Physics 139B Solutions to Homework Set 3 Fall 2009

Transcription

1 Physics 139B Solutions to Hoework Set 3 Fall Consider a particle of ass attached to a rigid assless rod of fixed length R whose other end is fixed at the origin. The rod is free to rotate about the origin. Classical echanics teaches us that the Hailtonian of this syste is given by H = I where I = R is the oent of inertia and = r p is the angular oentu. In quantu echanics is an operator and the Schrodinger equation for the energy levels of the rigid rotator is given by H ψ = E ψ. (a What are the possible energy eigenvalues of the syste? Since the eigenvalues of are l(l + 1 where l = it follows that the eigenvalues of H = /(I are l(l + 1/(I where l = (b Suppose that the particle of ass has no internal spin degree of freedo but carries an electric charge +e. It is placed in a unifor agnetic field B. Ug the principle of inial substitution write down the Schrodinger equation for the charged rigid rotator. According to the principle of inial substitution we replace p p e A in the Hailtonian. Thus r ( p e A ] H =. (1 I For a unifor agnetic field a convenient choice for A is given by: A = 1 r B ( which is consistent with the Coulob gauge condition A = 0. In the coordinate representation the oentu operator i given by p = i. Hence for any wave function ψ ] p Aψ = i ( Aψ = ih A ψ + ψ A = i A ψ = A p ψ. That is in the Coulob gauge A p = p A and we do not have to worry about the relative ordering of these two operators. We now proceed to siplify the Hailtonian given in eq. (1. Consider two operators C and D that do not coute. Ug ( C D i = jk ǫ ijk C j D k 1

2 it follows that: ( C D = jkl = jk ǫ ijk ǫ il C j D k C l D = (δ jl δ k δ j δ kl C j D k C l D jkl (C j D k C j D k C j D k C k D j = jk C j D k C j D k j C j ( D CD j where I have been careful to keep track of the order of the operators. Applying this result with C r and D p e A one obtains: r ( p e A ] = r j (p ea k r j (p ea k r j ( p e A r ] (p ea j. jk j If we now ake use of the expression for A given in eq. ( the second ter above can be siplified ce A r = 1 ( r B r = 0 ce r is perpendicular to r B. Hence ( p e A r = p r. Hence r ( p e A ] = ( r p e r j (A k r j p k +p k r j A k +e r j ( p ra j +e jk j jk r j A k r j A k. (3 Since A is a function of r (and not of p it follows that A coutes with r. However p does not coute with r. Instead r j p k ] = i δ jk or equivalently As a result r j ( p ra j = j jk r j p k r k A j = jk r j p k = p k r j + i δ jk. (p k r j +i δ jk r k A j = ( p r+i r A = 0 (4 ce r j r k ] = 0 and r A = A r = 0 as previously noted for a unifor agnetic field in the Coulob gauge. Next we exaine: r j p k r j A k = r j (r j p k i δ jk A k = r p A i r A = r p A (5 jk jk where we have again used r A = 0. As usual we denote r r r. Finally noting that A k r j ] = 0 it follows that: r j A k r j p k = r A p. (6 jk

3 Inserting the results of eqs. (4 (6 into eq. (3 one obtains: r ( p e A ] ( = ( r p er p A + A p + e r A = er A p + e r A (7 where in the last step we have used the fact that p A = A p in the Coulob gauge. We now substitute A = 1 r B (appropriate for a unifor agnetic field and note that: A p = 1 ( r B p = 1 ( B r p = 1 B ( r p = 1 B A = 1 4 ( r B = 1 4 r B ( r B ]. Inserting these results into eq. (7 we end up with H = 1 { er B I e r r B ( r B ]}. The tie-independent Schrodinger equation then reads: 1 { er B I e r r B ( r B ]} ψ = Eψ where H is given above and ψ i r ψ. (c Copute the energy levels of the syste described in part (b assuing that the agnetic field is weak (i.e. assue that the ter in the Hailtonian that is quadratic in B can be neglected. If the agnetic field is weak one can neglect the ter in H that is quadratic in the B-field. Thus one can use the approxiate Hailtonian H = 1 er B ]. I One is free to choose the z-axis to lie along B so that B = Bẑ. Moreover the length of the co-ordinate r is fixed r = R ce by assuption the rigid rotator has a fixed length. The oent of inertia of the rigid rotator is I = R. Hence H = R eb z. The eigenstates of H are siultaneous eigenstates of and z. Since the possible eigenvalues of z are l where l = l l l 1 l 3

