Quantum Transport :Atom to Transistor,Schrödinger Equation: Examples
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Transcript of Quantum Transport :Atom to Transistor,Schrödinger Equation: Examples
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