NMR Shielding and Coupling Constants – Derivation

NMR Shielding and Coupling Constants – Derivation

APPENDIX H NMR Shielding and Coupling Constants – Derivation This section is for those who do not fully believe the author, and want to check whether...

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APPENDIX H

NMR Shielding and Coupling Constants – Derivation This section is for those who do not fully believe the author, and want to check whether the final formulae for the shielding and coupling constants in nuclear magnetic resonance (NMR) are indeed valid (Chapter 4).

Shielding constants Let us begin from Eq. (4.88).

Applying vector identities We are going to apply some vector identities1 in the operators Bˆ 3 , Bˆ 4 , Bˆ 5 . The first identity is u·(v × w) = v·(w × u) = w·(u × v), which simply means three equivalent ways of calculating the volume of a parallelepiped. This identity applied to Bˆ 3 and Bˆ 4 gives ˆ Aj e   IA · L γA , Bˆ 3 = 3 mc rAj A j e  Bˆ 4 = H · Lˆ 0j . 2mc

(H.1) (H.2)

j

Let us transform the term Bˆ 5 by using the following identity: (u × v) · (w × s) = (u · w) (v · s) − (v · w) (u · s). We obtain  IA × rAj e2    Bˆ 5 = γ = H × r · A 0j 3 2mc2 rAj A j     e2   1 γA [ H · IA )(r0j · rAj − r0j · IA )(H · rAj ] · 3 . 2 2mc rAj A j 1 The reader may easily check each of them.

627

628 Appendix H

Putting things together We are now all set to put all this baroque furniture into its destination, i.e., into Eq. (4.88) for E,  EA , (H.3) E = A

where EA stands for the contribution of nucleus A, i.e.,   EA = − γA ψ0(0) | (IA · H) ψ0(0) + 

  1 (0)  e2 (0) (H · IA )(r0j · rAj ) − r0j · IA )(H · rAj · 3 ψ0 + + γA ψ 0 | 2mc2 rAj j ⎛ ⎡ ⎞ ⎛ ⎞

 IA · Lˆ Aj  e2 (0) (0) ⎠ Rˆ 0 ⎝ γA ⎣ ψ 0 | ⎝ H · Lˆ 0j ⎠ ψ0 + + 3 2m2 c2 r Aj j j ⎛ ⎞ ⎛ ⎞ 

⎤   ˆ I · L A Aj ⎠ (0) ⎦ (0) ψ0 | ⎝ H · Lˆ 0j ⎠ Rˆ 0 ⎝ . ψ0 3 r Aj j j

Averaging over rotations The expression for EA represents a bilinear form with respect to the components of vectors IA and H, EA = ITA CA H, where CA stands for a square matrix2 of dimension 3 and IA and H are vertical three-component vectors. A contribution to the energy such as EA cannot depend on our choice of coordinate system axes x, y, z, i.e., on the components of IA and H. We will obtain the same energy if we rotate the axes (orthogonal transformation) in such a way as to diagonalize CA . The resulting diagonalized matrix CA,diag has three eigenvalues (composing the diagonal) corresponding to the new axes x  , y  , z . The very essence of averaging is that none of these axes are to be privileged in any sense. This is achieved by constructing the averaged matrix       1  CA,diag x  x  + CA,diag y  y  + CA,diag z z 3      ¯ A,diag   = C ¯ A,diag   = C ¯ A,diag   ≡ CA , = C xx yy zz 2 We could write its elements from the equation for E , but their general form will turn out to be nonnecessary. A

NMR Shielding and Coupling Constants – Derivation 629   ¯ A,diag  = δqq  CA for q, q  = x  , y  , z . Note that since the transformation was orwhere C qq thogonal (i.e., the trace of the matrix is preserved), the number CA may also be obtained from the original matrix CA as follows: CA =

    1   1  CA,diag x  x  + CA,diag y  y  + CA,diag z z = CA,xx + CA,yy + CA,zz . 3 3 (H.4)

Then the averaged energy E becomes E¯ =



¯ A,diag H = ITA C

A



CA (IA · H) .