4 it follows that the energy levels of the charged rigid rotator in a weak unifor agnetic field are given by: E = l(l + 1 R l l = l = l l l 1 l Of course this is the Zeean effect. REMARK: The odel of the rigid rotator is a good one for diatoic olecules. Thus the above result provides a odel for the behavior of the rotational levels of diatoic olecules in the presence of a weak unifor agnetic field.. Positroniu is a bound state of two spin 1/ particles: an electron (e and a positron (e +. Consider the Hailtonian for the syste where we focus only on the spin degrees of freedo. In the presence of a unifor external agnetic field we ay take: H = A(I σ 1 σ + µ B ( σ 1 σ B where A is a constant I is the identity atrix in the direct product space of the two spin- 1 Hilbert spaces µ B e /(c and is the electron ass. The labels 1 and refer to the e and the e + respectively. (a Consider separately the case of zero agnetic field and the case of a unifor agnetic field pointing in the z-direction. In each case list the coplete set of siultaneously couting angular oentu operators that also coute with the Hailtonian H given above. Ug S i 1 σ i (i = 1 we can rewrite H as: H = A I 4 ] S 1 S + e ( S 1 S B. The total spin operator is S S 1 + S. Squaring this expression and solving for S 1 S one obtains: S 1 S = 1 S S 1 S ]. First suppose that B = 0. Then H coutes with S S z S 1 and S. Note that H does not coute with any of the coponents of S 1 and S. For exaple S 1i S 1 S ] = j S 1i S 1j ]S j = i jk ǫ ijk S 1k S j 0. Next if B Bẑ (without loss of generality we are free to choose the z-axis to lie in the direction of the B-field then H still coutes with S z S 1 and S. 4

5 However H no longer coutes with S. For exaple S (S 1 S i ] = S 1 + S + S 1 S (S 1 S i ] = j S 1j S j (S 1 S i ] = S j S 1j S 1i ] S 1j S j S i ] = i jk ǫ jik (S j S 1k S 1j S k = 4i jk ǫ ijk S 1j S k 0. (b In zero agnetic field a transition is observed to occur fro the S = 1 state to the S = 0 state (which is the ground state. The eitted photon is observed to have a frequency of 10 5 MHz. Copute the energy levels of the syste (for B = 0 and then evaluate the constant A that appears in the Hailtonian. If B = 0 then the eigenstates of H can be chosen to be siultaneous eigenstates of S S z S 1 and S. These eigenstates are denoted by s s where the possible values of s and s are: s = 1 and s = corresponding to the triplet spin state and s = 0 and s = 0 corresponding to the glet spin state. In the case of zero orbital angular oentu (l = 0 the triplet state is denoted by 3 S 1 and the glet state by 1 S 0. The latter is the ground state of the syste. Ug the results of part (a the Hailtonian in the case of zero external agnetic field can be written as: H = A I ( S S 1 S ]. Ug S s s = s(s + 1 s s S 1 s s = 3 4 s s S s s = 3 4 s s where the latter two results above are a consequence of the fact the the electron and positron are both spin- 1 particles it follows that H s s = A { 1 { s(s + 1 ]} 3 0 for s = 1 s s = 4A s s for s = 0. That is the energies of the 3 S 1 and 1 S 0 states are given by { 0 for the 3 S 1 state E = 4A for the 1 S 0 state. (8 5

