Note on: Methods of theoretical chemistry Prof. Roi Baer
Lesson 1: Molecular electric conduction
I. The density of states in a 1D quantum wire
Consider a quantum system with discrete en-
ergy levels πΈπ, π = 1,2, β¦
Take an energy (not necessarily an energy
level) πΈ and ask: how many states π(πΈ) have
an energy less than πΈ. One can use the Heavi-
side function to answer this (see box on right),
and then:
π(πΈ) = β π(πΈ β πΈπ)
π
(1.1)
The derivative of the accumulative number of
states is the density of states (DOS) at energy
πΈ: π(πΈ) = πβ²(πΈ). Clearly:
π(πΈ) = β πΏ(πΈ β πΈπ)
π
(1.2)
Since β« πΏ(π₯)ππ₯β
0= 1 the DOS has the prop-
erty that for any smooth function π(πΈ) of the
energy:
β« π(πΈ)π(πΈ)ππΈ = β π(πΈπ)
π
(1.3)
Thus we can replace a summation over the states by integration: βπ β β« π(πΈ)ππΈ. We
never really use delta-functions when we speak of DOS. Almost always there are decay
mechanisms that widen these delta spikes. Thus the delta functions are usually replaced
by lorentzian functions or gaussians.
The Heaviside π(π₯) and Dirac πΏ(π₯) functions.
These have the property that for any π(π₯):
β« π(π₯)π(π¦ β π₯)ππ₯β
ββ
= β« π(π₯)ππ₯π¦
ββ
β« π(π₯)πΏ(π¦ β π₯)ππ₯β
ββ
= π(π¦)
The first integral shows that the definition of the
Heaviside function is:
π(π₯) = 1 ππ π₯ > 0
π(π₯) = 0 ππ π₯ β€ 0
The second integral shows that πΏ(π₯) is zero for all
π₯ β 0 but, putting π(π₯) = 1:
β« πΏ(π¦ β π₯)ππ₯β
ββ
= 1
Taking the derivative of the 1st integral wrt π¦ of
shows the relation between the two functions:
πβ²(π₯) = πΏ(π₯)
Note on: Methods of theoretical chemistry Prof. Roi Baer
Let us compute the DOS of an important basic system: a particle on a ring of circumfer-
ence πΏ. As with any free particle the energy is πΈπ =ππ
2
2π where π = 0, Β±1, Β±2, β¦ and ππ
is the linear momentum of the particle corresponding to an angular momentum βπ. For a
particle on a ring of radius π the angular momentum π Γ π is simply ππ. Thus ππ =
βπ
πΏ/2π= π
β
πΏ. We see that the linear momentum ππ = π Γ Ξπ is equally spaced where the
spacing is Ξπ =β
π. Compute π(πΈ). We find the momentum corresponding to πΈ:
π(πΈ)2
2π= πΈ β π(πΈ) = Β±β2ππΈ
And thus the number of states with energy less than πΈ is the number of states π with
momentum ππ in the interval βπ(πΈ) < ππ < π(πΈ). So: π(πΈ) = 2 Γ π(πΈ) β 1 where
π(πΈ) = [π(πΈ)
Ξπ] and we use the notation that [π₯] is the largest integer smaller than π₯. If we
think of a long wire, we can imagine that the integer aspect is smoothed out, so we can
write:
π(πΈ) = 2πΏπ(πΈ)
β
The density of states is just the derivative of this. From π2 = 2ππΈ, and taking derivative
w.r.t. πΈ we have ππβ² = π so that: π(πΈ) = πβ²(πΈ). Thus, in terms of momentum:
π(πΈ) = 2ππΏ
βπ(πΈ)
(1.4)
Note that the density of states is proportional to the length of the wire and inverse prop-
tional to the square root of the energy.
