quantum transport by supriyo dadfasdgfasgtta.pdf

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1. QUANTUM TRANSPORT
BY SUPRIYO DATTA

2. OUR OBJECTIVE IS TO SOLVE SCHRODINGEREQUATION FOR THE
BELOW MODEL
The s-contact is a theoretical construct used to model the impact of scattering processes within the channel of
a nanoscale transistor
The source is the terminal through which carriers (electrons or holes) enter the channel of the transistor. It is
typically connected to a voltage supply or ground, depending on the type of transistor and its configuration.
Thedrainistheterminalthroughwhichcarriersexitthechannelofthetransistor.Itisusuallyconnectedtoa
loadoranotherpartofthecircuitwherethecurrentisneeded
Solving the schrodingerequation for
this model is our objective

3. RUTHERFORD MODEL OF ATOM
Rutherford model assumed the electrons are revolving around the nucleus in a circular orbit . This was a phenomenal
Because the previous model assumed both positive and negative particle was spread out uniformly inside the atom .
The above two basic equations developed form this model
But acceleration charges will lose energy in the form of electromagnetic radiation , therefore this theory
Cannot explain the stability of the atom

4. BOHR MODEL OF ATOM

5. BOHR’S ATOM MODEL
Bohr postulated that the electrons are moving in orbits whose radius are
The energy and radius of stable orbits given by
The energy difference between the orbits is given by

6. QUANTUM MECHANICAL MODEL OF ATOM

7. QUANTUM MECHANICAL MODEL OF ATOM
For hydrogen atom the schrodingerequation can be directly solved
The variable separation method will lead to a solution given below

8. Well, it’s the same way with the Schrodinger equation. If there is no confining potential (U = 0), we can write the
solutions to the 1D Schrodinger equation:
To keep things simple let us consider the vibrations u(x, t) of a one-dimensional (1D) string described by the 1D
wave equation:

9. Another kind of box that we will often use is a ring (Fig. 2.1.4) where the end point at x = L is connected back to
the first point at x = 0 and there are no ends. Real boxes are seldom in this form but this idealization is often used
since it simplifies the mathematics.
Electron density and probability current density: An electron with a wavefunction (x, t) has a probability of ∗dVof
being found in a volume dV. When a number of electrons are present, we could add up ∗for all the electrons to
obtain the average electron density n(x, t). What is the corresponding quantity we should sum to obtain the
probability current density J(x, t)

10. METHOD OF FINITE DIFFERENCES
The Schrodinger equation for a hydrogen atom can be solved analytically, but most other practical problems require a
numerical solution. In this section I will describe one way of obtaining a numerical solution to the Schrodinger equation.
Most numerical methods have one thing in common –they use some trick to convert the
We can represent the wavefunction (x, t) by a column vector
{ψ1(t) ψ2(t) ··· ···}T (“T” denotes transpose) containing its
values around each of the lattice points at time t. Suppressing
the time variable t for clarity, we can write. This representation
becomes exact only in the limit a →0, but as long as a is
smaller than the spatial scale on which varies, we can expect it
to be reasonably accurate

11. The next step is to obtain the matrix representing the Hamiltonian operator

12. Eigenvalues and eigenvectors: Now that we have converted the Schrodinger equation into a matrix equation.
We can solve that using standard procedure.
For example-consider first the “particle in a box” problem the Hamiltonian is given by

13. SELF CONSISTENT FIELD
In general, the ionization levels for multielectron atoms can be calculated approximately from the Schrodinger equation
by adding to the nuclear potential Unuc(r), a self-consistent field USCF(r) due to the other electrons
Step 1. Guess electronic potential USCF(r).
Step 2. Find eigenfunctions and eigenvalues from Schrodinger equation.
Step 3. Calculate the electron density n(r).
Step 4. Calculate the electronic potential USCF(r).
Step 5. If the new USCF(r) is significantly different from last guess, update USCF(r) and go back to Step 2. If the new
USCF(r) is within say 10 meVof the last guess, the result has converged and the calculation is complete
The nuclear potential for atoms is derived from the nuclear charge at the origin. The self-consistent field (SCF) potential
comes from the other electrons. To find the SCF potential, we need the electron wavefunctions, which are determined by
solving the Schrödinger equation with the SCF potential. This process is iterative to ensure consistency.

