Unformatted text preview: PH 216, Problem set 3, Due 4/20 1. Consider the hydrogen atom in an excited n = 2 state, which is subjected to an external uniform electric field E. Do not neglect the spin of the electron. Assume that the field E is sufficiently weak so that eEa0 is small compared to the fine structure, but such that the "Lamb shift" ( = 1057 MHz) cannot be neglected. This "Lamb shift" is an effect from QED that breaks the degeneracy between the 2S1/2 and 2P1/2 states: if the Lambshift is neglected, then the 2s1/2 and 2p1/2 states are degenerate with energy, 1 5 E (0) =  2 me c2 2  128 me c2 4 , as computed in class. Including the Lamb shift, 2s1/2 and 2p1/2 are no longer degenerate. Instead, we have E (0) (2s1/2 )  E (0) (2p1/2 ) > 0 , which defines . Assume here that the Lamb shift can be included as a term HLS in the unperturbed Hamiltonian. Thus, for present purposes you can treat the problem as a twolevel system consisting of the the 2S1/2 and 2P1/2 states of hydrogen. In particular, you may ignore the 2P3/2 state of hydrogen and the hyperfine interactions. (a) Compute the Stark effect for the 2S1/2 and 2P1/2 states of hydrogen by solving the twolevel system exactly (i.e. find the four matrix elements of H between the two states, and diagonalize.) HINT: When electron spin is included, the hydrogen atom energy eigenstates are twocomponent wave functions given (in the coordinate representation) by:
j () = Rn (r)Ym (, ) , r where Rn (r) is the radial wave function of the hydrogen atom, and the spin spherical harmonics are defined by:2 Ym where 1 j = 1 , m = 1 ( + 1 m)1/2 , 1 ; m  1 , 1 + ( + 1 m)1/2 , 1 ; m + 2 ,  1 . 2 2 2 2 2 2 2 2 2 + 1 (17) (b) Show that for eEa0 h, the energy shifts due to the external electric field are quadratic in E, whereas for eEa0 h, they are linear in E. Determine the (perturbed) energy eigenstates in both limiting cases.
2 j= 1 2 (, ) , j = 1 , m , 2 (16) Note that the usual spherical harmonics can be written as: Ym (, ) ,  , m. 14 (c) The critical field is defined as: h Ec , 3ea0 where the factor of 3 is conventional. The linear or quadratic behavior of the energy shifts obtained in part (b) depend on the magnitude of E as compared to Ec . Determine Ec in volts/cm. Solution: (a) The Hamiltonian governing the system is: H= p2 e2  + HFS  eEz + HLS , 2m r where HFS is responsible for the finestructure and HLS is responsible for the Lamb shift. Since only the 2s1/2 and 2p1/2 states matter in this calculation, all we need to do is compute the eigenvalues of the 2 2 matrix, 2s1/2 H2s1/2 2s1/2 H2p1/2 H . (18) 2p1/2 H2s1/2 2p1/2 H2p1/2 We shall treat eEz as the perturbation. That is, we denote: e2 p2  + HFS + HLS , H (1) = eEz . 