Currents from quantum mechanics
Physics for Semiconductors
Debdeep Jena (djena@nd.edu)
Department of Electrical Engineering
University of Notre Dame
(Spring 2013)
1
Introduction
In semiconductor devices, we are often concerned with currents. A current is a measure of the
flow of objects from one point in space to another. The flow of electric charges constitutes an
electric current, leading to the notion of electrical conductivity. In this chapter we develop
the recipe to understand current flow from a quantum-mechanical viewpoint. Since the
physical state of particles in quantum mechanics is represented by its wavefunction Ψ(x, t),
the current must be obtained from the wavefunction.
1.1
Probability current
Since |Ψ(x, t)|2 = Ψ⋆ Ψ is the probability density, let’s examine how it changes with time.
We obtain
∂|Ψ(x, t)|2
∂Ψ ∂Ψ⋆
= Ψ⋆
+
Ψ,
(1)
∂t
∂t
∂t
where we use the time-dependent Schrodinger equation ih̄∂Ψ/∂t = (p̂2 /2m + V )Ψ and
its complex conjugate −ih̄∂Ψ⋆ /∂t = (p̂2 /2m + V )Ψ⋆ to obtain
∂|Ψ(x, t)|2
(p̂2 /2m + V )Ψ⋆
(p̂2 /2m + V )Ψ
= Ψ⋆
+Ψ
,
∂t
ih̄
−ih̄
which simplifies to
1
∂|Ψ(x, t)|2
=
(Ψ⋆ p̂2 Ψ − Ψp̂2 Ψ⋆ ).
∂t
2mih̄
Since p̂ = −ih̄∇r , we recognize the resulting equation
(2)
(3)
∂|Ψ(x, t)|2
1
= −∇r · [
(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ )]
(4)
∂t
2m
as the familiar ‘continuity’ equation in disguise. A continuity equation is of the form
∂ρ/∂t = −∇r · j, where ρ is the particle ‘density’ and j is the current density. We read off
the quantum mechanical current density as
1
(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ ).
(5)
2m
This equation provides us the required recipe for calculating the probability density flow,
or current flow directly from the quantum mechanical wavefunctions of states. We make a
j=
1
√
few observations. If Ψ is real, j = 0. Since Ψ has dimension of 1/ V ol, the dimension of j
is per unit area per second. For 3D, volume is in m3 and j is then in 1/(m2 · s). For 2D j is
in 1/(m · s), and it is simply 1/s for 1D.
We also note that
I
Z
Z
d
d3 r∇ · j = − j · dS = 0.
(6)
d3 r|Ψ|2) = −
(
dt space
space
The conversion of the integral from volume to a closed surface uses Gauss’ theorem. The
value of the integral is zero because Ψ and consequently j goes to zero at infinity, and the
equality must hold for all space. This equation is a statement of the indestructibility of the
R
particle, which follows from space d3 r|Ψ|2 = 1. If the number of particles is not conserved,
then one needs to add recombination (‘annihilation’) and generation (‘creation’) terms to
the continuity equation. It then looks as ∂ρ/∂t = −∇ · j + (G − R) where R and G are
recombination and generation rates.
We also note that in the presence of a magnetic field B = ∇×A, the quantum-mechanical
momentum operator p̂ → p̂ + qA where q is the magnitude of the electron charge. This
leads to an additional term in the expression of the current density
1
qA ⋆
(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ ) +
Ψ Ψ.
(7)
2m
m
The additional term depending on the magnetic vector potential A is useful to explain
current flow in magnetic materials, magnetotransport properties, and superconductivity.
j=
1.2
Charge current
Lets focus on determining the electric current. To account for the flow of charge, the probability density current j is modified simply to J = qj, where q is the charge (in Coulombs) of
the charge particle. We assume these charge particles are electrons and q = 1.6 × 10−19 C
and free mass me = 9.1 × 10−31 kg. In the absence of a magnetic field, the electric current
density is then given by
J=
q
(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ ),
2me
(8)
which is now in A/m2 for 3D, A/m for 2D, and A for 1D current flow, where A=C/s is
the unit of current in Amperes. The current density is expressed in terms of the electron
wavefunctions. We wish to make the expression more ‘usable’.
