Quantum Mechanics
ECE 663-1, Fall ‘08
Why do we need it?
QM interference creates bandgaps and separates
metals from insulators and semiconductors
ECE 663-1, Fall ‘08
But how can electrons (particles)
create interference?
Lessons on http://hyperphysics.phy-astr.gsu.edu/hbase/quacon.html#quacon
or http://www.colorado.edu/physics/2000/quantumzone/ECE 663-1, Fall ‘08
Solar system model of atom
mv2/r = Zq2/4pe0r2
Centripetal
force
Electrostatic
force
Continuous radiation from orbiting electron
Pb1: Atom would
be unstable!
(expect nanoseconds
observe billion years!)
Spectrum of Helium
Transitions E0(1/n2 – 1/m2) (n,m: integers)
Pb2: Spectra of
atoms are discrete!
ECE 663-1, Fall ‘08
Why waves?
Only certain modes allowed (like a plucked string)
nl = 2pr (fit waves on circle)
Momentum ~ 1/wavelength
(DeBroglie)
p = mv = h/l
(massive classical particles  vanishing l)
This means angular momentum is quantized
mvr = nh/2p = nħ
From 2 equations, rn = (n2/Z) a0
a0 = h2e0/pq2m = 0.529 Å (Bohr radius)
ECE 663-1, Fall ‘08
Bohr’s suggestion
E = mv2/2 – Zq2/4pe0r
Using previous two equations
En = (Z2/n2)E0
E0 = -mq4/8ħ2e0 = -13.6 eV
= 1 Rydberg
Transitions E0(1/n2 – 1/m2) (n,m: integers)
Explains discrete atomic spectra
So need a suitable Wave equation so that imposing boundary
conditions will yield the correct quantized solutions
ECE 663-1, Fall ‘08
What should our wave equation look like?
∂2y/∂t2 = v2(∂2y/∂x2)
String
y
w
x
k
Solution: y(x,t) = y0ei(kx-wt)
w2 = v2k2
What is the dispersion (w-k) for a particle?
ECE 663-1, Fall ‘08
What should our wave equation look like?
Quantum theory:
E=hf = ħw (Planck’s Law)
p = h/l = ħk (de Broglie Law)
and E = p2/2m + U (energy of a particle)
w
Thus, dispersion we are looking for is
w  k2 + U
So we need one time-derivative
and two spatial derivatives
k
2y/∂tX
2 = v2(∂2y/∂x2)
∂X
ECE 663-1, Fall ‘08
Wave equation (Schrodinger)
iħ∂Y/∂t = (-ħ22/2m + U)Y
Kinetic
Energy
Potential
Energy
Makes sense in context of waves
Eg. free particle U=0
Solution Y = Aei(kx-wt) = Aei(px-Et)/ħ
w
k
We then get E = p2/2m = ħ2k2/2m
ECE 663-1, Fall ‘08
For all time-independent problems
iħ∂Y/∂t = (-ħ22/2m + U)Y = ĤY
Separation of variables for static potentials
Y(x,t) = y(x)e-iEt/ħ Oscillating solution in time
Ĥy = Ey, Ĥ = -ħ22/2m + U
BCs :
Ĥyn = Enyn (n = 1,2,3...)
En : eigenvalues (usually fixed by BCs)
yn(x): eigenvectors/stationary states
ECE 663-1, Fall ‘08
What does it all mean?
ECE 663-1, Fall ‘08
What does Y(x,t) represent?
• Probability amplitude of finding particle at x at time t
(Like electric field in phasor notation E, a complex #)
Y itself hard to measure so overall phase irrelevant
• Probability density P(x,t) = y*(x,t)y(x,t)
(Like intensity E*E  easier to detect)
• Must have ∫P(x,t)dx = 1 at all times
• Charge density r(x,t) = qP(x,t)
• Current density J(x,t) = qvP(x,t) (“v” to be defined later)
ECE 663-1, Fall ‘08
Averages
O = ∫|y(x,t)|2O(x)dx
O = ∫y*(x,t)Ô(x)y(x,t)dx
Symmetrized!
