Chapter 2
Green’s functions
2.1
Introduction
The Green’s functions are a powerful concept that allows us to evaluate the response at any point (inside or outside the conductor) due to an exci-tation at any other. In this chapter we will introduce the idea behind the Green’s functions, and we will explain how it can be applied to the problem of transport in a mesoscopic system and how we can discretize this approach in order to be able to use this approach when writing the code for our program. In the last paragraph of this chapter, the specific problem that the pro-gram needs to solve will be examined.
2.2
Mathematical interpretation of Green’s function
To introduce rigorously the concept of Green’s function a lot of mathe-matics would be needed; we will instead use an intuitive approach, based on an example that will illustrate what it is and what properties it has.
First of all, our example problem. We need to know the solution to the following problem:
d2u(x) dx2 + k
more precisely we want it solved for u subject to the boundary conditions:
u(0) = u(l) = 0. (2.2)
We can solve the problem employing the method of variation of parameters, assuming that if the solution exists it is of the form:
u(x) = A(x) cos kx + B(x) sin kx
by using the the simplifying hypotesis (for the only sake of easy math) dA(x) dx cos kx + dB(x) dx sin kx = 0 we finally find dA(x) dx = f(x) sin kx k dB(x) dx = −f (x) cos kx k Thus, formally we can write
u(x) = cos kx k
Z x
c1
f(y) sin kydy − sin kx k
Z x
c2
f(y) cos kydy
where c1 and c2 are constants that must be so chosen as to ensure that the
boundary conditions are satisfied.
Since f (y) is arbitrary, and since when we apply the condition u(0) = 0 we have
u(0) = Z 0
c1
it is obvious that we must choose c1 = 0.
By applying the other boundary condition, u(l) = 0, we require that
u(l) = cos kl k
Z l
0
f(y) sin kydy − sin kl k
Z l
c2
f(y) cos kydy = 0
After some slight manipulation we can rewrite it in the form
−sin kl k
Z 0
c2
f(y) cos kydy + 1 k
Z l
0
f(y) sin k(y − l)dy = 0
Combining the equations obtained from the boundary conditions with the general solution of the differential equation, we find the specific solution we were looking for:
u(x) = 1 k
Z x
0
f(y) sin k(y − x)dy − sin kx ksin kl
Z l
0
f(y) sin k(y − l)dy = Z x 0 f(y)sin kyP k(l − x) ksin kl dy+ Z l x f(y)sin kxP k(l − y) ksin kl dy = Z l 0 f(y)G(x, y)dy. (2.3)
Where, in equation (2.1), we introduced the function G(x, y), defined as:
G(x, y) = sin ky sin k(l−x)ksin kl , 0 ≤ y ≤ x G(x, y) = sin kx sin k(l−y)ksin kl , x≤ y ≤ l
(2.4)
This is what we call the Green’s function for the equation (2.1) and the boundary conditions (2.2). Its existance is assured in this particular problem, provided that sin kl 6= 0.
Equation (2.3) rephrases the problem stated by equations (2.1) and (2.2) in quite a convenient way for us. In the first place it is an integral equation, rather than a differential one, which is much more easily handled by numerical analysis. In second place we have to handle a function, G(x, y), which is independent of the forcing term f (x).
The fact that the Green’s function depends only upon the particular differential equation being examined and the boundary conditions imposed is really useful if we think of a possible physical interpretation of the problem we analyzed. In example, a taut string of length l whose ends are kept fixed is a problem ruled by equations (2.1) and (2.2). In such a case f (x) represents the way we act upon the system, but thanks to (2.3), regardless of the way we do so, the only thing we need to do is calculate (2.4) once, and then apply (2.3) for every different kind of stress we want to apply to the system.
By analyzing (2.4) we can deduce five properites of the Green’s function: 1. It satisfies the homogeneous form of the given differential equation:
d2G(x, y) dx2 + k 2G(x, y) = 0, ∀y, y 6= x 2. It is continuous at y = x since lim y→x−G(x, y) = sin kx sin k(l − x)
3. The derivative of G(x, y) with respect to y is discontinuous at y = x. This can be seen as follows:
G′(x, x−) = lim y→x− dG(x, y) dy = cos kx sin k(l − x) sin kl G′(x, x+) = lim y→x+ dG(x, y) dy = − sin kx cos k(l − x) sin kl G′(x, x−) − G′(x, x+) = 1
4. It satisfies the boundary conditions of the problem since it satisfies the relations
G(x, 0) = G(x, l) = 0 5. It is symmetric in its arguments, hence
G(x, y) = G(y, x)
Of course the properties are a little different when we consider a generic Green’s function. The first, second and fourth point are unchanged, the fifth property changes from simmetric to hermitian in case the Green’s function is complex instead of real. In the third point, the difference between the two derivatives is a constant factor c dependent on the specific problem.
