/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q12P Prove the commutation relation i... [FREE SOLUTION] | 91影视

91影视

Prove the commutation relation in Equation 9.74. Hint: First show that

[L2,z]=2ih(xLy-yLx-ihz)

Use this, and the fact that localid="1657963185161" r.L=r.(rp)=0, to demonstrate that

[L2,[L2,z]]=2h2(zL2+L2z)

The generalization from z to r is trivial.

Short Answer

Expert verified

The communication relation in Equation 9.74 is[L2,[L2,z]]=2h2(zL2+L2z)

Step by step solution

01

Definition of commutation relation

Fundamental quantum mechanical relationships that link successive operations on the wave function or state vector of two operators (L1 and L2) in opposing orders, i.e., L1 L2 and L2 L1. The operators' algebra is defined by the commutation relations.

02

Proving the commutation relation in Equation 9.74.

[L2,z]=[Lx2,z]+[Ly2,z]+[Lz2,z]=Lx[Lx,z]+[Lx,z]Lx+Ly[Ly,z]+[Ly,z]Ly+Lz[Lz,z]+[Lz,z]Lz.

But{[Lx,z]=[ypzzpy,z]=[ypz,z][zpy,z]=y[pz,z]=iy[Ly,z]=[zpxxpz,z]=[zpx,z][xpz,z]=x[pz,z]=ix[Lz,z]=[xpyypx,z]=[xpy,z][ypx,z]=0

厂辞:鈥夆赌夆赌[L2,z]=Lx(iy)+(iy)Lx+Ly(ix)+(ix)Ly=i(LxyyLx+Lyx+xLy).

But{Lxy=LxyyLx+yLx=[Lx,y]+yLx=iz+yLxLyx=LyxxLy+xLy=[Ly,x]+xLy=iz+xLy

So:[L2,z]=i(2xLyiz2yLxiz)[L2,z]=2i(xLyyLxiz)[L2,[L2,z]]=2i{[L2,xLy][L2,yLx]i[L2,z]}=2i{[L2,x]Ly+x[L2,Ly][L2,y]Lxy[L2,Lx]i(L2zzL2)}[L2,Ly]=[L2,Lx]=0

so,

[L2,Lx]=0,鈥夆赌夆赌[L2,Ly]=0,鈥夆赌夆赌[L2,Lz]=0(4.102).

[L2,[L2,z]]=2i{2i(yLzzLyix)Ly2i(zLxxLziy)Lxi(L2zzL2)}[L2,[L2,z]]=2i,or[L2,[L2,z]]=22(2yLzLy2zLy22zLx22z(Lx2+Ly2+Lz2)+2zLz22ixLy+2xLzLx+2iyLxL2z+z

=22(2yLzLy2ixLy+2xLzLx+2iyLx+2zLz22zL2L2z+zL2)=22(zL2+L2z)42[(yLzix)LzyLy+(xLz+iy)LzxLx+zLzLz]=22(zL2+L2z)42(LzyLy+LzxLx+LzzLz)Lz(rL)=0=22(zL2+L2z)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve Equation 9.13 to second order in perturbation theory, for the general case ca(0)=a,cb(0)=bca=-ihH'abe-颈蝇0tcb,cb=-ihH'bae-颈蝇0tca

(9.13).

The first term in Equation 9.25 comes from the eit/2, and the second from e-it/2.. Thus dropping the first term is formally equivalent to writing H^=(V/2)e-it, which is to say,

cbl-ihvba0tcos(t')ei0t'dt'=-iVba2h0tej(0+)t'+ej(0-)t'dt'=--iVba2hej(0+)t'-10++ej(0-)t'-10-(9.25).Hba'=Vba2e-it,Hab'=Vab2eit(9.29).

(The latter is required to make the Hamiltonian matrix hermitian鈥攐r, if you prefer, to pick out the dominant term in the formula analogous to Equation 9.25 forca(t). ) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 9.13 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the field.

c.a=-ihHab'e-i0tcb,c.b=-ihHba'e-i0tca,

(a) Solve Equation 9.13 in the rotating wave approximation (Equation 9.29), for the usual initial conditions: ca(0)=1,cb(0)=0. Express your results (ca(t)andcb(t))in terms of the Rabi flopping frequency,

r=12(-0)2+(Vab/h)2 (9.30).

(b) Determine the transition probability,Pab(t), and show that it never exceeds 1. Confirm that.

ca(t)2+cb(t)2=1.

(c) Check that Pab(t)reduces to the perturbation theory result (Equation 9.28) when the perturbation is 鈥渟mall,鈥 and state precisely what small means in this context, as a constraint on V.

Pab(t)=cb(t)2Vab2hsin20-t/20-2(9.28)

(d) At what time does the system first return to its initial state?


Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio at rest in a static magnetic fieldB0k^ precesses at the Larmor frequency0=纬叠0 (Example 4.3). Now we turn on a small transverse radiofrequency (rf) field,Brf[cos(t)^sin(t)j^]$, so that the total field is

role="math" localid="1659004119542" B=Brfcos(t)^Brfsin(t)j^+B0k^

(a) Construct the 22Hamiltonian matrix (Equation 4.158) for this system.

(b) If (t)=(a(t)b(t))is the spin state at time t, show that

a=i2(惟别颈蝇tb+0a):鈥夆赌夆赌b=i2(惟别颈蝇ta0b)

where 纬叠rfis related to the strength of the rf field.

(c) Check that the general solution fora(t) andb(t) in terms of their initial valuesa0 andb0 is

role="math" localid="1659004637631" a(t)={a0cos('t/2)+i'[a0(0)+b0]sin('t/2)}e颈蝇t/2b(t)={b0cos('t/2)+i'[b0(0)+a0]sin('t/2)}e颈蝇t/2

Where

'(0)2+2

(d) If the particle starts out with spin up (i.e. a0=1,b0=0,), find the probability of a transition to spin down, as a function of time. Answer:P(t)={2/[(0)2+2]}sin2('t/2)

(e) Sketch the resonance curve,

role="math" localid="1659004767993" P()=2(0)2+2,

as a function of the driving frequency (for fixed 0and ). Note that the maximum occurs at=0 Find the "full width at half maximum,"螖蝇

(f) Since 0=纬叠0we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude gauss. What will the resonant frequency be? (See Section for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.)

A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the 鈥渇loor鈥 of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent:V0(t),withV0(0)=V0(T)=0.

(a) Solve for the exact cm(t), using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" (T), in terms of the function V0(t)

(b) Analyze the same problem in first-order perturbation theory, and compare your answers. Compare your answers.
Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.8.

Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that

. Comment: Ostensibly, this system oscillates between 鈥溾 Doesn鈥檛 this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian鈥攁 measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T 鈥攖his doesn鈥檛 affect the calculations, but it allows for a more sensible interpretation of the result.

ca=-ihHabeigtcb,cb=-ihHbaeigtca 鈥(9.13).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.