Joel Broida UCSD Fall 2009 Phys 130B QM II Homework Set 5 Solutions 1. From E = p2 /2m + V we have p2 = 2m(E − V ) so that Hrel = − p4 (E − V )2 1 = − =− (E 2 − 2EV + V 2 ) 3 2 2 8m c 2mc 2mc2 and hence 1 (E 2 − 2EhV i + hV 2 i) . 2mc2 For the harmonic oscillator we have 1 En = ~ω n + 2 (1) Erel = hHrel i = − and 1 1 2 kx = mω 2 x2 . 2 2 To evaluate hV i and hV 2 i where the expectation values are taken with respect to the harmonic oscillator states, we use the creation and annihilation operators ∓ip + mωx a± = √ 2~mω to write r ~ (a+ + a− ) . x= 2mω Since the orthonormal harmonic oscillator states |ni satisfy √ √ and a− |ni = n |n − 1i a+ |ni = n + 1 |n + 1i V = we have hV i = 1 ~k khx2 i = hn|(a2+ + a+ a− + a− a+ + a2− )|ni 2 4mω = ~k hn|a+ a− + a− a+ |ni 4mω ~k ~k [n + (n + 1)] = 4mω 2mω ~ω 1 = n+ . 2 2 = 1 1 n+ 2 Note that the only nonzero terms in the expectation value are those with an equal number of creation and annihilation operators. With this in mind, we then also have hV 2 i = k2 4 ~2 k 2 hn|(a2+ + a+ a− + a− a+ + a2− )2 |ni hx i = 4 16m2 ω 2 = = = ~2 k 2 hn|(a2+ a2− + a+ a− a+ a− + a+ a− a− a+ 16m2 ω 2 + a− a+ a+ a− + a− a+ a− a+ + a2− a2+ )|ni ~2 k 2 [n(n − 1) + n2 + n(n + 1) 16m2 ω 2 + n(n + 1) + (n + 1)2 + (n + 1)(n + 2)] ~ω 4 2 [6n2 + 6n + 3] If we note that 2hV i = E, it is then easy to substitute the result for hV 2 i to obtain 3~2 ω 2 (1) Erel = − [2n2 + 2n + 1] . 32mc2 2. From equation 2.59 in the notes, we (0) En n (1) Efs = − 2 α2 − n j + 1/2 have 13.6 eV 2 n 3 3 =− α − 4 n4 j + 1/2 4 where e2 1 ≈ . ~c 137.036 (a) The n = 2 level has l = 0, 1 so the possible j values are j = 1/2 and j = 3/2, 1/2. Thus the n = 2 level will split into two levels. We have 2 3 13.6 eV 2 (1) = −5.66 × 10−5 eV α Efs (j = 1/2) = − − 24 1/2 + 1/2 4 2 3 13.6 eV 2 (1) = −1.13 × 10−5 eV α − Efs (j = 3/2) = − 24 3/2 + 1/2 4 α= (b) The n = 3 level has l = 0, 1, 2 so the possible j values are j = 1/2, j = 3/2, 1/2 and j = 5/2, 3/2. Thus the n = 3 level will split into three levels. The fine structure corrections are 13.6 eV 2 3 3 (1) Efs (j = 1/2) = − = −2.01 × 10−5 eV α − 34 1/2 + 1/2 4 3 3 13.6 eV 2 (1) = −0.67 × 10−5 eV α − Efs (j = 3/2) = − 34 3/2 + 1/2 4 2 (1) Efs (j = 5/2) = − 13.6 eV 2 3 3 = −0.22 × 10−5 eV α − 34 5/2 + 1/2 4 (c) The energy level diagram showing all n = 3 to n = 2 transitions looks like this (very much not to scale): n=3 j = 5/2 j = 3/2 j = 1/2 n=2 j = 3/2 j = 1/2 1 2 3 4 5 6 (d) The difference in energy levels for a transition is given by (0) (1) (0) (1) (0) (0) (0) (0) (1) (1) (E3 + E3,fs ) − (E2 + E2,fs ) = (E3 − E2 ) + (E3,fs − E2,fs ) = (E3 − E2 ) + ∆E where (0) =− E3 so that 13.6 = −1.51 eV 32 and (0) (0) E3 − E2 (0) E2 =− 