We still need to understand the behaviour of G(x, y) for y = x. Actually, truth to be told, the rigorous way to write the first point is:
d2G(x, y) dx2 + k
2G(x, y) = −δ(x − y)
where δ(t) is Dirac delta function (symbolic function, or distribution). We can rewrite our problem in a more general way by introducing the differential operator L:
Lu= −f where, for our problem it is:
L≡ d
2
dx2 + k 2
Defining V (0, l) the space of functions defined and continuous together with their first and second derivatives in the interval 0 ≤ x ≤ l, and which satisfies the Dirichlet boundary conditions in the interval [0, l] with value 0 in both cases, and defining C(0, l) the space of continuous functions defined in 0 ≤ x ≤ l, then we can consider the operator L as defining a mapping from the space V (0, l) to the space C(0, l) (a map is a continuous transformation between two metric space, which are essentially a set with an associated notion of distance, its metric).
With this terminology the specific boundary problem given by equations (2.1) and (2.2) is equivalent to the following abstract problem:
“Given f ∈ C(0, l), find those functions u ∈ V (0, l) which are such that Lu= −f .”
In order to solve this abstract problem, we can think to associate with the operator L an inverse operator L−1
which is such that
LL−1 = L−1L = I
where I is the identity operator. If such inverse operator does exist, than our solution can be written as
u = −L−1f
If we introduce the intregal operator G defined over 0 ≤ x ≤ l by
Gf = Z l
0
f(x)G(x, y)dx
we can see that, in some sense, −G provides an interpretation of the inverse operator L−1 . Lu(x) = Z LGf dt= Z δ(x − t)f (t)dt = f (x) L−1f(x) = Z G(x, t)f (t)dt
2.3
An interpretation of the Green’s function for
solid state physics
Starting exactly from where we left in the previous chapter: whenever in physics a response R is related to the excitation S by a differential operator L, such that LR = S, the problem can be solved by defining a Green’s function and expressing the response in the form
R= L−1S = GS, G≡ L−1
In our specific problem we need to solve the Schr¨odinger equation:
[E − H]Ψ = S
where H is the Hamiltonian operator describing our system, Ψ is the wave function and S is an equivalent excitation term due to a wave incident from one of the leads. The corresponding Green’s function can be written as
G= [E − H]−1
Once again we will proceeed to explain using an example, actually the same example as in the previous paragraph (from a mathematical point of view), but with a different perspective. In this case the physical problem is a one dimensional wire with a costant potential energy U0 and zero vector
potential. We approach the problem by looking for the Green’s function:
G= E− U0+ ¯ h2 2m ∂2 ∂x2 −1
E− U0+ ¯h 2 2m ∂2 ∂x2 G(x, x′) = δ(x − x′)
It can be noted that the last expression is formally equivalent to the Schr¨odinger equation, except for the source term δ(x − x′
) (the forcing term, as we called it in the previous paragraph):
E − U0+ ¯h 2 2m ∂2 ∂x2 Ψ(x) = 0
Thus it comes to mind the idea to see the Green’s function G(x, x′
) as the wave function at x resulting from a unit excitation appplied at x′
. Such an excitation can be expected to give rise to two waves travelling outwards from the point of excitation, with amplitudes A+ and A−
: G(x, x′) = A+exp ik(x − x′), x > x′ G(x, x′) = A−exp −ik(x − x′), x < x′ k ≡ [2m (E − U0)] 1 2 ¯h
So we once again have a solution defined over all the space but for one point, x= x′
in this case. We know our Green’s function must be coninuous, while its first derivative must have a discontinuity in x = x′
, which in this case turns out to be:
∂G(x, x′ ) ∂x x=x′ + − ∂G(x, x ′ ) ∂x x=x′ − = 2m ¯ h2
By using the equation of continuity for the function and of discontinuity for the first derivative we can calculate A+ and A−
, and so we find the equation of our waves: GR(x, x′) = − i ¯hν exp [+ik |x − x ′ |] (2.5) ν ≡ ¯hk m
Altough there is another solution, corresponding to the boundary conditions for which the waves are moving torward x = x′
instead of the opposite. GA(x, x′) = + i
¯hν exp [−ik |x − x
′
|] (2.6)
The Green’s function of equation (2.5), corrisponding to outgoing waves, is referred to as the retarded function, while the one in equation (2.6), function for the incoming waves, is called the advanced function.