II. Theory of electric conduction through molecules
Consider an electron on the ring discussed above and ask: for a given state π what is the
electric current? Since the velocity is π£π =ππ
π, the time the electron completes a rotation
Note on: Methods of theoretical chemistry Prof. Roi Baer
is ππ =πΏ
π£π and so πΌπ =
π
ππ=
ππ£π
πΏ=
πππ
ππΏ. For a long wire we can write these quantities as
functions of πΈ
πΌ(πΈ) =π
ππΏπ(πΈ) (2.1)
Now, let us try to connect the current and the density of states π. Using (1.4), we obtain a
relation between the current going left (say) and the density of states of the βwireβ:
πΌ(πΈ)π(πΈ) =π
β (2.2)
Note we βlostβ a factor 2 because for each energy we consider only the βleftβ going
states. This relation is remarkable: when an electron is transported through an energy lev-
el πΈ of a device, the product of the current by the density of states is independent of πΈ, of
the electron mass or the wire length! In fact the result is a constant of nature π
β and is a
purely quantum effect (due to the presence of β)! What does this inverse proportionality
of the current to the density of states mean? It means that wen you calculate the current
resulting from all energy levels in the interval ΞπΈ ΞπΈ = πΈ2 β πΈ1 then the current is the
simple result:
πΌ(πΈ2, πΈ1) =π
ββ« πΌ(πΈ)π(πΈ)ππΈ
πΈ2
πΈ1
=π
β(πΈ2 β πΈ1) =
π
βΞπΈ
The total current of electrons in an energy range ΞπΈ is just π
βΞπΈ: indeed a remarkably
simple rule! This rule is now going to be used to compute the current in a molecular junc-
tion.
Now, consider a wire connected to two electrodes. We focus on βnon-interacting elec-
tronsβ. This means that each electron is independent (except for Pauli principle) from
each other electron. The electrodes inject electrons into the wire and after the electron
passes through the wire it is absorbed by the electrodes. The total current in the wire is
the difference between currents going from the left to the right and from the right to the
left:
πΌ = πΌπΏβπ β πΌπ βπΏ (2.3)
Note on: Methods of theoretical chemistry Prof. Roi Baer
To get the current from left to right we have to sum over all states with positive momen-
tum: πΌπΏβπ = 2 β π€πππΏβπ π (πΈπ)π,π>0 where π enumerates the states in the left lead and
ππΏβπ π (πΈπ) is the electric current in the right lead as a result of an impingement of an elec-
tron of energy πΈπ coming from the left lead. Note that such an electron coming from the
left can either pass to the right side and contribute to ππΏβπ π or to be reflected backwards.
Clearly there is a probability π(πΈπ) that the impinging electron passes to the right. Be-
cause the number of electrons impinging on the wire coming from left is:
ππΏβπ π (πΈ) = π€(πΈ)ππΏβπ (πΈ) (
π
βππΏ(πΈ))
ππΏβπ (πΈ) is a transmission coefficient. π€(πΈ) is the joint probability that the left lead actu-
ally has an electron in state π and that on the right the corresponding level of the same
energy is vacant (because of the Pauli principle, if the level is occupied the electron can-
not flow there since a level cannot be occupied by 2 electrons or more). The factor 2
comes from the two possible spin states. Clearly, because electrons are fermions the
population of levels is determined from the Fermi Dirac distributions in the two elec-
trodes, thus:
π€(πΈ) = ππΏ(πΈ)(1 β ππ (πΈ)) (2.4)
Where ππ(πΈ) =1
1+ππ½(πΈβππ)
, π = πΏ, π is the Fermi-Dirac distribution, ππ is the chemical
potential of the left or right electrode and (ππ΅π½)β1 is the temperature (ππ΅ is Boltzmannβs
constant). All these give:
πΌπΏβπ = 2 β ππΏ(πΈπ)(1 β ππ (πΈπ))ππΏβπ π (πΈπ)
π,π>0
= 2 β« ππΏ(πΈ)(1 β ππ (πΈ))ππΏβπ π (πΈ)ππΏ(πΈ)ππΈ
=2π
ββ« ππΏ(πΈ)(1 β ππ (πΈ))ππΏβπ (πΈ)ππΈ
(2.5)
ππΏβπ (πΈ) is a transmission coefficient. It answers the question: what is the probability that
an electron of energy πΈ moving to the right in the left lead ends up in the right lead.
Posed this way, such a question is ill-defined, but it is intuitively clearer.
Note on: Methods of theoretical chemistry Prof. Roi Baer
A similar expression will be used for πΌπ βπΏ. Thus
πΌπ βπΏ =2π
ββ« ππ (πΈ)(1 β ππΏ(πΈ))ππ βπΏ(πΈ)ππΈ
(2.6)
We will show in the next section that ππ βπΏ(πΈ) = ππΏβπ (πΈ) and we thus call both βthe
transmission coefficientβ π(πΈ). From Eq. (2.3) we then find:
πΌ =2π
ββ«[π(πΈ β ππΏ) β π(πΈ β ππ )]π(πΈ)ππΈ
(2.7)
This is Landauerβs equation for the current in a junction. It is very simple. The electronic
structure of the leads and the constriction go in only through the transmission coefficient.
