Chapter 3
Building the Green’s function of
a complex system
3.1
Introduction
Now that the concept of Green’s function has been introduced, we can concentrate on applying such method to a real, complex, system. We will introduce a technique to find the Green’s function for a system for which we are not able to evaluate it directly.
In doing so we will also describe two methods to handle the problem of applying a probe to the system.
3.2
The recursive Green’s function technique
The purpose of this method is to find the Green’s function of a system by decomposing the problem into smaller ones, for which the Green’s function can be easily found. The Green’s function for the original problem is then calculated by assembling togheter the solutions for the reduced problems.
The first step is to discretize the problem by filling the domain with a tight binding lattice. For a 2D problem like ours, a rectangular lattice will be needed. In the original method devised for the analysis of disordered
conductors we calculate analytically the Green’s function for a semiinfinite lead at the output, than at every step of the procedure the next layer of the discretized lattice is added. If, however, there are sections of the device characterized by a longitudinally costant transverse potential, they can be treated as a single section, for which the Green’s function can be derived analytically.
Therefore our approach will be to divide the structure we are working on into slices characterized by a longitudinally costant transverse potential, as shown in Fig. 3.1, each of these sections, or slices, is supposed to have Dirich-let boundary conditions (so they are “sealed”) and to overlap the neighboring sections by one lattice unit.
0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 00000000 00000000 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 11111111 11111111 000000000000000 000000000000000 000000000000000 111111111111111 111111111111111 111111111111111 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 000 000 000 000 111 111 111 111 000 000 000 000 111 111 111 111
If we consider two neighboring sections, the overall Hamiltonian can be writ-ten as
H = H0+ V
where H0 is the Hamiltonian for the two decoupled sections and V is the
perturbation due to coupling them togheter (or rather opening up the facing ends of the two slices). Such perturbation is too large to be treated only with low order terms, since it corresponds to a substantial variation of the system geometry. We will than resort to Dyson’s equation for the exact calculation of the Green’s function G for the final section, starting from the Green’s functions of the decoupled slices G0.
G= G0V G
In order to obtain an explicit expression for G we need to do some calcula-tions that require we choose a specific representation for it. Specifically, we will adopt a mixed representation: in real space in the longitudinal direction, in the space of transverse eigenmodes in the transverse direction. Thus the matrix representing the Green’s function Gij between two locations i and j
along the device has elements hi, s |G| t, ji where we denoted with s a trans-verse eigenmode at location i and with t a transtrans-verse eigenmode at location j.
As we previously stated, such sections are supposed to be affected by a longitudinally costant transverse potential, so there can not be mode mixing and therefore the matrix representing their Green’s function will be diagonal: hi, s |G| t, ji = 0 for s 6= t.
Each element hi, s |G| s, ji can be treated as the Green’s function for a 1D discrete chain extending from i to j with an energy corresponding to the longitudinal energy component available for that specific mode. Due to the separability of the Schr¨odinger equation along the longitudinal and the transverse direction, we can subdivide the total energy of the impinging particles into a transverse component, equal to the difference between the total energy and the just mentioned transverse component. For each section of the problem is thereby transformed into a collection of non interacting 1D problems.
3.2.1 Coupling two slices togheter
What we need to do now is to calculate the Green’s function in an explicit way. To do so, we will analyze the coupling of two adjacent slices. Let the first slice, on the left, be defined between points d and c, and the second slice, on the right, be defined between points b and a. We want the Green’s function for the “sum” slice ranging from d to a, which we obtain by coupling the first two slices by the c and b ends.
As we already discussed, we will now proceed to apply Dyson’s equation: hd |G| ai =d G0 a + d G0V G a
which, using the completeness relationships P
i|ii hi| = I, with |ii being the
elements of a complete basis, can be rewritten as hd |G| ai =d G0 a + X m,n d G0 m hm |V | ni hn |G| ai
Since the coupling potential is only present between c and b, it is obvious that the only terms of the summation that are different from zero are those corresponding to the application of the potential between them, and thus we have hd |G| ai =d G0 a +d G0 b hb |V | ci hc |G| ai+d G0 c hc |V | bi hb |G| ai
it should be noted, by the way, that there is no connection, before the cou-pling, between the first and the second slice, so the terms with superscript 0 that range from a point of the first slice to one of the second are bound to be zero.