6 A photon with the a frequency of ν 10 5 MHz is observed in a transition fro the 3 S 1 state to the 1 S 0 ground state. Since it follows that E E( 3 S 1 E( 1 S 0 = hν = 4A A = 1 4 hν = 1 4 ( ev sec( sec 1 = ev. 3. iboff proble 1.5 on page 584. iboff defines the ter ultiplicity to be the nuber of possible j values for a given l and s. As shown on pp of iboff the ultiplicity is equal to s + 1 if s l and l + 1 for s l (the definitions agree in the case of l = s. (a 3 D eans l = s = 1 and j =. The ultiplicity is s + 1 = 3. (b 4 P 5/ eans l = 1 s = 3/ and j = 5/. The ultiplicity is l + 1 = 3. (c F 7/ eans l = 3 s = 1/ and j = 7/. The ultiplicity is s + 1 =. (d 3 G 3 eans l = 4 s = 1 and j = 3. The ultiplicity is s + 1 = iboff proble 13.4 on page 689. (a The unperturbed Hailtonian is { H (0 0 for x / (x = for x / and the perturbation is H (1 (x = { V 0 for x a/ 0 for x a/ where 0 < a <. First we need to solve for the energy eigenvalues and the corresponding eigenfunctions of H (0. These are given in eq. (6.100 on p. 179 of iboff: E n (0 = π n and n (0 = ( nπx for n = ( nπx cos for n = (9

7 The first order energy shift for the n = energy level is given by E (1 = (0 H (1 (0 = V a/ ( 0 πx dx = V 0a 1 (πa/ ] (πa/ a/ That is the corrected second eigenenergy to first order in the perturbation is given by: E = π + V 0a 1 (πa/ ]. (πa/ The corrected eigenfunction to first order in the perturbation (0 + (1 is obtained ug (1 = n (0 n (0 H (1 (0. n E (0 E n (0 Thus we copute n (0 H (1 (0 = V 0 V 0 a/ a/ a/ a/ ( πx ( πx cos ( nπx for n even ( nπx for n odd where the case of n = is excluded. The above integrals are easily evaluated: ( ( (n πa (n + πa V a/ ( 0 πx ( nπx = V 0 π n n + a/. a/ a/ ( πx cos ( nπx = 0. In evaluating the first integral above we set u = πx/ and eployed the trigonoetric identity: ] (u (nu = 1 cos(n u] cos(n + u]. The second integral above vanishes ce the integrand is an odd function with respect to x x. Hence ( ( (1 = V (n πa (n + πa ( 0 nπx πe (0 n n + 4 n 1 n=46... where E (0 1 π /( and the su is taken over all positive even nubers excluding n =. There is no siple closed-for expression for the su above. 7

8 (b In order for the approxiation to be valid the energy shift should be paraetrically saller than the unperturbed energy. Since the ratio a/ need not be particularly sall we siply require that V 0 π where we have dropped constants such as factors of or factors of a/. The condition above also ensures that the correction to the unperturbed eigenfunction is also sall. (c The parity of the the nth unperturbed energy level is ( 1 n corresponding to the behavior ψ n ( x = ( 1 n ψ n (x which follows fro eq. (9. Noting that the unperturbed Hailtonian and the perturbation H (1 are both even functions under x x it follows that both the unperturbed and the exact Hailtonians coute with the parity operator. Hence the exact energy eigenfunctions ust have definite parity. In particular if the unperturbed energy eigenfunction is odd under parity then the perturbed wave function will also be odd under parity. This is indeed the case for the n = energy eigenstate as the explicit results of part (a deonstrate. 5. Consider the Hailtonian for positroniu in the presence of a unifor external agnetic field given in proble above. Assue that the agnetic field points in the z-direction. (a Treating the agnetic field as a perturbation copute the energy eigenvalues to second order in B and the energy eigenstates to first order in B. Sketch the energy levels as a function of B. We take B = Bẑ. The unperturbed Hailtonian is given by H (0 = A I ( S S 1 S ] and the perturbation is given by H (1 = eb (S 1z S z. As shown in proble the unperturbed eigenstates are given by s s with s = 1 for the 3 S 1 triplet states and s = 0 for the 1 S 0 glet state. First consider the first-order energy shifts. For the 1 S 0 glet state E (1 00 = eb 0 0 S 1z S z

