22.51 Quantum Theory of Radiation Interactions

Transcription

1 .51 Quantum Theory of Radaton Interactons Fna Exam - Soutons Tuesday December 15, 009 Probem 1 Harmonc oscator 0 ponts Consder an harmonc oscator descrbed by the Hamtonan H = ω(nˆ + ). Cacuate the evouton of the expectaton vaue of J the poston of the harmonc oscator x = (a + a ) n the foowng cases: a) The harmonc oscator s ntay prepared n a superposton of number states: mω 1 ψ(t = 0)) = c a ) + c b ) where c a, c b are coeffcents such that the state s normazed (here for exampe take c a = cos(ϑ/) and c b = e ϕ sn (ϑ/)) We can use the Schrödnger pcture to fnd the evouton of the state: ψ(t)) = cos(ϑ/)e ωt ) + e ϕ sn (ϑ/)e ωt ) (Notce that I ve aready emnated the common phase factor e 1 ωt ). Then we can cacuate the expectaton vaue of x: ( ) ( ) (x(t)) = cos(ϑ/)e ωt ( + e ϕ sn (ϑ/)e ωt ( (a + a ) cos(ϑ/)e ωt ) + e ϕ sn (ϑ/)e ωt ) mω Ony the terms ( a ) = and ( a ) = survve, yedng (x(t)) = sn(ϑ) cos(ωt ϕ) mω It coud have been maybe smper to use the Heseberg pcture, rememberng that a(t) = a(0)e ωt. Then: (x(t)) = and the same resut as above s drecty obtaned. ( ) ( ) cos(ϑ/)( + e ϕ sn (ϑ/)( (ae ωt + a e ωt ) cos(ϑ/) ) + e ϕ sn (ϑ/) ) mω b) The nta state of the harmonc oscator s a superposton of coherent states: where c a, c b are coeffcents such that the state s normazed. In ths case t s convenent to use the Hesenberg pcture: ψ(t = 0)) = c a α) + c b β) ωt (x(t)) = (c a(α + c ωt b (β ) (ae + a e )(c a α) + c b β)) mω The mportant pont here was to remember that athough the coherent states are normazed, they are not orthogona, thus (α β) = 0, but ( α + β α β)/ (α β) = O α,β = e We then have (x(t)) = ( ca (αe ωt + α e ωt ) + c b (βe ωt + β e ωt ) + c c b (α e ωt + βe ωt )O α,β + c a c b (αe ωt + β e ωt )O ) a α,β mω c) Woud the choce c a = cos(ϑ/) and c b = e ϕ sn(ϑ/) normaze the above state? Wth the above choce ( ) (ψ ψ) = cos(ϑ/) + sn(ϑ/) ϕ + sn (ϑ/)cos(ϑ/) + O ϕ = 1 + sn(ϑ)e ( α + β )/ Re[e α β O e ϕ α,β e ] = 1 α,β e 1

