Physics 115C Homework 4

Physics 115C Homework 4 Problem 1 a In the Heisenberg picture, the ynamical equation is the Heisenberg equation of motion: for any operator Q H, we have Q H = 1 t i [Q H,H]+ Q H t where the partial erivative is efine as Q H e iht/ Q S t t e iht/ where Q S is the Schröinger operator. If we re intereste in the evolution of the lowering operator of the simple harmonic oscillator, we let Q = a, an we get a H = 1 t i [a H,H] To evaluate the commutator, let s express the Hamiltonian in terms of the Heisebnerg raising an lowering operators. To o so, recall that in the Schröinger picture, the Hamiltonian for the simple harmonic oscillator was H = ω a S a S + 1 Now, let s multiply the Hamiltonian by unity in the form 1 = e iht/ e iht/ Using the fact that the Hamiltonian commutes with any function of itself, we get H = 1 H = e iht/ e iht/ H = e iht/ He iht/ = ωe iht/ = ω e iht/ a S a S + 1 e iht/ a S a Se iht/ + 1 1

Cleverly inserting another factor of unity between the laer operators, we get H = ω e iht/ a S e iht/ e iht/ a S e iht/ + 1 = ω a H ta Ht+ 1 where we recognize the relationship between the Schröinger an Heisenberg operators a H t = e iht/ a S e iht/ an likewise for a. The next thing we ll nee are the commutation relations for a H an a H. In the Schröinger picture, we know that [a S,a S ] = 1 Multiplying on the left by e iht/ an the right by e iht/, we get e iht/ [a S,a S ]e iht/ = e iht/ e iht/ e a iht/ S a S a S a S e iht/ = 1 Again inserting unity between the operators, we get e iht/ a S e iht/ e iht/ a S e iht/ e iht/ a S e iht/ e iht/ a S e iht/ = 1 a H ta H t a H ta Ht = 1 Thus we obtain the so-calle equal-time commutation relation [a H t,a H t] = 1 it is crucial that the times be equal, for otherwise our arguments above wouln t have worke!. We are now reay to calculate the explicit time-evolution of a H. We ha the equation of motion a H = 1 t i [a H,H] Writing H = ω a H ta Ht+ 1 an using our equal-time commutation relations, we fin that [ [H,a H ] = ω a H t,a H ta Ht+ 1 ] [ ] = ω = ω a H t,a H ta Ht a H t[a Ht,a H t]+ = ωa H t ] [a H t,a H t a H t

The first commutator in the secon-to-last line vanishes trivially, since a H t commutes with itself. Thus the Heisenberg equation of motion gives a H t = 1 i ωa Ht = iωa H t This is a simple ifferential equation: its solution is a H t = a H 0e iωt Togettheinitialconitiona H 0,wegobacktotheefinitionoftheHeisenbergoperator: We see easily then that a H t = e iht/ a S e iht/ a H 0 = a S an therefore we have the simple relationship a H t = a S e iωt b To get the corresponing time-evolution of a H, we coul simply take the ajoint of a Ht, but let s work it out explicitly for practice. This time, we have a H t = 1 ] [a H i,h = 1 [ i ω a H t,a H ta Ht+ 1 ] [ ] = iω a H t,a H ta Ht [ ] ] = iω a H t a H t,a Ht + [a H t,a H t a H t = iωa H t Thus we fin that or as expecte. a H t = a H 0eiωt a H t = a S eiωt c That the Hamiltonian is time-inepenent in the Heisenberg picture follows trivially from our work in part a above, where we showe that the Hamiltonian is in fact the 3

same in both pictures since it commutes with both e iht/ an e iht/. However, we can show it explicitly using the results we just foun: H = ω a H ta Ht+ 1 = ω a S eiωt a S e iωt + 1 = ω a S a S + 1 So inee, the iniviual time epenencies in a H an a H a time-inepenent Hamiltonian. cancel themselves out to give 4