Our equation thus becomes hd |G| ai =d
G0 c hc |V | bi hb |G| ai
We now have an explicit expression for the Green’s function hd |G| ai, but we have a new variable: the Green’s function hb |G| ai. We now need to
apply Dyson’s equation to this new variable in order to be able to calculate hd |G| ai.
As we will see, there will be quite a bit of calculations involved, so it may be useful to switch to a more handy formalism. From now on we will refer to a general Green’s function hq |G| pi simply by typing Gq,p. Analogously,
we will use Vq,p when we want to refer to hq |V | pi.
Back to our calculations, we were about to apply Dyson’s equation to Gb,a:
Gb,a = G0b,a+ G0b,bVb,cGc,a + G0b,cVc,bGb,a
This time the first term of the right operand is not zero, since when the two slices are isolated a Green’s function between b and a can be defined and it will be different from zero. As we previously said, V is different from zero only between b and c, so only the terms Vb,cand Vc,b are not zero, so only two
other terms appear in the right operand. However, of those two terms the second include G0b,c which is zero since when the slices are decoupled (zero
superscript!) there is no connection between b and c.
Now, without further need to remember which terms are zero and why, we can proceed to apply Dyson’s equation to Gc,a and to recap the equations
we have found up to now:
Gb,a = G0b,a+ G0b,bVb,cGc,a (3.2)
Gc,a = G0c,cVc,bGb,a (3.3)
With this system of three equation and three variables we can finally find a solution. We will start by using equation (3.3) on equation (3.2), thus obtaining
Gb,a = G0b,a+ G0b,bVb,cG0c,cVc,bGb,a
which after some manipulation takes the form
Gb,a =I − G0b,bVb,cG0c,cVc,b
−1
G0b,a (3.4)
By applying (3.4) to (3.1) we finally find
Gd,a= G0d,cVc,bI − G0b,bVb,cG0c,cVc,b
−1
G0b,a (3.5)
We will now define, for our own convenience, α and ∆ as follows:
α ≡ G0d,cVc,b (3.6)
∆ ≡I − G0
b,bVb,cG0c,cVc,b
−1
(3.7) The final form for Gd,a is given by
Let us verify what data we need to know in order to calculate Gd,a. We used
G0d,c, G0
c,c, G0b,b and G0b,a, plus Vb,c and Vc,b. During the following step our
newly found Gd,awill play the role of G0b,a, the terms for the new slice to add
(G0d,c, G0c,c) will be calculated easily since they belong to an isolated slice.
The potential Vb,c (and its symmetrical expression Vc,b will be also known).
What is left is G0b,b, which corresponds to Gd,d of the current step. So, in
order to be able to perform the calculations in the following step, and thus apply recursively this method, we still need to find Gd,d.
We will use a procedure completely analogous to the one used up to now. We start from the implicit expression for Gd,d obtained from the Dyson
equation:
Gd,d = G0d,d+ G0d,cVc,bGb,d, (3.9)
and we then obtain the expression for Gd,d
Gb,d = G0b,bVb,cGc,d (3.10)
and for Gc,d
Gc,d = G0c,d+ G0c,cVc,bGb,d (3.11)
Following steps analogous to the ones of the previous case, we substitute equation (3.11) into (3.10), thus obtaining
We can notice that
Vb,cG0c,d =G0d,cVc,b
T
= αT (3.13)
where T denotes the operation of transposition. So by slightly manipulating (3.12), and in virtue of (3.13) and (3.7)
Gb,d = ∆G0b,bα T
(3.14)
By means of (3.14) and (3.6) we can thus write
Gd,d = G0d,d+ α∆G0b,bα T
(3.15)
By checking which Green’s function are needed to calculate Gd,d we see that
we need no new one, thus no other Green’s functions are needed to be able to recursively apply this procedure during the following steps. A graphic of the Green’s function needed for the calculations is supplied in Fig. 3.2.