Since the effect of the boundary conditions can be included in the equa-tion defining the Green’s funcequa-tion by simply adding a term ±iη, the advanced and retarded Green’s function are usually defined as:
G+(x, x′; E) = GR(x, x′; E) = lim η→0+G(x, x ′ ; E+iη) = lim η→0+[(E + iη) − H] −1 G−(x, x′; E) = GA(x, x′; E) = lim η→0+G(x, x ′ ; E+iη) = lim η→0+[(E − iη) − H] −1
These two solutions are different, but related to each other. We can easily notice that
but since the energy is real, it turns out that G(x, x′ ; E) is hermitian, and so it follows: G−(x, x′) =G+(x′ , x)∗ .
Thus the real part of the advanced and retarded Green’s functions are the same, while their imaginary parts are equal but opposite.
From a formal point of view, we can write the Green’s function as a function of the eigenvalues of E − H. In the case of a continuous spectrum of eigenvalues, an integral summation would be involved, and in case of E = λ with λ eigenvalues of H, such integral would not be well defined, and as a result, the problem would need to be approached by a limit procedure (defining G+ and G−
). For the discrete part of the spectrum, the Green’s function has poles in corrispondence of the discrete eigenvalues λn. Thus, in
a general problem of the kind
[E − H(r)] u(r) = f (r)
with r being the position in space, and where u(r) satisfies the same boundary conditions as G(r, r′
; E), the general solution to such problem is:
u(r) = R G(r, r′ ; E)f (r′ )dr′ , E 6= {λn} R G± (r, r′ ; E)f (r′ )dr′ + φ(r), E = {λ}
2.4
Tight binding model
For an arbitrarly shaped conductor the Green’s function problem can be written by using the effective mass equation for potential energy U (r) (including the conduction band energy value), and vector potential A:
[E + iη − H(r)] GR(r, r′) = δ(r − r′)
H(r) = (i¯h∇ + eA)
2
2m + U (r)
To solve the problem we discretize the spatial coordinate, thus turning the Green’s function into a matrix whose indexes denote points on a discrete lattice. The differential equation is thus turned into the matrix equation
[(E + iη) I − H] GR = I
where I is the identity matrix and H is the matrix representation of the Hamiltonian operator. Such discretized Hamiltonian is tri-diagonal, with the upper and lower diagonal being the the same, and it is similar to the tight-binding Hamiltonian widely used to model electronic transfer in con-densed matter and molecules. In the tight-binding model, the wave function is expressed in terms of localized atomic orbitals, one at each site. Orbitals on neighboring sites are connected by a “hopping matrix element” (or an “overlap integral”). The term on the diagonal in our model plays the role
of the energy of the orbital localized at site j, while the term on the upper or lower diagonal plays the role of the overlap integral between orbitals on neighboring sites.
We can partition the overall Green’s function into submatrices: GR = [(E + iη) I − H]−1 Gp GpC GCp GC ≡ (E + iη) I − Hp τp τp+ EI − HC −1
The matrix [(E + iη) I − Hp] represents the isolated lead, while [EI − HC]
represents the isolated conductor (no infinitesimal imaginary term iη is needed since the coupling with the leads will give rise to a finite imaginary term). The coupling matrix τp is non zero only for adjacent points, we will
denote with t their value.
Since we are interested in the sub-matrix GC we can proceed to
left-multiply the partitioned Green’s function matrix by the inverse of its equiv-alent representation and consider the equations that arise from the multipli-cation with the second column of the Green’s function matrix:
[(E + iη) I − Hp] GpC + [τp] GC = 0
[EI − HC] GC +τp+ GpC = I
The solution of the first equation leads to
GpC = −gRpτpGC, gRp ≡ [(E + iη) I − Hp] −1
The term gR
p is the Green’s function for the isolated semi-infinite lead. By
using this equation and the second equation of our system we can finally obtain
GC =EI − HC − τp+gpRτp
−1
(2.7) Matrices in (2.7) are finite and of dimension CxC, C being the number of points inside the conductor. The infite lead is taken into account exactly thanks to the last term in the left operand, τ+
p gRpτp, which is still, by the
way, a function of the inverse of an infinite dimensional matrix. What does the trick is that gR
p is the Green’s function for an isolated lead, which usually
can be determined analytically. Also, since the coupling matrix τp is zero fo
all points in the lead except those (pi, pj) adjacent to points (i, j) inside the
conductor, we can write
τ+ p gRpτp ij = t 2gR p (pi, pj)
Assuming that different leads are independent, so that their effect is additive, we finally can write
GC =EI − HC − ΣR
−1
ΣR =X
p
ΣRp, ΣRp(i, j) = t2gRp(pi, pj)
This term GC is actually the only component of the overall Green’s function
two points inside the conductor, taking the effect of the leads into account through the term ΣR.