13.6 = −3.4 eV 22 = 1.89 eV . We have ∆E1 = (−2.01 − (−1.13)) × 10−5 eV = −0.88 × 10−5 eV ∆E2 = (−0.67 − (−1.13)) × 10−5 eV = 0.46 × 10−5 eV ∆E3 = (−0.22 − (−1.13)) × 10−5 eV = 0.91 × 10−5 eV ∆E4 = (−2.01 − (−5.66)) × 10−5 eV = 3.65 × 10−5 eV ∆E5 = (−0.67 − (−5.66)) × 10−5 eV = 4.99 × 10−5 eV ∆E6 = (−0.22 − (−5.66)) × 10−5 eV = 5.44 × 10−5 eV Note that the 1/2 → 3/2 transition has an energy spacing that is less than the spacing between the unperturbed n = 3 and n = 2 levels. 3 (e) Using ∆Ei = hνi and h = 4.136 × 10−15 eV-sec, we find the following frequencies (in Hz): ν1 = 2.13 × 109 ν2 = 1.11 × 109 ν4 = 8.82 × 109 ν5 = 12.06 × 109 ν3 = 2.20 × 109 ν6 = 13.15 × 109 Therefore the spacings of adjacent transitions are as follows. (Note that the spacings within each n level are much closer than the spacing between n = 3 and n = 2.) ν2 − ν1 = (∆E2 − ∆E1 )/h = 3.24 × 109 Hz ν3 − ν2 = (∆E3 − ∆E2 )/h = 1.09 × 109 Hz ν4 − ν3 = (∆E4 − ∆E3 )/h = 6.62 × 109 Hz ν5 − ν4 = (∆E5 − ∆E4 )/h = 3.24 × 109 Hz ν6 − ν5 = (∆E6 − ∆E5 )/h = 1.09 × 109 Hz 3. For the n = 3 level of hydrogen we have the possible values l = 0, 1, 2 and the (0) unperturbed energy E3 = −13.6/32 = −1.51 eV. (a) In the strong-field regime we have the first-order energy correction due to the magnetic field (equation (2.70)) (1) Enlml ms = µB B(ml + 2ms ) := β(ml + 2ms ) and this is in addition to the fine structure energy (equation (2.72)) l(l + 1) − ml ms 1 A α2 (1) − eV := −1.51α2 eV Efs = −1.51 3 l(l + 1)(l + 1/2) 4 3 where for l = 0 we take the term in square brackets to be equal to 1. The total energy of each state is then given by (in eV) (0) (1) (1) E3 = E3 + Efs + Enlml ms A 2 = −1.51 1 + α + β(ml + 2ms ) . 3 4 Tabulating these results, we obtain the following table: state |n l ml ms i ml + 2ms A total energy E3 (eV) 1 3/4 −1 3/4 −1.51(1 + α2 /4) + β l = 1 |3 1 1 1/2i 2 1/4 l = 1 |3 1 1 −1/2i 0 7/12 l = 1 |3 1 0 1/2i 1 5/12 l = 1 |3 1 0 −1/2i −1 5/12 0 7/12 l = 1 |3 1 −1 −1/2i −2 1/4 l = 2 |3 2 2 1/2i 3 1/12 l = 2 |3 2 2 −1/2i 1 l = 2 |3 2 1 1/2i 2 13/60 −1.51(1 + 13α2 /180) + β l = 2 |3 2 1 −1/2i 0 l = 2 |3 2 0 1/2i 1 l = 2 |3 2 0 −1/2i −1 l = 2 |3 2 −1 1/2i 0 l = 0 |3 0 0 1/2i l = 0 |3 0 0 −1/2i l = 1 |3 1 −1 1/2i l = 2 |3 2 −1 −1/2i −2 l = 2 |3 2 −2 −1/2i −3 l = 2 |3 2 −2 1/2i −1 7/60 −1.51(1 + α2 /4) + β −1.51(1 + α2 /12) + 2β −1.51(1 + 7α2 /36) −1.51(1 + 5α2 /36) + β −1.51(1 + 5α2 /36) − β −1.51(1 + 7α2 /36) −1.51(1 + α2 /12) − 2β −1.51(1 + α2 /36) + 3β −1.51(1 + 7α2 /180) + 2β 11/60 −1.51(1 + 11α2 /180) 3/20 3/20 −1.51(1 + α2 /20) + β −1.51(1 + α2 /20) − β 11/60 −1.51(1 + 11α2 /180) 7/60 −1.51(1 + 7α2 /180) − 2β 13/60 −1.51(1 + 13α2 /180) − β 1/12 −1.51(1 + α2 /36) − 3β (b) For the weak-field regime we have (equation (2.73)) av Emag = βmj gJ where gJ = 1 + = j(j + 1) + s(s + 1) − l(l + 1) 2j(j + 1) 3 3/4 − l(l + 1) + 2 2j(j + 1) and also (equation (2.59)) (1) Efs = −1.51 α2 A 1 1 eV := −1.51α2 eV . − 3 j + 1/2 4 3 5 Thus the total energy is given by (0) (1) av E3 = E3 + Efs + Emag A 2 = −1.51 1 + α + βmj gJ eV . 