiσ yx
ε =⎜
ω
⎜
0
⎜⎜
⎝
4πiσ xy
iω p2ω c
ω
ω p2
1− 2
ω
3
0
r
r
D = εE
⎞
⎟
⎟
⎟
0 ⎟
⎟
ω p2 ⎟
1− 2 ⎟
ω ⎠
becomes
0
⎛
iω p2ω c ⎞
ω p2
⎜
⎟ E
3
⎛ Dx ⎞ ⎜ 1 − ω 2
ω
⎟⎛⎜ x ⎞⎟
⎜⎜ ⎟⎟ =
2
2 ⎜
⎟
⎝ D y ⎠ ⎜ − iω pω c 1 − ω p ⎟⎝ E y ⎠
⎜
⎟
ω3
ω2 ⎠
⎝
The two eigen modes are determined by
⎛
⎛ ω2 ⎞
iω p2ω c ⎞
⎟
⎜ ε − ⎜1 − p ⎟
−
3
⎜ ω2 ⎟
⎟
⎜
ω
⎝
⎠
det ⎜
⎟=0
2
⎛ ω p2 ⎞ ⎟
⎜ iω pω c
ε − ⎜⎜1 − 2 ⎟⎟
3
⎟
⎜
ω
⎝ ω ⎠⎠
⎝
2
⎡ ⎛ ω p2 ⎞⎤ ⎡ ω p2ω c ⎤
⎢ε − ⎜⎜1 − 2 ⎟⎟⎥ − ⎢ 3 ⎥ = 0
⎣⎢ ⎝ ω ⎠⎦⎥ ⎣ ω ⎦
ω p2 ω p2ω c
ε = 1− 2 ± 3
ω
ω
Note ω = ck / ε . We then have c 2 k 2 = ω 2ε = ω 2 − ω p2 ± ω cω p2 / ω
ω p2 ω p2ω c
−
, solving the original equation gives E y = iE x , D y = iDx .
ω2
ω3
ω p2 ω p2ω c
For ε = 1 − 2 +
, solving the original equation gives E y = −iE x , D y = −iDx .
ω
ω3
For ε = 1 −
These two modes then correspond to left and right circularly polarized light.
3. Static magnetoconductivity tensor.
For the drift velocity theory of
r r r
r
r
r
m(dv / dt + v / τ ) = −e( E + B × v / c ) , with B = Bzˆ , show that the static current
density can be written in matrix form as
0
⎛ jx ⎞
⎛ 1 − ω cτ
⎞⎛ E x ⎞
⎜ ⎟
⎜
⎟⎜ ⎟
σ0
ωτ
1
0
⎜ jy ⎟ =
⎟⎜ E y ⎟ .
2 ⎜ c
1
(
ω
τ
)
+
c
2
⎜j ⎟
⎜ 0
0
1 + (ω cτ ) ⎟⎠⎜⎝ E z ⎟⎠
⎝ z⎠
⎝
In the high magnetic field limit of ω cτ >> 1 , show that
nec
= −σ xy
B
= 0 , to order 1 / ω cτ . The quantity σ yx is called the Hall
σ yx =
In this limit σ xx
conductivity.
Solution:
From the result of the last problem, we have
ω p2τ (1 − iωτ )
σ xx = σ yy =
4π [(1 − iωτ ) 2 + (ω cτ ) 2 ]
σ xy = −σ yx
σ zz =
ω p2ω cτ 2
=−
4π [(1 − iωτ ) 2 + (ω cτ ) 2 ]
ω p2τ
4π (1 − iωτ )
At ω = 0 , the result is
σ xx = σ yy
ω p2τ
σ0
=
=
2
4π [1 + (ω cτ ) ] 1 + (ω cτ ) 2
σ xy = −σ yx = −
ω p2ω cτ 2
σ 0ω cτ
=−
2
4π [1 + (ω cτ ) ]
1 + (ω cτ ) 2
ω p2τ
= σ0
σ zz =
4π
ω p2τ ne 2τ
=
.
Here σ 0 =
4π
m
r
r
Then the j = σE can be written as
0
⎞⎛ E x ⎞
⎛ 1 − ω cτ
⎛ jx ⎞
⎟⎜ ⎟
⎜
⎜ ⎟
σ0
ω
τ
1
0
⎟⎜ E y ⎟ .
⎜
⎜ jy ⎟ =
c
2
2 ⎟⎜
⎜ j ⎟ 1 + (ω cτ ) ⎜ 0
0
1 + (ω cτ ) ⎠⎝ E z ⎟⎠
⎝
⎝ z⎠
In the high magnetic field limit of ω cτ >> 1 ,
σ xx = σ yy ≈
σ0
(ω cτ ) 2
σ xy = −σ yx ≈ −
σ 0ω cτ
σ0
ne 2τ / m
nec
=
−
=
−
=−
2
ω cτ
eBτ / mc
B
(ω cτ )
4. Maximum surface resistance. Consider a square sheet of side L, thickness d, and
electrical resistivity ρ. The resistance measured between opposite edges of the sheet
is called the surface resistance: Rsq = ρL / Ld = ρ / d , which is independent of the
area L2 of the sheet. ( Rsq is called the resistance per square and is expected in ohms
per square, because ρ / d has the dimensions of ohms.) If we express ρ by
ρ = m / ne 2τ , then Rsq = m / nde 2τ . Suppose now that the minimum value of the
collision time is determined by scattering from surfaces of the sheet, so that
τ ≈ d / vF , where vF is the Fermi velocity. Thus the maximum surface resistivity is
Rsq ≈ mv F / nd 2 e 2 . Show for a monatomic metal sheet one atom thickness that
Rsq ≈ h / e 2 = 4.1kΩ , where 1kΩ is 103 ohms.
Solution:
For 2D system,
εF
S 2πpdp
Sπ 2mdε S 4πmε F
=∫ 2
=
2
0
0
0
h
h2
h2
S 4πm mv F2 S 2πm 2 v F2 Sm 2 v F2
=
⋅
=
=
2
2πh 2
h2
h2
h 2πN
.
vF =
m
S
N =∫
εF
Ddε = ∫
εF
2
Rsq ≈ mv F / nd 2 e 2 = m
h
m
1
h
N
⋅ 2 2 =
2πS nd e
nd 2 e 2
2πN
.
S
1
N 1
~ , n ~ 3 , and d ~ a , we have
a
S
a
1.05 × 10 −34
h
h
Rsq ~
= 2 =
Ω = 4.1 × 10 3 Ω = 4.1kΩ .
−19 2
3
2 2
(1 / a )a e a e
(1.6 × 10 )
With