1. ECE 663-1, Fall ‘08 Quantum Mechanics

2. ECE 663-1, Fall ‘08 Why do we need it? QM interference creates bandgaps and separates metals from insulators and semiconductors

3. 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/

4. ECE 663-1, Fall ‘08 Spectrum of Helium Continuous radiation from orbiting electron Pb1: Atom would be unstable! (expect nanoseconds observe billion years!) Pb2: Spectra of atoms are discrete! Solar system model of atom mv 2 /r = Zq 2 /4  0 r 2 Centripetal force Electrostatic force Transitions E 0 (1/n 2 – 1/m 2 ) (n,m: integers)

5. ECE 663-1, Fall ‘08 Only certain modes allowed (like a plucked string) n  = 2  r (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, r n = (n 2 /Z) a 0 a 0 = h 2  0 /  q 2 m = 0.529 Å (Bohr radius)

6. ECE 663-1, Fall ‘08 Bohr’s suggestion E = mv 2 /2 – Zq 2 /4  0 r Using previous two equations E n = (Z 2 /n 2 )E 0 E 0 = -mq 4 /8ħ 2  0 = -13.6 eV = 1 Rydberg Transitions E 0 (1/n 2 – 1/m 2 ) (n,m: integers) Explains discrete atomic spectra So need a suitable Wave equation so that imposing boundary conditions will yield the correct quantized solutions

7. ECE 663-1, Fall ‘08 What should our wave equation look like? ∂ 2 y  ∂t 2 = v 2 (∂ 2 y/∂x 2 ) Solution: y(x,t) = y 0 e i(kx-  t)  2 = v 2 k 2 y x  k String What is the dispersion (  -k) for a particle?

8. ECE 663-1, Fall ‘08 What should our wave equation look like? Thus, dispersion we are looking for is   k 2 + U Quantum theory: E=hf = ħ  (Planck’s Law) p = h/  = ħk (de Broglie Law) and E = p 2 /2m + U (energy of a particle)  k So we need one time-derivative and two spatial derivatives ∂ 2 y  ∂t 2 = v 2 (∂ 2 y/∂x 2 ) XX

9. ECE 663-1, Fall ‘08 Wave equation (Schrodinger) iħ∂   ∂t = (-ħ 2  2 /2m + U)  Makes sense in context of waves Eg. free particle U=0 Solution  = Ae i(kx-  t) = Ae i(px-Et)/ħ We then get E  = p 2 /2m  = ħ 2 k 2 /2m Kinetic Potential Energy  k

10. ECE 663-1, Fall ‘08 For all time-independent problems iħ∂   ∂t = (-ħ 2  2 /2m + U)  = Ĥ  Separation of variables for static potentials  (x,t) =  (x)e -iEt/ħ Ĥ  =  E , Ĥ = -ħ 2  2 /2m + U BCs : Ĥ  n = E n  n (n = 1,2,3...) E n : eigenvalues (usually fixed by BCs)  n (x): eigenvectors/stationary states Oscillating solution in time

11. ECE 663-1, Fall ‘08 What does it all mean?

12. 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)

13. ECE 663-1, Fall ‘08 Averages  = ∫|  (x,t)| 2 O(x)dx One can show that averages follow ‘classical rules’ d /dt = md /dt =  = ∫  * (x,t)Ô(x)  (x,t)dx Symmetrized!  x.  p ≥ ħ/2 (Uncertainty Principle)

14. ECE 663-1, Fall ‘08 Examples of meaningful averages n =    (Electron Density) v = p/m = -iħ  /m (check KE) J = q = q∫  *(x)[-iħ  /m]  (x)dx (Symmetrize !!!) J = iqħ/2m(  */  x –  *  /  x) For plane waves  =  0 e ikx J = n(ħk/m) = nqv Also check:  n/  t +  J/  x = 0 (charge conservation) ˆ ˆˆ

15. 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 E n 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 a n  n (x)e -iE n t/ħ

16. 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) = -Zq 2 /4  0 r

17. ECE 663-1, Fall ‘08   e ikx doesn’t satisfy boundary conditions (should be zero at both ends) But can superpose allowed solutions  = Asin(kx) is zero at x = 0 What about x = L ? Simplest eg of a confined system U =  U = 0 Particle in a Box