At zero temperature, with no barriers (π(πΈ) = 1) we get the almost similar rule as we
had above:
πΌ =2π
β(ππΏ β ππ ) =
2π2
βΞπ
(2.8)
Where Ξπ = πβ1(ππΏ β ππ ) is the voltage bias between the two electrodes. The conduct-
ance is the derivative of the current by the bias. The conductance in the above case is thus
πΊ =ππΌ
πΞπ=
2π2
β.
III. 1D model of a molecular junction
We now build a model for a molecular junction. In the junction there are 3 entities: the
molecule, and the left/right metallic leads. Our model will assume that electrons are non-
interacting. This may sound strange, since electrons interact quite strongly, however, it is
well known that a good approximation for the behavior of electronic systems is that by
modifying the overall potential suitably they can be approximately assumed to be non-
interacting. In some phenomena interactions are extremely important, however, such
phenomena are beyond the scope of our treatment.
The model for the molecule is a well, for the present we will assume a square well, but
the methods we develop below are suitable for any shape. So our molecule is schemati-
cally shown in this diagram:
Note on: Methods of theoretical chemistry Prof. Roi Baer
The molecule is a potential well with several discrete energy levels up to a vacuum level.
The left lead has the following schematics:
When a molecular junction is formed one has:
E
Vacuum level
E1
E2
E3
E
Vacuum level
ΞΌL
E
E1
E2
E3 ππΏ
ππ
Note on: Methods of theoretical chemistry Prof. Roi Baer
Due to the double barrier tunneling phenomenon, we will discuss in the next section, one
can think of conduction as "going through" the energy levels of the molecule (this is a
good assumption when the molecule is weakly coupled to the metal). At zero temperature
one can see in the figure that conductance through molecular level 1 is impossible since
the corresponding energy levels on both sides are occupied (the Fermi-Dirac difference
terms at this energy give zero). In fact due to this, this level of the molecule will itself be
occupied by 2 electrons. I say it is nearly impossible because one can still imagine a pro-
cess in which one electron in level 1 of the molecule will jump up in energy to the lowest
unoccupied state of the right lead (energy ππ ) and an electron from the left lead at the
same energy will replace it. But as we shall see in calculations such events have very
small probability since they are essentially tunneling events (violate the energy conserva-
tion for a short period of time). Conduction through level 3 is also nearly impossible for
the same but opposite reason: there are no electrons of energy πΈ3 in the metals and thus
conductance will occur only through tunneling which has a small probability. However
conduction may occur more readily through level 2 since the left lead can supply elec-
trons that go through the molecule and come out in the vacant levels of the right lead.
IV. Calculating T(E): Transfer Matrix Method
In order to estimate the conductance we need a way to compute the transmission coeffi-
cient in the Landauer formula π(πΈ). We develop such a method, for 1D systems now.
Consider a potential step, going from potential ππΏ, ππ at π₯ = π. The wave functions with
energy πΈ on the left and right side of this point are:
ππΏ(π) = πΏ+ππππ + πΏβπβπππ ππΏβ² (π) = ππ(πΏ+ππππ β πΏβπβπππ) (3.1)
ππ (π) = π +ππππ + π βπβπππ ππ β² (π) = ππ(π +ππππ β π βπβπππ) (3.2)
Where β2π2
2π= πΈ β ππΏ,
2 2
2R
qE V . ππΏ(π) is composed of two states with exact mo-
mentum π: one going to the right with amplitude πΏ+ and the other goinf to the left, with
amplitude πΏβ. Similarly, the other wave functions.
Note on: Methods of theoretical chemistry Prof. Roi Baer
Consider the βwave functionβ π(π) = ππππ. It is certainly non-square integrable, so it is
not a usual wave function. We can always imagine it as the limit of a very wide square
integrable wave function, for example ππ,π(π) = πππππβ
π2
2π2/ββππ which is normalized
when is extremely large this function looks like a plane wave in the vicinity of the
origin. All calculations can be done with finite and the limit can be taken af-
ter the proper observables are calculated. For example, suppose we want to compute the
density of the particle at a point π0. Then:
ππ,π(π0) = |π(π0)|2 =π
βπ0
2
π2
βππ (3.3)
Keeping π0 fixed and increasing π indefinitely, we obtain:
ππ,π(π0) =1
βππ (3.4)
The fact that π becomes zero in the limit of π β β is expected since we have one particle
spread out all over space. Now we want to compute the current density ππ,π(π0) β‘
β
πImπβ(π0)πβ²(π0) as follows:
ππ,π(π0) =β
πIm {πβππππ
πβ
π02
π2
ββπππππππ
πβ
π02
π2
ββππ(ππ β
π0
π2)} =βπ
π
πβ
π02
π2
βππ (3.5)
Keeping π0 fixed and increasing π we obtain:
ππ,π(π0) =βπ
πππ,π(π0) (3.6)
While each quantity goes to zero when π β 0, the ratio of the current density to the den-
sity is βπ
π which is independent of π: it is a physical result.