Consider free electrons in 1D with periodic boundary conditions between
x = (0, L). The
√ ikx
wavefunction for a state |ki of definite energy E(k) is ΨE (x, t) = (1/ L)e e−iE(k)t/h̄ . In
the QM expression for current, the time evolution portion is not affected by the momentum
operator, and therefore factors to 1. It is another illustration of the virtues of working
with states of definite energy. The current carried by state |ki is then obtained as J(k) =
I(k) = qh̄k/me L. The current density and current are the same in 1D. The current I(k) =
qh̄k/me L = qv(k)/L connects to the classical notion of current carried by a particle with
velocity v(k) = h̄k/me traversing a distance L. Another way to picture the same current is
to split it as I = q × v(k) × n, where n = 1/L is the ‘volume density’ of particles.
2
To find the total current carried by multiple k-states, we must sum the contribution from
each state. We use the velocity picture since it carries over to wave packets and electrons
in crystals. We define the group velocity vg (k) = ∇k E(k)/h̄. We also generalize to any
dimension d. The expression for the current then becomes
Jd =
q X
vg (k)f (k),
Ld k
(9)
where we have now included the Fermi-Dirac occupation probability of state |ki. We next
split off the spin degeneracy gs = 2, and allow a valley degeneracy gv for each k-state. Picture
the current flow from a ‘left’ contact to a ‘right’ contact. Let us also allow for scattering,
which are quantum-mechanical reflections in going from the left to the right contact. Let
the quantum-mechanical transmission coefficient from the left to the right contact be T (k).
Then, the expression for the net current density flowing from the left to the right becomes
Jd =
qgs gv X
vg (k)T (k)[fL (k) − fR (k)].
Ld k
(10)
The Fermi-Dirac function difference results because fL (k)[1 −fR (k)] −fR (k)[1 −fL (k)] =
fL (k) − fR (k). The transmission coefficient T (k) does not depend on the direction of current
flow.
The sum over k states is typically always converted into an integral. The recipe for this
step uses the fact that the allowed states in the k-space are discrete ‘boxes’ of each side 2π/L
and volume (2π/L)d in d dimensions. Then, the sum converts to an integral via
X
k
(...) =
Z
dd k
(...).
( 2π
)d
L
(11)
Note the cancellation of the dependence on the macroscopic dimension L. The kcoordinate system is chosen to be either cartesian, cylindrical, or spherical based on the
specific problem. The expression for the current density is then
qgs gv
Jd =
(2π)d
Z
dd k × vg (k)T (k)[fL (k) − fR (k)].
(12)
The unit of current density is in A/md−1 for current flow in d-dimensions. This expression
for the current density is applicable in a wide range of situations ranging from ballistic
transport, to scattering limited drift or diffusion, and to tunneling transport, and in multiple
(1, 2, or 3) dimensions. The group velocity term vg (k) is locked down by the electron band
structure E(k), and external forces modify the Fermi-Dirac functions. For example, the
response of f (k) to electric fields, or concentration gradients are tracked by a Boltzmann
transport equation.
1.3
Charge current in semiconductor crystals
We now turn our attention to electric current flow by electrons in crystals. To evaluate the
charge-current density by electrons in crystals, we must identify their group velocities vg (k).
The definite energy wavefunctions of electrons in crystals are Bloch functions Ψ(k, r, t) =
3
ψE (k, r)e−iE(k)t/h̄ , where ψE (k, r) = eik·r u(k, r) with k the Bloch wavevector in the reduced
zone and u(k, r + a) = u(k, r). E(k) are the eigenvalues of the Bloch states, and constitute
the electron bandstructure. To obtain the group velocity, we first derive the following useful
identity for Bloch states:
Z
Z
me
∇k E(k) = d3 rΨ⋆ p̂Ψ = − d3 rΨp̂Ψ⋆
(13)
h̄
The identity is evidently dimensionally correct since both sides have units of momentum.
