Ch12

CHAPTER 12. 1. With a potential of the form 1 2 2 V(r) = mω r 2 the perturbation reduces to 2 1 1 dV (r) ω H1 = 2 2 S • L = 2 (J2 − L2 − S2 ) 2m c r dr 4mc 2 (hω ) = 2( j( j + 1) − l(l + 1) − s(s + 1)) 4mc

where l is the orbital angular momentum, s is the spin of the particle in the well (e.g. 1/2 for an electron or a nucleon) and j is the total angular momentum. The possible values of j are l + s, l + s – 1, l + s –2, …|l – s|. 3 hω (2n r + l +). Each of the The unperturbed energy spectrum is given by E n r l = 2 levels characterized by l is (2l + 1)-fold degenerate, but there is an additional degeneracy, not unlike that appearing in hydrogen. For example nr =2, l = 0. nr =1, l = 2 , nr = 0, l = 4 all have the same energy. A picture of the levels and their spin-orbit splitting is given below.

2.

The effects that enter into the energy levels corresponding to n = 2, are (I) the basic Coulomb interaction, (ii) relativistic and spin-orbit effects, and (iii) the hyperfine structure which we are instructed to ignore. Thus, in the absence of a magnetic field, the levels under the influence of the Coulomb potential consist of 2n2 = 8 degenerate levels. Two of the levels are associated with l = 0 (spin up and spin down) and six

levels with l = 0, corresponding to ml = 1,0,-1, spin up and spin down. The latter can be rearranged into states characterized by J2, L2 and Jz. There are two levels characterized by j = l –1/2 = 1/2 and four levels with j = l + 1/2 = 3/2. These energies are split by relativistic effects and spin-orbit coupling, as given in Eq. (12-16). We ignore reduced mass effects (other than in the original Coulomb energies). We therefore have 1 1 1 3 ∆E =− mec 2α4 3  − 2 n  j +1/ 2 4n 

(b)

1 5 =− me c 2α4   2 64 

j =1 / 2

1 1  =− me c 2α4 64  2

j =3 / 2

The Zeeman splittings for a given j are ehB 2  ∆ EB = m 2me j3

j =1 / 2

ehB 4  = m 2me j 3 

j =3 / 2

ehB 1 5 4 5 = 14.47 × 10− eV , me c 2α ≈ 1.132 × 10 − eV , while for B = 2.5T 2me 128 so under these circumstances the magnetic effects are a factor of 13 larger than the relativistic effects. Under these circumstances one could neglect these and use Eq. (1226). Numerically

3.

The unperturbed Hamiltonian is given by Eq. (12-34) and the magnetic field interacts both with the spin of the electron and the spin of the proton. This leads to S I S• I H= A 2 + a z+ b z h h h

Here  m  I •S 4 A = α4 me c 2 gP  e  2 3  MP  h ehB 2me ehB b =−gP 2M p a=2

Let us now introduce the total spin F = S + I. It follows that S •I 1  2 3 2 3 2 2 = 2 h F(F +1) − h − h h 2h  4 4  1 = for F =1 4 3 =− for F =0 4 bI z for eigenstates of F2 and Fz . We next need to calculate the matrix elements of aSz + These will be exactly like the spin triplet and spin singlet eigenstates. These are 1 〈 1,1| aSz + bI z |1,1〉= 〈χ ξ bIz | χ ξ 〉= (a + b) + +| aSz + + + 2 2

1  〈 1,0 | aSz + bI z |1,0〉=   〈χ ξ+χξ| aS +bI z | χ ξ χ ξ 〉=0 + −+ + −  2  +− − + z 1 〈 1,− 1 | aSz + bIz |1,− 1〉= 〈χ ξ bIz | χ ξ 〉=− (a + b) − −| aSz + − − 2 And for the singlet state (F = 0) 2

1 1  〈1,0 | aSz +bI z | 0,0〉=  〈χ ξ χ ξ bI z | χ ξ χ+ξ−〉= (a −b) + −+ − +| aSz + + −−  2 2 2

1  〈0,0 | aSz +bI z | 0,0〉=  〈χ ξ−χξ | aS +bIz | χ ξ χ ξ 〉=0 + −− + −  2  +− −+ z

Thus the magnetic field introduces mixing between the |1,0> state and the |0.0> state. We must therefore diagonalize the submatrix A/4 (a − b) / 2 − A/4 0  A /2 (a − b) / 2    = +        (a − b) / 2 − 3A / 4   0 − A / 4  (a − b) / 2 − A /2  The second submatrix commutes with the first one. Its eigenvalues are easily determined to be ± A2 / 4 + (a − b) 2 / 4 so that the overall eigenvalues are − A /4 ± A2 / 4 + (a − b) 2 / 4 Thus the spectrum consists of the following states:

F = 1, Fz = 1

E= A /4 + (a + b) / 2

F =1, Fz = -1

E= A /4 − (a + b) / 2

F = 1,0; Fz = 0

E= − A/4± (A 2 / 4 + (a − b)2 / 4

We can now put in numbers. For B = 10-4 T, the values, in units of 10-6 eV are 1.451, 1.439, 0(10-10), -2.89 For B = 1 T, the values in units of 10-6 eV are 57.21,-54.32, 54.29 and 7 x 10-6. 4. According to Eq. (12-17) the energies of hydrogen-like states, including relativistic + spin-orbit contributions is given by E n, j

