Answers to Exam II (Nov. 4-7 2005)

Prob 1

LS coupling

For LS coupling of identical particles, we have the restriction L+S is even. Therefore,
(3s3s) → \(TraditionalForm\`\(\^1\) S\)                                        1 substate
(3  p3p) (→ ^1)S,  \(TraditionalForm\`\^3\)P,  \(TraditionalForm\`\^1\)D                        1+3×3+5  = 15 substates
(3  d3d) (→ ^1)S,  \(TraditionalForm\`\^3\)P,  \(TraditionalForm\`\^1\)D, \(\(\ \)\(\^3\)\)F ((, )^1) G     1+3×3+5+3×7+9 = 45 substates
(3  s3d) (→ ^1)D,  \(TraditionalForm\`\^3\)D                                5+3×5 = 20 substates
Total number of substates = 1+15+45+20=81

jj coupling

For jj coupling of identical particles J is even. Therefore,
(3  (Subscript[s, 1/2]) 3  (Subscript[s, 1/2])) →  [0]                             1 substate
(3  (Subscript[p, 1/2]) 3  (Subscript[p, 1/2])) → [0]                             1 substate
(3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) → [1], [2]                      3+5 = 8  substates
(3  (Subscript[p, 3/2]) 3  (Subscript[p, 3/2])) → [0], [2]                      1+5 = 6  substates 
(3  (Subscript[d, 3/2]) 3  (Subscript[d, 3/2])) → [0], [2]                      1+5 = 6  substates
(3  (Subscript[d, 3/2]) 3  (Subscript[d, 5/2])) → [1], [2], [3], [4]        3+5+7+9 = 24  substates
(3  (Subscript[d, 5/2]) 3  (Subscript[d, 5/2])) → [0], [2], [4]               1+5+9 = 15  substates
(3  (Subscript[s, 1/2]) 3  (Subscript[d, 3/2])) → [1], [2]                       3+5 = 8 substates
(3  (Subscript[s, 1/2]) 3  (Subscript[d, 5/2])) → [2], [3]                       5+7 = 12  substates
Total number of substates = 1+(1+8+6)+(6+24+15)+(8+12)=1+15+45+20=81

A mathematica version of the abpbe tables is constructed as follows:

\(tLS\  := 
      TableForm[{{"LS", "", "", "", "", "", "Sum"}, 
{"(3s3s)", \*""\!\(\(\(\\ \)\(\^1\)\)\)S"", "", "", "", "", 
            1}, {"(3p3p)", \*""\!\(\(\(\\ \)\(\^1\)\)\)S"", \ \*""\!\(\(\(\\ \)\(\^3\)\)\)P"", \*""\!\(\(\(\\ \)\(\^1\)\)\)D"", "", "", 
            1 + \ 3\ 3 + 
              5}, \ {"(3d3d)", \*""\!\(\(\(\\ \)\(\^1\)\)\)S"", \ \*""\!\(\(\(\\ \)\(\^3\)\)\)P"", \*""\!\(\(\(\\ \)\(\^1\)\)\)D"", \ \*""\!\(\(\(\\ \)\(\^3\)\)\)F"", \*""\!\(\(\(\\ \)\(\^1\)\)\)G"", 
            1 + 3\ 3\  + 5 + \ 3\ 7\  + \ 
              9}, {"(3s3d)", \*""\!\(\(\(\\ \)\(\^1\)\)\)D"", \ \*""\!\(\(\(\\ \)\(\^3\)\)\)D"", "", "", "", 
            5 + 3\ 5}}]\)

\(tjj\  := 
      TableForm[{{"jj", "", "", "", "", "Sum"}, 
{\*""(3(Subscript[s, 1/2])3(Subscript[s, 1/2]))"", "[0]", "", "", "", 
            1}, {\*""(3(Subscript[p, 1/2])3(Subscript[p, 1/2]))"", "[0]", \ "", "", "", 
            1}, {\*""(3(Subscript[p, 1/2])3(Subscript[p, 3/2]))"", "[1]", \ "[2]", "", "", 
            3 + 5}, \ 
{\*""(3(Subscript[p, 3/2])3(Subscript[p, 3/2]))"", "[0]", \ "[2]", "", "", 
            1 + \ 5}, \ \ {\*""(3(Subscript[d, 3/2])3(Subscript[d, 3/2]))"", "[0]", \ "[2]", "", \ "", 
            1 + 5}, {\*""(3(Subscript[d, 3/2])3(Subscript[d, 5/2]))"", "[1]", \ "[2]", "[3]", "[4]", 
            3 + 5 + 7 + 
              9}, 
{\*""(3(Subscript[d, 5/2])3(Subscript[d, 5/2]))"", "[0]", \ "[2]", "[4]", \ "", 
            1 + 5 + 9}, 
{\*""(3(Subscript[s, 1/2])3(Subscript[d, 3/2]))"", "[1]", \ "[2]", "", "", 
            3 + 5}, {\*""(3(Subscript[s, 1/2])3(Subscript[d, 5/2]))"", "[2]", \ "[3]", "", "", 5 + 7}}]\)