3 Since the possible values of j are l + s, l + s − 1, . . . , |l − s|, and mj = j, j − 1, . . . , −j, we have the following table. state |3 l j mj i gJ A total energy E3 (eV) l = 0; j = 1/2 |3 0 1/2 1/2i 2 3/4 2 3/4 −1.51(1 + α2 /4) + β l = 1; j = 3/2 |3 1 3/2 3/2i 4/3 1/4 4/3 1/4 l = 1; j = 3/2 |3 1 3/2 −1/2i 4/3 1/4 l = 1; j = 3/2 |3 1 3/2 −3/2i 4/3 1/4 l = 1; j = 1/2 |3 1 1/2 1/2i 2/3 3/4 l = 1; j = 1/2 |3 1 1/2 −1/2i 2/3 3/4 l = 2; j = 5/2 |3 2 5/2 5/2i 6/5 1/12 −1.51(1 + α2 /36) + 3β l = 2; j = 5/2 |3 2 5/2 1/2i 6/5 1/12 −1.51(1 + α2 /36) + 3β/5 l = 0; j = 1/2 |3 0 1/2 −1/2i l = 1; j = 3/2 |3 1 3/2 1/2i −1.51(1 + α2 /4) − β −1.51(1 + α2 /12) + 2β −1.51(1 + α2 /12) + 2β/3 −1.51(1 + α2 /12) − 2β/3 −1.51(1 + α2 /12) − 2β −1.51(1 + α2 /4) + β/3 −1.51(1 + α2 /4) − β/3 6/5 1/12 −1.51(1 + α2 /36) + 9β/5 l = 2; j = 5/2 |3 2 5/2 3/2i l = 2; j = 5/2 |3 2 5/2 −1/2i 6/5 1/12 −1.51(1 + α2 /36) − 3β/5 l = 2; j = 5/2 |3 2 5/2 −3/2i 6/5 1/12 −1.51(1 + α2 /36) − 9β/5 l = 2; j = 5/2 |3 2 5/2 −5/2i 6/5 1/12 −1.51(1 + α2 /36) − 3β l = 2; j = 3/2 |3 2 3/2 3/2i 4/5 1/4 l = 2; j = 3/2 |3 2 3/2 1/2i 4/5 1/4 l = 2; j = 3/2 |3 2 3/2 −1/2i 4/5 1/4 l = 2; j = 3/2 |3 2 3/2 −3/2i 4/5 1/4 −1.51(1 + α2 /12) + 6β/5 −1.51(1 + α2 /12) + 2β/5 −1.51(1 + α2 /12) − 2β/5 −1.51(1 + α2 /12) − 6β/5 (c) Now for the intermediate-field regime. Note that there are 18 degenerate unperturbed states. (The hydrogen atom energies En are n2 -fold degenerate, and this becomes 2n2 if we include spin. Alternatively, the l = 0 level has one state, the l = 1 level has 2l + 1 = 3 states, and the l = 2 level has 2l + 1 = 5 states, for a total of 9. Multiplying this by 2 for spin and we have a total of 18 states.) As in the notes, we choose to work in the |j mi basis because both Hrel and Hso are already diagonal in this basis, and hence so is Hfs . This means that we can immediately use equation 6 (2.59) to write (1) Efs (0) 2 E α 3 3 1 =− eV − 34 j + 1/2 4 4 13.6α2 − 1 eV =− 108 j + 1/2 4 = −3γ − 1 eV j + 1/2 where it is convenient to define γ= 13.6 2 α . 324 Thus we have (1) Efs = −9γ for j = 1/2 = −3γ for j = 3/2 = −γ for j = 5/2 . (1) Efs (1) Efs The magnetic Hamiltonian is Hmag = eB µB B β (Lz + 2Sz ) = (Lz + 2Sz ) = (Lz + 2Sz ) . 