Physics 115C Homework 3

Transcription

1 Physics 115C Homework 3 Problem 1 In this problem, it will be convenient to introduce the Einstein summation convention. Note that we can write S = i S i i where the sum is over i = x,y,z. In the Einstein summation convention, we leave out the overall summation sign, and assume the convention that if an index is repeated in an expression, there is an implied summation over it, like so: i S i i i S i This will make our expressions easier to work with (and will save some typing for me. Now, let s start. Recall that the commutation relations for angular momentum are [ i, j ] = iǫ ijk k [, i ] = where ǫ ijk is the evi-civita symbol, and identical expressions hold for S and J (and there is an implied sum over the index k!. Using these commutation relations along with the definition J = +S, let s compute the commutator of S with J,, S, J z, z, and S z : [ S, ] = [ S,S ] = These two are trivial, since and S commute with all of their components, and therefore also with S. Next, [ S,J ] = [ S,J J] = [ S,(+S (+S] = [ S, +S + S] = [ S, ]+[ S,S ]+[ S, S] = [ S, ]+[ S,S ] = 1

2 To get the the fourth line, I used the fact that since S and commute, S = S ; to get to the fifth line, I used the fact that S commutes with itself; and to get to the last line, I used the fact that commutes with all the components of S and. Continuing, [ S, z ] = [ i S i, z ] = i [S i, z ]+[ i, z ]S i = +iǫ izk k S i = iǫ zki k S i = i( S z To get to the third line, I used the fact that the components of S and commute with each other; to get to the fourth line, I used the antisymmetry of the evi-civita symbol under exchange of any two indices (e.g. ǫ ijk = ǫ jik = ǫ jki, etc.; and to get to the last line I used the expression for the dot product Moving on, Finally, (A B i = ǫ ijk A j B k [ S,S z ] = [ i S i,s z ] = i [S i,s z ]+[ i,s z ]S i = iǫ izk i S k + = iǫ zik i S k = i( S z [ S,J z ] = [ S, z +S z ] = [ S, z ]+[ S,S z ] = i( S z i( S z = where I made use of our results above. Therefore, J,,S, and J z commute with S, but not z and S z. Since the spin-orbit coupling perturbation is proportional to S, this is why the good basis states are those labeled by n,l,s,j, and m j, but not those labeled by n,l,s,m l, and m s.

3 Problem (Griffiths 6.1 From Griffiths equations 6.75 and 6.76, we have that the first-order corrections to the energy in the weak-field Zeeman effect are E (1 Z = µ Bg J B ext m j where µ B is the Bohr magneton, B ext is the external electric field (along which we align our z-axis, and g J is the andé g-factor: g J = 1+ j(j +1 l(l+1+3/4 j(j +1 (incidentally, Griffiths derivation of the andé g-factor gives a good physical idea of what s going on, but isn t very mathematically rigorous. A more rigorous derivation of equation 6.73 makes use of the Wigner-Eckart theorem. Now, the eight n = states of the hydrogen atom have s = 1/ and l = or l = 1. For l =, the only possible value of j is 1/; for l = 1, the possible values of j are j = 1/ and j = 3/. Thus there are two states with l = (corresponding to quantum numbers m j = ±1/ and six states with l = 1 (j = 1/ and m j = ±1/, and j = 3/ and m j = ±1/,±3/. et s label these states as follows: using the notation ljm j, we define the eight states as 1 =,1/,1/ =,1/, 1/ 3 = 1,1/,1/ 4 = 1,1/, 1/ 5 = 1,3/,3/ 6 = 1,3/,1/ 7 = 1,3/, 1/ 8 = 1,3/, 3/ Now, let s calculate the g-factors for these states. For 1 and, we have g J = 1+ 1/(1+1/ +3/4 (1/(1+1/ = For 3 and 4, we have g J = 1+ 1/(1+1/ 1(1+1+3/4 (1/(1+1/ = 3 For the last four states, we have g J = 1+ 3/(1+3/ 1(1+1+3/4 (3/(1+3/ = 4 3 3

4 To find the energies of these eight states, we ll need to include fine structure. From Griffiths equation 6.67, we have that the energies of the first four states (including fine structure but not yet the Zeeman perturbation are E fs = E ( [1+ α n n n j +1/ 3 ] 4 = E ( [1+ α 4 4 1/+1/ 3 ] 4 = E ( α where E = 13.6 ev is the (unperturbed Hydrogen ground-state energy. The energies of the last four states (again, including just fine structure are E fs = E ( [1+ α 4 4 3/+1/ 3 ] 4 = E ( α Including the first-order perturbation from the Zeeman effect, the energy of each state is thus E = E fs +µ B g J B ext m j Using our expressions for E fs and the g-factor, we can write down the energies of all eight states: E 1 = E ( α +µ B B ext E = E ( α µ B B ext E 3 = E ( α µ BB ext E 4 = E ( α 1 3 µ BB ext E 5 = E ( α +µ B B ext E 6 = E ( α + 3 µ BB ext E 7 = E ( α 3 µ BB ext E 8 = E ( α µ B B ext 4