Problem a The Biot-Savart law states that in magnetostatics, the magnetic fiel create by an infinitesimal current element Il at position r is Br = µ 0 Il r r 4π r r 3 Now, this formula is only true in magnetostatics i.e. for a steay line current, so technically, it oes not apply to our case. The reason is that for a changing current, we nee to take into account retaration: since information can t travel faster than the spee of light, it takes time for the information about the changing current to propagate from r to r. However, assuming that this effect can be neglecte i.e. if we work in the nonrelativistic regime, then for a moving point charge we can approximate the moving current element as Il = qv, an we fin that the magnetic fiel create by a point charge moving with velocity v at position r is Br µ 0 qv r r when v/c 1 4π r r 3 Incientally, even if we wante the full relativistic formula, we couln t get it from the given information. To get the exact answer, we nee to know the trajectory of the particles, but all we re given are their position an velocity at some instant, so the nonrelativistic formulas are all we can get. Using this formula, we can calculate the magnetic fiel each particle feels ue to the other. First, we have v 1 = v 1 ˆx v = v ŷ r 1 r = ŷ The magnetic fiel that particle 1 feels ue to particle is thus B 1 = µ 0 qv r 1 r 4π r 1 r 3 = µ 0 qv 4π ŷ ŷ = 0 Does this make sense? Sure! Particle effectively creates a current pointing up along the y-axis, so the magnetic fiel lines by the right-han rule shoul be going aroun the y-axis. But particle 1 is right on the y-axis, where there is no magnetic fiel from particle. 5

Next, the magnetic fiel that particle feels ue to particle 1 is B 1 = µ 0 qv 1 r r 1 4π r r 1 3 qv 1 = µ 0 4π = µ 0 4π ˆx ŷ qv 1 ẑ This makes sense too: particle 1 looks like a current traveling along the x-axis, so the magnetic fiel lines go aroun the x-axis, an so along the negative y-axis where particle is, the magnetic fiel shoul be pointing in the negative z-irection, as we foun. Finally, we nee to calculate the electric fiel between the charges. This one s easy, since in the nonrelativistic regime, we just use Coulomb s law: E 1 = 1 4πǫ 0 q ŷ E 1 = 1 4πǫ 0 q ŷ To fin the force on each particle, we just use the Loretz force law: The force on particle 1 is then F = qe +v B F 1 = qe 1 +v 1 B 1 1 q = q 4πǫ 0 ŷ +0 Next, F 1 = 1 4πǫ 0 q ŷ F = qe 1 +v B 1 = q 1 q 4πǫ 0 ŷ µ 0 qv 1 v ŷ ẑ 4π F = 1 4πǫ 0 q ŷ µ 0 4π q v 1 v ˆx Clearly, these forces are in violation of Newton s thir law: the elecric forces between the particles are inee equal an opposite, but the magnetic forces are not. 6

b In special relativity, kinetic energy actually, total energy, but they just iffer by a constant is the time component of the momentum four-vector. The spatial components are no surprise the spatial momentum. Thus the momentum four-vector is p µ = E/c,p where the inex µ takes the values 0, 1,, 3 or t,x,y,z, if you prefer. The extra factor of c is simply neee so that all the components of p µ have units of momentum. Likewise, the scalar potential φ is the time component of a four-vector whose spatial components are the vector potential A. The four-vector potential is thus A µ = φ/c,a Again, the factor of c is necessary to give all the components units of A. At this point, we recognize that the conservation equation 1 t mv +qφ = 0 is simply the time-component of the four-vector conservation equation τ pµ +qa µ = 0 I switche from time coorinate t to proper time τ to emphasize the covariant nature of the equation; in the nonrelativistic limit, τ t an therefore /τ /t. All we have to o now is extract the spatial components of the conservation equation to get τ p+qa = 0 In the nonrelativistic limit, p mv, /τ /t, an thus we get mv +qa = 0 t Of course, this wasn t at all a proof; we simply motivate the above equation base on relativistic covariance. However, it is relatively simple to prove it from the Hamiltonian formalism. Recall that the classical Hamiltonian for a particle moving in an electromagnetic fiel is H = p canon m +qφ where p canon is the canonical momentum: p canon = mv +qa 7