18. ECE 663-1, Fall ‘08 Particle in a box  = Asin(kx) is zero at x = L only for special values of k k n L = n  (n = 1, 2, 3, …) Quantization condition k n L = n  (n = 1, 2, 3, …) ie, L = n n /2 (exactly like acoustic waves) U =  U = 0

19. ECE 663-1, Fall ‘08 Particle in a box  n = Asin(k n x) k n L = n  (n = 1, 2, 3, …) Fixed k’s give fixed E’s E n = ħ 2 k n 2 /2m = ħ 2 n 2  2 /2mL 2 Coeff A fixed by normalization A=√2/L U =  U = 0 Full Solution  n = 2/L sin(n  x/L) exp(-iħn 2  2 /2mL 2 t)  k k2k2 k3k3 k1k1 E1E1 E2E2 E3E3

20. ECE 663-1, Fall ‘08 Particle in a box E n = ħ 2 k n 2 /2m = ħ 2 n 2  2 /2mL 2  n =` 2/L sin(n  x/L) exp(-iħn 2  2 /2mL 2 t)  k k2k2 k3k3 k1k1 E1E1 E2E2 E3E3 Find J J = iqħ/2m(  */  x –  *  /  x)

21. ECE 663-1, Fall ‘08 Extracting Physics from Pictures !! Since Kinetic energy ~  ∂ 2  /∂x 2 Lowest energy wavefunction must have smallest curvature It must also vanish at ends and be normalized to unity 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

22. ECE 663-1, Fall ‘08 Extracting Physics from Pictures !! This increases curvature and thus energy levels and their separation Notice this from exact results,  E n  1/L 2 (Uncertainty: Localizing particle increases its energy!) U =  U = 0 Small boxes (atoms), energies discrete and well separated Large boxes (metal contacts), they are bunched up U =  U = 0 If we decrease box size, but keep area under modes same, then each mode must peak more

23. ECE 663-1, Fall ‘08 Constant, non-zero potential k n L = n  still But dispersion k n = √ 2m(E n -U 0 )/ħ 2 U =  U = U 0

24. ECE 663-1, Fall ‘08 Finite potential walls: thinking ‘outside the box’ U = U 0 U = 0 Solve piece-by-piece, and match boundary condns (Match , d  /dx) Wavefunction penetrates out (“Tunneling”) Asinkx + Bcoskx, k =  2mE/ħ 2 exp(±ik’x), k’ =  2m(E-U 0 )/ħ 2 exp(±  x),  =  2m(U 0 -E)/ħ 2

25. ECE 663-1, Fall ‘08 Particle at a Step U0U0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 x 0 Boundary Conditions:  (0 - ) =  (0 + ) d  /dx| 0 - = d  /dx| 0 + k =  2mE/ħ 2 k’ =  2m(E-U 0 )/ħ 2 E > U 0 k,k’ real

26. ECE 663-1, Fall ‘08 Particle at a Step U0U0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 x 0 Boundary Conditions: 1 + r = t k(1-r) = k’t t = 2k/(k+k’) r = (k-k’)/(k+k’)

27. ECE 663-1, Fall ‘08 Particle at a Step U0U0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 x 0 Transmission = curr transmitted/curr incident Reflection Coeff = curr reflected/curr incident

28. ECE 663-1, Fall ‘08 Particle at a Step U0U0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 x 0 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)

29. ECE 663-1, Fall ‘08 Particle at a Step U0U0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 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’)

30. ECE 663-1, Fall ‘08 Repeat for E < U 0 U0U0 x 0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 k =  2mE/ħ 2 k’ =  2m(E-U 0 )/ħ 2 E < U 0 k’ imaginary = i 

31. ECE 663-1, Fall ‘08 Repeat for E < U 0 U0U0 x 0  (x) = e ikx + re -ikx, x < 0  = te ik’x, x > 0 k =  2mE/ħ 2  =  2m(U 0 -E)/ħ 2 t = 2k/(k+i  ) r = (k-i  )/(k+i  ) T = Re(i  |t| 2 /k) = 0 R = |r| 2 = |k-i  | 2 /|k+i  | 2 = 1 Expected?

32. ECE 663-1, Fall ‘08 Summary: Particle at a step U0U0 x 0 E U0U0 T Classical Quantum 1 0 $$$ Question: How do we make the curves approach each other?

33. ECE 663-1, Fall ‘08 Particle at a Barrier U0U0 x 0 Boundary Conditions:  (0 - ) =  (0 + ) d  /dx| 0 - = d  /dx| 0 + k =  2mE/ħ 2 k’ =  2m(E-U 0 )/ħ 2 E > U 0 k,k’ real L  (L - ) =  (L + ) d  /dx| L - = d  /dx| L + e ikx + re -ikx Ae ik’x + Be -ik’x te ikx

34. ECE 663-1, Fall ‘08 Particle at a Barrier U0U0 e ikx + re -ikx Ae ik’x + Be -ik’x te ikx x 0 1+ r = A + B k(1-r) = k’(A-B) L Ae ik’L + Be -ik’L = te ikL k’(Ae ik’L – Be -ik’L ) = kte ikL 2 = (1+k’/k)A + (1-k’/k)B 0 = (1-k’/k)Ae ik’L + (1+k’/k)Be -ik’L

35. ECE 663-1, Fall ‘08 Particle at a Barrier U0U0 e ikx + re -ikx Ae ik’x + Be -ik’x te ikx x 0 L 2 = (1+k’/k)A + (1-k’/k)B 0 = (1-k’/k)Ae ik’L + (1+k’/k)Be -ik’L A = 2e -ik’L (1+k’/k)/[(1+k’/k) 2 e -ik’L – (1-k’/k) 2 e ik’L ] B = 2e ik’L (1-k’/k)[(1-k’/k) 2 e ik’L -(1+k’/k) 2 e -ik’L ]

36. ECE 663-1, Fall ‘08 Particle at a Barrier U0U0 e ikx + re -ikx Ae ik’x + Be -ik’x te ikx x 0 L A = 2e -ik’L (1+k’/k)/[(1+k’/k) 2 e -ik’L – (1-k’/k) 2 e ik’L ] B = 2e ik’L (1-k’/k)[(1-k’/k) 2 e ik’L -(1+k’/k) 2 e -ik’L ] te ikL = 2(2k’/k)/ [(1+k’/k) 2 e -ik’L – (1-k’/k) 2 e ik’L ] = 2kk’/[-i(k 2 +k’ 2 )sink’L + 2kk’cosk’L]

37. ECE 663-1, Fall ‘08 Particle at a Barrier U0U0 e ikx + re -ikx Ae ik’x + Be -ik’x te ikx x 0 L te ikL = 2(2k’/k)/ [(1+k’/k) 2 e -ik’L – (1-k’/k) 2 e ik’L ] = 2kk’/[-i(k 2 +k’ 2 )sink’L + 2kk’cosk’L] T = 4k 2 k’ 2 /[(k 2 +k’ 2 ) 2 sin 2 k’L + 4k 2 k’ 2 cos 2 k’L] Resonances: Maximum if cosk’L =  1, sink’L = 0 ie, k’L = , , 2 , ...  L = 0, /2, 3 /2,....

38. ECE 663-1, Fall ‘08 More obvious if it’s a well U0U0 e ikx + re -ikx Ae ik’x + Be -ik’x te ikx x 0L ie, k’L = , , 2 , ...  L = 0, /2, 3 /2,.... Represent Resonances, with E > U 0 (They would be bound states if E < U 0 ) Here k =  2m(E-U 0 )/ħ 2 k’ =  2mE/ħ 2

39. ECE 663-1, Fall ‘08 Back to Barrier but lower Energy U0U0 e ikx + re -ikx Ae -  x + Be  x te ikx x 0 L T = 4k 2  2 /[(k 2 -  2 ) 2 sinh 2  L + 4k 2  2 cosh 2  L] Large or wide barriers:  L >> 1, sinh(  L) ~ cosh(  L) ~ e  L /2 T ≈ 16 k 2  2 e -2  L /(k 2 +  2 ) 2 ~ [16E(U 0 -E)/U 0 2 ]e -2  L k’ = i  te ikL = 2ik  /[(k 2 -  2 )sinh  L + 2ik  cosh  L]