One can show that averages follow ‘classical rules’
Dx.Dp ≥ ħ/2
(Uncertainty Principle)
d<x>/dt = <v>
md<v>/dt = <F>
l
ECE 663-1, Fall ‘08
Examples of meaningful averages
n = Y*Y
(Electron Density)
ˆ
v=ˆ
p/m = -iħ/m (check KE)
J = q<v> = q∫y*(x)[-iħ/m]y(x)dx (Symmetrize !!!)
ˆ
J = iqħ/2m(YY*/x – Y*Y/x)
For plane waves Y = Y0eikx
J = n(ħk/m) = nqv
Also check: n/t + J/x = 0 (charge conservation)
ECE 663-1, Fall ‘08
Relation between Y and yn ?
yn s are allowed solutions (like mode shapes of a fixed string)
Their energies (‘frequencies’) are the eigenvalues En
Aside: They are orthogonal (independent), like modes of a string
∫y*n(x)ym(x)dx = 0 if n ≠ m
and normalized
∫y*n(x)yn(x)dx = 1
Y shows the actual solution (superposition of allowed ones)
In general, Y(x,t) = Sn an yn(x)e-iEnt/ħ
ECE 663-1, Fall ‘08
Why are levels quantized as Bohr suggested?
Due to confinement, like acoustic waves on a string
Same for H-atom
Vacuum
U(r) = -Zq2/4pe0r
-0.85 eV
-1.51 eV
-3.4 eV
-13.6 eV
ECE 663-1, Fall ‘08
Simplest eg of a confined system
U=
U=
Particle in a Box
U=0
y  eikx doesn’t satisfy boundary conditions
(should be zero at both ends)
But can superpose allowed solutions
y = Asin(kx) is zero at x = 0
What about x = L ?
ECE 663-1, Fall ‘08
Particle in a box
U=
U=
y = Asin(kx) is zero at x = L only
for special values of k
knL = np (n = 1, 2, 3, …)
U=0
Quantization condition
knL = np (n = 1, 2, 3, …)
ie,
L = nln/2 (exactly like acoustic waves)
ECE 663-1, Fall ‘08
Particle in a box
U=
w
yn = Asin(knx)
knL = np (n = 1, 2, 3, …)
U=
E3
E2
E1
k
k 1 k2 k3
U=0
Fixed k’s give fixed E’s
En = ħ2kn2/2m = ħ2n2p2/2mL2
Coeff A fixed by normalization
A=√2/L
Full Solution Yn = 2/L sin(npx/L) exp(-iħn2p2/2mL2 t)
ECE 663-1, Fall ‘08
Particle in a box
Yn=` 2/L sin(npx/L) exp(-iħn2p2/2mL2 t)
w
En = ħ2kn2/2m = ħ2n2p2/2mL2
E3
E2
E1
k
k1 k2 k3
Find J
J = iqħ/2m(YY*/x – Y*Y/x)
ECE 663-1, Fall ‘08
Extracting Physics from Pictures !!
U=
U=
Ground State y1(x) 
Smoothest curve with no kinks
Next mode y2(x) should also minimize energy
but be orthogonal to the first mode
(since modes must be independent!)
U=0
Hence the single kink!
Next one y3(x) must be orthogonal to
the other two
Since Kinetic energy ~ ∂2y/∂x2
Lowest energy wavefunction must have smallest curvature
It must also vanish at ends and be normalized to unity
ECE 663-1, Fall ‘08
Extracting Physics from Pictures !!
U =U= 
UU==
U=0
If we decrease box size, but keep
area under modes same, then each mode
must peak more
This increases curvature and thus energy
levels and their separation
Notice this from exact results, DEn  1/L2
U=0
(Uncertainty: Localizing particle increases
its energy!)