9 To evaluate this we convert fro the total spin basis to the product basis. Recall that: 0 0 = 1 ] ( = 1 + ]. (11 Then H (1 0 0 = eb ( ] ] S 1z Sz = ] + = 1 0. (1 Since 0 0 and 1 0 are orthogonal states it follows that E (1 00 = 0. Next consider the 3 S 1 triplet states. These are three degenerate states with respect to H (0 so we ust use degenerate perturbation theory. Thus we shall evaluate the atrix W s s 1 s H(1 1 s which is a 3 3 atrix. To evaluate these atrix eleents we again rewrite the unperturbed eigenstates in ters of the product basis ug eq. (11 and We copute: 1 1 = 1 1 =. H (1 1 1 = eb (S 1z S z = 0 H (1 1 1 = eb (S 1z S z = 0 and H (1 1 0 = eb ( ] ] S 1z + Sz + = ] = 0 0. Since the s s states are orthonoral it follows that W s s = 0 for all s s = Thus we have proven that all first-order energy shifts vanish. 9

10 Next we consider the perturbed energy eigenstates at first order in the perturbation expansion. In class we derived: (0 (0 H (1 n (0 n (1 = n (0 + n E n (0 E (0 (13 where we exclude the degenerate states fro the su. Since H (1 1 1 = 0 and H (1 1 1 = 0 it follows that there is no first order shift in the wave functions for 1 1 and 1 1. On the other hand there are first-order shifts in the wave functions for 1 0 and 0 0. Ug eq. (13 it follows that ψ (1 = H( A ψ (1 = H( A 0 where we have used the unperturbed energies obtained in eq. (8. Evaluating these expressions ug results obtained above one finds: ψ (1 = ( A ψ (1 = ( A The second order energy shifts can now be deterined ug E ( = n (0 H (1 n (1. (16 Due to H (1 1 1 = 0 and H (1 1 1 = 0 it follows that there are no secondorder energy shifts for 1 1 and 1 1. On the other hand there are secondorder energy shifts for 1 0 and 0 0. Ug eq. (16 one finds: 10 1 = 0 H (1 ψ (1 10 E ( 00 0 = 0 H (1 ψ (1 00 E ( = 4A 1 0 H(1 0 0 = 1 4A = 4A 0 0 H(1 1 0 = 1 4A The results obtained above are suarized in the following table: state energy A 4A A + 4A 4A 10 ( ( ( (.

11 where A < 0 as shown in proble (b. A sketch of the energy levels as a function of B is shown below. REMARK: Since the degeneracy was not resolved at first order one ight worry that we have not yet identified the correct unperturbed degenerate eigenstates. In class I indicated that to resolve the degeneracy at second order one should solve the eigenvalue proble for the atrix W ( which is defined by: where n (0 i W ( = n (0 n j H (1 (0 k n (0 n k H (1 (0 i E k n n (0 E k (0 i are the degenerate states. Applying this result to the present proble the su over k consists of a su over one unperturbed state 0 0 which is not degenerate with the set of degenerate states 1 s s = Thus the 3 3 atrix W ( is given by: W ( ji = n (0 j 0 H (1 n H (1 (0 0 4A i = 1 4A ( n (0 j after ug H (1 0 0 = (/ 1 0 obtained in eq. (1. Since the states 1 s s = are orthonoral it follows that W ( is a diagonal 11 n (0 i