LStot  :=  
      Sum[tLS[[1, k, Length[tLS[[1, 2]]]]], {k, 2, 
          Length[tLS[[1]]]}]

jjtot  :=  
      Sum[tjj[[1, k, Length[tjj[[1, 2]]]]], {k, 2, 
          Length[tjj[[1]]]}]

tLS

tjj

LStot

jjtot == LStot

Prob 2

The ground-state energy is  E[ (((1  s))^2) \(\^1\)(Subscript[S, 0])] = 2(Subscript[ϵ, 1  s])+ (Subscript[R, 0])(1s1s1s1s) - 2  (Subscript[U, 1  s1s]),  where U = (5/16)(1/r)

ϵ1s  =  -Zs^2/2

P1s[r_]  :=  2 Zs^(3/2) r Exp[-Zs r]

U1s1s  =  (5/16) Integrate[P1s[r]^2 /r , {r, 0, Infinity}, 
          Assumptions ->  Zs > 0]

Y0s[r_]  := 
      Integrate[P1s[x]^2, {x, 0, r}, Assumptions ->  Zs > 0]/r  + 
        Integrate[P1s[x]^2/x, {x, r, Infinity}, 
          Assumptions ->  Zs > 0]

R0s  =  
      Integrate[ P1s[r]^2 Y0s[r], {r, 0, Infinity}, 
        Assumptions ->  Zs > 0]

The final result for the ground-state energy is  -(Zs^2) (as should be expected from first section of Chap 4)

e1S0  =  2 ϵ1s  + R0s  - 2  U1s1s

e1S0  /.  Zs -> 5  -  5/16 // N

Compare to NIST  :  IP = 2091960 (cm^(-1))  [ground state energy = (Subscript[E, ion]) -  IP = -((Z^2)/2)-IP(a.u.)]

eNist  =  -12.5  -  2091960/219474.624

del  =  
      100 ((e1S0 /.  Zs -> 5 - 5/16) - eNist)/eNist

The ground state energy differs from measured value by 0.27%

Now let's look at the (1s2p) singlet and triplet P states. The energy of the 2P level is 
E  [(((1  s2p)  )^(2  S + 1)) 
          P]  =  (Subscript[ϵ, 1  s])+(Subscript[ϵ, 2  p])+(Subscript[R, 0])(1  s2p1s2p) + ((((-1))^S)/3)(Subscript[R, 1])(1s,2p,2p,1s) -U(1s1s)-U(2p2p).  
Note that the counter potential U for the 2p state is (1/r)
As a first step , we calculate the integrals (Subscript[R, 0])(1s2p1s2p)  & (Subscript[R, 1])(1s,2p,2p,1s)

P2p[r_]  :=  
      Zp^(5/2)/(2  Sqrt[6]) r^2  Exp[-Zp r/2]

R0 = 
      Integrate[ P2p[r]^2 Y0s[r], {r, 0, Infinity}, 
        Assumptions ->  {Zp > 0, Zs > 0}]

R0  /.  {Zp -> 4,  Zs -> 5 - 5/16}  // N

Y1[r_]  := 
      Integrate[P2p[x] P1s[x] x, {x, 0, r}, 
            Assumptions ->  {Zp > 0, Zs > 0}]/r^2  + 
        r Integrate[P2p[x] P1s[x]/x^2, {x, r, Infinity}, 
            Assumptions ->  {Zp > 0, Zs > 0}]

R1  =  
      Integrate[ P2p[r] P1s[r] Y1[r], {r, 0, Infinity}, 
        Assumptions ->  {Zp > 0, Zs > 0}]

R1  /.  {Zp -> 4,  Zs -> 5 - 5/16}  // N

Now, we evaluate the counter term U(2p2p)

U2p2p  =  
      Integrate[P2p[r]^2 /r , {r, 0, Infinity}, 
        Assumptions ->  Zp > 0]

