QUANTUM MECHANICS II PHYS 517. Solutions to Problem Set # 1

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1 QUANTUM MECHANICS II PHYS 57 Solutions to Problem Set #. The hamiltonian for a lassial harmoni osillator an be written in many different forms, suh as use ω = k/m H = p m + kx H = P + Q hω a. Find a anonial transformation from the familiar oordinate x and momentum p to their dimensionless ounterparts Q, P. b. Chek that the Hamilton equations of motion are satisfied for both sets of oordinates.. To go from lassial to quantum mehanis, Shrödinger tells us to replae x ˆx and p ˆp and require that these operators satisfy the ommutation relations [ˆp, ˆx] = h/i. What ommutation relations does this fore the operators ˆQ, ˆP to have? Solution.a. Equate the two expressions for H: H = p m + k x = P + Q hω Now ompare the kineti energy terms and the potential energy terms: p m = hω P P = k x = hω Q Q = p m hω k hω x = mω h x Dimensional analysis shows [ m hω ] = MLT = [p], also [ h/mω ] = L = [x], verifying that P and Q are dimensionless.

2 b. The Hamilton equations of motion in the two ases are: dt = H p = p m dq dt = H P = hωp dp dt = H x = kx dp dt = H Q = hωq. Plug the expressions for P and Q into the ommutator: [ ] p mω [P, Q] =, m hω h x = h [p, x] = i. The Lorentz fore on a partile of harge q moving with veloity v in the presene of eletri and magneti fields E, B is F = q E + v B a. Express E and B in terms of the vetor and salar potential A and φ. b. Find an expression for the total derivative da/dt in terms of its partial derivatives and v. Explain exatly how this expression is derived.. Expand v A. d. Plug the expressions from b. and. into a. to find an expression of the form F = U+ the total time derivative of something. e. What is U? What is something and why isn t it important? f. Construt the Lagrangian for the motion of a harged partile in the presene of an eletri and magneti field: L = mv v U. g. Construt the momentum onjugate to x in the usual way: p = L/ v. h. Use the standard Legendre transformation to onstrut the Hamiltonian from L: H = p v L, and express H as a funtion of x, p, A. i. Verify that the Hamiltonian equations of motion give the orret result. Solution.a. From our disussion of eletromagneti potentials in lass B = A and E = φ b. The A vetor potential is omputed at the loation of the partiles on whih the field ats, so that A = Axt, t. As a result, the total

3 time derivative ontains an expliit term, as well as impliit dependene through the dependene of the partile oordinates on time. Result: da dt = xt, t + xt, t j t x j dt = + v A. Use the identity for vetors from high shool: a b = a b a b, set a = v/, b =, and = A, also reall that the differential operator ats only on A, to find v A = v A v A Keep this result in mind beause it will be used again in part i. to onstrut the seond of the Hamilton equations of motion and the Lorentz fore. d. Plugging the expressions for E and B in terms of the potentials A and φ into the expression for the Lorentz fore gives F = q φ + q v A v A v = q φ + A q q v A [ = q φ v ] A q da dt e. The potential U is qφ v A. Something is q da. This term an be dt negleted beause the addition of a total time derivative to a Lagrangian from whih dynamis is onstruted via the the standard Calulus of Variations proedure that leads to the Euler-Lagrange equations without hanging the dynamial equations of motion. f. L = T U = mv v qφ v A. g. It is a standard proedure to onstrut the momentum that is anonially onjugate to a oordinate by taking the derivative of the Lagrangian with respet to the veloity oordinate. The anonial momentum is used in Hamilton s anonial formalism, whih transforms n Newton s seond order equations of motion to n first order equations of motion. p = L v = mv + q A 3