2me c ~ ~ To evaluate the matrix elements of Hmag , we need to write our 18 basis |j mi states in terms of |ml ms i states using Clebsch-Gordan coefficients. For l = 0 and l = 1 the |j mi states are just the 8 states constructed in the notes, Section 2.5.3, where we worked out the intermediate-field case for n = 2. Therefore the β terms in the matrix of H ′ = Hfs + Hmag are the same as in the notes, and the Hfs contributions are −9γ for j = 1/2 and −3γ for j = 3/2 (as compared with −5ξ and −ξ in the notes). Thus this part of H ′ consists of the blocks " (−9γ + β); (−9γ − β); −3γ + 23 β − − √ 2 3 β √ 2 3 β −9γ + 31 β # (−3γ + 2β); ; " −3γ − 23 β − √ 2 3 β (−3γ − 2β) − √ 2 3 β −9γ − 31 β # . So now we need only construct the remaining 10 states corresponding to l = 2, or j = 5/2, 3/2. Since these are the result of adding a spin 2 state to a spin 1/2 state, we use the table of Clebsch-Gordan coefficients to write the various |j mi states as linear combination of |m1 m2 i = |ml ms i states. Note that we group these so that linear combinations of the same |ml ms i 7 states are kept together. This will put the matrix of H ′ = Hfs + Hmag into block diagonal form. | 25 5 2i = |2 12 i | 25 − 25 i = | −2 − 12 i q q = 15 |2 − 21 i + 45 |1 21 i | 25 32 i q q | 32 32 i = 45 |2 − 21 i − 15 |1 21 i q q = 25 |1 − 21 i + 35 |0 21 i | 25 12 i q q | 32 12 i = 35 |1 − 21 i − 25 |0 21 i q q | 25 − 21 i = 35 |0 − 21 i + 25 | −1 12 i q q | 23 − 21 i = 25 |0 − 21 i − 35 | −1 12 i q q | 25 − 23 i = 45 | −1 − 12 i + 51 | −2 21 i q q | 32 − 23 i = 15 | −1 − 12 i − 54 | −2 21 i We already showed what the matrix elements of Hfs are for these states, so we need only evaluate the matrix elements of Hmag . Each of the above pair of equations yields a 2 × 2 matrix since each pair is orthogonal to any other pair. For example, we have ! r r 5 3E 1E 1 4 1E = β(Lz + 2Sz ) + Hmag 2 − 1 2 2 5 2 5 2 ! r r 1E 1 4 1E +2 2 − 1 5 2 5 2 =β and hence 5 3E D5 3 Hmag 2 2 2 2 ! ! r r r D r D 1 1 1E 1 4 1 4 1E 2 − + 1 +2 =β 2 − 1 5 2 5 2 5 2 5 2 =β 1 4 +2· 5 5 = 9 β 5 (1) so that together with Efs = −γ (since this is the diagonal element for j = 5/2) we would have the H ′ matrix element −γ + (9/5)β. And the other nonvanishing matrix element for the | 52 32 i state is 5 3E D3 3 2 2 2 =− β −2· =β Hmag 2 2 2 2 5 5 5 8 but this is off-diagonal so there is no additional term due to Hfs . Continuing in this manner, we find the following blocks corresponding to these 10 states: (−γ + 3β); (−γ − 3β) √ " # # " −γ + 95 β − 52 β − 56 β −γ + 35 β ; √ − 52 β −3γ + 56 β −3γ + 25 β − 56 β √ " # # " −γ − 53 β −γ − 95 β − 56 β − 52 β . ; √ −3γ − 65 β − 52 β −3γ − 52 β − 56 β Thus we see that the matrix of H ′ is the direct sum of six 1 × 1 blocks and six 2 × 2 blocks. The 1 × 1 blocks are already the eigenvalues, so we need to find the eigenvalues