PHYS Quantum Mechanics I - Fall 2011 Problem Set 7 Solutions

Transcription

1 PHYS Fall PHYS Quantum Mechanics I - Fall Problem Set 7 Solutions Joe P Chen / joepchen@gmailcom For our reference, here are some useful identities invoked frequentl on this problem set: J j,m jj + j,m J z j,m m j,m J ± J x ±ij J ± j,m j mj ±m+ j,m± Angular momentum a There are at least two was to approach this problem: Using the ladder operators, J x J + + J and J i J + J When J x resp J acts on j,m, it produces a linear combination of two ket states j,m+ and j,m, both of which are orthogonal to j,m So if we put the bra j,m with the ket J x j,m resp J j,m together, the bracket must vanish: that is, the expectation value of J x resp J in the state j,m is It is also possible to exploit the commutation relations of the J i alone without invoking the ladder operators Since J,J z ij x, and J z j,m m j,m, we can compute the expectation value of J x in j,m as follows: j,m J x j,m i j,m J,J z j,m i j,m J J z j,m j,m J z J j,m i m j,m J j,m m j,m J j,m Essentiall the same arguments go to show that j,m J j,m b For this part we use: The commutation relation J i,j j iǫ ijk J k i,j,k,3 The Jacobi identit of commutatorslie brackets: A,B, C B,C, A+C,A, B Thus if an operator O commutes with both J i and J j i j, then it must commute with the third component: O,J k ǫ ijk i O,J i,j j ǫ ijk i {J i, J j,o+j j, O,J i }

2 PHYS Fall Spin precession We re given a spin-/ particle ψ in a B field Bt Bcosωtî+Bsinωtĵ+B ˆk which consists of a static z-component and an oscillating component in the x-plane In this lab frame, ψ evolves according to the Schrödinger equation i d ψt Ht ψt γs Bt ψt dt Due to the time-dependence of B or H, it is more complicated to write down the solution ψt in the lab frame So instead we shifts to a frame which co-rotates with the oscillating field, and introduce ψ r t e iωszt/ ψt, which should see a time-independent B field What is the effective Hamiltonian H r in the rotating frame? A direct calculation shows that i d dt ψ rt i d dt e iωszt/ ψt i iωs z /e iωszt/ ψt +ie iωszt/ d dt ψt ωs z ψ r t +e iωszt/ Ht ψt ωs z +e iωszt/ Hte iωszt/ ψ r t }{{} H r To go on we must unravel the operator e iωszt/ Hte iωszt/ This can be done b considering its matrix representation in the eigenbasis of S z, { +, }: Clearl e iωszt/ e iωt/ e iωt/ and e iωszt/ e iωszt/ Meanwhile, Ht γs Bt γ γ γ 3 σ j B j j B Bcosωt+isinωt Bcosωt isinωt B B Be iωt Be iωt B Thus B Be iωt e iωszt/ Hte iωszt/ γ e iωt/ e iωt/ e iωt/ Be iωt B B e iωt/ Be iωt/ e iωt/ e iωt/ γ e iωt/ B B Be iωt/ B e iωt/ γ B B γb S z +BS x

3 PHYS Fall Putting it together, we get i d dt ψ rt γs B r ψ r t where B r Bîr + B ω ˆk γ Now that the effective Hamiltonian H r γs B r is time-independent, we ma easil write down the time evolution of an state in the rotating frame as ψ r t e ihrt/ ψ r e iγs Brt/ ψ r ToexplicitlcomputetheactionoftheunitarevolutionoperatorUt e iγs Brt/ onanarbitrar state, it helps to exploit the following identities: Define θ θ ˆn;+ cos + +e iφ sin θ θ ˆn; sin + +e iφ cos, where ˆn is the unit vector in R 3 with azimuthal angle θ from +ˆk and polar angle φ Then S ˆn ˆn;± ± ˆn;± In the current problem, the operator S B r B r S ˆn, where B r B + B ω B and ˆn has associated angles θ sin, φ γ B r Thereforeithaseigenkets ˆn;± witheigenvalues± B r / Bthefunctionalcalculusofoperators see #, PS, the operator e iγs Brt/ has the same eigenkets ˆn;± with eigenvalues e ±iγ Br t/ This means that ψ r t e iγ Br t/ ˆn;+ ψ r ˆn;+ +e iγ Br t/ ˆn; ψ r ˆn; Now suppose the initial ket is ψ r ψ +, per the problem Then in the rotating frame its time evolution is given b ψ r t e iγ Br t/ ˆn;+ + ˆn;+ +e iγ Br t/ ˆn; + ˆn; θ θ e iγ Br t/ cos ˆn;+ e iγ Br t/ sin ˆn; θ θ θ e iγ Br t/ cos cos + +sin θ θ θ e iγ Br t/ sin sin + +cos γ Br t γ Br t γ Br t cos +isin cosθ + +isin sinθ ωr t ω ω ωr t γb ωr t cos +i sin + +i sin, where we have used the shorthands ω B /γ and ω r B r /γ ω r ω r 3