α

1 me c 2 (Z ) 2 1 1 1 1 3 = − mec 2 (Z )4 3  −  2 − 2 (1+ me / M p ) n 2 n  j+ 1/ 2 4n 

α

The wavelength in a transition between two states is given by

λ=

2πhc ∆E

where ∆E is the change in energy in the transition. We now consider the transitions n =3, j = 3/2  n = 1, j = 1/2 and n = 3 , j = 1/2  n = 1, j = 1/2.. The corresponding energy differences (neglecting the reduced mass effect) is

α

α

(3,3/21,1/2)

1 1 1 1 1 ∆ E= mec 2 (Z )2 (1− )+ me c 2 (Z ) 4 (1− ) 2 9 2 4 27

(3,1/21,1/2)

1 1 1 1 3 me c 2 (Z ) 2 (1− )+ me c 2 (Z ) 4 (1− ) 2 9 2 4 27

α

We can write these in the form (3,3/21,1/2)

13 ∆ E 0 (1 + (Zα )2 ) 48

(3,1/21,1/2)

1 ∆ E 0 (1 + (Zα ) 2) 18

where 1 8 ∆ E 0 =mec 2 (Zα )2 2 9 The corresponding wavelengths are

α

(3,3/21,1/2)

13 λ (1− (Zα ) )= 588.995 × 10 m 48

(3,1/21,1/2)

1 λ (1− (Zα ) )= 589.592 × 10 m 18

2

0

− 9

2

0

− 9

We may use the two equations to calculate λ0 and Z. Dividing one equation by the other we get, after a little arithmetic Z = 11.5, which fits with the Z = 11 for Sodium. (Note that if we take for λ0 the average of the two wavelengths, then , using 2 hc / ∆ E0 = 9 h / 2mc(Z )2 , we get a seemingly unreasonably small value of Z = 0 = 0.4! This is not surprising. The ionization potential for sodium is 5.1 eV instead of Z2(13.6 eV), for reasons that will be discussed in Chapter 14)

λ π πα

2

1  p2  4. The relativistic correction to the kinetic energy term is −   . The energy 2mc 2 2m shift in the ground state is therefore 2

 1 p2  1 1 ∆ E= −2 〈 0 |  | 0〉 = −2 〈 0 | (H − m 2 r 2 ) 2 | 0〉 2mc  2m 2mc 2

ω

To calculate < 0 | r2 | 0 > and < 0 | r4 | 0 > we need the ground state wave function. We know that for the one-dimensional oscillator it is mω   − mω x 2 / 2h u0 (x) = e π h 1/4

so that for the three dimensional oscillator it is mω  − r 2 / 2h u0 (r) = u0 (x)u0 (y)u0 (z) = e mω   π h 3/ 4

It follows that 3/2

mω 2 −mωr2 / h 4πr dr re =  πh 

∞

〈0 | r | 0〉=∫0 2

2

3/ 2

5/2

mω  h  =4π    πh mω 3h = 2mω

∞

∫

0

dyy 4 e −y

2

We can also calculate

〈0 | r | 0〉=∫0

3/ 2

mω 4 −mωr2 / h 4πr dr re =  πh 

∞

4

2

3/ 2

7/2

mω  h  =4π    πh mω

∞

∫

0

dyy 6e −y

2

15  h 2 = 4 mω We made use of Thus

∞ n − z

dzz e ∫ 0

= Γ (n + 1) = nΓ (n) and

1 Γ ( )= 2

π

2 2 1  3  3  15    2 3h  1 2 4  h  ∆ E= − 2  hω− hωmω +m ω   2mω  4 4    2mc  2   2  mω 

15 (hω )2 = − 32 mc 2 6.

(a) With J = 1 and S = 1, the possible values of the orbital angular momentum, such that j = L + S, L + S –1…|L – S| can only be L = 0,1,2. Thus the possible states are 3 S1 ,3 P1,3 D1 . The parity of the deuteron is (-1)L assuming that the intrinsic parities of the proton and neutron are taken to be +1. Thus the S and D states have positive parity and the P state has opposite parity. Given parity conservation, the only possible 3 admixture can be the D1 state. (b)The interaction with a magnetic field consists of three contributions: the interaction of the spins of the proton and neutron with the magnetic field, and the L.B term, if L is not zero. We write H= − Mp • B− Mn • B− ML • B Mp =

eg p eh S p  S p =(5.5792)   2M 2M  h 

egn eh Sn  Sn =(−3.8206) 2M 2M  h  e ML = L 2M red

where M n =

We take the neutron and proton masses equal (= M ) and the reduced mass of the two3 particle system for equal masses is M/2. For the S1 stgate, the last term does not contribute.

If we choose B to define the z axis, then the energy shift is S pz    S  eBh 3 − 〈 S1 | gp   + gn nz  |3 S1 〉   2M h  h We write S pz   gn S pz + Snz g p − gn S pz − Snz  Snz g p + gp  + gn   = +  h  h 2 h 2 h It is easy to check that the last term has zero matrix elements in the triplet states, so S 1 that we are left with (g p +gn ) z , where Sz is the z-component of the total spin.. 2 h Hence gn 3Bh g p + 3 〈 S1 | H1 |3 S1 〉 = − ms 2M 2 where ms is the magnetic quantum number (ms = 1,0,-1) for the total spin. We may therefore write the magnetic moment of the deuteron as gn e gp + e µ − S= − (0.8793) S eff = 2M 2 2M The experimental measurements correspond to gd = 0.8574 which suggests a small 3 admixture of the D1 to the deuteron wave function.