4 As a result, v = p q A/m. h. Carry out the steps indiated to find: H = mv+ q A v mv v+qφ v A = mv v+qφ = p q +qφ m A i. The first of the Hamilton equations of motion is a piee of ake to ompute: dt = H p = p q m A = v The seond is not so simple, as it involves gymnastis to extrat the Lorentz fore from the vetor and salar potentials. Here goes: dp j = H = q j φ q p q A dt x j x j m m dv dt + q da dt = q φ + q v A q + q The last two terms have been introdued in the spirit of the empty poket trik. The idea is to use one of the partials with φ to onstrut an expression for the eletri field E and the other to introdue the total time derivative of A. m dv dt + q da = q φ + q dt + q v A Now bring the total time derivative over to the right hand side of the equation. The differene between the total and partial derivatives is da = dt v A. This yields m dv dt = q φ + q v A v A The terms in the first braket are the representation of E in terms of the vetor and salar potentials. The terms in the seond pair of parentheses are the expansion of q v A, whih is the representation for q v B. The final result is m dv dt = q E + v B = F Lorentz 4

5 3. Compute the thermal expetation value for a quantum harmoni osillator with angular frequeny ω. Reall that i. The oupation probability of a state with energy E n is proportional to the Boltzmann fat: P n e βen. ii. Show P n = e βen /Z, where Z is the partition funtion. iii. Use E n = n + hω. a. Show E = n + hω. What is n? b. How many modes of the eletromagneti field exist in the angular frequene range ω to ω + dω?. Compute the mean energy per unit volume in the eletromagneti field when it is in thermal equilibrium at temperature T β = /kt with its surroundings. Show that this energy onsists of two terms, one of whih is proportional to T 4 and the other of whih diverges. Solution 3.a. The probability that a state of energy E n is oupied is proportional to the Boltzmann fator: P n e βen. To turn the proportionality into an equality multiply the Boltzmann fator by a normalization onstant: P n =. Requiring that the sum of the probabilities is + Z e βen defines Z as the partition funtion: Z = n e βen. Now ompute the expetation of the energy using these probabilities: E = n E n P n = n E n e βen /Z = β e βen /Z = n β Zβ Zβ an be omputed as a geometri sum sine the energies are equally spaed: /Zβ = β logzβ Z = n e n+ β hω = e β hω/ n=0 e β hω n = e β hω/ e β hω The logarithm of Zβ is easily taken log Zβ == β hω log e β hω The negative gradient with respet to β follows: hωe β hω logzβ = + β hω e β hω Cleaning up the fration gives the standard result 5

6 E = n + hω n = e β hω b. The modes of the eletromagneti field without soures subjet to periodi boundary onditions in a ube of length L are defined by triples of integers aording to k = πn L, n, n 3, where the n i are integers. The number of integer triples inside a sphere of radius N is about 4πN 3, where 3 N = n + n + n 3. The number of integer triples in a spherial shell with inside radius N and outer radius N + dn is approximately 4πN dn. N, k, and ω are related by As a result k = π L n + n + n 3 = ω L 3 4πN dn = 4π ω dω π There are two polarization modes for eah spatial k mode, so this mode ounting funtion is multiplied by.. The total energy is obtained by ounting the average energy over all modes of the eletromagneti field: E = 0 4π L 3 ω π e β hω + hω dω The energy density, energy per unit volume, is obtained by dividing out by L 3 = V. The integral onsists of two terms. One is finite beause the asymptoti ω dependene is exponentially deaying e β hω, killing off the polynomial growth. This integral an be done in losed form. E /V = 0 4π 3 ω hω dω = π kt 4 π e β hω 5 h 3 The T 4 dependene was first derived from an elegant thermodynami argument involving a Carnot heat engine using eletromagneti radiation as a working fluid. The seond ontribution to the average energy diverges as ω. It is at the ore of many deep problems in physis: 6