4 PHYS Fall Back in the lab frame, the state would read ψt e iωszt/ ψ r t ωr t ω ω ωr t γb cos +i sin e iωszt/ ωr t + +i sin ω r ω r ωr t ω ω ωr t γb cos +i sin e iωt/ ωr t + +i sin ω r ω r e iωszt/ e iωt/ Note that ψ +, as required It follows that S z ψ S z ψ / If ω ω on resonance, the effective field B r in the rotating frame consists of the transverse component Bîr onl, so the spin state precesses about the îr axis at angular frequenc ω r ω γb From the perspective of the lab frame, the state evolves as ω t ψt cos e iωt/ ω t + +isin e iωt/ e iω t/ cos ω t e iω t/ ˆnt;+, + +e iπ/ sin ω t e iωt where ˆnt is the unit vector with associated angles θt ω t and φt π/ ω t The orientations of the spin state trace out a figure-8 on the Bloch sphere Fig Note that the up state + flips into the down state in a duration T π/ω, and vice versa Figure : Time evolution of a spin-/ state in a field Bt Bcosωtî+Bsinωtĵ+B ˆk when on resonance ω γb Initial state is the up state + ẑ;+ Dots on the Bloch sphere indicate the successive spin orientations in the lab frame For general ω, the z-magnetization of the state ψt at time t is S z t ψt S z ψt 4

5 PHYS Fall ωr t ω ω ωr t γb cos i sin e iωt/ ωr t + i sin ω r ω r ωr t ω ω ωr t γb S z cos +i sin e iωt/ ωr t + +i sin ω r ω r cos ωr t ω ω ωr t γb +i sin sin ωr t ω r ω r cos ωr t + ω ω γb ωr sin ωr t +cosωr t + ω ω γb cosωr t ωr ω ω ωr + γb ωr cosω r t ω ω S z ω ω +γb + γb ω ω +γb cosω rt From this we deduce that the up state reverses orientation in a duration T π ω r π ω ω +γb e iωt/ e iωt/ 3 Angular momentum of an unknown particle Sakurai 35 a It helps to rewrite ψx in spherical coordinates: ψx rfrsinθcosφ+sinθsinφ+3cosθ To check whether ψ is an eigenfunction of L, we ma carr out a direct computation First note that the operator L can be explicitl written in spherical coordinates eg Sakurai Eq 365: L ψx sin θ φφ + sinθ θsinθ θ ψx φφ ψx rfrsinθcosφ+sinθsinφ θ ψx rfrcosθcosφ+cosθsinφ 3sinθ θ sinθ θ ψx rfr θ sinθcosθcosφ+sinφ 3sin θ rfrcosθcosφ+sinφ 3sinθ So L ψx sin θ sinθcosφ+sinφ+ sinθ cosθcosφ+sinφ 3sinθ rfr rfr sinθcosφ+sinφ 3cosθ ψx Thus ψx is an eigenfunction of L with eigenvalue ll+, or l 5

6 PHYS Fall b Alternativel, we can re-express ψx as a linear combination of spherical harmonics Using the normalized spherical harmonics 3 Y ± 3 cosθ, Y 4π 8π sinθe±iφ, we find e iφ +e iφ ψx rfr sinθ + eiφ e iφ +3cosθ i rfr isinθeiφ + +isinθe iφ +3cosθ π π rfr 3 iy + +iy + 3πY 3 In one fell swoop, we ve shown that ψx is an eigenfunction of L with eigenvalue +, and expanded ψx in the eigenbasis of the j Hilbert space, ie, ψx where rfr m c my m π c 3 i, c π 3π, c 3 +i It ought to be clear that the probabilit of ψ being found in the state,m is given b Since c 9 c 9 c, we have Pm c m m c m P, P 9, P c Recall that the Laplacian in R 3 can be written as r r r r + r sin θ φφ + sinθ θsinθ θ r r r r L So the time-independent Schrödinger equation in R 3 takes the form { mr r r L } r + mr +Vr Ψx EΨx Now suppose the energ eigenstate Ψx is the known wavefunction ψx rfrζω, where ζω sinθcosφ+sinθsinφ+3cosθ From Part a we alread saw that L ψx ψx Meanwhile, r ψx fr+rf rζω r r r ψx rfr+rf rζω+r f r+rf rζω Plugging the various terms into ields rfr+4rf r+r f rζω mr fr+4rf r+r f rζω+ mr Thus Vr E + rfr mr rfrζω+vrrfrζω ErfrζΩ 4rf r+r f r E + mr 4rf r+r f r fr 6