of the six 2 × 2 blocks. There are three distinct characteristic equations (and another three that result from letting β → −β). These are 2 1 2 −3γ + β − λ −9γ + β − λ − β 2 = 0 (1a) 3 3 9 9 6 4 −γ + β − λ −3γ + β − λ − β 2 = 0 (1b) 5 5 25 3 2 6 −γ + β − λ −3γ + β − λ − β 2 = 0 . (1c) 5 5 25 Multiplying these out we have the three equations λ2 + λ(12γ − β) + γ(27γ − 7β) = 0 33 2 2 2 =0 λ + λ(4γ − 3β) + 3γ − γβ + 2β 5 11 2 λ + λ(4γ − β) + γ 3γ − β = 0 . 5 Using the quadratic formula, these have the roots r 1 1 (a) λ± = −6γ + β ± 9γ 2 + γβ + β 2 2 4 r 3 3 1 (b) λ± = −2γ + β ± γ 2 + γβ + β 2 2 5 4 r 1 1 1 (c) λ± = −2γ + β ± γ 2 + γβ + β 2 2 5 4 9 (2a) (2b) (2c) (3a) (3b) (3c) Combining our results, we have the following nine energy levels: (0) E1 = E3 − 9γ + β (0) E2 = E3 − 3γ + 2β (0) E3 = E3 − γ + 3β E4 = (0) E3 − 6γ + (0) E5 = E3 − 6γ + (0) E6 = E3 − 2γ + (0) E7 = E3 − 2γ + (0) E8 = E3 − 2γ + (0) E9 = E3 − 2γ + r 1 1 β + 9γ 2 + γβ + β 2 2 4 r 1 1 β − 9γ 2 + γβ + β 2 2 4 r 3 3 1 β + γ 2 + γβ + β 2 2 5 4 r 3 3 1 β − γ 2 + γβ + β 2 2 5 4 r 1 1 1 β + γ 2 + γβ + β 2 2 5 4 r 1 1 1 β − γ 2 + γβ + β 2 2 5 4 plus another nine levels that result from letting β → −β. What about the limiting cases of a very strong field (i.e., γ ≪ β) and very weak field (i.e., β ≪ γ)? In the strong-field case, we ignore√terms of order γ 2 , and expand the square root to first order in γ/β using 1 + x ≈ 1 + x/2. For example, r r 1/2 1 1 β β γ 9γ 2 + γβ + β 2 ≈ γβ + β 2 = ≈ +γ. 1+4 4 4 2 β 2 Continuing in this manner yields the strong-field limits (0) E4 ≈ E3 − 5γ + β (0) E5 ≈ E3 − 7γ 7 (0) E6 ≈ E3 − γ + 2β 5 13 (0) E7 ≈ E3 − γ + β 5 9 (0) E8 ≈ E3 − γ + β 5 11 (0) E9 ≈ E3 − γ 5 Since 324 = 9(36), we note that γ= 13.6 2 α2 α = 1.51 324 36 10 and hence it is not hard to see that these limits reproduce the table in part (a). Similarly, in the weak-field limit β ≪ γ we ignore terms of order β 2 , expand the square root to first order in β/γ, and find that 2 (0) E4 ≈ E3 − 3γ + β 3 1 (0) E5 ≈ E3 − 9γ + β 3 9 (0) E6 ≈ E3 − γ + β 5 6 (0) E7 ≈ E3 − 3γ + β 5 3 (0) E8 ≈ E3 − γ + β 5 2 (0) E9 ≈ E3 − 3γ + β 5 Again, we see that these reproduce the results of part (b). In the figure below we plot E1 – E9 . The three points on the E-axis (1) represent the three possible values of Efs that we would have if B = β = 0. E 25 20 15 10 5 Β 2 4 6 8 10 -5 The slope of the curves for E1 – E3 are constant, but the slope of the curves for E4 – E9 varies as β increases. For example, at small β the slope of E4 is 2/3, but at large β it is 1. Similarly, the slope of E5 starts out as 1/3, but then decreases to 0. Also note that this is only half of the energies. There are another nine curves that result from letting β → −β, and hence they have a negative (downwards) slope. 