Answers to Exam IIa (Nov.  2005)

Prob 1

LS coupling

For LS coupling of identical particles, we have the restriction L+S is even.  As a first step, couple two particles
(2s2s) → \(TraditionalForm\`\^1\)S   (even parity)
(2s2p) → \(TraditionalForm\`\^1\)P,  \(TraditionalForm\`\^3\)P   (odd-parity)                          
(2  p2p) (→ ^1)S,  \(TraditionalForm\`\^3\)P,  \(TraditionalForm\`\^1\)D   (even-parity)   
                
For an even parity 3-particle state  couple two 2p states (2p2p)
then couple the resulting state with the 2s  (2p2p)[LS]2s 
The possible combinations are
(2p2p) [ \(TraditionalForm\`\^1\)S]2s  \(TraditionalForm\`\^2\)S    2×1 substates
(2p2p) [ \(TraditionalForm\`\^3\)P]2s  \(TraditionalForm\`\^2\)P     2×3 substates
(2p2p) [ \(TraditionalForm\`\^3\)P]2s  \(TraditionalForm\`\^4\)P     4×3 substates
(2p2p) [ \(TraditionalForm\`\^1\)D]2s  \(TraditionalForm\`\^2\)D   2×5 substates
Total number of substates = 2+6+12+10=30

jj coupling

For jj coupling of identical particles J is even. The (2p2p) states are
(3  (Subscript[p, 1/2]) 3  (Subscript[p, 1/2])) → [0]                            
(3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) → [1], [2]                     
(3  (Subscript[p, 3/2]) 3  (Subscript[p, 3/2])) → [0], [2]
      
Combining these with a 2s state  leads to

 [( 3  (Subscript[p, 1/2]) 3  (Subscript[p, 1/2])) [0] 2  (Subscript[s, 1/2])]    [1/2]                          2                  substates
 [( 3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) [1] 2  (Subscript[s, 1/2])]    [1/2]  & [3/2]           2 + 4  = 6    substates
 [( 3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) [2] 2  (Subscript[s, 1/2])]    [3/2]  & [5/2]           4 + 6  = 10  substates 
 [( 3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) [0] 2  (Subscript[s, 1/2])]    [1/2]                          2                  substates
 [( 3  (Subscript[p, 1/2]) 3  (Subscript[p, 3/2])) [2] 2  (Subscript[s, 1/2])]    [3/2]  & [5/2]           4 + 6  = 10  substates 
  Total number of substates = 2+6+10+2+10=30

Prob 2

The  energy of the (1s3p) state  is 
E[ (1s3p) \(\^3\)P] =  (Subscript[ϵ, 1  s])+(Subscript[ϵ, 3  p]) + (Subscript[R, 0])(1s3p1s3p) - (1/3)(Subscript[R, 1])(1s3p3p1s) -  (Subscript[U, 3  p3p]) 
where U = (1/r) accounts for the screening of the 3p (or 2s) electron. 
The energy of the (1s2s) state is
 E[ (1s2s) \(TraditionalForm\`\(\(\^3\)\(S\)\)\)] =  (Subscript[ϵ, 1  s])+(Subscript[ϵ, 2  s]) + (Subscript[R, 0])(1s2s1s2s) - (Subscript[R, 0])(1s2s2s1s) -  (Subscript[U, 2  s2s]).
As a first step we need lowest order energies and Coulomb wave functions:

ϵ1s  =  -Z^2/2

Let Zs = Z-1 be the screened charge.

ϵ2s  =  -Zs^2/(2 2^2)

ϵ3p  =  -Zs^2/(2 3^2)

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

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

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

Now evaluate the screening potentials and Slater Integrals

v01s[
        r_]  :=   Integrate[P1s[x]^2, {x, 0, r}, 
            Assumptions ->  Z > 0]/r  + 
        Integrate[P1s[x]^2/x, {x, r, Infinity}, 
          Assumptions ->  Z > 0]

v01s2s[r_]  :=  
      Integrate[P1s[x] P2s[x], {x, 0, r}, 
            Assumptions ->  {Z > 0, Zs > 0}]/r  + 
        Integrate[P1s[x] P2s[x]/x, {x, r, Infinity}, 
          Assumptions ->  {Z > 0, Zs > 0}]

v11s3p[r_] := 
      Integrate[P1s[x] P3p[x] x, {x, 0, r}, 
            Assumptions ->  {Z > 0, Zs > 0}]/
          r^2  + r Integrate[
            P1s[x] P3p[x]/x^2, {x, r, Infinity}, 
            Assumptions ->  {Z > 0, Zs > 0}]

G01s2s  :=  
      Integrate[P2s[r]^2 v01s[r], {r, 0, Infinity}, 
        Assumptions ->  {Z > 0, Zs > 0}]