14. Step 1: Initial Guess for Electron Wavefunctions
•Start with an initial guess for the electron wavefunctions, often based on hydrogen-like orbitals or another reasonable
approximation.
•Use these initial wavefunctions to calculate the initial electronic charge distribution and potential.
Step 2: Solving the Schrödinger Equation
•Use a similar method as for the hydrogen atom, though an analytical solution is usually not possible.
•The potential USCF(r)U_{\text{SCF}}(r) is generally not isotropic, but for atoms, it can be assumed to be isotropic
without significant error.
•The dependence on r is complex, making an analytical solution impossible. However, numerically solving the Schrödinger
equation with any U(r)U(r) is as straightforward as solving it for the hydrogen atom.
Step 3: SummingProbabilityDistributions
Sum the probability distributions for all occupied eigenstates:
If the charge distribution is assumed to be isotropic, we can write:

15. Step 4: Calculating the SCF Potential
Use electrostatics to show that:
•The two terms arise from the contributions due to the charge within a sphere of
radius r and the charge outside of this sphere.
•The first term is the potential at r outside a sphere of charge, and the second term
is the potential at r inside a sphere of charge
•The charge density for each eigenstate should exclude the eigenstate under
consideration, as no electron feels any repulsion due to itself.
•For example, in silicon with 14 electrons, the self-consistent field includes all but
one of these electrons for each level.
•It is more convenient to scale the total charge density by the factor (Z−1)/Z(Z-1)/Z,
preserving spherical symmetry and usually not causing significant differences

16. RELATION TO MULTI-ELECTRON PICTURE
Multi-electron Schrodinger equation: It is important to recognize that the SCF method is really an approximation that is
widely used only because the correct method is virtually impossible to implement. For example, if we wish to calculate
the eigenstates of a helium atom with two electrons, we need to solve a two-electron Schrodinger equation of the form
How would we use our results (in principle, if not in practice) to construct a one-electron energy level diagram like the
ones we have been drawing

17. For the ionization levels, the one-electron energies εnrepresent the difference between the ground state energy EG(N)
of the neutral N-electron atom and the nth energy level En(N −1) of the positively ionized (N −1)-electron atom:
To calculate the affinity levels we should look at the difference between the ground state energy EG(N) and the nth
energy level En(N + 1) of the negatively ionized (N + 1)-electron atom:
Note that if we want the energy levels to correspond to optical transitions then we should look at the difference
between the ground state energy EG(N) and the nth energy level En(N) of the N-electron atom, since visible light does
not change the total number of electrons in the atom, just excites them to a higher energy:

18. How do we choose this effective potential? If we use Uee(N) to denote the total electron–electron interaction energy
of an N-electron system then the appropriate USCF for the ionization levels is equal to the change in the interaction
energy as we go from an N-electron to an (N −1)-electron atom:
Similarly the appropriate USCF for the affinity levels is equal to the change in the interaction energy between an N-
electron and an (N + 1)-electron atom:
The electron–electron interaction energy of a collection of N electrons is proportional to the number of distinct pairs:
The self consistent field for the first two cases are given below

19. Hartree approximation: In large conductors (large R) U0 is negligible and the distinction between Z and (Z −1) can be
ignored. The self-consistent potential for both ionization and affinity levels is essentially the same and the expression
can be generalized to obtain the standard expression used in density functional theory (DFT)
which tells us that the self-consistent potential at any point r is equal to the change in the electron–electron interaction
energy due to an infinitesimal change in the number of electrons at the same point. If we use the standard expression for
Ueefrom classical electrostatic
It is solution to poison’s equation
It is then more convenient to solve a modified form of the
Poisson equation that allows a spatially varying relative
permittivity:

20. Correlation energy: The actual interaction energy is less than that predicted by Eq. (3.2.9) because electrons can
correlate their motion so as to avoid each other –this correlation would be included in a many-electron picture but is
missed in the one-particle picture. One way to include it is to write
where g is a correlation function that accounts for the fact that the probability of finding two electrons simultaneously
atrand r is not just proportional to n(r)n(r ), but is somewhat reduced because electrons try to avoid each other
(actually this correlation factor is spin-dependent, but we are ignoring such details). The corresponding selfconsistent
potential is also reduced
The basic effect of the correlation energy is to add a negative term Uxc(r) to the Hartree term UH(r) discussed above
local density approximation (LDA) -

21. Formation of Na+Cl−from individual Na and Cl atoms with a 3s electron from Na “spilling over” into the 3p levels of Cl
thereby lowering the overall energy.