2m r By definition, 2s1/2 and 2p1/2 are eigenstates of H (0) . If the Lambshift is neglected, then the 2s1/2 and 2p1/2 states are degenerate with energy, H (0) = as computed in class. Including the Lamb shift, 2s1/2 and 2p1/2 are no longer degenerate. Instead, we have E (0) (2s1/2 )  E (0) (2p1/2 ) > 0 , which defines . We now compute the matrix H [cf. eq. (18)]. Note that parity considerations imply that 2s1/2 H (1) 2s1/2 = 2p1/2 H (1) 2p1/2 = 0 . Moreover, H (0) is diagonal with respect to the 2s1/2 and 2p1/2 states. That is, 2s1/2 H (0) 2p1/2 = 2p1/2 H (0) 2s1/2 = 0 . Thus, all we need to evaluate is 2s1/2 H (1) 2p1/2 . In order to compute this matrix element, we shall make use of eq. (17) to express the 2s1/2 and 2p1/2 states in terms of eigenstates of L 2 , Lz , S 2 and Sz . In the coordinate representation,
j 1  j = 2 , mj = Rn (r)Ymj (, ) , r E (0) =  1 me c2 2  2 5 m c2 4 128 e , 15 where Rn (r) is the radial wave function of the hydrogen atom, and the spin spherical harmonics are defined by:3 1 mj + 2 Y,m  1 (, ) 1 j= 1 j 2 Ymj 2 (, ) , j = 1 , mj = , 2 1 2 + 1 mj + 2 Y,m + 1 (, )
j 2 where we have used [cf. eq. (17)]: 1 1 1 1/2 1 1 1/2 1 1 1 1 1 j = , mj = , 2 ; m  2 , 2 + (m+ 2 ) , 2 ; m + 2 ,  2 . (m+ 2 ) 2 2 + 1 Due to the WignerEckart theorem, it is sufficient to evaluate the matrix element 2s1/2 H (1) 2p1/2 for a fixed value of mj . Thus, for mj = 1 , 2 Y10 (, ) Y00 (, ) 1  2s1/2 = R20 (r) r ,  2p1/2 = R21 (r) r , 3 0 2 Y11 (, ) where R20 (r) = 1 2a0 3/2 r 2 a0 e
r/(2a0 ) , 1 R21 (r) = 3 1 2a0 3/2 r r/(2a0 ) e , a0 and a0 is the Bohr radius. Hence, using H (1) = eEr cos , eE (1) 2s1/2 H 2p1/2 = d3 r Y00 (, )R20 (r)Y10 (, )R21 (r) r cos 3 3 eE 1 1 3 4 r/a0 2 = r (2a0  r)e dr cos2 d 3 2a0 4 3 3a0 0 =  3 eEa0 , where we have used, 2 cos d = 2 4 cos d cos = , 3 1
2 1 and xn ex dx = n! , 0 with x r/a0 . Thus, the matrix H is given by: H= where Es  3 eEa0  3 eEa0 Ep , Es 2s1/2 H2s1/2 ,
3 Ep 2p1/2 H2p1/2 , Note that the usual spherical harmonics can be written as: Ym (, ) ,  , m. 16 and Es  Ep = . Since H is a real symmetric matrix, it can be diagonalized by a real orthogonal matrix. In an appendix to these solutions, I provide details of the computation of the eigenvalues and eigenvectors. Denoting the two eigenvalues by E , we have: 1 2 + 12e2 E 2 a2 . E = 2 Es + Ep 0 The corresponding eigenvectors are: cos 2s1/2 + sin 2p1/2 ,  sin 2s1/2 + cos 2p1/2 , where 0 < (by convention) with: 2 3eEa0 sin 2 = , 2 + 12e2 E 2 a2 0 sin = 1  cos 2 , 2 cos 2 = , 2 + 12e2 E 2 a2 0 1 + cos 2 . 2 which imply that 1 < < . Then, it follows that: 2 cos =  (b) If eEa0 , then we can expand the square root, The Lamb shift corresponds to /h = 1057 Mhz. 6e2 E 2 a2 0 2 + 12e2 E 2 a2 + . 0 1 2 It follows that Putting Es  Ep then yields: E+ Es + 3e2 E 2 a2 0 , 6e2 E 2 a2 0 E Es + Ep + . 3e2 E 2 a2 0 , E Ep  As advertised, the energy shifts due to the external electric field are quadratic in E. The corresponding eigenstates are determined by: 2 3eEa0 6e2 E 2 a2 0 sin 2 , cos 2 1  , 2 Hence, it follows that: sin 3eEa0 , cos 1 + O e2 E 2 a2 0 2 . 