2 Probem Coupng of a spn to an harmonc oscator 0 ponts Consder the system n fgure 1. A cantever wth a magnetc tp s postoned cosed to a spn- 1 (of gyromagnetc rato γ) n a strong externa magnetc fed B aong the z-drecton. The magnetc tp creates a magnetc gradent G z such that the fed fet by the spn depends on the poston of the tp tsef, B tot = B + G z z. In the mt of sma dspacements, the cantever can be modeed as an harmonc oscator of mass m, oscatng aong the z drecton at ts natura frequency ω c. Cantever Spn Fgure 1: A cantever couped to a spn. Adapted from P. Rab, P. Cappearo, M.V. Gurudev Dutt, L. Jang, J.R. Maze, and M.D. Lukn, Strong magnetc coupng between an eectronc spn qubt and a mechanca resonator, Phys. Rev. B 79, 0410 R 0 (009) a) What s the tota Hamtonan of the system (spn+harmonc oscator)? 1 1 H spn = γbs z = γbσ z = ωσ z 1 H h.o. = ω c (nˆ + ) The coupng between the cantever and the spn s gven by the extra fed G z z(t) actng on the spn: σz λ V = γg z zs z = γg z (a + a ) = σ z (a + a ) mω c J wth λ = γg z mω c. The tota Hamtonan s thus 1 1 λ Htot = H 0 + V = ωσ z + ωc(nˆ + ) + σ z (a + a ) b) The magnetc gradent s usuay sma, thus the coupng term between the spn and the harmonc oscator can be consdered a sma perturbaton. Use perturbaton theory to cacuate the energy and egenstates to the owest non-vanshng order. The egenstatates of H 0 are egenstates of the σ z and nˆ operators: (0) (0) ψ 0,n ) = 0) n), ψ 1,n ) = 1) n) wth energes: (0) 1 1 (0) 1 1 E 0, n = ω + ω c (n + ), E 1,n = ω + ω c (n + ) The frst order correcton s cacuated as Δ (1) = (ψk 0 V ψk 0 ). Here: (1) (1) λ Δ 0,n = ( 0 ( n [ σ z(a + a )] 0) n) = 0 and Δ 1,n = 0 as we. Thus we need to cacuate the second order energy shft. Frst we cacuate the frst order egenstates. (1) (0, m V 0, n ) 0, m ) ( 1, m V 0, n ) 1, m ) ψ 0,n ) = + (0) ( 0) ( 0) ( 0) m=n E 0,n E 0,m m E 0,n E 1,m λ(0 σ z 0)( m (a + a ) n) λ 1 σz/ 0 m (a + a ) n = 0, m ( )( ) ) + 1, m) ω (n m) (0) (0) c m=n m E 0,n E 1,m

3 fnay, Smary, we obtan (1) λ ( ψ 0,n ) = n 0, n 1 ) n + 1 0, n + 1) ) ω c (1) λ 1, ωc ψ n ) = The second order energy shft can be cacuated from Δ = (ψ 0 V ψ 1 k k ): ( n 1, n 1) n + 1 1, n + 1) ) λ λ Δ 0,n = [n (n + 1)] = ω ω Probem Tme-dependent perturbaton theory: harmonc perturbaton 5 ponts Use tme-dependent perturbaton theory to derve the transton rate for a perturbaton Hamtonan V (t) = V 0 cos(ωt). You can use the foowng steps: a) The unperturbed Hamtonan s H 0, wth egenvectors and egenvaues: H 0 k) = ω k k). In the nteracton pcture defned by H 0, the state evoves under the propagator U I (t): ψ(t)) I = U I (t) ψ(0))). What s the dfferenta equaton descrbng the evouton of U I (t)? wth V (t) = e H0 t V (t)e H0 t I. du I dt = V I (t)u I (t) b) Wrte an expanson for U I (t) to frst order ( Dyson seres). Integratng the equaton above: U I (t) = 11 0 t dt V I (t ) c) Cacuate the transton amptude c k (t) = (k U I (t) ) from the nta state ) to the egenstate k) to frst order, as a functon of ω k = ω k ω, V k = (k V 0 ) and ω. Hnt: the foowng ntegra mght be usefu: From the expresson n b) and the defnton of V I : Takng k = we have: c (t) = (k U (t) ) = δ k I k t dt e ω 1 t sn ((ω 1 ω )t/) e ±ω t = e (ω1±ω)t/ ± 0 ω 1 ± ω t t dt (k V (t ) ) = δ I k 0 0 t dt (k V (t ) )e (ω k ω ) t ck(t) = dt (k V I (t ) ) = (k V0 ) dt cos(ωt)e (ω ω k ) Settng ω k = ω k ω and usng the gven formua, we have: 0 V k c (t) = e (ω k+ω)t/ sn ((ω k + ω)t/) (ω k ω)t/ sn ((ω k ω)t/) k + e ω k + ω ω k ω 0 d) Cacuate the probabty of transton p k (t) = c k (t) n the ong tme mt, wth the foowng approxmatons: sn (Ωt/) π m = tδ(ω) t Ω and sn (Ω 1 t/)sn (Ω t/) m = 0 t Ω 1 Ω

