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..”

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

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

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

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)

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

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 ?

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)

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)

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

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

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”)

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

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

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

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)

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’)

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

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?

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

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

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

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

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]

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]

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

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]

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

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)

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

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

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

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)

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)

Finite Difference Method y 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]

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

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

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 =

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

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)

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

Add a field

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

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...

Add asymmetry

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([-5 5 -2 10])

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

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.