ProblemSet7

Prob 2  Li  Isotope Shift

Specific Mass Shift

We use the formula  T = - (∑_a) (([v])^(-1)) <v|| C1 \(TraditionalForm\`\(\(||\)\(\(a\)\( > \^2\)\)\)\)  (Pva^2) to evaluate the the SMS. Since Li has only one core shell  (((1  s))^2), the sum reduces to a single term. Moreover, since the reduced matrix element <v|| C1 ||a> vanishes between two s states, the SMS vanishes for the 2s state. (This statement is true only in the independent particle approximation and is modified when correlation corrections are considered.)

Introduce the Coulomb wave function Pa[r] for the 1s core orbital and Pv[r] for the 2p valence orbital. Check that the wave functions are properly normalized.

Pa[r_]  :=  2 Za^(3/2) r Exp[-Za r]

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

Na  =  
      Integrate[Pa[r]^2, {r, 0, Infinity}, 
        Assumptions  ->  Za > 0]

Na  =  
      Integrate[Pv[r]^2, {r, 0, Infinity}, 
        Assumptions  ->  Zv > 0]

Work out the valence-core matrix element of the momentum operator Pva

Pva  =  
      Integrate[ Pv[r] (D[Pa[r], r] - Pa[r]/r), {r, 0, Infinity}, 
        Assumptions  ->  {Za > 0, Zv > 0}]

Check the above by working out the matrix element in the reverse order Pav. This sould be the adjoint of the above.  Since we are ignoring the factor 1/i, the reverse order Pva ans Pav should have opposite signs.

Pav  =   Integrate[  Pa[r] (D[Pv[r], r] + Pv[r]/r), {r, 0, 
          Infinity}, Assumptions  ->  {Za > 0, Zv > 0}]

Substitute numerical values for Za and Zv

Za  =  3 - 5/16 // N

Zv  =  1  + 1/8 // N

Pva // N

Pav//N

Now evaluate the reduced matrix element <v|| C1 ||a>

RME  =  - 
          Sqrt[3] ThreeJSymbol[{1, 0}, {1, 0}, {0, 0}]

Put the above together to obtain the SMS (up to a factor 1/Ma)

T1  =  - N[(3^(-1)) RME^2 Pva^2]

Clear[Za, Zv]

Now put in the isotope masses:  \(TraditionalForm\`\(\ \^7\)\)Li → 6.015 u and \(TraditionalForm\`\(\ \^7\)\)Li → 7.016 u.  Note that the electron mass is 0.00054858 u.  If we express electron mass in u, the conversion factor from a.u. to MHz  is 3.60948×(10^6)  For the 2p state, we find:

SMS6= -0.0405946 3.60948 (10^6) /6.015

SMS7= -0.0405946 3.60948 (10^6) /7.016

SMS67  =  SMS6 - SMS7

Normal Mass Shift

The energy of the 2s state in the NIST data base is 0; therefore there is no NMS for this state. The 2p state energy is 0.067905 a.u. The corresponding NMS is:

NMS6= -0.067905 3.60948 (10^6) /6.015

NMS7= -0.067905 3.60948 (10^6) /7.016

NMS67  =  NMS6 - NMS7

Total Isotope Shift

The resulting shift NMS+SMS in MHz is

ISMHz  =  SMS67 + NMS67

This value differs from the experimental result -10533 MHz primarily because the lowest-order calculation of the 2s SMS vanishes, whereas a higher-order MBPT calculation gives -1100 MHz.  Adding this to the previous result gives -10389 MHz, in much closer agreement with experiment.

To convert the above energies to (cm^(-1)), simply divide by c.  To convert energies expressed in (cm^(-1 ))to wavelengths, take the reciprocal.

c  =  2.997525 10^10

Thus, The isotope shift of the 2p-2s line in ((cm^(-1))) is

IScm  =  ISMHz 10^6/ c

The corresponding wavlength difference in (cm) is the inverse of the previous vakue

dλ  =  1/Abs[IScm]