11 4. As Rin Example 3.1 of the notes, the potential energy is given by −e E · dr so that 2 2 H ′ (t) = −eE e−t /τ z . R F · dr = The transition probability is given by Z 2 1 ∞ ′ Pif (t) = 2 Hf i (t)eiωf i t dt ~ −∞ where the hydrogen atom wave functions ψnlm are ψi = ψ100 and ψf = ψ2lm . Since the matrix element is proportional to h2 l m|z|1 0 0i, we see that parity considerations tell us that the only nonzero term is when l = 1. This is because z is an odd function, and the parity of the spherical harmonics goes like (−1)l . Since Y00 is even, we must have a transition to an odd state, and hence l = 1 is the only possibility for n = 2. Thus the potential final states are ψ210 , ψ211 and ψ21 −1 . The wave functions are given by ψnlm = Rnl Ylm , and the relevant functions are −5/2 a0 re−r/2a0 R21 = √ 24 Y00 Y11 −3/2 −r/a0 R10 = 2a0 1 = √ 4π r =− Y10 3 iφ e sin θ 8π Y1−1 e = r 3 cos θ 4π = r 3 −iφ e sin θ 8π Since z = r cos θ and the volume element depends on sin θ dθ, the fact that π Z π Z π sin3 θ 2 2 cos θ sin θ dθ = sin θ d sin θ = =0 3 0 0 0 tells us that the only nonzero matrix element Hf′ i ∼ h2 1 m| cos θ|1 0 0i is h2 1 0| cos θ|1 0 0i. So we have Z 2 2 Hf′ i = −eE e−t /τ R21 Y10 [r cos θ]R10 Y00 r2 sin θ drdθdφ r Z 1 −4 1 a r4 e−3r/2a0 cos2 θ drd cos θdφ = −eE e 2π 8 0 r Z 1 −4 ∞ 4 −3r/2a0 −t2 /τ 2 2 = −eE e a0 dr . r e 3 8 0 −t2 /τ 2 The integral was done in HW4, Problem 3, and is 5 Z ∞ 2a0 4 −3r/2a0 . dr = 4! r e 3 0 12 This leaves us with Hf′ i = −eE e −t2 /τ 2 r 1 8 2 9 a0 35 and therefore Pif (t → ∞) = eE ~ 2 215 a20 310 Z ∞ e −t2 /τ 2 iω21 t e −∞ 2 dt . But this integral was evaluated in Example 3.1 of the notes and is equal to √ 2 2 πτ 2 e−ω21 τ /4 . Thus we finally arrive at Pif (t → ∞) = eE ~ 2 215 a20 310 2 πτ 2 e−ω21 τ 2 /2 . 5. (a) For a spin one-half particle at rest, the Hamiltonian is simply H = −µ · B = − gq γ~ S · B = −γS · B = − σ · B 2mc 2 where the Pauli matrices are given by " # " # 0 1 0 −i σx = σy = 1 0 i 0 σz = " 1 0 0 −1 # . For the given field, this becomes H=− =− =− γ~ (Brf cos ωt σx − Brf sin ωt σy + B0 σz ) 2 # " B0 Brf cos ωt + iBrf sin ωt γ~ 2 γ~ 2 " Brf cos ωt − iBrf sin ωt # B0 Brf eiωt . Brf e−iωt −B0 −B0 (b) The time-dependent Schrödinger equation is i~ which becomes " i~ ȧ(t) ḃ(t) # γ~ =− 2 ∂ χ(t) = Hχ(t) ∂t " B0 Brf eiωt Brf e−iωt −B0 13 #" a(t) b(t) # . Equating components on both sides of this equation we have ȧ = iγ iγ B0 a + Brf eiωt b 2 2 ḃ = iγ iγ Brf e−iωt a − B0 b 2 2 Defining ω0 = γB0 and Ω = γBrf these equations become ȧ(t) = ḃ(t) = i Ωeiωt b + ω0 a 2 i Ωe−iωt a − ω0 b . 