3.2.2 Green’s function for a finite slice and semiinfinite chain
In the previous chapter we found the Green’s function for an infinite chain, but in order to apply the recursive Green’s function method we need two different types of Green’s functions: those for a semiinfinite chain (to account for the leads) and those for a finite segment (to account for the slices of the system).
G
d,d d,cG
G
c,cG
b,aG
a,aG
d,dB
C
D
A
G
0 0 0 0 0 d,aFig. 3.2 Scheme of the Green’s function needed for our calculations.
In particular, for the semiinfinite chain we need to find the Green’s function from a generic site q back to itself, Gs
q,q, and from a generic site q
back to the first site of the lead, Gs
q,1. For a finite chain of N sites we need
instead the Green’s function from the its first site back to itself, Gf1,1, and between the ends, GfN,1.
B A B A
Fig. 3.3 Infinite chain split into two semiinfinite sections.
We start from the Green’s function between two generic sites, q and p, for a discretized infinite chain. We thus have
Gq,p=
iei|q−p|θ
2V sin θ (3.16)
where we defined θ = ka, with k being the wave vector for propagation and a is the discretization step.
In order to find the Green’s function for a semiinfinite chain we start from the Green’s function for the infinite chain, and apply Dyson’s equation backwards. As shown in Fig. 3.3, we will split the infinite chain into two semiinfinite chain, whose “finite ends” will be called a and b. If we define G the Green’s function for the infinite chain, we can infact write the Dyson equation we would have considering that we are adding togheter the two
semiinfinite chains of the previous figure. In the following expression we are defining Gs the Green’s function for the semiininite chain from a to infinity.
G= Gs+ GsV G
focusing on the pair of sites q and p, both to the left of a, it can be rewritten as
hq| G |pi = hq| Gs|pi + hq| GsV G|pi
Inserting the completeness relation and considering that the perturbation V acts only between b and a we obtain
hq| G |pi = hq| Gs|pi + hq| Gs|ai ha| V |bi hb| G |pi + hq| Gs|bi hb| V |ai ha| G |pi
Since the semiinfinite chain ends in a, it is straightforward that hq| Gs|bi = 0
and thus the previous equation reduces to
hq| G |pi = hq| Gs|pi + hq| Gs|ai ha| V |bi hb| G |pi
which becomes, with the already used shorthand notation,
Gq,p= Gsq,p+ G s
q,aVa,bGb,p (3.17)
Let us now assume that p is coincident with a. In such a case we have
Gq,a = Gsq,a+ G s
or, after some manipulation,
Gsq,a = Gq,a[I + Va,bGb,a]−1 (3.19)
Since the Green’s function from two sites on an infinite chain, Gb,a, is known,
we can now find the Green’s function for a semiinfinite chain from one site back to its first site, Gs
q,a: Gsq,a = ie i|q−a|θ 2V sin θ 1 + V ie i|a−b|θ 2V sin θ −1 = ie i|q−a|θ 2V sin θ 1 + ie iθ 2 sin θ −1
which after some more manipulation reads
Gsq,a = 1 V e
iθ
ei|q−a|θ (3.20)
From equation (3.20) it is strightforward to find the Green’s function for a semiinfinite chain from the first location back to itself, since we just need to set q = a to find
Gsa,a = 1 V e
iθ (3.21)
while for the one between the q-th site and the first site we have
Gsq,a = 1 V e
iqθ
(3.22)
We are still looking for the Green’s functions for a segment of finite length. In order to find them we will use a procedure analogous to the one just used: we
will apply the Dyson equation backwards starting from a semiinfinite chain beginning at point a, which we can see as the composition of a finite segment, ranging from q to a, and another semiinfinite chain starting at q + 1. This setup is shown in Fig. 3.4.
