ece 663-1, fall ‘08 quantum mechanics. ece 663-1, fall ‘08 why do we need it? qm interference...
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ECE 663-1, Fall ‘08
Quantum Mechanics
ECE 663-1, Fall ‘08
Why do we need it?
QM interference creates bandgaps and separatesmetals 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
Spectrum of Helium
Continuous radiation from orbiting electron
Pb1: Atom wouldbe unstable!(expect nanosecondsobserve billion years!)
Pb2: Spectra ofatoms are discrete!
Solar system model of atom
mv2/r = Zq2/40r2
Centripetalforce
Electrostaticforce
Transitions E0(1/n2 – 1/m2) (n,m: integers)
ECE 663-1, Fall ‘08
Only certain modes allowed (like a plucked string)
n= 2r (fit waves on circle)
Momentum ~ 1/wavelength (DeBroglie)p = mv = h/(massive classical particles vanishing )
This means angular momentum is quantized
mvr = nh/2= nħ
Why waves?
From 2 equations, rn = (n2/Z) a0
a0 = h20/q2m = 0.529 Å (Bohr radius)
ECE 663-1, Fall ‘08
Bohr’s suggestion
E = mv2/2 – Zq2/40r
Using previous two equations
En = (Z2/n2)E0
E0 = -mq4/8ħ20 = -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)
Solution: y(x,t) = y0ei(kx-t)
2 = v2k2
y
x
k
String
What is the dispersion (-k) for a particle?
ECE 663-1, Fall ‘08
What should our wave equation look
like?
Thus, dispersion we are looking for is k2 + U
Quantum theory:
E=hf = ħ (Planck’s Law) p = h/= ħk (de Broglie Law)
and E = p2/2m + U (energy of a particle)
k
So we need one time-derivativeand two spatial derivatives
∂2y∂t2 = v2(∂2y/∂x2) X X
ECE 663-1, Fall ‘08
Wave equation (Schrodinger)
iħ∂∂t = (-ħ22/2m + U)
Makes sense in context of waves
Eg. free particle U=0Solution = Aei(kx-t) = Aei(px-Et)/ħ
We then get E= p2/2m= ħ2k2/2m
Kinetic PotentialEnergy Energy
k
ECE 663-1, Fall ‘08
For all time-independent problems
iħ∂∂t = (-ħ22/2m + U)= Ĥ
Separation of variables for static potentials
(x,t) = (x)e-iEt/ħ
Ĥ=E, Ĥ = -ħ22/2m + U
BCs : Ĥn = Enn (n = 1,2,3...)
En : eigenvalues (usually fixed by BCs)
n(x): eigenvectors/stationary states
Oscillating solution in time
ECE 663-1, Fall ‘08
What does it all mean?
ECE 663-1, Fall ‘08
What does (x,t) represent?
• Probability amplitude of finding particle at x at time t (Like electric field in phasor notation E, a complex #)
itself hard to measure so overall phase irrelevant
• Probability density P(x,t) = *(x,t)(x,t) (Like intensity E*E easier to detect)
• Must have ∫P(x,t)dx = 1 at all times
• Charge density (x,t) = qP(x,t)
• Current density J(x,t) = qvP(x,t) (“v” to be defined later)
ECE 663-1, Fall ‘08
Averages
= ∫|(x,t)|2O(x)dx
One can show that averages follow ‘classical rules’
d<x>/dt = <v>
md<v>/dt = <F>
= ∫*(x,t)Ô(x)(x,t)dxSymmetrized!
x.p ≥ ħ/2(Uncertainty Principle)
ECE 663-1, Fall ‘08
Examples of meaningful averages
n = (Electron Density)
v = p/m = -iħ/m (check KE)
J = q<v> = q∫*(x)[-iħ/m](x)dx (Symmetrize !!!)
J = iqħ/2m(*/x – */x)
For plane waves = 0eikx J = n(ħk/m) = nqv
Also check: n/t + J/x = 0 (charge conservation)
ˆ
ˆ ˆ
ECE 663-1, Fall ‘08
Relation between and n ?
n 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
∫*n(x)m(x)dx = 0 if n ≠ m
and normalized ∫*n(x)n(x)dx = 1
shows the actual solution (superposition of allowed ones) In general,(x,t) =n an n(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
-13.6 eV
-3.4 eV
-1.51 eV-0.85 eV
Vacuum
U(r) = -Zq2/40r
ECE 663-1, Fall ‘08
eikx doesn’t satisfy boundary conditions(should be zero at both ends)
But can superpose allowed solutions
= Asin(kx) is zero at x = 0What about x = L ?