2 (4a) (4b) (c) Taking the derivative of (4a) and substituting (4b) for ḃ we have −2i ä = iωΩeiωt b + Ωeiωt ḃ + ω0 ȧ = iωΩeiωt b + Ωeiωt i Ωe−iωt a − ω0 b + ω0 ȧ 2 i i = iωΩeiωt b + Ω2 a − ω0 Ωeiωt b + ω0 ȧ 2 2 i ω0 + ω0 ȧ + Ω2 a . = iΩeiωt b ω − 2 2 But from (4a) we can write iΩeiωt b = 2ȧ − iω0 a so that i ω0 + ω0 ȧ + Ω2 a −2i ä = (2ȧ − iω0 a) ω − 2 2 i 2 = 2ω ȧ + Ω − ω0 (2ω − ω0 ) a 2 i 2 = 2ω ȧ + Ω − ω 2 + (ω − ω0 )2 a . 2 Rearranging, this becomes 1 2 Ω − ω 2 + (ω − ω0 )2 a = 0 . 4 This is now just a second-order differential equation with constant coefficients, and its solutions are of the form ert where r is a root of the polynomial 1 2 r2 − iωr + Ω − ω 2 + (ω − ω0 )2 = 0 . 4 ä − iω ȧ + 14 From the quadratic formula, the roots of this are r= iω iω iω ′ ip 2 Ω + (ω − ω0 )2 := ± ± 2 2 2 2 and hence the general solution for a can be written in the form ω′t ω′t iωt/2 + β sin . a(t) = e α cos 2 2 Clearly we have α = a(0) := a0 . To find β, we take the derivative of a(t) to write iω iωt/2 ω ′ iωt/2 ω′t ω′t ω′t ω′t ȧ(t) = − e . α cos α sin e + β sin − β cos 2 2 2 2 2 2 Then (using α = a0 ) ω′ iω a0 + β 2 2 2 iω β = ′ ȧ(0) − a0 . ω 2 ȧ(0) = or Using (4a) to evaluate ȧ(0) and letting b(0) = b0 , this becomes 2 i i iω β= ′ (Ωb0 + ω0 a0 ) − a0 = ′ [Ωb0 + (ω0 − ω)a0 ] . ω 2 2 ω Thus the solution for a(t) is ω′t ω′t i a(t) = a0 cos eiωt/2 . + ′ [Ωb0 + (ω0 − ω)a0 ] sin 2 ω 2 To solve for b(t) you could repeat all this work starting from (4b). However, it is a lot easier to realize that equations (4a) and (4b) are symmetric under the exchange a ↔ b along with ω → −ω and ω0 → −ω0 . This means that we can immediately write down the solution for b(t) as ω′t i ω′t e−iωt/2 . + ′ [Ωa0 + (ω − ω0 )b0 ] sin b(t) = b0 cos 2 ω 2 (d) If the particle is initially in the spin up state, then a0 = 1 and b0 = 0. Therefore i ω ′ t −iωt/2 b(t) = ′ Ω sin e ω 2 so the probability of being in the spin down state is 2 Ω ω′t 2 Pdown (t) = |b(t)| = sin2 . ′ ω 2 15 (e) The resonance curve satisfies the equation P (ω) = Ω2 . (ω − ω0 )2 + Ω2 To plot this in Mathematica so the whole curve would show, I simply let Ω = 2 and ω0 = 2. Then schematically, the resonance curve looks like this. 1.0 0.8 0.6 0.4 0.2 -10 5 -5 10 No matter what Ω is, the maximum value of P (ω) is 1. Therefore, the width at half-maximum comes from solving 1 Ω2 = 2 (ω − ω0 )2 + Ω2 which implies that (ω − ω0 )2 = Ω2 or ω = ω0 ± Ω . Thus the width at half-maximum is ∆ω = 2Ω . (f) For the proton we have gp e ω0 = γB0 = B0 2mp c = 5.586(4.803 × 10−10 statC) 104 gauss 2(1.673 × 10−24 g)(2.998 × 1010 cm/s) = 2.674 × 108 rad/s so that ω0 = 4.257 × 107 Hz . 2π By definition, Ω = γBrf . Then ν0 = ∆ω = 2Ω = 2(2.674 × 104 )10−2 = 534.8 rad/s or ∆ν = 534.8 = 85.12 Hz . 2π 16