C
C B A
B A
Fig. 3.4 Semiinfinite chain as composition of a finite segment and another semiinfinite chain.
Denoting the Green’s function for the finite segment as G0, we will now proceed to apply Dyson’s equation in order to find hq| Gs|ai:
Gsq,a = G0q,a+ G0q,qVq,q+1Gsq+1,a (3.23)
which holds since the coupling potential is meant to connect q and q + 1, so the second term of the right operand is the only one left after applying the completeness relation and getting rid of the zero terms of the summation.
In order to further proceed, we will need to focus onto the Green’s function for a segment of unitary length, by setting a = q. Thus equation (3.23) becomes
Gsq,q = G0q,q + G0q,qVq,q+1G s q+1,q
which once inverted takes the form
G0q,q = Gsq,qI + Vq,q+1G s q+1,q
−1
(3.24)
By substituting (3.21) and (3.22) into (3.24) we obtain
G0q,q = 1 2V cos θ =
sin θ
V sin 2θ (3.25)
In order to further proceed in our study, we can recursively add one point to another, and starting from a slice of unitary length we will find the formula for a slice of arbitrary length N . So, by means of equations (3.5) and (3.25) we obtain G0q+1,q = 1 2V cos θV 1 − 1 2V cos θV 1 2V cos θV −1 1 2V cos θ = sin θ V sin 3θ
This last equation seems promising, and it spawns the idea that a similar formula holds for a generic slice of length N . Since the formula is true for our “zero case” (Green’s function with identical slice indexes), in order to
prove by induction we need to start from a slice of lenght N , characterized by the following Green’s function
G0q+N,q = sin θ
V sin (N + 1) θ(3.26) If we can prove that such equation implies that
G0q+(N +1),q = sin θ
V sin (N + 2) θ (3.27)
than we have proven a general formula to find the Green’s function for a slice of length N .
By applying (3.5) to (3.26) we indeed find, after some manipulation, equation (3.27) so the formula holds true for every N .
Actually, in order to combine the Green’s function for a one-site chain with that for a N -site chain we also need the Green’s function for a finite chain from one end back to itself. It can be shown, by using an analogous recursive procedure, that for a chain of length N
G0q+N,q+N = G0q,q =
sin N θ
V sin (N + 1) θ (3.28)
As far as the coupling potential V is concerned, as we have seen in chapter 2, it is given by (2.9)
V = ¯h
2
where a is the discretization step. Actually in the program we use a different coordinate system, such that our coupling potential will be denoted by
V = − ¯h
2
2ma2 (3.29)
3.3
Probe iteration and the Metalidis-Bruno
ap-proach
In order to evaluate the effect of the probe, the standard strategy is to apply the probe to the system (or rather, to every isolated slice that is affected by it, in its current position, when we take into account the whole system). We will than start the recursive procedure just explained and obtain the Green’s function for the entire device.
In their paper [7] Metalidis and Bruno suggest a different approach to the problem. Their method is to apply the probe as a perturbation to the system “whole device, uneffected by the probe”. In the following lines we need to keep in mind that in their paper the model of the probe they adopt is that of an ideal probe, a delta function, so it adds a repulsive potential v to a single point of the system.
Also, it should be noted that they do not work with slices, but with single columns (slices of a single mesh point), so their math gives results a bit different from those found in the previous paragraphs of this chapter.
Without further ado, we can start presenting their procedure. The first step is to start from the rightmost mesh column and add to it the adjacent mesh column, thus obtaining the Green’s function from the second column back to the first, G0,R2,1. Where 0 refers to the fact that the Green’s function describes a system uneffected by the probe, and R stands for “right” (use of suc a superscript is not really needed, it is just a reminder that we will need to use this function to keep into account the right part of the system).