Simplest eg of a confined system
U = U =
U = 0
Particle in a Box
ECE 663-1, Fall ‘08
Particle in a box
= Asin(kx) is zero at x = L only for special values of k
knL = n(n = 1, 2, 3, …)
Quantization conditionknL = n(n = 1, 2, 3, …)
ie,L = nn/2 (exactly like acoustic waves)
U = U =
U = 0
ECE 663-1, Fall ‘08
Particle in a box
n = Asin(knx)
knL = n(n = 1, 2, 3, …)
Fixed k’s give fixed E’s
En = ħ2kn2/2m = ħ2n22/2mL2
Coeff A fixed by normalization A=√2/L
U = U =
U = 0
Full Solution n = 2/L sin(nx/L) exp(-iħn22/2mL2 t)
kk2 k3k1
E1
E2
E3
ECE 663-1, Fall ‘08
Particle in a box
En = ħ2kn2/2m = ħ2n22/2mL2
n=` 2/L sin(nx/L) exp(-iħn22/2mL2 t)
kk2 k3k1
E1
E2
E3
Find J
J = iqħ/2m(*/x – */x)
ECE 663-1, Fall ‘08
Extracting Physics from Pictures !!
Since Kinetic energy ~∂2/∂x2
Lowest energy wavefunction must have smallest curvatureIt must also vanish at ends and be normalized to unity
U = U =
U = 0
Ground State 1(x)
Smoothest curve with no kinks
Next mode 2(x) should also minimize energy
but be orthogonal to the first mode(since modes must be independent!)
Hence the single kink!
Next one 3(x) must be orthogonal to
the other two
ECE 663-1, Fall ‘08
Extracting Physics from Pictures !!
This increases curvature and thus energylevels and their separation
Notice this from exact results, En 1/L2
(Uncertainty: Localizing particle increasesits energy!)
U = U =
U = 0
Small boxes (atoms), energies discrete and well separated
Large boxes (metal contacts), they are bunched up
U = U =
U = 0
If we decrease box size, but keeparea under modes same, then each modemust peak more
ECE 663-1, Fall ‘08
Constant, non-zero potential
knL = n still
But dispersion kn = √2m(En-U0)/ħ2
U = U =
U = U0
ECE 663-1, Fall ‘08
Finite potential walls: thinking ‘outside the box’
U = U0 U = U0
U = 0
Solve piece-by-piece, and match boundary condns (Match , d/dx)
Wavefunction penetrates out (“Tunneling”)
Asinkx + Bcoskx, k = 2mE/ħ2
exp(±ik’x), k’ = 2m(E-U0)/ħ2exp(±x), = 2m(U0-E)/ħ2
ECE 663-1, Fall ‘08
Particle at a Step
U0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
x0
Boundary Conditions: (0-) = (0+)d/dx|0- = d/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
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
x0
Boundary Conditions: 1 + r = tk(1-r) = k’t
t = 2k/(k+k’)r = (k-k’)/(k+k’)
ECE 663-1, Fall ‘08
Particle at a Step
U0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
x0
Transmission = curr transmitted/curr incidentReflection Coeff = curr reflected/curr incident
ECE 663-1, Fall ‘08
Particle at a Step
U0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
x0
Free particle J = ħk/m|0|2, 0: amplitude
Thus, T = Re(k’|t|2/k), R = |r|2
(More correctly, J = Re(ħk/m|0|2), in case
k is complex)
ECE 663-1, Fall ‘08
Particle at a Step
U0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
x0
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
x0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
E < U0
k’ imaginary = i
ECE 663-1, Fall ‘08
Repeat for E < U0
U0
x0
(x) = eikx + re-ikx , x < 0 = teik’x, x > 0
k = 2mE/ħ2
= 2m(U0-E)/ħ2
t = 2k/(k+i)r = (k-i)/(k+i)
T = Re(i|t|2/k) = 0R = |r|2 = |k-i|2/|k+i|2 = 1
Expected?
ECE 663-1, Fall ‘08
Summary: Particle at a step
U0
x0
EU0
T
Classical
Quantum
1
0
$$$ Question: How do we make the curves approach each other?
