1. The strange world of Quantum Mechanics
“It is impossible for both The Ultimate Answer and
the Ultimate Question to be known about in the same
universe, as they will cancel each other out and take
the Universe with them, to be replaced by something
even more bizarre..”

2. QM of solids QM interference creates bandgaps and separates
metals from insulators and semiconductors

3. Wave equation (Schrodinger)
iħ∂Y/∂t = (-ħ22/2m + U)Y
Kinetic Potential
Energy Energy

4. 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
Ĥy = Ey, Ĥ = -ħ22/2m + U
Oscillating solution in time
BCs : Ĥyn = Enyn (n = 1,2,3...)
En : eigenvalues (usually fixed by BCs)
yn(x): eigenvectors/stationary states

5. 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)

6. Why are levels quantized as Bohr suggested?
Due to confinement, like acoustic waves on a string
Same for H-atom
Vacuum
-0.85 eV
-1.51 eV
U(r) = -Zq2/4pe0r
-3.4 eV
-13.6 eV

7. Simplest eg of a confined system
U = 
U = 0
Particle in a Box
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 ?

8. Particle in a box U =  y = Asin(kx) is zero at x = L only
for special values of k
knL = np (n = 1, 2, 3, …)
Quantization condition
knL = np (n = 1, 2, 3, …)
ie,
L = nln/2 (exactly like acoustic waves)

9. Particle in a box yn = Asin(knx) knL = np (n = 1, 2, 3, …) U =  w
Fixed k’s give fixed E’s
En = ħ2kn2/2m = ħ2n2p2/2mL2
Coeff A fixed by normalization
A=√2/L k3 E2 k2 E1 k1
Full Solution Yn = 2/L sin(npx/L) exp(-iħn2p2/2mL2 t)

10. Particle in a box Yn=` 2/L sin(npx/L) exp(-iħn2p2/2mL2 t) wk k2 k3 k1 E1 E2 E3
En = ħ2kn2/2m = ħ2n2p2/2mL2
Find J
J = iqħ/2m(YY*/x – Y*Y/x)

11. Constant, non-zero potential
U = 
U = U0
knL = np still
But dispersion kn = √2m(En-U0)/ħ2

12. Finite potential walls: thinking ‘outside the box’
U = U0
U = U0
exp(±ik’x), k’ = 2m(E-U0)/ħ2
exp(±kx), k = 2m(U0-E)/ħ2
Asinkx + Bcoskx, k = 2mE/ħ2
U = 0
Solve piece-by-piece, and match boundary condns
(Match y, dy/dx)
Wavefunction penetrates out (“Tunneling”)

13. Particle at a Step U0 x y(x) = eikx + re-ikx , x < 0 k = 2mE/ħ2
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
Boundary Conditions:
y(0-) = y(0+)
dy/dx|0- = dy/dx|0+
E > U0
k,k’ real

14. Particle at a Step U0 x y(x) = eikx + re-ikx , x < 0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
t = 2k/(k+k’)
r = (k-k’)/(k+k’)
Boundary Conditions:
1 + r = t
k(1-r) = k’t

15. Particle at a Step U0 x y(x) = eikx + re-ikx , x < 0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Transmission = curr transmitted/curr incident
Reflection Coeff = curr reflected/curr incident

16. Particle at a Step U0 x y(x) = eikx + re-ikx , x < 0
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
Free particle J = ħk/m|y0|2, y0: amplitude
Thus, T = Re(k’|t|2/k), R = |r|2
(More correctly, J = Re(ħk/m|y0|2), in case
k is complex)

17. Particle at a Step U0 x Y(x) = eikx + re-ikx , x < 0
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’)

18. Repeat for E < U0 U0 x y(x) = eikx + re-ikx , x < 0 k = 2mE/ħ2
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
E < U0
k’ imaginary
= ih

19. Repeat for E < U0 U0 x y(x) = eikx + re-ikx , x < 0 k = 2mE/ħ2
y(x) = eikx + re-ikx , x < 0
= teik’x, x > 0
k = 2mE/ħ2
h = 2m(U0-E)/ħ2
T = Re(ih|t|2/k) = 0
R = |r|2 = |k-ih|2/|k+ih|2 = 1
t = 2k/(k+ih)
r = (k-ih)/(k+ih)
Expected?