Small boxes (atoms), energies discrete and well separated
Large boxes (metal contacts), they are bunched up
ECE 663-1, Fall ‘08
Constant, non-zero potential
U=
knL = np still
U=
U = U0
But dispersion kn =
√2m(En-U0)/ħ2
ECE 663-1, Fall ‘08
Finite potential walls: thinking ‘outside the box’
U = U0
U = U0
22
exp(±ik’x),kk’= =2m(U
exp(±kx),
2m(E-U
0-E)/ħ
0)/ħ
U=0
Asinkx + Bcoskx, k = 2mE/ħ2
Solve piece-by-piece, and match boundary condns
(Match y, dy/dx)
Wavefunction penetrates out (“Tunneling”)
ECE 663-1, Fall ‘08
Particle at a Step
U0
x
0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Boundary Conditions:
y(0-) = y(0+)
dy/dx|0- = dy/dx|0+
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
E > U0
k,k’ real
ECE 663-1, Fall ‘08
Particle at a Step
U0
0
x
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Boundary Conditions:
1+r=t
k(1-r) = k’t
t = 2k/(k+k’)
r = (k-k’)/(k+k’)
ECE 663-1, Fall ‘08
Particle at a Step
U0
x
0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Transmission = curr transmitted/curr incident
Reflection Coeff = curr reflected/curr incident
ECE 663-1, Fall ‘08
Particle at a Step
U0
0
x
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Free particle J = ħk/m|y0|2, y0: amplitude
(More correctly, J = Re(ħk/m|y0|2), in case
k is complex)
Thus, T = Re(k’|t|2/k), R = |r|2
ECE 663-1, Fall ‘08
Particle at a Step
U0
0
x
Y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
T = Re(k’|t|2/k) = 4kk’/(k+k’)2
R = |r|2 = (k-k’)2/(k+k’)2
T+R=1
t = 2k/(k+k’)
r = (k-k’)/(k+k’)
ECE 663-1, Fall ‘08
Repeat for E < U0
U0
x
0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
E < U0
k’ imaginary
= ih
ECE 663-1, Fall ‘08
Repeat for E < U0
U0
0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
T = Re(ih|t|2/k) = 0
R = |r|2 = |k-ih|2/|k+ih|2 = 1
Expected?
x
k = 2mE/ħ2
h = 2m(U0-E)/ħ2
t = 2k/(k+ih)
r = (k-ih)/(k+ih)
ECE 663-1, Fall ‘08
Summary: Particle at a step
U0
x
0
T
Classical
1
Quantum
0
U0
E
$$$ Question: How do we make the curves approach each other?
ECE 663-1, Fall ‘08
Particle at a Barrier
eikx + re-ikx
Aeik’x + Be-ik’x
0
teikx
U0
L
x
Boundary Conditions:
y(0-) = y(0+)
dy/dx|0- = dy/dx|0+
k = 2mE/ħ2
y(L-) = y(L+)
dy/dx|L- = dy/dx|L+
E > U0
k,k’ real
k’ = 2m(E-U0)/ħ2
ECE 663-1, Fall ‘08
Particle at a Barrier
eikx + re-ikx
Aeik’x + Be-ik’x
0
1+ r = A + B
k(1-r) = k’(A-B)
Aeik’L + Be-ik’L = teikL
k’(Aeik’L – Be-ik’L) = kteikL
U0
L
teikx
x
2 = (1+k’/k)A + (1-k’/k)B
0 = (1-k’/k)Aeik’L + (1+k’/k)Be-ik’L
ECE 663-1, Fall ‘08
Particle at a Barrier
eikx + re-ikx
Aeik’x + Be-ik’x
0
teikx
U0
L
x
A = 2e-ik’L(1+k’/k)/[(1+k’/k)2e-ik’L– (1-k’/k)2eik’L]
B = 2eik’L(1-k’/k)[(1-k’/k)2eik’L-(1+k’/k)2e-ik’L]
2 = (1+k’/k)A + (1-k’/k)B
0 = (1-k’/k)Aeik’L + (1+k’/k)Be-ik’L
ECE 663-1, Fall ‘08
Particle at a Barrier
eikx + re-ikx
Aeik’x + Be-ik’x
0
teikx
U0
L
x
A = 2e-ik’L(1+k’/k)/[(1+k’/k)2e-ik’L– (1-k’/k)2eik’L]
B = 2eik’L(1-k’/k)[(1-k’/k)2eik’L-(1+k’/k)2e-ik’L]
teikL = 2(2k’/k)/ [(1+k’/k)2e-ik’L– (1-k’/k)2eik’L]
= 2kk’/[-i(k2+k’2)sink’L + 2kk’cosk’L]
ECE 663-1, Fall ‘08
Particle at a Barrier
eikx + re-ikx
Aeik’x + Be-ik’x
0
teikx
U0
L
x
T = 4k2k’2/[(k2+k’2)2sin2k’L + 4k2k’2cos2k’L]
Resonances: Maximum if cosk’L = 1, sink’L = 0
ie, k’L = 0, p, 2p, ...  L = 0, l/2, 3l/2, ....
teikL = 2(2k’/k)/ [(1+k’/k)2e-ik’L– (1-k’/k)2eik’L]
= 2kk’/[-i(k2+k’2)sink’L + 2kk’cosk’L]
ECE 663-1, Fall ‘08
More obvious if it’s a well
eikx + re-ikx
Aeik’x + Be-ik’x
0
teikx
U0
L
x
ie, k’L = 0, p, 2p, ...  L = 0, l/2, 3l/2, ....
