Quantum Transport:Quantum Transport:Atom to Transistor

Prof. Supriyo DattaECE 659Purdue University

Network for Computational Nanotechnology

01.31.2003

Lecture 8: Schrödinger Equation: ExamplesRef. Chapter 2.3

Summary:

? - became a column vector U - became a diagonal NxN matrix

- became a tri-diagonal NxNmatrix

• And the corresponding eigenvectors and eigenvalues form a set by which wave functions may be represented as a linear combination of eigen vectors.

• Recall, the analytical form of the Schrödinger Equation:

where

• In numerical form:

With the eigenvalue equation [H]{a}= Ea{a}

)(2

22

rUm

H oprh

+∇−

=

),(),( trHtrt

i oprrh Ψ=Ψ

∂∂

{ } [ ]{ }?? Η=∂∂t

ih

22

2∇−

mh

{ } { } htiEeC ααα

α−∑=Ψ

? n

x1 2 3 n-1 n n+1 N

a

Last time we looked at the Schrödinger Equation in one dimension. Along ‘x’ a discrete set of lattice points were placed.

With the discrete set of lattice points shown on the right, the Schrödinger Equation was converted into a difference equation.

Retouch on Concepts00:00

Computationally it is rather easy to find matrix eigenvectors and eigenvalues. For example in Matlab we simply call “[V,D]=eig(H)”.

Finally, it is important to emphasize that eigenvalues correspond to energies and eigenvectors to wavefunctions.

Matrix D provides the eigenvalues in a diagonal matrix and Matrix V provides the corresponding eigenvectors as columns of a square matrix.

D= V=

E2

E1

E3

Wave Functions and Energies

Retouch on Concepts05:21

• The main advantage of the numerical method is that once it is solved for a particular dimensional structure, that same method may then be applied with little difficulty to other scenarios of the same nature.

Example: The slanted well has a complicated analytical solution (involving Airy Functions) but numerically the solution is the same as that implemented for the infinite square well.

U=0

• So, how do we go beyond the one dimensional numerical solution seen thus far?• In two dimensions the Hamiltonian operator looks like:

The lattice for this operator is shown below .

It is straight forward to extend the finite difference method to two dimensions, the main problem is size. i.e

),(2 2

2

2

22

yxUyxm

Hop +

∂∂

+∂∂

−=h

100x100 pts=10,000 pts

Slanted Well

Separating Variables08:39

• One solution is to separate the 2 -D problem into two 1-D. However, this is only possible if the potential is separable:

• Analytically this is done by separation of variables.Let,

Thus,

WhereFrom,

time independent form

)()(),( yUxUyxU yx +=

)()(),( yYxXyx ×=Ψ

)(X)(2

)( 2

22

xxUxm

xXE xx

+

∂∂−= h

)()(2

)( 2

22

yYyUyxm

yYEy

+

∂∂−

=h

yx EEE +=

),(),(2

),( 2

2

2

22

yxyxUyxm

yxE Ψ

+

∂∂+

∂∂−=Ψ h

• Numerically this translates into two 1 -D lattices (one for the x-axis and one for the y-axis).

• For example, take a two dimensional infinite square well:

y

x

Y(y)

X(x)

0

2-D Infinite Square Well

Reducing the Lattice14:32

• Numerically if we split the x and y lattices by 100 points. How many eigenvectors will result?

y

•Ans: Total=100x100=10000 eigenvectors . We combine X(x) and Y(y) functions such that? nm (x,y)= Xn (x) Ym (y)

x01

1 2

2

100

100

2

2

1

1

• Notice that two eigenvectors with the same m+n value do not necessarily have the same energy:• Take eigenvectors {? 22 } and {? 13 }, which has more energy?

Recall:

(E13= E1+E3) > (E22= E2+E2)

• Now what about a 3-D infinite square well or a 3-D box? The same thing, separation of variables gives eigenfunctions of ? 111 , ? 112,….etc.

2

222

2mLn

En

πh=

∴

Infinite Square Well

2-D Infinite Square Well22:00

• Consider the Hydrogen Atom, can it be handled by separation of variables?

• The Hydrogen atom cannot be separated in Cartesian coordinates

But it can be separated in spherical coordinates such that U becomes

∂∂+

∂∂+

∂∂−= 2

2

2

2

2

22

2 zyxmHop

h

222

2

xpe4 0 zy

q

++

−

rpe4 0

2q−

• Note: A similar approach can be taken with all atoms but not with molecules because you don’t have this spherical symmetry.

• So separation of variables in spherical coordinates gives the radial Hamiltonian as:

• In general ? becomes

rpe4 0

2q−

+

∂∂

+∂∂−

22

22 )?,(22 r

Lrrrm

φh

)(?)()(f )?,(r, φΦΘ=Ψ rR

Hydrogen Atom27:16

Time – 34:40

• Note: The radial equation

can be simplified by letting R(r) = f (r) /r

Thus,

R(r)rpe4

)1(2 0

2

2

2

−

+−+

qrll

mh

2

)()(')(rrf

rrf

drrdR

−=

3222

2 )(2)()()()(r

rfr

rfrrf

rrf

drrRd

+′

−′

−′′

=

rrf

rRdrd

rdrd )("

)(2

2

2

=

+∴

• Hence, the entire radial equation simplifies to:

• Furthermore, the radial equation gives the s,p,d …. levels of the hydrogen atom for different values of “l”:s: l=0p: l=1D: l=2,etc

f(r)4

)(2

)(0

2

2

22

rq

drrfd

mrEf

πε−

−=

h

f(r))1(

2 2

2

rll

m+

+h

)(R2

2)( 2

22

rdrd

rdrd

mrER

+

−=

h

Hydrogen Atom

• How do we get a numerical solution to the hydrogen atom?

• Ans: We set up a discrete axis along a radial axis and solve the appropriate radial Hamiltonian matrix for l=0,1,2….

• Back to the hydrogen Hamiltonian, where does l(l+1) come from?

•Ans: l(l+1) forms the eigenvalues of the spherical harmonics or angular part of

• Where,

L(?,F )=

and

• Note: The angular term, , is the same for all atoms which results in isotropic behavior. It is only the radial term that changes.

,??sin

1

??sin

??sin1

2

2

2 ∂∂+

∂∂

∂∂

Y

Y

mmll ll Y)1()Y,L( +−=φθ

mlY

r0

)(?)()(f )?,(r, φΦΘ=Ψ rR

)(?)(Y φmlm

l ΦΘ=

Hydrogen Atom40:54

• Closing Notes:

• For higher dimensions use separation of variables whenever poss ible.

• Separable solutions are of the form

•The angular term is the same for all separable spherical solutions.

• For hydrogen-like atoms the radial term can be treated as a 1-D numerical problem with coulomb potential

and angular “potential”

f )?,()(f )?,(r, mlYrR=Ψ

rpe4 0

2Zq−

2

2

2)1(

mrll ++ h

Summary49:01