7 3 E /V = 4π ω hω dω 0 π h ω 4 π 3 8 ω The zero point energy ontribution to the energy density and mass density of the universe is by far too large to be ompatible with the universe as we now know it. 4. The nonrelativisti hamiltonian that desribes the interation of a harged partile with the eletromagneti field is H = m p q A + qφ In quantum mehanis p h/i and the hamiltonian ats on a wavefuntion ψx, t. a. Assume that the wavefuntion is hanged by a onstant phase ψ ψ = e iα ψ. Show that ψ satisfies the original Shrödinger equation with the original vetor potential A. b. Assume that the wavefuntion is hanged by a nononstant phase ψ ψ = e iαx,t ψ. Show that ψ does not satisfy the original Shrödinger equation with the original vetor potential A.. Show that ψ does satisfy the original Shrödinger equation, but with a new vetor potential A = A + χ. How is the gauge term χx, t related to the phase term αx, t? d. How does the salar potential hange: φ φ +?? Solution 4. The produt rule from elementary alulus is d f g = df dg g +f or more simply fg = f g +fg. When one of these funtions is an exponential this rule an be expressed rather elegantly: d df ef g = e f f g + e f g = e f + d g We will use this result in the following form e f d g = d df e f g The original time-dependent Shrödinger equation is 7

8 h m i q A + qφx, t ψx, t = i h ψx, t We modify this by moving the potential term to the right hand side: m h i q A ψx, t = i h qφx, t ψx, t a. If we multiply ψ by a onstant phase fator, ψ ψ = e iα ψ, and plug this into Shrödinger s time dependent equation, we an move the onstant phase fator through the derivatives, and then finally multiply both sides of the equation by its inverse = e iα to show that ψ satisfies the same equation. N.B.: this is alled a gauge transformation of the first kind. b.-d. Now multiply both sides of this equation by a phase fator and move it through the derivatives using the result above: e f d g = d df ef g. First apply this to the right hand side of the rewritten time-dependent Shrödinger equation: Now multiply ψ by a phase fator whih is not onstant, ψ ψ = e iαx,t ψx, t, and plug this into Shrödinger s time dependent equation. As we try to move the phase fator through the derivative terms, we pik up extra terms as follows. On the right hand, time derivative side: e iαx,t i h qφx, t ψx, t = i h qφx, t i hi α e iαx,t ψx, t We an do similar on the left hand side, where spae derivatives our. We do this in two steps: m eiαx,t = h i q A ψx, t h m i q A h i i α e hi iαx,t q A ψx, t = m h i q A h i i α e iαx,t ψx, t 8

9 These two expresisons, one involving time derivatives, the other involving spae derivatives, an now be equated. The result is the time-dependent Shrödinger equation for the wave funtion ψ x, t = e iαx,t ψx, t: m h i q A h i i α ψ x, t = i h qφx, t i hi α ψ x, t If αx, t is a onstant, the derivatives of α disappear, and the equation satisfied by ψ = e iα ψ is the same as that satisfied by ψ. If αx, t is not a onstant, the equation satisfied by ψ = e iα ψ is the different from that satisfied by ψ. They differ by a gauge transformation. A = A + h α = A + χ q φ = φ h q α = φ χ We identify the phase hange αx, t with a gauge transformation χx, t as follows: χ = h q α or α = q h χ so that the phase hange of the fermion field is related to the gauge transformation of the boson field by e iq h χ. N.B.: this is alled a gauge transformation of the seond kind. This alulation serves as a model for later desriptions of the interations between fermions and bosons. The first extension was done by C. N. Yang and R. L. Mills Physiists now believe that every orret theory of physis partiularly physial interations must be guage invariant. An aessible early desription of gauge theory and its relation to group theory is by R. Utiyama http: //prola.aps.org/abstrat/pr/v0/i5/p597_. This presents E&M gauge theory and Yang-Mills gauge theory as the U and SU implementations of this general mahine. The Bosons that are predited to exist are all massless. We know many are not, and the Higgs mehanism/higgs boson has been invented to solve this problem. The Higgs is a prinipal driver for the LHC. If it isn t found this time around very many people will be desolate, while a few will be very happy. 9