17 We conclude that the corresponding eigenstates are:4 3eEa0 3eEa0 2s1/2  2p1/2 , 2s1/2 + 2p1/2 . We see that there is an admixture of the 2p1/2 state in the 2s1/2 state when an external uniform electric field is present. If eEa0 , then we can simply set = 0 to first approximation. Then, E 1 (Es + Ep ) 3eEa0 , 2 which is linear in the external electric field as advertised. In this limit, cos 2 = 0, in which case = 3/4. Hence, sin =  cos = 1/ 2, and we find that the corresponding eigenvectors are: 1 2s1/2 2p1/2 . 2 h Ec , 3ea0 where the factor of 3 is conventional. The linear or quadratic behavior of the energy shifts obtained in part (b) depend on the magnitude of E as compared to Ec . Determine Ec in volts/cm. We finish off with some numerics. Putting hc = 12, 400 eV a0 = 0.53 A, A, = 1057 106 s1 and c = 3 1010 cms1 , it follows that: (12, 400)(1057 106 ) Ec = volts/cm = 476 Vcm1 . 10 )(0.53) 3(3 10 2. (a) Consider a twolevel system with E1 < E2 , and a Hamiltonian H = H 0 +V , where 1 and 2 are the energy eigenstates of H 0 . V is a timedependent potential that connects the two levels as follows: V11 = V22 = 0, V12 = eit , V21 = eit ( real) , (19) (c) The critical field is defined as: where Vij = iV j. At time t = 0, it is known that only the lower level is populatedthat is, c1 (0) = 1 and c2 (0) = 0. We saw in class that if (t)I = cn (t) n, (20)
n=1
4 For convenience, I have multiplied both eigenstates by an overall factor of 1, as the overall phase of the state is a matter of convention. 18 where (t)I eiH 0 t/ then the cn (t) satisfy a differential equation i (t)S , where Vnm (t) nV (t)m and kn Ek  En . By solving the above system of differential equations exactly, find c1 (t)2 and c2 (t)2 for t > 0. HINT: It is convenient to define new coefficients, c1 (t) ei(21 )t/2 c1 (t) , c2 (t) ei(21 )t/2 c2 (t) . dcn = Vnm (t)einm t cm , dt m=1 (21) Then, show that eq. (21) reduces to a matrix differential equation of the form d c1 (t) c (t) =A 1 i , (22) c2 (t) dt c2 (t) where A is a timeindependent 2 2 traceless hermitian matrix. Verify that the solution to eq. (22) is c1 (t) iAt/ c1 (0) =e . c2 (t) c2 (0) By writing A = (where the vector is uniquely determined), it is straighta a iAt/ forward to compute e and complete part (a) of the problem. (b) Do the same problem using timedependent perturbation theory to lowest nonvanishing order. Compare the two approaches for small values of . Treat the following two cases separately: (i) very different from 21 , and (ii) close to 21 . Solution: (a) The full Hamiltonian is given by H = H (0) + V , where the matrix elements of V are given in eq. (19). The timedependent Schrodinger equation is given by i (t) = H (0) + V (t) . t Inserting the expansion for (t) given in eq. (20) into the above equation, and using H (0) n = En n, it follows that5