Represent Resonances, with E > U0
(They would be bound states if E < U0)
Here
k = 2m(E-U0)/ħ2
k’ = 2mE/ħ2
ECE 663-1, Fall ‘08
Back to Barrier but lower Energy
k’ = ik
eikx + re-ikx
Ae-kx + Bekx
0
U0
L
teikx
x
T = 4k2k2/[(k2-k2)2sinh2kL + 4k2k2cosh2kL]
Large or wide barriers: kL >> 1, sinh(kL) ~ cosh(kL) ~ ekL/2
T ≈ 16 k2k2e-2kL/(k2+k2)2 ~ [16E(U0-E)/U02]e-2kL
teikL = 2ikk/[(k2-k2)sinhkL + 2ikhcoshkL]
ECE 663-1, Fall ‘08
Back to Barrier but lower Energy
eikx + re-ikx
Ae-kx + Bekx
0
U0
L
teikx
x
T ≈ [16E(U0-E)/U02]e-2kL
Even though E < V0, T > 0 (tunneling)
U(x)
E
x1
x2
More generally, WKB approximation
x2
T ~ exp[-2∫dx 2m[U(x)-E]/ħ2]
x1
ECE 663-1, Fall ‘08
Example: Tunneling
x2
T ~ exp[-2∫dx 2m[U(x)-E]/ħ2]
x1
Well
Barrier
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•
•
Alpha particle decay from nucleus
Source-Drain tunneling in MOSFETs
Single Electron Tunneling Devices (SETs)
Resonant Tunneling Devices (RTDs)
ECE 663-1, Fall ‘08
Example: Tunneling
Quantum states
(Speer et al, Science ’06)
Needed for designing
Heterojunctions/superlattices/
Photonic devices, etc
Upswing in
Current due
To Tunneling
Gloos
ECE 663-1, Fall ‘08
Barrier problem: Summary
U0
L
0
T
1
x
Classical
Quantum
0
Tunneling
T ~ e-2kL, k ~ (U0-E)
U0
E
Resonances
k’L = np, E = U0 + ħ2k’2/2m
ECE 663-1, Fall ‘08
Matlab plots
As barrier width increases, we recover particle
on a step
ECE 663-1, Fall ‘08
Matlab code
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subplot(2,2,4); % vary this from plot window to plot window
m=9.1e-31;hbar=1.05e-34;q=1.6e-19;
L=1e-9; %m, vary this from plot window to plot window!
U0=1; %Volts
Ne=511;E=linspace(0,5,Ne);
k=sqrt(2*m*E*q/hbar^2);%/m
eta=sqrt(2*m*(E-U0)*q/hbar^2);%/m
T=4.*k.^2.*eta.^2./((k.^2+eta.^2).^2.*sin(eta.*L).*sin(eta.*L) + 4.*k.^2.*eta.^2.*cos(eta.*L).*cos(eta.*L));
plot(E,T,'r','linewidth',3)
title('L = 15 nm','fontsize',15) % vary this from plot window to plot window!
grid on
hold on
tcl=2.*k./(k+eta);tcl=tcl.*conj(tcl);
Tcl=real((eta./k).*tcl);
plot(E,Tcl,'k--','linewidth',3)
gtext('step','fontsize',15)
ECE 663-1, Fall ‘08
Can we solve for arbitrary Potentials?