7 PHYS Fall 4 Rotated angular momentum Sakurai 3? A state ψ rotated b an angle β about the -axis becomes e ijβ/ ψ So the probabilit for the new state to be in,m m,±,± is given b the modulus squared of the projection of e ijβ/ l,m onto the subspace l,m, ie, D m α,β,γ,m e ijβ/,, where α, β, and γ are the Euler angles At this stage we ma invoke Sakurai Eq 365: D l 4π m α,β,γ l+ Y l m β,α Using the expressions Y m θ,φ in Appendix A, we find D 4π 4π 5 3, α,β,γ 5 Y β, 5 3π sin β 8 sin β, D 4π 4π 5 3, α,β,γ 5 Y β, 5 8π sinβcosβ sinβcosβ, D 4π 4π 5, α,β,γ 5 Y β, 5 6π 3cos β 3cos β, D, D, Thus α,β,γ D, α,β,γ, α,β,γ D, α,β,γ D ±, α,β,γ D ±, α,β,γ D, α,β,γ It is straightforward to check that m D 3 8 sin4 β, 3 sin βcos β, 4 3cos β m α,β,γ 5 Rotation matrix for j states Sakurai 3 a Since J i J + J, it is clear that the matrix element j,m J j,m vanishes for an m, m where m m Also j,m J j,m j,m J j,m b hermiticit of J Soforj, itisenoughtocomputethematrixelements, J, and, J, From J, i J +, J, i, i, ; J, i J +, J, i, i,, Please read Sakurai Eqs 3646 through 365 and the accompaning text for the derivation 7

8 PHYS Fall we deduce that, J, i/ and, J, i/ So the matrix representation of J in the {,,,,, } basis reads J j i i i i i i i i b A direct computation shows that J j i i i i i i i i 4 and J j 3 3 In other words, i i i i J j 4 3 i i i i 3 j / J /, which means that for positive integers n, J j n / J j J j /, n odd, / n even Therefore e ij β/ iβ n+ + n+! n n β n+ i n+! n J i sinβ + J n+ + J J + n n iβ n n! n β n n! cosβ J n J c The matrix representation d β of e ij β/ reads d β isinβ i i i +cosβ i +cosβ sinβ cosβ sinβ cosβ sinβ cosβ sinβ +cosβ 4 8

9 PHYS Fall 6 Neutrino oscillations B assumption, the initial state of the neutrino is the weak eigenstate ψ ν e cosθ ν +sinθ ν, where ν j j, are the mass eigenstates, and θ is the mixing angle Since the neutrinos are assumed free, the mass eigenstates evolves in time according to ν j t e iht/ ν j e iejt/ ν j where E j m j c +pc The assumption that ν e is a momentum eigenstate allows us to replace the operator ˆp with the scalar p As a result, ψ evolves in time according to ψt e iht/ ψ e ie t/ ν ψ ν +e ie t/ ν ψ ν e ie t/ cosθ ν +e ie t/ sinθ ν Thus the probabilit of the sstem being in ν µ at time t is ν µ ψt sinθ ν +cosθ ν e iet/ cosθ ν +e iet/ sinθ ν sinθcosθ e ie t/ +e ie t/ sin θ ie ie +E t/ e ie E t/ +e ie E t/ i sin θsin E E t Knowing that the mass of the neutrino is ver small, ie, m j c pc, we ma approximate E j in the usual wa: mj E j m j c +pc c pc + pc + mj c mj c +O 4 pc pc pc So E E pc m m c4 pc +O mj c 4 m m c4 pc pc +higher-order terms Using the shorthand m m m, we deduce that ν µ ψt m sin θsin c 3 t, 4p which shows that neutrino oscillation occurs at period T 4πp/ m c 3 9