2 n=1
5 i dcn iEn t/ e n = cn eiEn t/ V n . dt n=1 2 For ease in notation, we denote the eigenvalues of the unperturbed Hamiltonian H (0) by En rather than (0) En . 19 d i dt c1 (t) c2 (t) = V11 V12 ei12 t V22 V21 ei21 t c1 (t) c2 (t) , where 21 = 12 = (E2  E1 )/ > 0, and the matrix elements of V are given by eq. (19). That is, 0 ei(21 )t c1 (t) d c1 (t) , i = dt c2 (t) c2 (t) ei(21 )t 0 Multiplying out the matrix and vector above, one obtains coupled differential equations for c1 (t) and c2 (t): i dc1 = ei(21 )t c2 , dt i dc2 = ei(21 )t c1 . dt (23) At this point, it is convenient to define new coefficients, c1 (t) ei(21 )t/2 c1 (t) , c2 (t) ei(21 )t/2 c2 (t) . (24) We now can express eq. (23) in terms of c1 (t) and c2 (t): i d i(21 )t/2 e c1 = ei(21 )t ei(21 )t/2 c2 , dt i Expanding out the derivatives, one sees that the exponential factors cancel. The resulting equations for c1 (t) and c2 (t) are: dc1 dc2 1 1 i 2 i(  21 )c1 + = c2 , i  2 i(  21 )c2 + = c1 . (25) dt dt In matrix form, eq. (25) is given by: c1 (t) d c1 (t) i =A , dt c2 (t) c2 (t) where A is the timeindependent 2 2 traceless hermitian matrix, 1 (  21 ) 2 A .  1 (  21 ) 2 Denoting the column vector (c1 , c2 )T , eq. (26) is of the form c d i(21 )t/2 e c2 = ei(21 )t ei(21 )t/2 c1 . dt (26) (27) d c iA =  . c dt The solution to this equation is: (t) = eiAt/ 0 , c c 20 where 0 (t = 0) . c c (28) This is easily verified by inserting the above solution back into eq. (28). To compute the matrix exponential, a simple strategy is to write writing A = , where the vector is uniquely determined. Such a relation holds for any a a 2 2 traceless hermitian matrix. It is easy to see that for A given by eq. (27), A = x + 1 (  21 )z . 2 Using the wellknown result: exp 1
2 6 i^ = I cos(/2)+i^ sin(/2) = n n cos(/2) + inz sin(/2) (inx  ny ) sin(/2) (inx + ny ) sin(/2) cos(/2)  inz sin(/2) where I is the 2 2 identity matrix, we can identify it 1 exp(iAt/) = exp  x + 2 (  21 )z = exp 1 i^ , n 2 where7 1 n= ^ 2 , 0 , (  21 ) , 2 (  21 )2 + 4 2 = t 2 (  21 )2 + 4 2 . (29) (30) and (31) We conclude that c1 (t) cos(/2) + inz sin(/2) inx sin(/2) c1 (0) = , c2 (t) inx sin(/2) cos(/2)  inz sin(/2) c2 (0) where is defined by eq. (31) and 2 nx , 2 (  21 )2 + 4 2 c1 (0) = c1 (0) = 1
6 In light of eq. (24), it follows that the relevant initial conditions are and c2 (0) = c2 (0) = 0 , (  21 ) nz = . 2 (  21 )2 + 4 2 (32) If this result is not wellknown to you, please derive it as follows. First note that for any nonnegative integer k, (^ )2k = I and (^ )2k+1 = n , where n is a unit vector and I is the 2 2 identity matrix. n n ^ ^ Then, using the Taylor series definition of the matrix exponential, exp 1 n 2 i^ = k=0 1 n 2 i^ k! k =I k even 1 k 1 i k 2 i 2 + n ^ = I cos(/2) + i^ sin(/2) . n k! k!