Approximation Techniques




Graphical solutions (e.g. particle in a finite box)
Special functions (harmonic oscillator, tilted well, H-atom)
Perturbation theory (Taylor expansion about known solution)
Variational Principle (assume functional form of solution
and fix parameters to get minimum energy)
Numerical Techniques (next)
ECE 663-1, Fall ‘08
Finite Difference Method
y
yn-1
xn-1
y=
yn-1
yn
yn+1
Uy =
One particular
mode
yn y
n+1
xn
xn+1
Un-1
Un-1yn-1
Unyn
Un+1yn+1
=
yn-1
Un
Un+1
= [U][y]
yn
yn+1
ECE 663-1, Fall ‘08
What about kinetic energy?
y
yn-1
xn-1
y=
yn-1
yn
yn y
n+1
xn
xn+1
(dy/dx)n = (yn+1/2 – yn-1/2)/a
(d2y/dx2)n = (yn+1 + yn-1 -2yn)/a2
yn+1
ECE 663-1, Fall ‘08
What about kinetic energy?
y
yn-1
xn-1
y=
yn-1
yn
yn y
n+1
xn
xn+1
-ħ2/2m(d2y/dx2)n = t(2yn - yn+1 - yn-1)
yn+1
t = ħ2/2ma2
ECE 663-1, Fall ‘08
What about kinetic energy?
y
yn-1
xn-1
yn y
n+1
xn
xn+1
-ħ2/2m(d2y/dx2)n = t(2yn - yn+1 - yn-1)
y=
yn-1
-t 2t -t
yn
yn+1
Ty =
-t 2t -t
-t 2t -t
yn-1
yn
yn+1
ECE 663-1, Fall ‘08
What about kinetic energy?
y
yn-1
xn-1
y=
yn-1
yn
yn y
n+1
xn
xn+1
[H] = [T + U]
yn+1
ECE 663-1, Fall ‘08
What next?
y
yn-1
xn-1
yn y
n+1
xn
xn+1
Now that we’ve got H matrix, we can
calculate its eigenspectrum
>> [V,D]=eig(H); % Find eigenspectrum
>> [D,ind]=sort(real(diag(D))); % Replace eigenvalues D by sorting, with index ind
>> V=V(:,ind); % Keep all rows (:) same, interchange columns acc. to sorting index
(nth column of matrix V is the nth eigenvector yn plotted along the x axis)
ECE 663-1, Fall ‘08
Particle in a Box
Results agree with analytical results E ~ n2
Finite wall heights, so waves seep out
ECE 663-1, Fall ‘08
Add a field
ECE 663-1, Fall ‘08
Or asymmetry
Incorrect, since we need
open BCs which we didn’t
discuss
ECE 663-1, Fall ‘08
Harmonic Oscillator
Shapes change from box: sin(px/L)  exp(-x2/2a2)
Need polynomial prefactor to incorporate nodes (Hermite)
E~n2 for box, but box width increases as we go higher up
 Energies equispaced E = (n+1/2)ħw, n = 0, 1, 2...
ECE 663-1, Fall ‘08
Add asymmetry
ECE 663-1, Fall ‘08
Matlab code
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t=1;
Nx=101;x=linspace(-5,5,Nx);
%U=[100*ones(1,11) zeros(1,79) 100*ones(1,11)];% Particle in a box
U=x.^2;U=U;%Oscillator
%U=[100*ones(1,11) linspace(0,5,79) 100*ones(1,11)];%Tilted box
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%Write matrices
T=2*t*eye(Nx)-t*diag(ones(1,Nx-1),1)-t*diag(ones(1,Nx-1),-1); %Kinetic Energy
U=diag(U); %Potential Energy
H=T+U;
[V,D]=eig(H);
[D,ind]=sort(real(diag(D)));
V=V(:,ind);
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% Plot
for k=1:5
plot(x,V(:,k)+10*D(k),'r','linewidth',3)
hold on
grid on
end
plot(x,U,'k','linewidth',3); % Zoom if needed
axis([-5 5 -2 10])
ECE 663-1, Fall ‘08
Grid issues
For Small energies, finite diff. matches exact result
Deviation at large energy, where y varies rapidly
Grid needs to be fine enough to sample variations
ECE 663-1, Fall ‘08
Summary
• Electron dynamics is inherently uncertain. Averages of
observables can be computed by associating the electron
with a probability wave whose amplitude satisfies the
Schrodinger equation.
•
Boundary conditions imposed on the waves create quantized
modes at specific energies. This can cause electrons to
exhibit transmission ‘resonances’ and also to tunnel through
thin barriers.
• Only a few problems can be solved analytically. Numerically,
however, many problems can be handled relatively easily.
ECE 663-1, Fall ‘08