Download Semiclassical motion in a perpendicular uniform electric

Semiclassical motion in a perpendicular
uniform electric and magentic field
Sankalpa Ghosh, Physics Department, I I T Delhi
September 4, 2008
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Semiclassical Motion in a crossed electric
and magnetic field
Let us recall the semiclassical motion again. It describes then the Bloch
wave packets localized over several lattice spacing moving in a slowly varying
external field. Understanding this is the route to understanding electronic or
any other transport in an otherwise weakly interacting electronic system. The
motion of the electrons in the periodic crystal potential is treated quantum
mechanically. That is how we got wave packets of Bloch waves. But then
)
the rest of the motions was treated almost classically (only ṙ 6= ~k
m
We consider semiclassical motion in mutually perpendicular electric and
magentic field. We already know that in the presence of a uniform field the
k-space orbits are in the plane perpendicular to the magentic field H.
The set of equations are again the same ones
ṙ =
1 ∂E
~ ∂k
1
~k̇ = (−e)[E + v(k) × H]
c
with E · H = 0 ( since they are mutually perpendicular)
We note that E = E Ê, where Ê is a unit vector along the direction of
the electric field. Again we multiply both side of the second equation with
1
Ĥ. This gives us
e
Ĥ × ~k̇ = −eĤ × E − Ĥ × ṙ × H)
c
eH
(ṙ − Ĥ(Ĥ · ṙ)
= −eE[Ĥ × Ê] −
c
eH
= −eE[Ĥ × Ê] −
r˙⊥
c
Now integrating over time we obtain
(1)
Z
Z
dk
d
r˙⊥ = −eE[Ĥ × Ê] dt − Ĥ × ~ dt
dt
dt
~c
r⊥ (t) − r⊥ (0) = −
Ĥ × [k(t) − k(0)] + wt
eH
eH
c
Z
Here
E
(Ê × Ĥ)
(2)
H
As compared to the motion in presence of the uniform magentic field
alone, namely
w=c
~c
Ĥ × [k(t) − k(0)]
eH
this motion has an additional drift velocity w. This velocity is the velocity
of the reference frame in which the electric field actually vanishes.
Home assignment;
Show explicitly the above result.
Hint:
Recall that the Lorenz transformation of the electric field is given by
r⊥ (t) − r⊥ (0) = −
Ek0 = Ek
1
E⊥0 = q
1−
v2
c2
[E⊥ + (v × H)⊥ ]
(3)
Here v is the relative velocity between two frames.
The direction of this velocity is perpendicular to the electric as well as
the magentic field . Thus the motion in the real space perpendicular to H
is the superposition of
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• the k-space orbit rotated and scaled just as it would be if only the
magnetic field were present.
• a uniform drift with velocity w
To determine the k-space orbit we note that when E and H are perpendicular then
|Ê × Ĥ| = sin
π
=1
2
Therefore we can write
1
~k̇ = (−e)[E + v(k) × H]
c
e cE
e 1 ∂E(k)
~k̇ = −
H Ê −
×H
cH
c ~ ∂k
e 1 ∂E(k)
e1
×H
~k̇ = − ~|w|H Ê −
c~
c ~ ∂k
e1 ∂
~k̇ = −
(E − ~k · w) × H
c ~ ∂k
(4)
The last line is possible since
w × H = wH
By definition w is perpendicular to H.
Thus the equation of motion can be written in a fashion such that it is
the equation of motion an electron would have if only the magentic field H
is present and if the band structure is given by
Ē(k) = E(k) − ~k · w
For a free electron Ē is simply the electrons energy in the frame moving
with velocity w.
We can therefore conclude from this analysis that k-space orbits are given
by the intersections of surfaces of constant Ē with planes perpendicular to
the magnetic field.
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