From there, a recursive procedure starts, until we have found, and saved, every single Green’s function from G0,R1,1 to G0,RN,1, where N is the number of columns which form the system (or rather, the number of discretization points along the longitudinal direction). We can see a graphical representa-tion of this step on the left picture of Fig. 3.5.
During this first step we also need to calculate the Green’s function from the end back to itself: G0,L1,1, G0,L2,2, G0,Ln,n and so on.
The second step is not much different from the first: we start from the last column and recursively attach to it the other columns of the system, one by one, saving every Green’s function calculated, as shown on the right of Fig. 3.5. These Green’s function will be written as G0,LN,N, G0,LN,N−1 and so on. In this case the superscript is L since we are working on the left part of the system (and it helps indeed in not mistaking G0,LN,N for G0
n n−1 1 N n+2 n+1 G G G Gn,1 n−1,1 N,n+2 N,n+1
+
+
Fig. 3.5 On the left: first step of the Metalidis-Bruno method. On the right: second step of the Metalidis-Bruno method.
being the Green’s function for the isolated N -th column, the second being the Green’s function for the whole system from the end back to itself). During this step the Green’s function G0,LN−1,N −1,G0,LN−2,N −2,G0,Ln+1,n+1 and so on are calculated, too.
The third step is to use the left and right Green’s function to find the Green’s functions for the whole system. In particular, we need to find G0
G0n,n and G0
n,1 for every n. To do so, we use the Green’s functions from the
left side from column 1 to n, and those from the right from column n + 1 to N. Since these calculation are done between two slices, and not between a slice and a column, we can use equation (3.8) and (3.15) to find G0n,n and
G0n,1, remembering that in the previous paragraphs the slice on the left was the one from c to d, and the one on the right ranged from a to b.
The Green’s functions G0N,ncan be found by substituting equation (3.10)
in (3.11).
During the last step we apply Dyson’s equation to the whole system. Like we have already said, the application of the probe is the perturbation represented by the V . Since the probe, or rather the tip, to use the same term used by the authors (and shorten our notation in the following calculations), affects only one point, and since they use real space both longitudinally and transversally, the potential V from the Dyson equation only affects the column n where the tip is applied.
Since we are interested in the Green’s function for the whole system, we will apply the Dyson equation as follows:
hN | G |1i = hN | G0VtipG|1i
GN,1 = G0N,1 + G0N,nV tip
The term G0
N,1 is not zero, of course, since it represents the Green’s function
for the whole system, from an end back to the other, when no tip is applied to it.
Anyway, once again we need to apply the Dyson equation, since we ignore the term Gn,1:
Gn,1 = G0n,1+ G0n,nV tip n,nGn,1
which by straightforward manipulation becomes
Gn,1 =I − G0n,nV tip n,n
−1
G0n,1 (3.31)
Thanks to (3.31) we can rewrite (3.30) in its final form:
GN,1 = G0N,1+ G0N,nV tip n,nI − G0n,nV tip n,n −1 G0n,1 (3.32)
Due to the real-space representation along both dimensions, and due to the tip being ideal, the matrix Vtip
n,n has only one element different from zero,
thus the inversion of the matrix I − G0 n,nVn,ntip
is pretty simple, since it is reduced to the simple inversion of a scalar if we consider the entire term Vn,ntipI − G0n,nVn,ntip−1. So, when using this method, the application of the tip is essentially a costless application from an algorithmical point of view, since in the recursive procedure the inversions are what defines the computational cost. The standard method, instead, involves the calculation of the entire
structure from the beginning for every application of the probe. If we work with M xM matrices, with a system of N columns, the standard method requires M4N2 inversions, while the Metalidis-Bruno approach only M3N, since the first three steps require each N − 1 inversions, as the evaluation of the Green’s functions for the isolated columns do, and the inversion of the matrices scales like M3.
As it can easily be noted, between M4N2 and M3N there is a factor
M N, which is the size of the system, or rather the number of places where we need to apply the probe and iterate the calculations.