24. BAND STRUCTURE
This chapter will demonstrate how this method can be applied to calculate energy eigenvalues for an infinite periodic
solid, using a few simple examples to show that band structure can be determined by solving a matrix eigenvalue equation.
The matrix [h(k )] is (b ×b) in size, b being the number of basis orbitals per unit cell. The summation over m runs over all
neighboringunit cells (including itself) with which cell n has any overlap (that is, for which Hnmis non-zero). The sum can
be evaluated choosing any unit cell n and the result will be the same because of the periodicity of the lattice. The
bandstructurecan be plotted out by finding the eigenvalues of the (b ×b) matrix [h(k )] for each value of k and it will
have b branches, one for each eigenvalue. This is the central result which we will first illustrate using toy exampl

25. Toy examples
Let us start with a toy one-dimensional solid composed of N atoms (see Fig. 5.1.1). If we use one orbital per atom we can
write down a (N ×N) Hamiltonian matrix using one orbital per atom (the off-diagonal element has been labeledwith a
subscript “ss,”
We have used what is called the periodic boundary condition (PBC), namely, that the Nth atom wraps around and overlaps
the first atom as in a ring. This leads to non-zero values for the matrix elements H1,N and HN,1 which would normally be
zero if the solid were abruptly truncated. The PBC is usually not realistic, but if we are discussing the bulk properties of a
large solid then the precise boundary condition at the surface does not matter and we are free to use whatever boundary
conditions make the mathematics the simplest, which happens to be the PBC.
eigenvalues of the matrix [H]

26. Why is it that we can write down the eigenvalues of this matrix so simply? The reason is that because of its periodic
nature, the matrix equation E(ψ) = [H](ψ) consists of a set of N equations that are all identical in form and can all be
written as (n = 1, 2,... N)
This set of equations can be solved analytically by the ansatz:
This is a result of two factors. Firstly, periodic boundary conditions require the wavefunction to be periodic with a period of
Na and it is this finite lattice size that restricts the allowed values of k to the discrete set kαa = α2π/N (see Eq. (5.1.2)).
Secondly, values of ka differing by 2π do not represent distinct states on a discrete lattice. The wavefunctions

27. represent the same state because at any lattice point xn= na,
They are NOT equal between two lattice points and thus represent distinct states in a continuous lattice. But once we
adopt a discrete lattice, values of kαdiffering by 2π/a represent identical states and only the values of kαa within a range
of 2π yield independent solutions. In principle, any range of size 2π is acceptable, but it is common to restrict the values of
kαa to the range (sometimes called the first Brillouin zone)

28. Lattice with a basis: Consider next a one-dimensional solid whose unit cell consists of two atoms as shown
Using one orbital per atom we can write the matrix representation of [H] as

30. General result
matrix [Hnm] of size (b ×b), b being the number of basis functions per unit cell. We can write the overall matrix
equation in the form
The important insight is the observation that this set of equations can be solved by the ansatz The important
insight is the observation that this set of equations can be solved by the ansatz
It is a consequence of the periodicity of the lattice and it ensures that when we substitute our ansatz. we obtain

32. Brillion zone
The Brillouin zone is the Wigner-Seitz cell in the reciprocal lattice. It is the region in reciprocal space that is closer to
a given reciprocal lattice point than to any other.