k odd 7 A common mistake made by students is to neglect the fact that n appearing in eq. (29) must be a unit ^ vector. Note that n in eq. (30) is properly normalized so that n n = 1. ^ ^ ^ 21 and c1 (t)2 = c1 (t)2 = cos(/2) + inz sin(/2)2 = cos2 (/2) + n2 sin2 (/2) , z c2 (t)2 = c2 (t)2 = n2 sin2 (/2) . x An important check of the above result is: c1 (t)2 + c2 (t)2 = cos2 (/2) + (n2 + n2 ) sin2 (/2) = 1 , x z where we have used the fact that n2 + n2 = n2 + n2 + n2 = 1, keeping in mind x z x y z that ny = 0 and n is a unit vector. Using the explicit forms of nx , nz and given ^ in eqs. (31) and (32) we arrive at: 1/2 4 2 2 2 2 1 c2 (t) = 2 sin + 4 (  21 ) t , 2 4 + 2 (  21 )2
2 c1 (t)2 = 1  c2 (t)2 (33) Note that the initial conditions c1 (0) = 1 and c2 (0) = 0 are indeed satisfied. ALTERNATIVE DERIVATION: Starting from eq. (23), it immediately follows that d i it(21 ) dc2 i e = eit(21 ) c2 . dt dt After evaluating the derivative on the lefthand side of the above equation, the exponential factors cancel out, resulting in dc2 2 c2 d2 c2 + i(  21 ) + 2 = 0. dt2 dt The solution is obtained by solving the auxiliary equation: x2 + i(  21 )x + where 2 =0 2 2 2 x = i 1
2 (  21 ) . (34) + 1 (  21 )2 . 4 It follows that: c2 (t) = A exp it 1 (  21 ) +  exp it 1 (  21 )  , 2 2 (35) after imposing the initial condition c2 (0) = 0. To determine the overall coefficient, we make use of eqs. (23) and (35) to obtain: 1 it(21 ) dc2 c1 = e = iA 1 (  21 ) + exp it 2 (  21 ) + 2 i dt 1 +iA 1 (  21 )  exp it 2 (  21 )  . 36) ( 2 22 Setting t = 0 and using c1 (0) = 1, one obtains: A= Thus, eqs. (35) and (27) yield: c2 (t)2 = = = = 1 2 2  2 Re exp it 2 (  21 ) + exp it 1 (  21 )  2 2 2 4 2 2  2 Re e2it 42 2 2 [1  cos(2t)] 22 2 2 sin2 (t) . 2 2 . 2 (37) Finally, we take the absolute square of eq. (36), which yields: 2 2 2 c1 (t)2 = A2 1 (  21 ) + + A2 1 (  21 )   2A2 1 (  21 )2  2 Re(e2it ) 2 2 4 2 1 = 2A2 4 (  21 )2 + 2  2A2 1 (  21 )2  2 cos(2t) 4 = A2 (  21 )2 sin2 (t) + 4A2 2 cos2 (t) 2 2 2 = 4A  2 sin2 (t) + 4A2 2 cos2 (t) 2 2 2 2 = 4A  2 sin (t) , where in the penultimate step we used (  21 )2 = 4(2  2 /2 ), which follows from the definition of [cf. eq. (34)]. Noting that 4A2 = 2 /(2 2 ) [cf. eq. (37)], we end up with c1 (t)2 = 1  2 sin2 (t) = 1  c2 (t)2 . 2 2 Thus, we have correctly reproduced the results of eq. (33). (b) Using the results of timedependent perturbation theory, we identify cn (t) of part (a) as: cn (t) = nUI (t, 0)i , where UI (t, 0) is the timeevolution operator in the interaction representation. The firstorder perturbation theory expression for cn (t) obtained in class is: i t ini t e cn (t) = ni  Vni (t ) dt . 0 23 Thus, to first order in perturbation theory, c1 (t) 1 and i t i21 t it 1  ei(21 )t c2 (t) =  e e . dt = 0 (21  ) Hence, c2 (t)2 = 2 2 [1  cos(21  )t] 4 2 = 2 sin2 1 (21  )t . 2 2 (  )2 2 21 (21  ) (38) We observe that eq. (38) agrees with the exact formula obtained in eq. (33) in the limit of 1 21  . Note that the latter inequality is never satisfied near 2 resonance when 21 . However, if 21 then the exact formula, eq. (33), reduces to c2 (t)2 = sin2 (t/) , for = 21 . In contrast, the corresponding firstorder perturbative result exhibited in eq. (38) is: 2 t2 2 c2 (t) 2 , for = 21 . Thus, near the resonance, the exact result for c2 (t)2 agrees with the firstorder perturbative result in the limit of t /. To summarize, the firstorder perturbative result is a good approximation to the exact result in the limit of a weak perturbation (which corresponds to small ). Away from the resonance, small means 1 21  , whereas near the 2 resonance (where 21 ) small means /t. For a fixed value of , the latter can be satisfied only at early times t. 24 ...
View
Full
Document
 Spring '09

Click to edit the document details