4 (for Ω 1 = Ω ) The probabty p k (t) = c ( k t) w have contrbutons from terms ke (ω k +ω)t/ sn ((ω k + ω)t/) e ω k + ω and e (ω k+ω)t/ sn ((ω k + ω)t/) (ω k ω)t/ sn ((ωk ω)t/) e ω k + ω ω k ω Ths ast term goes to zero by the second reatonshp provded, whe the frst terms gve: p k = π Vk t [δ(ω k + ω) + δ(ω k ω)] e) Fnay, you shoud wrte down the transton rate W k = dp k. The transton rate s just the probabty per tme: dt π Vk W k = [δ(ω k + ω) + δ(ω k ω)] Probem 4 Rayegh ght scatterng 5 ponts Consder the eastc scatterng of ght from a moecue n the atmosphere. We want to cacuate the frequency dependence of the crosssecton, to understand why the sky s bue and the sunset s red. The system of nterest s descrbed by a moecue, wth egenstates m k ) and energes E k and two modes of the radaton fed, k and k k, λ wth energes ωk and ω k and poarzatons λ and λ. For convenence the system s encosed n a cavty of voume V = L. The nteracton between the radaton fed and the moecue s descrbe by k, λ Moecue the hamtonan V = d E E n the dpoe approxmaton, where L πωh h R h R E = ahξe + a e Eǫ hξ Fgure : Rayegh scatterng, showng the ncomng and hξ V outgong photon nto the voume of nterest. h,ξ wth R s the poston of the center of mass of the moecue. Wf You can use the foowng steps to cacuate the scatterng cross secton dσ =, wth W f = π (f T ) Φ ρ(e f ), where T s the nc transton matrx and ρ(e f ) the fna densty of states. a) Wrte a forma expresson for the transton matrx eement (f T ), to the owest non-zero order n the perturbaton V. ( f V )( V ) (f T ) = ( f V ) E As V does not aow transtons nvovng two photons, the frst order term s zero and we have: ( f V )( V ) (f T ) = E E E b) What are the possbe ntermedate (vrtua) states that we need to consder n ths scatterng process? Use them to smpfy the expresson n a). The nta state s m, 1 k,λ, 0 k,λ ) and fna state m, 0 k,λ, 1 k,λ ). Intermedate states are such that there s ony 1-photon transton, ether m, 0 k,λ, 0 k,λ ) or m, 1 k,λ, 1 k,λ ). Thus: (mf, 0 k,λ, 1 k,λ V m, 0 k,λ, 0 k,λ )(m, 0 k,λ, 0 k,λ V m, 1 k,λ, 0 k ) ( ),λ f T = ( E + ω k ) E (mf, 0 k,λ, 1 k,λ V m, 1 k,λ, 1 k,λ )(m, 1 k,λ, 1 k,λ V m, 1 k,λ, 0 k,λ + ) (E + ωk) (E + ω k + ω 4 k )

5 Usng the expct expresson for V, we have: π J E E E E (k k) R) (Eǫ d f )(Eǫ k d ) (Eǫ k d f )(Eǫ d ) ( f T ) = ω ω k k k k V e E + E + ω k E E ω k c) What s the fux of ncomng photons and the densty of states of the outgong photons? Φ = c/l nc and L ωk ρ(ef) = dω π c d) Fnd an expresson for the dfferenta cross secton dσ (where dω s the sod ange nto whch the photon s scattered) From dσ = Wf π (f T ) = Φ nc Φ nc ρ(e f ) we fnd: dω dσ π L L ω E E E E k 4π (Eǫ k d f)(eǫ k d ) (Eǫ k d f )(Eǫ k d ) = (ω k ωk dω c π ) + c L 6 E E + ω k E E ω k smpfyng the expresson: dσ ω E E E E k ω k (Eǫ k d f )(Eǫ k d ) (Eǫ k d f )(Eǫ k d ) = + dω c 4 E E + ωk E E ω k e) In the case of eastc scatterng m f ) = m ) and ω k = ω k. What s the scatterng cross-secton dependence on the photon frequency ω k? How does that hep expanng why the sky s bue and the sunset red? We can further smpfy the cross secton to dσ ω 4 E E k (Eǫ k d )(Eǫ k d )(E E ) = dω c 4 (E E) (ω k) As (E E ) ω k, the cross secton depends on the frequency as ωk 4, thus bue ght s scattered more than red ght, gvng the coor of the sky. 5

6 MIT OpenCourseWare Quantum Theory of Radaton Interactons Fa 01 For nformaton about ctng these materas or our Terms of Use, vst: