The meaning of it all..

1. The meaning of it all..

2. 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 #) • 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 An uncertain world

3. What does Y(x,t) represent? Y itself hard to measure so overall phase irrelevant Probabilities allow us to compute expectations !! <O> = ΣnOnPn • Charge density r(x,t) = qP(x,t) • Current density J(x,t) = qvP(x,t) (“v” to be defined later)

4. Averages d<x>/dt = <v> md<v>/dt = <F> Dx.Dp ≥ ħ/2 (Uncertainty Principle) l 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’

5. Current Density ˆ ˆ n = Y*Y (Electron Density) v = p/m = -iħ/m (check KE) J = q<v> = q∫y*(x)[-iħ/m]y(x)dx (Symmetrize !!!) J = iħ/2m(YY*/x – Y*Y/x) For plane waves Y = Y0eikx J = n(ħk/m) = nv Also check: n/t + J/x = 0 (Continuity Equation) ˆ

6. Relation between Y and fn ? fn 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 ∫f*n(x)fm(x)dx = 0 if n ≠ m and normalized ∫f*n(x)fn(x)dx = 1 Y shows the actual solution (superposition of allowed ones) In general, Y(x,t) = Sn an fn(x)e-iEnt/ħ

7. Back to radiating atom Why don’t the ‘Bohr levels’ radiate? A Bohr level is an eigenstate (‘mode’) Y(x,t) = fn(x)e-iEnt/ħ Since P(x,t) = |fn(x)|2 : time-independent that’s why electrons in Bohr orbits don’t radiate! What about acceleration? Average position <r> = ∫|Y(r,t)|2rd3r (‘Dipole’) This is time-independent for an eigenstate (since the average position of a plane wave is everywhere!), So there’s no classical ‘motion’ (remember classical rules only apply for averages!) and thus no radiation

8. Time-dependence (Radiation) Back to radiating atom So when do they radiate? When you’re in a superposition of modes Y(x,t) =cmfm(x)e-iEmt/ħ + cnfn(x)e-iEnt/ħ P(x,t) = cm2|fm(x)|2 + cn2|fn(x)|2 + 2cmcnfm*fncos(Em-En)t/ħ Strongest radiation occurs between states only when dipole term ∫fm*fnrd3r between them is non-zero In that case, dipole emits radiation at the energy Emn=Em-En and oscillates (accelerates) at the rate of wmn = Emn/ħ

9. How are the transitions executed? How do we know when to jump between states? This needs a time-dependent U(x,t) so that the coefficients of the mode-mixing become time-dependent Y(x,t) =cm(t)fm(x)e-iEmt/ħ + cn(t)fn(x)e-iEnt/ħ But these give oscillations – what about decay? Good question! If you treat the electromagnetic field itself as a quantum object, as a bunch of oscillators, then absorption terms  Nw and emission  Nw+1. The ‘1’ accounts for spontaneous emission

10. 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

11. Simplest eg of a confined system U =  U =  U = 0 f  eikx doesn’t satisfy boundary conditions (should be zero at both ends) But can superpose allowed solutions f = Asin(kx) is zero at x = 0 What about x = L ?

12. Particle in a box U =  U =  U = 0 f = 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)

13. Particle in a box U =  U =  k1 k2 k3 w U = 0 k Full Solution Yn = 2/L sin(npx/L) exp(-iħn2p2/2mL2 t) fn = Asin(knx) knL = np (n = 1, 2, 3, …) E3 E2 E1 Fixed k’s give fixed E’s En = ħ2kn2/2m = ħ2n2p2/2mL2 Coeff A fixed by normalization A=√2/L

14. Extracting Physics from Pictures !! Ground State f1(x)  Smoothest curve with no kinks U =  U =  Next mode f2(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 f3(x) must be orthogonal to the other two Since Kinetic energy ~∂2f/∂x2 Lowest energy wavefunction must have smallest curvature It must also vanish at ends and be normalized to unity

15. In other words.. At intermediate energies between eigenvalues, BCs not met U = 0 The culprit is the boundary condition !! We can find more solutions to the equation, but not with the right BCs. Either the curvature is too large or too small.

16. Extracting Physics from Pictures !! U =  U =  U =  U =  U = 0 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 (Uncertainty: Localizing particle increases its energy!) Small boxes (atoms), energies discrete and well separated Large boxes (metal contacts), they are bunched up

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

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

19. Simplest box U =  U =  k1 k2 k3 w U = 0 k Full Solution Yn = 2/L sin(npx/L)exp(-iħn2p2/2mL2t) E3 E2 E1 Eigenvalues En = ħ2kn2/2m = ħ2n2p2/2mL2 Eigenstates fn = Asin(knx) knL = np (Quantization)

20. Extracting Physics from Pictures !! Ground State f1(x)  Smoothest curve with no kinks U =  U =  Next mode f2(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 f3(x) must be orthogonal to the other two Since Kinetic energy ~∂2f/∂x2 Lowest energy wavefunction must have smallest curvature It must also vanish at ends and be normalized to unity

21. In other words.. At intermediate energies between eigenvalues, BCs not met U = 0 The culprit is the boundary condition !! We can find more solutions to the equation, but not with the right BCs. Either the curvature is too large or too small.

22. Extracting Physics from Pictures !! U =  U =  U =  U =  U = 0 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 (Uncertainty: Localizing particle increases its energy!) Small boxes (atoms), energies discrete and well separated Large boxes (metal contacts), they are bunched up

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

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

25. Superposition: dealing with an uncertain world + |Y> = a|fCATalive> + b|fCATdead> |a|2 + |b|2 = 1