ECE 663-1, Fall ‘08
Particle at a Barrier
U0
x0
Boundary Conditions: (0-) = (0+)d/dx|0- = d/dx|0+
k = 2mE/ħ2
k’ = 2m(E-U0)/ħ2
E > U0
k,k’ real
L
(L-) = (L+)d/dx|L- = d/dx|L+
eikx + re-ikx Aeik’x + Be-ik’x teikx
ECE 663-1, Fall ‘08
Particle at a Barrier
U0
eikx + re-ikx Aeik’x + Be-ik’x teikx
x0
1+ r = A + Bk(1-r) = k’(A-B)
L
Aeik’L + Be-ik’L = teikL
k’(Aeik’L – Be-ik’L) = kteikL
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
U0
eikx + re-ikx Aeik’x + Be-ik’x teikx
x0 L
2 = (1+k’/k)A + (1-k’/k)B
0 = (1-k’/k)Aeik’L + (1+k’/k)Be-ik’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]
ECE 663-1, Fall ‘08
Particle at a Barrier
U0
eikx + re-ikx Aeik’x + Be-ik’x teikx
x0 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]
ECE 663-1, Fall ‘08
Particle at a Barrier
U0
eikx + re-ikx Aeik’x + Be-ik’x teikx
x0 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]
T = 4k2k’2/[(k2+k’2)2sin2k’L + 4k2k’2cos2k’L]
Resonances: Maximum if cosk’L = 1, sink’L = 0ie, k’L = , , 2,... L = 0, /2, 3/2, ....
ECE 663-1, Fall ‘08
More obvious if it’s a well
U0
eikx + re-ikx Aeik’x + Be-ik’x teikx
x0 L
ie, k’L = , , 2,... L = 0, /2, 3/2, ....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
U0
eikx + re-ikx Ae-x + Bex teikx
x0 L
T = 4k22/[(k2-2)2sinh2L + 4k22cosh2L]
Large or wide barriers: L >> 1, sinh(L) ~ cosh(L) ~ eL/2T ≈ 16 k22e-2L/(k2+2)2 ~ [16E(U0-E)/U0
2]e-2L
k’ = i
teikL =2ik/[(k2-2)sinhL + 2ikcoshL]
ECE 663-1, Fall ‘08
Back to Barrier but lower Energy
U0
eikx + re-ikx Ae-x + Bex teikx
x0 L
T ≈ [16E(U0-E)/U02]e-2L
More generally, WKB approximation
T ~ exp[-2∫dx 2m[U(x)-E]/ħ2]x1
x2
U(x)
Ex1 x2
Even though E < V0, T > 0 (tunneling)
ECE 663-1, Fall ‘08
Example: Tunneling
T ~ exp[-2∫dx 2m[U(x)-E]/ħ2]x1
x2
Well
Barrier
• 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
Gloos
Upswing inCurrent dueTo Tunneling
ECE 663-1, Fall ‘08
Barrier problem: Summary
U0
x0 L
EU0
TClassical
Quantum
1
0Resonancesk’L = nE = U0 + ħ2k’2/2m
TunnelingT ~ e-2L, ~ (U0-E)
ECE 663-1, Fall ‘08
Matlab plots
As barrier width increases, we recover particleon a step
ECE 663-1, Fall ‘08
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)
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
xn-1 xn xn+1
n-1
n
n+1
=
n-1 n n+1
Un-1
Un
Un+1
=
n-1
n
n+1
Un-1n-1
Unn
Un+1n+1
U =
One particular mode
= [U][]
ECE 663-1, Fall ‘08
What about kinetic energy?
xn-1 xn xn+1
n-1
n
n+1
=
n-1 n n+1
(d/dx)n = (n+1/2 – n-1/2)/a
(d2/dx2)n = (n+1 + n-1 -2n)/a2
ECE 663-1, Fall ‘08
What about kinetic energy?
xn-1 xn xn+1
n-1
n
n+1
=
n-1 n n+1
-ħ2/2m(d2/dx2)n = t(2n - n+1 - n-1)
t = ħ2/2ma2
ECE 663-1, Fall ‘08
What about kinetic energy?
xn-1 xn xn+1
n-1
n
n+1
=
n-1 n n+1
-t 2t -t n-1
n
n+1
T = -t 2t -t
-t 2t -t
-ħ2/2m(d2/dx2)n = t(2n - n+1 - n-1)
ECE 663-1, Fall ‘08
What about kinetic energy?
xn-1 xn xn+1
n-1
n
n+1
=
n-1 n n+1
[H] = [T + U]
ECE 663-1, Fall ‘08
What next?
xn-1 xn xn+1
n-1 n n+1
Now that we’ve got H matrix, we cancalculate 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 n 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 needopen BCs which we didn’t discuss
ECE 663-1, Fall ‘08
Harmonic Oscillator
Shapes change from box: sin(x/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)ħ, n = 0, 1, 2...
ECE 663-1, Fall ‘08
Add asymmetry
ECE 663-1, Fall ‘08
• 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([-5 5 -2 10])
Matlab code
ECE 663-1, Fall ‘08
Grid issues
For Small energies, finite diff. matches exact resultDeviation at large energy, where varies rapidlyGrid 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.
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