12 atrix in the total spin basis. Thus the triplet states of the total spin basis are the correct unperturbed degenerate eigenstates. Moreover the eigenvalues of W ( are the correct second-order energy shifts two zero energy eigenvalues corresponding to 1 1 and 1 1 and a third energy eigenvalue given by 1/(4A](/. The corresponding energy eigenstate is also correctly coputed in eq. (14 as we have eployed the correct unperturbed degenerate eigenstates. (b Repeat the calculation of part (a but now solve the proble exactly. Expand out your solutions in a power series in B and verify that the results of part (a are indeed correct. This proble can be solved exactly by diagonalizing the full 4 4 atrix Hailtonian s s H(1 s s where s s = { } (17 indicates a convenient ordering of the entries in the rows and coluns of the atrix representation of H. In fact we have already coputed all the atrix eleents of H in part (a of this proble. For exaple H 1 1 = (H (0 + H (1 1 1 = 0 H 1 1 = (H (0 + H (1 1 1 = 0. Ug results already obtained in proble and in part (a of this proble the 4 4 atrix H with respect to the basis of eq. (17 is given by: s s H(1 s s = A This atrix is easy to diagonalize. All we have to do is focus on the lower block i.e. the subspace spanned by 1 0 and A The eigenvalues of this atrix are obtained by solving: ( E(4A E = 0. 1

13 The solutions to this quadratic equation are denoted by E ± where ( E ± = A ± 4A +. (18 To check that this reproduces the perturbative coputation of part (a we expand eq. (18 in a power series about B = 0. First rewrite: E ± = A A ( ] 1/. 4A Noting that A < 0 it follows that A = A and so E ± A A ( ] 8A 1 ( for the state 4A = 4A + 1 ( for the state 4A ψ (1 10 ψ (1 10 in agreeent with the results obtained in part (a. To obtain the exact energy eigenfunction corresponding to the first eigenvalue E + we solve: 0 cosθ ( cos θ = A + 4A + (0 4A θ θ ( cos θ ce is the ost general eigenvector of a real syetric atrix that is θ noralized to unity. (Without loss of generality one can take the energy eigenfunctions to be real. Eq. (0 yields ( θ = A + 4A + cosθ. It is convenient to solve for tan θ = θ/ cos θ. One then obtains: tanθ = ( A + 4A +. Ug the identities θ = tan θ 1 + tan θ cos θ = tan θ (19

14 one can copute: θ = cos θ = ( A + 4A + ( 4A + ( A + 4A + ( 4A + where we have used cos θ = 1 θ. The above forulae look sipler when expressed in ters of the energy eigenvalues deterined above. Ug the results of eq. (18 and noting that: it then follows that: θ = E + E = 4A + ( E + E + E cos θ = E E + E. (1 Since A < 0 one sees that E + > 0 and E < 0 in which case the above results are consistent with a real angle θ that can be chosen to lie in the range 0 θ π/. Thus the eigenstate corresponding to the eigenvalue E + is given by: ( E E + E 1/ ( E+ E + E 1/ 0 0. The eigenstate corresponding to the eigenvalue E ust be orthogonal to the eigenstate obtained above and is thus given by: ( 1/ ( 1/ E+ E E + E E + E To coplete the diagonalization of H we note that the other two energy eigenvalues and eigenstates are not shifted by the perturbation due to the zeros in the atrix. That is the other two eigenstates are 1 1 and 1 1 with corresponding zero eigenvalues to all orders in perturbation theory. 14

15 To suarize the exact energy eigenstates and eigenvalues are given in the following table: state energy ( E E + E 1/ ( E+ E + E 1/ 0 0 E ( E+ E + E 1/ ( E E + E 1/ 0 0 E where E ± A ± 4A + (. As a final check one can expand these results in a power series about B = 0 and show that one obtains the energy eigenfunctions at first order that atch the results obtained in part(a. Ug eqs. (19 and (1 one finds: θ = E + E + E 1 4A 4A 1 A ( ( = 1 16A Since 0 θ π/ by convention (and A < 0 it follows that θ 1 ( + O(B 3. 4A Finally and therefore cos θ = 1 θ = 1 O(B cosθ = 1 O(B. ( + O(B 4 That is we identify: ( ψ (1 10 = cosθ θ A ( ψ (1 00 = θ cos θ A which reproduces the results obtained in part (a by the first-order perturbation theory coputation. 15