The Hamilton-Jacobi equations of motion tell us that p canon t = H x = φ mv +qa = φ t In the presence of no external electric fiels as is require for conservation of momentum, φ = 0, an we obtain the esire conservation equation. c Our job is to evaluate the expression mv +qa t for each particle. Iniviually, these will not necessarily be equal to zero, because each particle feels an external electrical force the Coulomb force from the other particle. However, the combine two-particle system oes not feel any external electrical force, an therefore we expect that the total momentum conservation law shoul hol: Let s begin by evaluating t mv 1 +qa 1 + t mv +qa 1 = 0 t mv 1 +qa 1 = ma 1 +q A 1 t = F 1 +q ri 1 t = F 1 +qv i 1 A 1 r1 i r1 =0,r = ŷ A 1 r i 1 r1 =0,r = ŷ where I use Newton s secon law to write F = ma, an use the chain rule to write A/t = A/r i r i /t remember that there s an implie sum over the inex i = x,y,z; the erivative is to be evaluate at the positions r 1 = 0,r = ŷ. Since v 1 = v 1ˆx, we get t mv A 1 1 +qa 1 = F 1 +qv 1 x 1 r1 =0,r = ŷ The problem is now to calculate the vector potential felt by particle 1 ue to particle, A 1. There is no unique choice for the vector potential because of gauge freeom, but a hany choice for the nonrelativistic vector potential ue to a point charge with velocity v at position r is Ar = µ 0 4π 8 qv r r

Thus A 1 = µ 0 4π = µ 0 4π qv r 1 r qv r ŷ where I ve now left the istance between particles 1 an a variable r instea of fixe because I ll be ifferentiating with respect to it; in particular, r x 1 x +y 1 y +z 1 z We now have t mv A 1 1 +qa 1 = F 1 +qv 1 x 1 r1 =0,r = ŷ = 1 q 4πǫ 0 ŷ + µ 0 1 4π q v 1 v x 1 r = 1 4πǫ 0 q ŷ + µ 0 4π q v 1 v [ ] x x 1 r 3 r1 =0,r = ŷŷ r 1 =0,r = ŷ where I use the fact that 1 xr = x x 1 r 3 Now, note that at r 1 = 0 an r = ŷ, we have x x 1 = 0, an therefore the secon term just evaluates to zero. Thus t mv 1 +qa 1 = 1 4πǫ 0 q ŷ As promise, momentum is not conserve for this single particle, because of the electric force from the secon particle ha we ignore electric forces an only ealt with magnetic forces, then this expression woul inee have been zero. Now, on to the secon particle: t mv +qa 1 = ma +q A 1 t = F +q ri t = F +qv i A 1 r i r1 =0,r = ŷ A 1 r i Using v = v ŷ, we get t mv A 1 +qa 1 = F +qv y 9 r1 =0,r = ŷ r1 =0,r = ŷ ŷ

This time, we have A 1 = µ 0 4π Thus t mv A 1 +qa 1 = F +qv y = µ 0 4π qv 1 r r 1 qv 1 r ˆx r1 =0,r = ˆx = 1 q 4πǫ 0 ŷ µ 0 q v 1 v ˆx+ µ 0 1 4π 4π q v 1 v y r = 1 q 4πǫ 0 ŷ µ 0 q v 1 v ˆx+ µ 0 4π 4π q v 1 v But at r 1 = 0 an r = ŷ, y 1 y = an r =, so [ ] y1 y r 3 r1 =0,r = ŷ t mv +qa 1 = 1 q 4πǫ 0 ŷ µ 0 q v 1 v ˆx+ µ 0 4π 4π q v 1 v 3ˆx = 1 q 4πǫ 0 ŷ ˆx r 1 =0,r = ŷ Note that the contributions from the magnetic fiels o inee cancel out, leaving just the electric fiel stuff; again, if we were to ignore the Coulomb force an just focus on the magnetic fiels, we woul have foun that the complete momentum was conserve for just this one particle. Combining these results, we fin that t mv 1 +qa 1 + t mv +qa 1 = 0 So the complete momentum is inee conserve when consiering the entire twoparticle system, because there are no external electrical forces acting on it. ˆx 10