Let us go back to the wave function which on the left is ππΏ and on the right is ππ . What
are the conditions we must demand the the fragments meet? The function must be contin-
uous and so must the 1st derivative. Thus:
( ππππ πβπππ
πππππ βππβπππ) (
πΏβ
πΏ+) = (
ππππ πβπππ
πππππ βππβπππ) (π β
π +) (3.7)
Define:
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π(π; π β π) = (ππππ πβπππ
πππππ βππβπππ)β1
( ππππ πβπππ
πππππ βππβπππ) (3.8)
And so:
π(π; π β π) (πΏβ
πΏ+) = (
π β
π +) (3.9)
π(π; π β π) is the transfer matrix. It determines the way the wave with wave-number π
interacts with the interface at π, after which the wave must have wave-number π. Note
that:
(ππππ πβπππ
πππππ βππβπππ)β1
=1
2π(
ππβπππ πβπππ
πππππ βππππ) (3.10)
Thus, π(π; π β π) =1
2π(
ππβπππ πβπππ
πππππ βππππ) ( ππππ πβπππ
πππππ βππβπππ) and:
π(π; π β π) =1
2π(
(π + π)ππ(πβπ)π (π β π)πβπ(π+π)π
(π β π)ππ(π+π)π (π + π)πβπ(πβπ)π) (3.11)
The transfer matrix can be used as in "Lego-land": You simply calculate the matrix at
each junction and then multiply them. For example, a wave hitting two consecutive inter-
faces. One at π1 and the other at π2. The potential at left is π0, after π1 it is π1 and after
π2 it is π2. Then the TM of the entire "device" is:
π = π(π2; π1 β π2)π(π1; π0 β π1) (3.12)
To calculate the probability to transfer from left to right, we impose the boundary condi-
tions that we have an incoming wave from left (amplitude 1) and an outgoing wave from
the right. There is no incoming wave from the right thus: π β = 0. There may be a reflect-
ed wave, of amplitude πΏβ. Thus the scattering is described by:
π (1
πΏβ) = (
π +
0)
(3.13)
There are 2 equations with 2 unknowns. After rearrangement, we get:
(π11 + π12πΏβ
π21 + π22πΏβ) = (
π +
0)
(3.14)
From the second equation the πΏ_ amplitude is:
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πΏβ = βπ21
π22
(3.15)
And the right amplitude is:
π + = π11 βπ21π12
π22=
det π
π22
(3.16)
We now consider the meaning of these amplitudes. Our physical problem consists of an
electron coming in from the left with a wave ππππΏπ and this causes two scattering waves:
πΏβπβπππΏπ on the left and π +ππππ π on the right (ππΏ and ππ may not be equal if the left and
right potential difference is non-zero). The current density on the left is the sum of cur-
rents of the incoming and reflected currents:
π½πΏ =βππΏ
πβ
βππΏ
π|πΏβ|2 =
βππΏ
π(1 β |πΏβ|2)
(3.17)
While on the right is just one wave:
π½π =βππ
π|π +|2
(3.18)
Now, the two current densities must be equal, π½πΏ = π½π , since there is no accumulation of
particles anywhere, thus ππΏ = ππ |π +|2 + ππΏ|πΏβ|2 or:
1 =ππ
ππΏ
|π +|2 + |πΏβ|2 (3.19)
Since |πΏβ|2 is the probability to be reflected, this equation shows that the probability to
be transmitted is given by:
π(πΈ) = 1 β |πΏβ|2 =ππ
ππΏ
|π +|2 (3.20)
In the appendix at the end of this document we show that the probability to be reflected
|πΏβ|2 for the π β πΏ direction is the same as for the πΏ β π direction and similarly for the
transmittance:
π(πΈ) =ππ
ππΏ
|π +πΏβπ |2 =
ππΏ
ππ
|π +π βπΏ|2
(3.21)
Note on: Methods of theoretical chemistry Prof. Roi Baer
V. Numerical Integration
We want to discuss the way to compute a controlled approximation to the definite inte-
gral of a function π(π₯):
πΌ(π) = β« π(π₯)ππ₯π
π
(4.1)
Formulas which give approximations that can be used for any function are called a quad-
rature rule. To be controlled, they must give an estimate for the error.
A. The Rectangle Rule
The simplest way to perform numerical integration πΌ(π) is to divide the interval [π, π] to
many (large π) equally spaced intervals, with spacing β =πβπ
π:
π = π₯0, π₯π+1 = π₯π + β, π₯π = π (4.2)
And then:
πΌ(π) β β ππβ
πβ1
π=0
= (π0 + π1 + β― ππβ1)β
(4.3)
(we use: ππ β‘ π(π₯π)). The error associated with this procedure can be estimated using
Taylor expansion to first order:
πΌ(π) = β« π(π₯)ππ₯π
π
= β β« π(π₯)ππ₯π₯π+1
π₯π
πβ1
π=0
= β β« (π(π₯π) + π(β))ππ₯π₯π+1
π₯π
πβ1
π=0
= β ππ
πβ1
π=0
β + π(β2)π = β ππ
πβ1
π=0
β + π(β)
(4.4)
Thus, when N is increased 2 fold, making the use of (4.3) twice as expensive, the error is
decreased by a factor of 2. This is a first order formula.
B. The Trapeze Rule
There is a simple way to improve. For π₯ β [π₯π, π₯π+1]. Instead of using the Taylor expan-
sion to first π(π₯) = ππ + π(β). Let us use it to second order:
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π(π₯) = ππ + ππβ²(π₯ β π₯π) + π(β2) (4.5)
where ππβ² β‘ πβ²(π₯π), thus:
β« π(π₯)ππ₯π₯π+1
π₯π
= ππβ +1
2ππ
β²β2 + π(β3) (4.6)
We need a finite-difference expression for ππβ² which is at least π(β). This is taken again
from the same Taylor expansion:
ππβ² =
ππ+1 β ππ
β+ π(β)
(4.7)
Plugging this into Eq. (4.6) we obtain after rearrangement:
β« π(π₯)ππ₯π₯π+1
π₯π
=ππ + ππ+1
2β + π(β3)
(4.8)
The error in this formula can be shown to be bounded by β3
12|πβ²β²| where |πβ²β²| is the largest
value of πβ²β² in the interval. Using this this in πΌ(π), we arrive at the "Trapezoidal Rule":
πΌ(π) = [π0
2+ (π1 + π2 + β― + ππβ1) +
ππ
2] β + π(β2)
(4.9)
The error is opbviously bounded by πβ3
12|πβ²β²| =
β2(πβπ)
12|πβ²β²|. Thus, at negligible cost (just
2 divisions by 2), we obtain a quadrature rule that gives a higher order integration: twice
as much work reduces the error by a factor 4. This formula approximates the integral as a
sum of trapeze pieces:
Note on: Methods of theoretical chemistry Prof. Roi Baer
a
Figure 1: The Trapez Rule (left) where a line is stretched between consecutive points. Simpson's rule
(right) where a parabola is cast between three consecutive points.
C. Simpson's rule
Next, we can use the Taylor theorem to the 4th order around an odd indexed point:
π(π₯) = π2π+1 + π2π+1β² (π₯ β π₯2π+1) +
1
2π2π+1
β²β² (π₯ β π₯2π+1)2
+1
6π2π+1
β²β²β² (π₯ β π₯2π+1)3 + π(β4)
(4.10)
Upon integration in the interval [π₯2π, π₯2π+2] the odd derivatives cancel so we have up to
5th order (two orders better than the trapez rule!):
β« π(π₯)ππ₯π₯2π+2
π₯2π
= π(2π+1)2β +1
2π2π+1
β²β²2β3
3+ π(β5)
(4.11)
In order to continue, we must find an approximation to π2π+1β²β² which is good to order 2
h .
Using the same Taylor expression (4.10) twice, once sticking π₯2π and once π₯2π+2:
π2π = π2π+1 β π2π+1β² β +
1
2π2π+1
β²β² β2 β1
6π2π+1
β²β²β² β3 + π(β4)
π2π+2 = π2π+1 + π2π+1β² β +
1
2π2π+1
β²β² β2 +1
6π2π+1
β²β²β² β3 + π(β4)
(4.12)
Adding these two expressions and rearranging, we have:
π2π+1β²β² =
π2π β 2π2π+1 + π2π+2
β2+ π(β2)
(4.13)
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Plugging into Eq. (4.11) and rearranging gives:
β« π(π₯)ππ₯π₯2π+2
π₯2π
= (π2π + 4π(2π+1) + π2π+2)β
3+ π(β5)
(4.14)
In this case, the error term is bounded by (2β)5
2880|πβ²β²β²| =
β5
90|πβ²β²β²|. Summing over π, and as-
suming π is even (π = 2π) we have Simpson's rule:
πΌ(π) = β β« π(π₯)ππ₯π₯2π+2
π₯2π
πβ1
π=0
= β (π2π + 4π(2π+1) + π2π+2)
πβ1
π=0
β
3+ π(β4)
= [(π0 + 4π1 + π2) + (π2 + 4π3 + π4) + β― ]β
3+ π(β4)
= [π0 + 4π1 + 2π2 + 4π3 + 2π4 + β― ]β
3+ π(β4)
(4.15)
Thus, Simpson's rule is:
πΌ(π) = [π0 + 4 β π2πβ1
π
π=1
+ 2 β π2π
πβ1
π=1
+ π2π]β
3+ π(β4)
(4.16)
With minute amount of additional work we obtain a much higher degree of approxima-
tion. Now, increasing work by a factor of two improves the accuracy by a factor 16.
One can continue this procedure to higher orders, but one has to always make sure that
the Nth derivatives of the integrand do not grow faster than the order of the method. Since
on computer we have finite arithmetic, numerical noise grows as you take higher and
higher order derivatives.
Going one step further in this analysis gives Booleβs rule:
πΌ(π) = [7π0 + 32 β π4πβ3
π
π=1
+ 12 β π4πβ2
π
π=1
+ 32 β π4πβ1
π
π=1
+ 14 β π4π
πβ1
π=1
+ 7π4π]β
90+ π(β6)
(4.17)
Note on: Methods of theoretical chemistry Prof. Roi Baer
VI. Appendix:
Notice also that for a particle coming in from the opposite side the same method will in-
volve instead of π(π1, π1 β π2)π(π0, π0 β π1) the following matrix product:
π(βπ0, π1 β π0)π(βπ2, π2 β π1) (just redraw the junction using a mirror image). No-
tice the following:
π(π; π β π) =1
2π(
(π + π)ππ(πβπ)π (π β π)πβπ(π+π)π
(π β π)ππ(π+π)π (π + π)πβπ(πβπ)π)
π(βπ; π β π) =1
2π(
(π + π)ππ(πβπ)π β(π β π)ππ(π+π)π
β(π β π)πβπ(π+π)π (π + π)πβπ(πβπ)π)
(4.18)
Note the special structure of the π matrices:
π(π, π β π) =1
2π(
π΄ π΅π΅β π΄β)
(4.19)
When one multiplies two matrices one obtains:
(π΄ π΅π΅β π΄β) (
πΆ π·π·β πΆβ) = (
π΄πΆ + π΅π·β π΄π· + π΅πΆβ
π΅βπΆ + π΄βπ·β π΅βπ· + π΄βπΆβ) (4.20)
The product is also of the same structure. These then form a group. The mirror change
moves from:
(π΄ π΅π΅β π΄β) β (
π΄ βπ΅β
βπ΅ π΄β ) (4.21)
We define a βmirror of a matrixβ:
(π΄ π΅π΅β π΄β)
π
= (π΄ βπ΅β
βπ΅ π΄β ) (4.22)
Clearly (ππ)π = π. It is easy to check that:
(ππ)π = ππππ (4.23)
One further easily sees that
det π = det ππ
π11 = (ππ)11 π22 = (ππ)22
(4.24)
From this we have:
Note on: Methods of theoretical chemistry Prof. Roi Baer
π(π, π β π) =1
2π(
π΄ π΅π΅β π΄β)
π(βπ, π β π) =1
2π(
π΄ π΅π΅β π΄β)
π
=π
ππ(π, π β π)π
(4.25)
We thus have that the total TM of the mirror obeys:
ππ βπΏ =ππ
ππΏ
(ππΏβπ )π (4.26)
Thus one sees that the mirror barrier has:
π +π =
ππ
ππΏπ + πΏβ
π =π12
M22= βπΏβ
(4.27)
Thus:
(ππΏ
ππ ) |π +
π|2 = (ππ
ππΏ) |π +|2 πΏβ
π =π12
M22= βπΏβ
(4.28)