33. Reciprocal lattice: In general, if the direct lattice is not rectangular or cubic, it is not possible to construct the reciprocal
lattice quite so simply by inspection. We then need to adopt a more formal procedure as follows. We first note that any
point on a direct lattice in 3D can be described by a set of three integers (m, n, p) such that
It is easy to see that this formal procedure for constructing the reciprocal lattice leads to the lattice shown in Fig.
5.2.3b if we assume the real-space basis vectors to be a1 = xaˆ , a2 = ybˆ , a3 = zcˆ . Equation (5.2.9) then yields
Using Eq. we can now set up the reciprocal lattice shown in Fig.. Of course, in this case we do not really need the formal
procedure. The real value of the formal approach lies in handling non-rectangular lattices, as we will now illustrate with a
2D example

34. 2D Example: Graphene Layer
•Graphene Structure: Carbon atoms on the surface of a graphite sheet (graphene layer) are arranged in a
hexagonal pattern. This structure is not truly periodic as adjacent carbon atoms have different environments.
•Unit Cell: By grouping two carbon atoms into a unit cell, the lattice of unit cells becomes periodic, with
each site having an identical environment.
•Direct Lattice: The direct lattice shows the periodic arrangement of unit cells with basis vectors.
•Reciprocal Lattice: The reciprocal lattice is constructed using basis vectors determined by specific
conditions. The Brillouin zone is obtained by drawing perpendicular bisectors of lines joining the origin to
neighboringpoints on the reciprocal lattice.
•Energy Eigenvalues: For a given wave vector kk, the corresponding energy eigenvalues are obtained from
specific equations. The size of the matrix depends on the number of basis functions per unit cell.
•Graphene Band Structure: In graphene, the conduction and valence band states around the Fermi energy
are primarily formed from the 2pz orbitals of carbon atoms. This results in a simplified matrix for calculating
energy eigenvalues.

35. Common semiconductors
All the commonsemiconductors (like gallium arsenide) belong to the diamond structure which has ahe diamond structure
consists of two interpenetrating face-centeredcubic (FCC) lattices. For example, if we look at GaAs, we find that the gallium
atoms occupy the sites on an FCC lattice. The arsenic atoms occupy the sites of a different FCC lattice offset from the
previous one by a quarter of the distance along the body diagonal –that is, the coordinates of this lattice can be obtained by
adding (xˆ + yˆ + zˆ)a/4 to those of the first one. If a gallium atom is located at the origin (0 0 0)a/4 then there will be an
arsenic atom located at (xˆ + yˆ + zˆ)a/4, which will be one of its nearest neighbors. Actuallyit will also have three more
arsenic atoms as nearest neighbors.unitcell consisting of two atoms, a cation (like gallium) and an anion (like arsenic).
For elemental semiconductors like silicon, both cationic and anionic sites are occupiedby the same atom.

36. Spin orbit coupling
The bandstructurewe have obtained is reasonably accurate but does not describe the top of the valence band very well. To
obtain the correct bandstructure, it is necessary to include spin–orbit coupling
How could we modify the Schr¨odingerequation so that each level becomes two levels with identical energies? The
answer is simple. Replace
In simple terms, spin-orbit coupling occurs because the motion of an electron around a nucleus creates a magnetic field,
which then interacts with the electron's intrinsic magnetic moment (its spin). This interaction leads to a splitting of energy
levels, which can be observed in the spectral lines of atoms.

37. Typicallyin solids the velocities are not high enough to require this, but the electric fields are very high near the nuclei of
atoms leading to weak relativistic effects that can be accounted for by adding a spin–orbit correction Hsoto the
Schr¨odingerequation:
c being the velocity of light in vacuum. The spin–orbit Hamiltonian Hsois often written as

38. Usually the spin–orbit matrix elements are significant only if both orbitals are centeredon the same atom, so that we
expect a matrix of the form
Similarlywe can go on with the 21 and the 22 components. As it turns out, all these matrix elements can be shown to
be zero from symmetry arguments if we assume the potential U(r) to be spherically symmetric as is reasonable for
atomic potentials. The same is true for the s* orbitals as well. However, some of the matrix elements are non-zero
when we consider the X, Y, and Z orbitals. These non-zero matrix elements can all be expressed in terms of a single
number δafor the anionic orbitals:

40. Heavy hole, light hole, and split-off bands: The
nature of the valence band wavefunction near the
gamma point (kx= ky= kz= 0) plays a very
important role in determining the optical properties
of semiconductor nanostructures. At the gamma
point, the Hamiltonian matrix has a relatively simple
form because only g0 is nonzero, while g1, g2, and g3
are each equal to zero (see Eq. (5.3.3)). Including
spin–orbit coupling the Hamiltonian decouples into
four separate blocks at the gamma point:
The transformed Hamiltonian for Block III looks like