40. ECE 663-1, Fall ‘08 Back to Barrier but lower Energy U0U0 e ikx + re -ikx Ae -  x + Be  x te ikx x 0 L T ≈ [16E(U 0 -E)/U 0 2 ]e -2  L More generally, WKB approximation T ~ exp[-2∫dx  2m[U(x)-E]/ħ 2 ] x1x1 x2x2 U(x) E x1x1 x2x2 Even though E 0 (tunneling)

41. ECE 663-1, Fall ‘08 Example: Tunneling T ~ exp[-2∫dx  2m[U(x)-E]/ħ 2 ] x1x1 x2x2 Well Barrier Alpha particle decay from nucleus Source-Drain tunneling in MOSFETs Single Electron Tunneling Devices (SETs) Resonant Tunneling Devices (RTDs)

42. ECE 663-1, Fall ‘08 Example: Tunneling Quantum states (Speer et al, Science ’06) Needed for designing Heterojunctions/superlattices/ Photonic devices, etc Gloos Upswing in Current due To Tunneling

43. ECE 663-1, Fall ‘08 Barrier problem: Summary U0U0 x 0 L E U0U0 T Classical Quantum 1 0 Resonances k’L = n  E = U 0 + ħ 2 k’ 2 /2m Tunneling T ~ e -2  L,  ~  (U 0 -E)

44. ECE 663-1, Fall ‘08 Matlab plots As barrier width increases, we recover particle on a step

45. 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)

46. 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)

47. ECE 663-1, Fall ‘08 Finite Difference Method x n-1 xnxn x n+1  n-1 nn  n+1  = =   n-1 nn  n+1 U n-1 UnUn U n+1 =  n-1 nn  n+1 U n-1  n-1 UnnUnn U n+1  n+1 U =U = One particular mode = [U][  ]

48. ECE 663-1, Fall ‘08 What about kinetic energy? x n-1 xnxn x n+1  n-1 nn  n+1  = =   n-1 nn  n+1 (d  /dx) n = (  n+1/2 –  n-1/2 )/a (d 2  /dx 2 ) n = (  n+1 +  n-1 -2  n )/a 2

49. ECE 663-1, Fall ‘08 What about kinetic energy? x n-1 xnxn x n+1  n-1 nn  n+1  = =   n-1 nn  n+1 -ħ 2 /2m(d 2  /dx 2 ) n = t(2  n -  n+1 -  n-1 ) t = ħ 2 /2ma 2

50. ECE 663-1, Fall ‘08 What about kinetic energy? x n-1 xnxn x n+1  n-1 nn  n+1  = =   n-1 nn  n+1 -t 2t -t  n-1 nn  n+1 T =T = -t 2t -t -ħ 2 /2m(d 2  /dx 2 ) n = t(2  n -  n+1 -  n-1 )

51. ECE 663-1, Fall ‘08 What about kinetic energy? x n-1 xnxn x n+1  n-1 nn  n+1  = =   n-1 nn  n+1 [H] = [T + U]

52. ECE 663-1, Fall ‘08 What next? x n-1 xnxn x n+1   n-1 nn  n+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 (n th column of matrix V is the n th eigenvector  n plotted along the x axis)

53. ECE 663-1, Fall ‘08 Particle in a Box Results agree with analytical results E ~ n 2 Finite wall heights, so waves seep out

54. ECE 663-1, Fall ‘08 Add a field

55. ECE 663-1, Fall ‘08 Or asymmetry Incorrect, since we need open BCs which we didn’t discuss

56. ECE 663-1, Fall ‘08 Harmonic Oscillator Shapes change from box: sin(  x/L)  exp(-x 2 /2a 2 ) Need polynomial prefactor to incorporate nodes (Hermite) E~n 2 for box, but box width increases as we go higher up  Energies equispaced E = (n+1/2)ħ , n = 0, 1, 2...

57. ECE 663-1, Fall ‘08 Add asymmetry

58. 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

59. ECE 663-1, Fall ‘08 Grid issues For Small energies, finite diff. matches exact result Deviation at large energy, where  varies rapidly Grid needs to be fine enough to sample variations

60. 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.