Download Two dimensional electrons in a periodic potential in a uniform

Two dimensional electrons in a periodic potential in a uniform external magnetic field∗
(Dated: September 25, 2015)
I.
TWO DIMENSIONAL ELECTRONS IN A PERIODIC POTENTIAL IN A UNIFORM EXTERNAL
MAGNETIC FIELD
A.
Magnetic Translation Operators
Let us consider a system of non-interacting electrons under the effect of a periodic lattice potential, U(~r),
~ r). The Hamiltonian is then given by,
and in an external vector potential, A(~
~2
Π
+ U(~r), where,
2m
~ = ~p − q A(~
~ r).
Π
c
H =
(1)
(2)
Let the lattice vectors, ~a, ~b be such that ~a × ~b , 0 and U(~r) = U(~r + ~a) = U(~r + ~b). The question then is can
we take advantage of the discrete translational symmetry?
~ = 0, we know how to do this by using
In the absence of a vector potential, when the magnetic field B
Bloch’s theorem and defining a Brillouin-zone in the momentum-space. We can introduce the translation
operators, T w~0 = exp(−i~
w · ~p/~), such that,
~ ).
T w~0 f (~r) = f (~r − w
(3)
Note that the translation operators are unitary. Furthermore, they are such that,
[T a0 , H] = 0, [T b0 , H] = 0, [T a0 , T b0 ] = 0,
(4)
and it is possible to define simultaneous eigenfunctions (|ψi) of these operators, where
~
~~
T a0 |ψi = e−ik·~a |ψi, T b0 |ψi = e−ik·b |ψi.
(5)
This allows us to restrict ourselves to only the first Brillouin zone, −π < ~k · ~a < π and −π < ~k · ~b < π.
In the presence of a non-zero vector potential (and magnetic field), the above does not work even when
~ , 0 for general w
~ . However, in analogy with
the periodic lattice potential is absent. This is because, [T w~0 , Π]
the discussion of the zero-field case, we can try to find as set of “magnetic translation operators”, T w~ , such
∗
These notes are based on lectures originally given by Prof. B. I. Halperin at Harvard University during Fall, 2013. Transcribed
by Debanjan Chowdhury and edited by BIH.
2
that,
T w~ = eiφw~ (~r) T w~0 ,
~ ),
T w~ f (~r) = eiφw~ (~r) f (~r − w
(6)
(7)
~ = 0. The choice of φw~ (~r) depends on the
where we would like to choose φw~ (~r) in a way such that [T w~ , Π]
~ ·A
~ is linear in ~r, with ∇
~ = 0, so that we can write,
gauge. Let us choose to work with a gauge where A
A j (~r) = N jk rk + A0j .
(8)
B
N jk = − ε jk + b1 σ(1)
+ b3 σ(3)
jk ,
jk
2
(9)
~ = B.
~ Therefore,
We require ∇ × A
~
where ε jk = iσ(2)
jk . Note that ∇ × A = N21 − N12 = B, as required.
~ on eiφw~ (~r) f (~r − w
~ ),
Let us guess the following form for φw~ (~r) = w j M jk rk . Then consider the action of Π
~ ) = T w~0 f (~r), as follows,
where f (~r − w
q
~ ) = p j − N jk rk eiMmn wm rn f (~r − w
~ ),
Π j eiφw~ (~r) f (~r − w
c
q
~ ).
= eiMmn wm rn p j − N jk rk + ~δ jn Mmn wm f (~r − w
c
(10)
(11)
We want this to equal,
q
~ ).
eiMmn wm rn T w~0 Π j f (~r) = eiMmn wm rn p j − N jk (rk − wk ) f (~r − w
c
(12)
This requires,
q
N jm wm ,
c
q
q B
(3)
=
N jk =
− ε jk + b1 σ(1)
+
b
σ
3
jk
jk .
~c
~c
2
~wm Mm j =
(13)
⇒ Mk j
(14)
Example: Landau Gauge
~ = B(−y, 0). Then, from above,
Let us now take one specific example, namely the Landau gauge where A
b3 = 0, b1 = −B/2. Therefore,


 0 1 


,
N = −B 
 0 0 
(15)




qB  0 0 

 .
M=−
~c  1 0 
(16)
which implies that,
3
In the light of our discussion above,
φw~ (~r) = −
qB ~c
(17)
wy x.
Therefore, if wy , 0, T w~ translates the wavefunction and multiplies by a phase that depends linearly on x.
B.
Does this help us solve the Schrödinger equation in a periodic potential?
~ 2 ] = 0 for all w
~.
Going back to the discussion of the magnetic translation operators, we note that [T w~ , Π
~ = ~a, or, ~b, then [T w~ , U(~r)] = 0, so [T~a , H] = [T~b , H] = 0. We now still have to evaluate
Moreover, if w
[T~a , T~b ].
We state here the result, and leave the proof as an exercise:
T w~ T w~ 0 = e2iπϕ T w~ 0 T w~ ,
(18)
where,
2πϕ = wm Mmn (−w0n ) − w0m Mmn (−wn )
(19)
= w0m wn [Mmn − Mnm ]
~ · (~
~ 0)
qB
w×w
=
~c
~
~ 0)
B · (~
w×w
⇒ϕ =
.
Φ0
(20)
(21)
(22)
Here, Φ0 = 2π~/q is the quantum of magnetic flux, and ϕ is the magnetic flux, in units of the flux quantum,
~ and w
~ 0.
through the area of a parallelogram with sides w
Therefore, [T~a , T~b ] = 0 if and only if the area of the unit cell, ~a × ~b contains an integer number of flux
quanta, Φ0 . If this condition is satisfied, then we can use Bloch’s theorem to diagonalize the Hamiltonian
and find eigenstates |ψk,n i, such that,
~
~~
T~a |ψi = eik·~a |ψi~k,n , T~b |ψi = eik·b |ψi~k,n .
It is natural to ask how the above considerations get modified when ϕab ≡
(23)
~ a×~b)
B·(~
Φ0
is not an integer. In the
case where φab is a rational number,
ϕab =
~ · (~a × ~b) r
B
= ,
Φ0
s
(24)
we can choose a unit cell that is larger by a factor s than the original unit cell. The new cell has area s(~a × ~b),
which will contain an integer r flux quanta. We can then use Bloch’s theorem, where now the Brillouin zone
4
is smaller than the original cell, by a factor s, and the total number of bands is increased by a factor s, in
any energy interval. Solution of the Schrödinger equation for the eigenvalues and eigenfunctions becomes
correspondingly more difficult, as s increases, but we can imagine that the problem is solved in principle.
This leaves us with the case when ϕab is irrational. We shall address this situation later in the course.
The problem will be approached by trying to analyze solutions of the Schrödinger equation for a sequence
of rational values of ϕab approaching the irrational value we are interested in.