20. Summary: Particle at a stepx T 1
Classical
Quantum E U0
$$$ Question: How do we make the curves approach each other?

21. Particle at a Barrier eikx + re-ikx Aeik’x + Be-ik’x teikx U0 x L
Boundary Conditions:
y(0-) = y(0+)
dy/dx|0- = dy/dx|0+
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
y(L-) = y(L+)
dy/dx|L- = dy/dx|L+
E > U0
k,k’ real

22. Particle at a Barrier eikx + re-ikx Aeik’x + Be-ik’x teikx U0 x L
1+ r = A + B
k(1-r) = k’(A-B)
2 = (1+k’/k)A + (1-k’/k)B
Aeik’L + Be-ik’L = teikL
k’(Aeik’L – Be-ik’L) = kteikL
0 = (1-k’/k)Aeik’L + (1+k’/k)Be-ik’L

23. Particle at a Barrier eikx + re-ikx Aeik’x + Be-ik’x teikx U0 x L
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

24. Particle at a Barrier eikx + re-ikx Aeik’x + Be-ik’x teikx U0 x L
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]

25. Particle at a Barrier eikx + re-ikx Aeik’x + Be-ik’x teikx U0 x L
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]

26. More obvious if it’s a well
eikx + re-ikx Aeik’x + Be-ik’x 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

27. Back to Barrier but lower Energy
k’ = ik
eikx + re-ikx Ae-kx + Bekx teikx U0 L 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]

28. Back to Barrier but lower Energy
eikx + re-ikx Ae-kx + Bekx teikx U0 L x
T ≈ [16E(U0-E)/U02]e-2kL
U(x)
Even though E < V0, T > 0 (tunneling) E x1 x2
More generally, WKB approximation
T ~ exp[-2∫dx 2m[U(x)-E]/ħ2] x1 x2

29. Example: Tunneling T ~ exp[-2∫dx 2m[U(x)-E]/ħ2] Well Barrier
Alpha particle decay from nucleus
Source-Drain tunneling in MOSFETs
Single Electron Tunneling Devices (SETs)
Resonant Tunneling Devices (RTDs)

30. Example: Tunneling Quantum states (Speer et al, Science ’06)
Needed for designing
Heterojunctions/superlattices/
Photonic devices, etc
Upswing in
Current due
To Tunneling
Gloos

31. Barrier problem: Summaryx T
Classical 1
Quantum E U0
Tunneling
T ~ e-2kL, k ~ (U0-E)
Resonances
k’L = np, E = U0 + ħ2k’2/2m

32. Matlab plots
As barrier width increases, we recover particle
on a step

33. Matlab code
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)

34. 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)

35. Finite Difference Methody
One particular
mode
yn-1 yn
yn+1
xn-1 xn
xn+1
yn-1 yn
yn+1
y =
Un-1yn-1
Unyn
Un+1yn+1
Uy =
Un-1 Un
Un+1 =
yn-1 yn
yn+1
= [U][y]

36. What about kinetic energy?
yn-1 yn
yn+1
xn-1 xn
xn+1
yn-1 yn
yn+1
y =
(dy/dx)n = (yn+1/2 – yn-1/2)/a
(d2y/dx2)n = (yn+1 + yn-1 -2yn)/a2

37. What about kinetic energy?
yn-1 yn
yn+1
xn-1 xn
xn+1
yn-1 yn
yn+1
y =
-ħ2/2m(d2y/dx2)n = t(2yn - yn+1 - yn-1)
t = ħ2/2ma2

38. What about kinetic energy?
yn-1 yn
yn+1
xn-1 xn
xn+1
-ħ2/2m(d2y/dx2)n = t(2yn - yn+1 - yn-1)
yn-1 yn
yn+1
y =
-t 2t -t
yn-1 yn
yn+1
Ty =

39. What about kinetic energy?
yn-1 yn
yn+1
xn-1 xn
xn+1
yn-1 yn
yn+1
y =
[H] = [T + U]

40. What next? y yn-1 yn yn+1 xn-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)

41. Particle in a Box Results agree with analytical results E ~ n2
Finite wall heights, so waves seep out

42. Add a field

43. Or asymmetry
Incorrect, since we need
open BCs which we didn’t
discuss

44. 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...

45. Add asymmetry

46. Matlab code 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
%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);
% 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([ ])

47. 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

48. 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.