The lowest order energy (Subscript[ϵ, 2  p ])of  the 2p orbital is

ϵ2p  =  -Zp^2/8

Put together everything for singlets then triplets

e2pS  =  ϵ1s+ϵ2p + R0 + (1/3) 
          R1 -U1s1s -U2p2p

e2pT  =  ϵ1s + ϵ2p  +  R0  -  (1/3) R1  - 
        U1s1s  - U2p2p

The multiplet ine separation is e2pS-e2pT.  The NIST value  is 0.095935  (Theory gives a value about 5% too large)

e2pS - e2pT  /.  {Zp -> 4,  Zs -> 5 - 5/16}  // 
      N

Next, we set up the singlet and triplet energies relative to the  ground state  (These are the  excitation energies!)

e2pS  -  e1S0 /.  {Zp -> 4,  Zs -> 5 - 5/16}  // 
      N

The NIST value is

nistS  =  1658020/219474.624

For the triplet state

e2pT  -  e1S0 /.  {Zp -> 4,  Zs -> 5 - 5/16}  // 
      N

nistT  =  1636882/219474.624

The percentage differences with NIST are

ΔS  =  
      100 (e2pS - e1S0 - nistS)/nistS  /.  {Zp -> 4,  
            Zs -> 5 - 5/16}  // N

ΔT  =  
      100 (e2pT - e1S0 - nistT)/nistT  /.  {Zp -> 4,  
            Zs -> 5 - 5/16}  // N

The predicted excitation energies within 0.1% of NIST, while the predicted splitting is about 5% too large.

Prob 3

The reduced matrix element (Subscript[R, E1]) =<(((1  s))^2) (1^1)S || r || (2p1s) ((2 )^1) P> = (Sqrt[2]) <0 ||C1||1> (Subscript[R, 1  s 2  p]) , where 
 (Subscript[R, 1  s 2  p]) = ∫ (Subscript[P, 2  p])(r) r (Subscript[P, 1  s])(r) dr  is the radial dipole integral.

Rps  =  
      Integrate[P2p[r] r P1s[r], {r, 0, Infinity}, 
        Assumptions ->  {Zp > 0, Zs > 0}]

Let cps =  <0 ||C1||1>

cps  =  Sqrt[3] ThreeJSymbol[{0, 0}, {1, 0}, {1, 0}]

Then, we find that the reduced matrix element is:

RE1  =  Sqrt[2] cps Rps

The associated line strength is

SE1  =  RE1^2

The  transition energy from the NIST web site in 1/cm is

Eexp  =  1657980

The corresponding wavelength (Angstrom) is

λ  =  N[10^8/Eexp]

The transition rate is A = ((2.02613 (10^18))/(λ^3)) (SE1/(Subscript[g, a])) in 1/s, where (Subscript[g, a]) is the initial state degeneracy (3 for a singlet P state)

A = ((2.02613 (10^18))/(λ^3)) (SE1/(Subscript[g, a])) /. {Zs -> 5 - 5/16,  Zp ->  4,  
          (Subscript[g, a]) ->  3}

The lifetime is τ= 1/A (s)

τ  =  1/A

This compares favorably with the "exact" result 2.69 (10^(-12)) s.

Prob 4

H  3p state

The 3p level in H decays by E1 emission to both  2s and 1s levels.  The  A-coefficient is  ((2.02613 (10^18))/(λ^3))(SE1/3) for each transition. The total rate is the sum of the partial rates to each final state.  The measured wavelengths in Angstrom of the  transitions to the 2s and 1s states are 6564.5 and 1025.7,  respectively.

Details of a H 3p rate calculation.

P3p[r_]  :=  
      8  Z^(5/2)/(27 Sqrt[6]) r^2 (1 - 
            Z r/6) Exp[-Z r/3]

P2s[r_]  :=  
      Z^(3/2)/Sqrt[2] r (1 - Z r/2) Exp[-Z r /2]

The radial matrix elements are

R32  =  
      Integrate[ P2s[r] r P3p[r], {r, 0, Infinity}, 
        Assumptions ->  Z > 0]

R31  =  
      Integrate[( P1s[r] /. Zs -> Z)  r P3p[r], {r, 0, 
          Infinity}, Assumptions ->   Z > 0]

The angular factor is cps = <s || C1 || p>

cps  =  Sqrt[3] ThreeJSymbol[{0, 0}, {1, 0}, {1, 0}]

The line strengths for the 3p → 2s transitions is S32 = <s || C1 || \(TraditionalForm\`\(\(p\)\( > \^2\)\)\) (R32^2)

S32  =  cps^2 R32^2

S31  =  cps^2 R31^2

The transition energy (au) for the  3p → 2s transitions is ΔE32 = (Z^2)/(2 (2^2)) - (Z^2)/(2 (3^2))