G01s3p  :=  
      Integrate[P3p[r]^2 v01s[r], {r, 0, Infinity}, 
        Assumptions ->  {Z > 0, Zs > 0}]

F01s2s  :=  
      Integrate[P2s[r] P1s[r] v01s2s[r], {r, 0, Infinity}, 
        Assumptions ->  {Z > 0, Zs > 0}]

F11s3p  :=  
      Integrate[P3p[r] P1s[r] v11s3p[r], {r, 0, Infinity}, 
        Assumptions ->  {Z > 0, Zs > 0}]

Now evaluate the background potentials

U2s := Integrate[
        P2s[r]^2 /r , {r, 0, Infinity}, 
        Assumptions ->  Zs > 0]

U3p := Integrate[
        P3p[r]^2 /r , {r, 0, Infinity}, 
        Assumptions ->  Zs > 0]

Put things together: E[ (1s3p) \(\^3\)P] =  (Subscript[ϵ, 1  s])+(Subscript[ϵ, 3  p]) + (Subscript[R, 0])(1s3p1s3p) - (1/3)(Subscript[R, 1])(1s3p3p1s) -  (Subscript[U, 3  p3p])

e1s3p  =  ϵ1s +  ϵ3p  +  G01s3p  -  
        F11s3p/3  - U3p

e1s3p  =  
      e1s3p /.  {Z ->  5,  Zs ->  4} // N

E[ (1s2s) \(TraditionalForm\`\(\(\^3\)\(S\)\)\)] =  (Subscript[ϵ, 1  s])+(Subscript[ϵ, 2  s]) + (Subscript[R, 0])(1s2s1s2s) - (Subscript[R, 0])(1s2s2s1s) -  (Subscript[U, 2  s2s]).

e1s2s  =  ϵ1s +  ϵ2s  +  G01s2s  -  F01s2s  - 
        U2s

e1s2s  =  
      e1s2s  /.  {Z ->  5,  Zs ->  4} // N

ΔE  =  e1s3p-e1s2s

Energy in (cm^(-1))

decm  =  219474.62 ΔE

Wavelength in Angstrom

λ  =  (10^8)/decm

The NIST table  gives 337.07 Angstrom so the  difference with NIST is only 0.5 %

Prob 3

The reduced matrix element (Subscript[R, E1]) =<(1  s2s)\(TraditionalForm\`\(\ \^2\)\)S || r || (1s3p) \(TraditionalForm\`\(\(\ \^3\)\(\(P\)\(>\)\)\)\) = (Sqrt[3]) <0 ||C1||1> (Subscript[R, 1  s 3  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.

Rsp  =  
      Integrate[P3p[r] r P2s[r], {r, 0, Infinity}, 
        Assumptions ->  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[3] cps Rsp

The associated line strength is

SE1  =  RE1^2

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 (9 for a triplet P state) and λ= 335.269 from previous problem

A = ((2.02613 (10^18))/(λ^3)) (SE1/(Subscript[g, a])) /. {Zs -> 2,  (Subscript[g, a]) ->  9}

Prob 4

5  (Subscript[d, 5/2])

This state decays to the ground state by M1 emission 
λ = 62374 Angstrom

4  (Subscript[f, 5/2]) &  4  (Subscript[f, 7/2])

These states decay by E1 emission to the 5d levels
4  (Subscript[f, 5/2]) → 5  (Subscript[d, 3/2])  λ = 13898 Angstrom
4  (Subscript[f, 5/2]) → 5  (Subscript[d, 5/2])  λ = 17883 Angstrom
4  (Subscript[f, 7/2]) → 5  (Subscript[d, 5/2])  λ = 14100 Angstrom

6  (Subscript[s, 1/2])

The 6s decays to the lower 5d levels b E2 emission
6  (Subscript[s, 1/2]) → 5  (Subscript[d, 3/2])  λ = 7358  Angstrom
6  (Subscript[s, 1/2]) → 5  (Subscript[d, 5/2])  λ = 8342  Angstrom

Prob 5

The only lower state in Be is the \(TraditionalForm\`\(\(\^1\)\(S\_0\)\)\) ground state.  

The J=0 sublevel cannot decay by single-photon emission since the angular momentum triangle relations require  |0-0| ≤ J ≤ 0+0 and there is no multipole with J=0!.  

The J=1 sublevel could decay to the ground state by E1 emission; however, for this transition (Subscript[S, F])≠(Subscript[S, I]) so the transition is forbidden nonrelativistically.

The J=2 sublevel could decay by M2 emission.  As above, (Subscript[S, F])≠(Subscript[S, I]), so the transition is forbidden nonrelativistically.