Problem 3 Let s first think about this system physically. Before we change the magnetic fiel, the system consists of a stationary ring an static magnetic an electric fiels. In general, electromagnetic fiels can carry momentum linear an angular, but because of the rotational symmetry of the system, we expect the electromagnetic fiels to carry no total momentum. When we increase the magnetic fiel, there is a corresponing increase of magnetic flux through the ring. The conuctor will oppose this increase in flux by generating a current to opposeit; ifwetakethe+z irectionparalleltothemagneticfiel, thentheringwillgenerate a current rotating clockwise when viewe from above epening on whether the ring is conucting or insulating, it may remain stationary or begin to rotate, but either way the charge carriers will carry some mechanical angular momentum. This will be accompanie by a net angular momentum, which we expect will be cancelle out by a corresponing change in the angular momentum of the electromagnetic fiel. Looking back at our equations, our goal is to show that t r p canon = 0 Using our expression for the canonical momentum, we want to show that mr v +qr A = 0 t or t mr v = qr A t The left-han sie of the above equation represents the change in angular momentum of the ring, while the right-han sie represents the change in the angular momentum store in the electromagnetic fiel, as iscusse. To show that this equality is true, let s begin by working on the left-han sie, i.e. by working out the change in the angular momentum of the ring. From the angular version of Newton s secon law, L/t = τ, we have that mr v = τ t So the left-han sie is nothing more than the torque exerte on the ring uring the change in the magnetic fiel. Next, if we integrate Maxwell s equation E = B t over the area insie the ring an apply Stokes theorem, we get Faraay s law: E l = Φ t 11

where Φ is the magnetic flux through the loop, an the integral is taken aroun the loop. The area of the loop is πr, so the change in the flux is Φ t = πrδb δt I ll enote the change in the magnetic fiel by δb in time δt, rather than use B an t, just to make clear what s a erivative an what s just an orinary fraction. Thus from Faraay s law, we have E l = πr δb δt Now, the force on a segment l of the ring is F = Eq, where q = ql/πr is the charge of the segment the charge per unit length is q/πr. The torque on this small segment is thus The total torque is thus τ = r F = RF = REql πr = q π El τ = q π But Faraay s law tells us what the line integral of the electric fiel aroun the loop is; using our previous result an momentarily ignoring overall minus signs, which we ll restore later, we fin that the total torque is τ = q El π = qr δb δt One final issue: we nee the vector torque, not just its magnitue. To figure out which irection the torque points, recall that we ha set up our coorinate system so that the z- axis is parallel to B. Now, an increase in the magnetic fiel will inuce a current in the ring to try to oppose it: thus positive charges in the ring will travel clockwise as viewe from above. By the right-han rule, the torque necessary to inuce this motion of the charges must point in the negative z-irection; or, alternatively, in the δb irection, so the vector torque is El τ = 1 qrδb δt 1

Note that if the sign of the charges is changes or the fiel is ecrease instea of increase, the irection of τ changes appropriately. In conclusion, we have foun that the change in the mechanical angular momentum of the ring is t mr v = 1 qrδb δt Alrighty. Next, we move on to calculating the change in the angular momentum store in the fiels: qr A t To calculate this quantity, we ll nee to pick a gauge for A. For our constant magnetic fiel, a gauge choice with a nice rotational symmetry well-aapte to this problem is the Coulomb gauge A = 0. In this gauge, we can write the vector potential for a uniform an constant magnetic fiel as A = 1 B r Then we obtain t qr A = qr A t = 1 qr B t r = 1 qr δb r δt Since the raius of the ring oesn t change uring the process, I freely carrie time eriviatives through any rs. Now, δb points along the axis of the ring, while r points perpenicular to it; by the right-han rule, δb r will have magnitue RδB an point along a counterclockwise tangent to the circle again, as viewe from above. This, too, is perpenicular to r, so r δb r will have magnitue R δb an point along the axis of the circle, in the δb irection. Thus r δb r = R δb this can also be compute by evaluating the cross proucts by brute force, but with orthogonal vectors like we have here, it s easier to just figure out via the right-han rules. Then t qr A = 1 qrδb δt 13

an thus, lo an behol, we have shown explicitly that L canon t = mr v +qr A t = t mr v+ qr A t = 1 qrδb δt + 1 qrδb δt = 0 Remarkably, even though the mechanical angular momentum of the charges in the ring an the angular momentum store in the EM fiels were not conserve separately, the gran total canonical angular momentum was inee conserve. Cool! 14