5 Here s a depiction of how the energies vary with the perturbing magnetic field B ext (not to scale, of course: 5

6 Problem 3 If a system begins at time t = in an eigenstate i of the unperturbed Hamiltonian, then the probability of measuring the system to be in some other state f at a later time t (after the perturbation has been turned on is c f (t, where c f (t = δ fi i t e i(e( f E( i t / f H (t i dt where H (t is the perturbing Hamiltonian. In our particular case, we have H (t = βxe t/τ To give this perturbation a physical interpretation, we might imagine that this would be the perturbation the oscillator would experience if it were placed in an exponentially decaying electric field(which might arise, for instance, inside of a capacitor as it was being discharge in anrcircuit. Wearefurthergiventhattheoscillatorbeginsinthegroundstate,so i =. Since the (unperturbed energies of the harmonic oscillator are given by E n ( = (n+1/ω, we have c n (t = δ n i = δ n iβ t t [ (( exp i n+ 1 e inωt e t /τ n x dt ω 1 ω t ] n H (t dt Since we re interested only in the probability of measuring the system to be in the nth state at large time, we might as well take the limit t : c n ( = δ n iβ n x e (1/τ inωt dt = δ n iβ n x e (1/τ inωt dt = δ n iβ [ n x 1 1/τ inω e (1/τ inωt = δ n iβ (1/τ inω n x To evaluate the matrix element n x, let s use the ladder operators: recall that we can express x = mω (a +a ] 6

7 Thus and therefore n x = = = c n ( = δ n mω n (a +a mω n 1 mω δ n1 mω iβ (1/τ inω δ n1 Note that this expression is zero unless n = 1 or n =, i.e. the probability of the system transitioning to any state higher than n = 1 is zero (at least to first order in perturbation theory. To get the probability of transition, we take the norm squared: the probability of no transition at all is P = c ( = 1 = 1 The probability of transitioning to the first excited state is P 1 = c 1 ( = iβ mω (1/τ iω = = β mω (1/τ +ω β mω(1/τ +ω Thus the probability of measuring the system to be in the nth state at large time is 1 n = β P n = n = 1 mω(1/τ +ω otherwise 7

8 Problem 4 As in the previous problem, we have In this case, we have c f (t = δ fi i H (t = t e i(e( f E( i t / f H (t i dt { V T < t < T, < x < / otherwise If the well starts in the ground state, then we have i = 1 (for the infinite square well, it is conventional to start numbering the states at 1 rather than zero as in the harmonic oscillator. Since the energies of the states in the infinite square well go like E n = E 1 n, we have that the transition coefficients at large time (t > T are c n (t > T = δ n1 i T T = δ n1 i n H 1 exp [i ( ] E 1 n t E 1 n H (t 1 dt T For n = 1, the integral is just T; for n 1, we get T e i(n 1E 1 t / dt c n (t > T = δ n1 i [ n H 1 e i(n 1E 1 t / i(n 1E 1 = δ n1 i n H 1 (n 1E 1 sin ] T T ( (n 1 E 1T We only need to evaluate the matrix element. Recall that the stationary state wave functions of the infinite square well are ( nπ ψ n = sin x Thus n H 1 = ψnh ψ 1 dx / = V ψnψ 1 dx = V = V / / ( nπ ( π sin x sin x dx [ ( ( ] (n 1π (n+1π cos x cos x dx 8

9 (I used the identity sin(asin(b = (1/[cos(a b cos(a+b]. The integral has different behaviors for n = 1 and n 1. For the case n = 1, we have For n 1, we have n H 1 = V = V = V 1 H 1 = V / = V = V = V / [ ( ] π 1 cos x [x π ( ] / π sin x [ ( (n 1π cos ( (n+1π x cos dx ] x dx [ ( ( (n 1π (n 1π sin (n+1π x (n+1π sin [ ( ( ] (n 1π (n 1π sin (n+1π (n+1π sin ] / x Using the fact that sin((n 1π/ = sin((n+1π/ = for odd n, and sin((n 1π/ = ( 1 n/ and sin((n+1π/ = ( 1 n/ for even n, we get n H 1 = V [ ] (n 1π ( 1n/ (n+1π ( 1n/ [ ( 1 n/ 1 = V π n ] n+1 = V ( 1 n/ π n n 1 for even n, and n H 1 = for odd n. Consequently, for n even we have c n (t > T = i ( n H 1 sin (n 1 E 1T (n 1E 1 = iv ( n π ( 1n/ sin (n 1 E 1T n 1(n 1E 1 = iv ( ( 1 n/ 4n πe 1 (n 1 sin (n 1 E 1T 9

10 For n = 1, we instead get c 1 (t > T = 1 i T 1 H 1 T = 1 i V T = 1 iv T For n 1 and n odd, we have simply c n (t > T =. Thus the transition probabilities are 1+ ( V T n = 1 ( P n = V 16n πe 1 sin ( (n 1 E 1T even n (n 1 4 odd n where E 1 = π /m, as usual. In particular, the probability of measuring the well to be in the first excited state is P, or P = ( V πe 1 dt ( 16(4 (4 1 4 sin (4 1 E 1T ( ( 8V P = sin 3E1 T 9πE 1 1