ΔE32  =  Z^2/2/2^2  -  Z^2/2/3^2

ΔE31  =  Z^2/2/1^2  -  Z^2/2/3^2

The conversion from (a.u.) to 1/cm is 219474.62

con  =  219474.62

The (32) and (31) transition energies in 1/cm are

Δcm32  =  con ΔE32

Δcm31  =  con ΔE31

The transition wavelengths (angstrom) are

λ32  =  10^8/Δcm32

λ31  =  10^8/Δcm31

The Einstein A-coefficients  are  (Subscript[A, 31])=((2.02613 (10^18))/((Subscript[λ, 31])^3))(S31/3) and  (Subscript[A, 32])=((2.02613 (10^18))/((Subscript[λ, 32])^3))(S32/3)

A31  =  2.02631 10^(18)/λ31^3 S31/3

A32  =  2.02631 10^(18)/λ32^3 S32/3

The total rate is the sum of partial rates

A3  =  A31 + A32

The lifetime (s) of the 3p state in H is (Subscript[A, 3]^(-1)) evaluated with Z=1

τ  =  1/A3  /.  Z -> 1

(Al^(2+)) (Na-like) 3p state

The 3p state in (Al^(2+)) decays  to the 3s state by E1 emission. The  A-coefficient is ((2.02613 (10^18))/(λ^3))(SE1/3)
The average transition energy is (53684.1+2×53916.6)/3 (cm^(-1))and the corresponding average wavelength in Angstrom is

λ =  (10^8) 3/(53684.1 + 2 53916.6)

Al 3  (Subscript[p, 3/2]) state

The 3  (Subscript[p, 3/2]) state in Al decays to the 3  (Subscript[p, 1/2])ground  state by M1 emission. The A-coefficient is ((2.69735 (10^13))/(λ^3))((Subscript[S, M1])/4), where  (Subscript[S, M1])= | <(Subscript[p, 1/2])|| L + 2  S || (Subscript[p, 3/2 ])> \( | \^2\) The transition energy is 112.04 (cm^(-1)) and the corresponding wavelength in Angstrom is

λ =  (10^8)/112.04

\(TraditionalForm\`\(Ba\^+\)\) 5  (Subscript[d, 3/2])state

The 5  (Subscript[d, 3/2]) state in \(TraditionalForm\`\(Ba\^+\)\) decays to the 6  (Subscript[s, 1/2]) ground state by E2 emission. The A-coefficient is ((1.1198 (10^18))/(λ^5))((Subscript[S, E2])/4), where  (Subscript[S, E2])= | <5  (Subscript[d, 3/2])|| (Q^((2))) || 6  (Subscript[s, 1/2 ])> \( | \^2\) The transition energy is 4873.85 (cm^(-1)) and the corresponding wavelength in Angstrom is

λ =  (10^8)/4873.85

Prob 5

H 2s decay

The only  state in H below the 2  (Subscript[s, 1/2]) state is the 1  (Subscript[s, 1/2]) state.  The multipolarity J of the transition is limited by the angular momentum selections to 0= |1/2-1/2|⩽ J ⩽1/2+1/2 = 1. Therefore, since the parity of the two states is relatively even, only an M1 transition is possible. Since the nonrelativistic M1 amplitude is nonvanishing only for (Subscript[n, a])=(Subscript[n, b]), the rate  (Subscript[A, 1  s 2  s]^M1) vanishes and the lifetime to single-photon decay is infinite. (This state actually decays to the ground state by a two-photon transition and has a lifetime of about 1/7 s.)

He (1s2s) \(TraditionalForm\`\(\(\ \^1\)\(S\_0\)\)\)

This state cannot decay by single-photon emission to the  (((1  s))^2) \(TraditionalForm\`\(\(\ \)\(\^1\)\) S\_0\) ground state, since the triangle relation would require multipolarity J=0!  (The state actually decays by a two-photon transition with lifetime 19.6 ms.)

He  (1s2s) \(TraditionalForm\`\(\(\ \^3\)\(S\_1\)\)\)

This state could, in principle,  decay by single-photon M1 emission to the  (((1  s))^2) \(TraditionalForm\`\(\(\ \)\(\^1\)\) S\_0\) ground state. However, because of the requirement (Subscript[S, F])=(Subscript[S, I]) for transitions between two particle states, the M1 rate is forbidden nonrelativistically. The two-photon rate is very small 3.17×(10^(-9))(s^(-1)).  The dominant decay mode is through relativistic corrections to the M1 rate. The corresponding lifetime is about 7900 s.