Problem 4 To start, let s calculate the resonant frequency for a transition from the groun state to the first excite state of this infinite square well. This is given by ω 1 E E 1 = 4E 1 E 1 = 3E 1 = 3π ml Plugging in our numbers with m the mass of an electron, we get ω 1 1.714 10 15 ra/s This is very close to the three riving frequencies we re aske to consier, which means this perturbation shoul excite a resonance between these two energy levels. Now, let s get calculating: from first-orer perturbation theory, the probability of transitioning from the groun state to the first excite state is c 1, where c 1 = δ 1 i T 0 e ie E 1 t/ H 1 t where in our case T = 1 10 15 s, an the perturbing Hamiltonian is H = V 0 xcosωt Let s begin by calculating the matrix element: H 1 = V 0 cosωt x 1 0 = V 0 cosωt L L 0 xsin π π L x sin L x x The integral shoul be familiar to you from 115A I i it on Homework 4 this past fall, if you took it this year. It comes out to be L π π xsin L x sin L x x = 8L 9π an thus H 1 = V 0 cosωt 8L L9π = 16V 0L cosωt 9π 15

This will give c 1 = i 16V 0 L 9π = 16iV 0L 9π = 16iV 0L 9π = 8iV 0L 9π = 8iV 0L 9π = 8iV 0L 9π T 0 T 0 T 0 T 0 e ie E 1 t/ cosωtt e iω 1t cosωtt e iω 1t 1 e iωt +e iωt t e iω 1 ωt +e iω 1+ωt t [ 1 iω 1 ω e iω 1 ωt 1 + iω 1 +ω e iω 1+ωt e iω 1 ωt 1 + e iω 1+ωT 1 iω 1 ω iω 1 +ω At this point, we coul go ahea an square this expression exactly, but we can make our lives easier by remembering that we re close to resonance, i.e. that our riving frequency ω is close to the resonant frequency ω 1. In particular, ω 1 ω ω 1 +ω, so the first term in the above expression ominates, an we can go ahea an rop the secon term. Thus near resonance, c 1 8V 0L e iω1 ωt 1 9π ω 1 ω The transition probability is then P 1 = c 1 8V 0 Le iω1 ωt 1 9π ω 1 ω 8V0 L e iω 1 ωt 1 e iω1 ωt 1 = 9π ω 1 ω 8V0 L cosω 1 ωt = 9π ω 1 ω 16V0 L sin ω 1 ωt/ = 9π ω 1 ω Now, we ha V 0 = 1 10 ev/m, an L = 1 10 9 m, so 16V0 L.74 10 3 ra/s 9π so.74 10 3 ra/s P 1 = sin ω ω 1 ω 1 ωt/ 16 ] T 0

Now, let s start plugging in our numbers. First, we have ω = 1.537 10 15 ra/s, so.74 10 3 ra/s P 1 = sin 1.714 1.537 10 15 ra/s1.0 10 15 s/ 1.714 1.537 10 15 ra/s Next, for ω = 1.691 10 15 ra/s, we have P 1 1.87 10 4.74 10 3 ra/s P 1 = sin 1.714 1.691 10 15 ra/s1.0 10 15 s/ 1.714 1.691 10 15 ra/s Finally, for ω = 1.706 10 15 ra/s, we have P 1 1.877 10 4.74 10 3 ra/s P 1 = sin 1.714 1.706 10 15 ra/s1.0 10 15 s/ 1.714 1.706 10 15 ra/s P 1 1.877 10 4 Note that these probabilities are almost ientical. That s because near resonance an for small enough T, we can expan giving sin ω 1 ωt/ 1 4 ω 1 ω T P 1 8V0 LT 9π So near resonance, the transition probability attains its maximum value for fixe T, an epens only on T but not not ω. 17