The middle and far right are complex conjugate relations of each other, whereas the left side
is real. Note that the left side involves a gradient of E(k) in k−space, whereas the middle
and right sides involve gradients of Ψ, Ψ⋆ in the r−space.
To arrive at this identity, write the Schrodinger equation for the Bloch states as
h̄2 2
−
∇r ψE (k, r)e−iE(k)t/h̄ = [E(k) − V ]ψE (k, r)e−iE(k)t/h̄ .
(14)
2me
The time evolution cancels since Bloch states are stationary states. We now take a
gradient in the k−space. Since [∇r , ∇k ] = 0, and ∇k ψE = irψE + eik·r ∇k u = irψE + X we
have
h̄2 2
∇ (irψE + X) = [E(k) − V ](irψE + X) + [∇k E(k)]ψE .
−
2me r
We now use the identity for Bloch eigenstates:
∇2r (rψE ) = r∇2r ψE + 2∇r ψE .
(15)
(16)
in Equation 15 and rearrange to obtain
h̄2
ih̄2
h̄2 2
2
−
(ir∇r ψE ) − ∇r ψE = [E(k) − V ](irψE ) +[E(k) − (−
∇ + V )]X + [∇k E(k)]ψE .
|
{z
}
2me
me
2me r
|
{z
1
}
|
1
{z
Ĥ
}
(17)
The terms indicated as ‘1’ constitute the Schrodinger equation for Bloch eigenstates
multiplied by ir, and hence cancel. Multiplying by ψE⋆ on the left and integrating over all
space, the second term on the right vanishes since ψE⋆ is an eigenstate. Then, we are left
with
ih̄2
∇k E(k) d
= ∇k E(k) = −
d3 r(ψE⋆ ∇r ψE ),
(18)
me
which proves the identity in Equation 13 since p̂ = −ih̄∇r . The probability current
density carried by a Bloch state is given by
Z
3
Z
rψE⋆ ψE
1
(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ ),
(19)
2me
from which we obtain the group velocity of the Bloch state with help of the identity in
Equation 13 as
j=
4
vg (k) =
1
2me
Z
d3 r(Ψ⋆ p̂Ψ − Ψp̂Ψ⋆ ) =
∇r E(k)
.
h̄
(20)
The simplicity of the result vg (k) = ∇r E(k)/h̄ is one of the central reasons why the
k−space is where one should investigate transport properties. This velocity is used in Equation 12 for evaluating the charge current carried by Bloch states in d-dimensions:
Jd =
qgs gv
h̄(2π)d
Z
dd k[∇k E(k)]T (k)[fL (k) − fR (k)].
(21)
Note that the group velocity term is a first-order derivative in k. Therefore, for the
particular case of d =1 dimension, the current simplifies to
qgs gv
I=
h̄(2π)
Z
dE · T (E)[fL (E) − fR (E)] ≈ (gs gv
q2
)V
h
(22)
R
if we assume T (E) = 1 for perfect (ballistic) transmission and since dE[fL (E) −
fR (E)] = qV at T → 0K. The ballistic conductance of a 1D mode is therefore given by
G0 = gs gv × q 2 /h and does not depend on the exact bandstructure E(k).
1.4
Energy (heat) current
The flow of an electron of energy E(k) transports not just charge, but the associated energy
P
as well. The energy current density is given by JE (k) = k E(k) · vg (k)/Ld , and thus we
obtain
gs gv
JE =
(2π)d
Z
dd kE(k)vg (k)T (k)[fL (k) − fR (k)].
(23)
The units are in Watts/md−1 for transport in d-dimensions. Using the same approximations as earlier for a 1D conductor, we obtain the energy current carried by electrons to be
IE ≈ (gs gv q 2 /h) · V 2 in Watts if the transport is ballistic (T (k) = 1) at T → 0K.
1.5
Any current
Based on the above recipes, spin currents, polarization currents, and other related quantities
may be evaluated.
5