The term ΣR can be viewd as an effective Hamiltonian arising from
the interaction of the conductor with the leads, and since a similar term is often used to describe the interaction of the electrons with phonons and other electrons, by analogy it is called the “self-energy” due to the leads. Although usually the self energy only provides an approximate description, while our “self-energy” describes exactly the leads. We will not stress more over the concept of self energy, since we will use a different approach to include the effect of the leads, which will be explained in the next chapter.
2.5
Application of Green’s function to the problem
to be simulated
As we discussed in a previous paragraph, what we are interested in is the retarded Green’s function GR(x, y; x′
, y′), or rather the wave function at a point (x, y) due to an excitation at a point (x′
, y′). In the following figure an image of the wire we are analyzing and the coordinate system adopted is shown.
x=x’ y=y’ y x A m A m − +
Fig. 2.1An image of the system under study, including our choice for the coordinate system. Again, we are generalizing a previous example; by recalling the 1D case analyzed in paragraph 2.3 we can write the Green’s function in the form
GR(x, x′) =X
m
A±mχm(y) exp [ikm|x − x ′
|]
where A+m and A−
m are the amplitudes of the different modes that propagate
away from the source. The transverse mode wave functions χm(y) satisfy the
equation − ¯h 2 2m ∂2 ∂y2 + U (y) χm(y) = ǫm,0χm(y)
Following the same method explained in paragraph 2.3 for the 1D case we use the equation for the continuity of the function and the discontinuity of its first derivative to find
A+m= A−m= − i ¯hνm
χm(y ′
)
which shows how the amplitude of any mode m is proportional to the trans-verse wave function at the point of excitation. Our Green’s function thus is GR(x, y; x′, y′) =X m − i ¯hνm χm(y)χm(y ′ ) exp [ikm|x − x ′ |] km ≡ p2m (E − ǫm,0) ¯h , νm ≡ ¯hkm m
Now we need a discretized version of it, but keeping in mind that we are working on a wire infinite along x but bounded along y. In other words, our problem is to represent a 1D chain, and in order to do so we discretize it by applying it to any function of x, let us call our choice F (x), and choosing a discrete lattice whose points are located at x = ja, j being an integer and a being the step of the mesh. We can thus write
[HF ]x=ja = −¯h 2 2m d2F dx2 x=ja + UjFj, Fj = F (ja), Uj = U (ja)
Assuming a is small, by the method of finite differences we approximate the derivative operators. For the first derivate operator we have
∂F ∂x x=(j+1 2)a = 1 a [Fj+1− Fj] and thus we find the second derivative operator
∂2F ∂x2 x=ja = 1 a ( ∂F ∂x x=(j+1 2)a − ∂F ∂x x=(j−1 2)a ) = 1 a2 [Fj+1− 2Fj+ F j − 1]
We can then write
[HF ]x=ja = (Uj+ 2t) Fj − tF j − 1 − tF j + 1 (2.8)
t ≡ ¯h
2
2ma2 (2.9)
Equation (2.8) can be rewritten in the form
[HF (x)]]x=ja =X
i
H(j, i)Fi (2.10)
where
H(j, i) = Ui+ 2t if i = j
= −t if i and j are nearest neighbors
= 0 otherwise
We thus have, thanks to (2.9), our desired discretized Hamltonian operator:
H = · · · −t 0 0 0 −t U−1+ 2t −t 0 0 0 −t U0+ 2t −t 0 0 0 −t U1+ 2t −t 0 0 0 −t · · ·
One final note about equation (2.9) is in order. The term m represents the effective mass of the electron, which is supposed to be costant for the evaluation of the matrix elements. If the effective mass is not a costant along the transverse direction, then such evaluation must be done considering the mass m as a function of position, which leads to more complex calculations in order to find the transverse eigenfunctions.
Once we have found the Green’s function for the whole system we are finally able to find the transmision and recflection coefficients, and thus we can evaluate both the conductance and the shot noise in our device.
We will not discuss how to find the formula linking the Green’s functions to the transmission and reflection coefficients, but we will introduce it since it is an essential part of the method.
The formula for the transmission coefficients between locations q and p is
t′n,m = −i2V (sin θnsin θm)
1
2 ei(θmp−θnq)
hn |Gq,p| mi (1.14)
While for the reflection coefficients
r′n,m = − sin θn sin θm
12
ei2(θm+θn)p
(i2V sin θmhn |Gp,p| mi + δn,m) (1.15)
In both the equations θn and θm are the longitudinal wave vectors for the
is a Kronecker delta, which is defined as: δn,m = 0, n6= m δn,m = 1, n= m