A

Thus we obtain the sum of energy contributions over the nuclei, each contribution with its own coefficient averaged over rotations,3 

 2   2 e 1 (0) (0) (r0j · rAj ) 3 ψ0 ψ0 | (H.5) E¯ = − γ A IA · H 1 − 3 2mc2 r Aj A j ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ 

 2     ˆ ˆ LAj L 1 e Aj (0) ⎣⎝ (0) ˆ 0j ⎠ + ⎝ ⎠ Rˆ 0 ⎝ ⎠⎦ ψ ψ0 | − 2 2 L Lˆ 0j ⎠ Rˆ 0 ⎝ , 0 3 3 2m c 3 r r Aj Aj j j j j  3 Indeed, making C = 1 C A A,xx + CA,yy + CA,zz for the terms of Eq. (H.3) we have the following contribu3 tions (term by term):

• −γA 13 [1 + 1 + 1] = −γA ,

     1 ψ (0) + ψ (0) |  r · r 1 ψ (0) + ψ (0) |  r · r 1 ψ (0) = r · r j 0j Aj r 3 j 0j Aj r 3 j 0j Aj r 3 0 0 0 0 0 Aj  Aj Aj   2 (0) 1 e γ ψ (0) | j r0j · rAj r 3 ψ0 , 0 2mc2 A Aj        2 2 • − e 2 γA ψ0(0) | j 13 x0j xAj + y0j yAj + z0j zAj 13 ψ0(0) = − e 2 γA ψ0(0) | j 13 r0j · rAj 13 ψ0(0) ,



e2 γ 1 2mc2 A 3



(0) 

ψ0 |

2mc

2mc

rAj



rAj

 1 e2 1 γA × (0) (0) 3 2m2 c2 k E0 − Ek ⎤ ⎛ ⎡ ⎞



  Lˆ  Aj x  (0) (0) (0) (0) ⎠ψ ψk | + similarly y, z + cc⎦ = Lˆ 0j x ψ0 × ⎣ ψ0  ⎝ k 3 rAj j j ⎤ ⎛ ⎞ ⎡



ˆ Aj   1 ⎣ (0)  ⎝ L 1 1 e2 (0) (0) (0) ˆ 0j ψ ⎠ψ ψ0  ψk | + cc⎦ , γA × L k 0 (0) (0) 3 2m2 c2 3 r3 E −E +

k

0

k

j

Aj

where cc means the “complex conjugate” counterpart. This reproduces Eq. (H.5).

j

630 Appendix H with the matrix elements (Uˆ x , Uˆ y , Uˆ z ).

  ˆ U

kl

  ˆ (0) of the corresponding operators U ˆ = = ψk(0) |Uψ l

Finally, after comparing the formula with Eq. (4.81), we obtain the shielding constant for nucleus A (the change of sign in the second part of the formula comes from the change in the denominator) given in Eq. (4.89).

Coupling constants Averaging over rotations In each contribution on p. 323 there is a double summation over the nuclear spins, which, after averaging over rotations (similarly as for the shielding constant), gives rise to an energy  dependence of the kind A
We have



2  rAj · rBj (0) e (0) E¯ DSO = γ A γ B IA · IB ψ 0 | 3 3 ψ 0 2mc2 rAj rBj A,B j 

1 e2   (0) xAj xBj (0) − γA γB IA · IB ψ0 | 3 3 ψ0 + 3 2mc2 rAj rBj A,B j



  y z y z (0) Aj Bj (0) (0) Aj Bj (0) , ψ 0 | 3 3 ψ0 + ψ0 | 3 3 ψ0 rAj rBj rAj rBj

because the first part of the formula does not need any averaging (it is already in the appropriate form), the second part is averaged according to (H.4). Therefore, 

2  rAj · rBj (0) e (0) E¯ DSO = γ A γ B IA · IB ψ 0 | 3 3 ψ 0 . 3mc2 rAj rBj A,B j •

We have

  E¯ PSO = ψ0(0) |Bˆ 3 Rˆ 0 Bˆ 3 ψ0(0) = aver 

 × r I ie 2   × r I A Aj ˆ B Bl (0) (0) R 0 ∇l · γA γB ψ0 |∇j · ψ0 3 3 mc r r Aj Bl A,B j,l

= aver

NMR Shielding and Coupling Constants – Derivation 631 

 rAj × IA r × I ie 2   Bl B (0) (0) γA γB ψ0 |∇j · ψ0 = Rˆ 0 ∇l · 3 mc rBl r Aj A,B j,l aver !  "

  rAj r e 2   Bl (0) (0) ψ0 γA γB ψ0 |IA · ∇j × 3 Rˆ 0 IB · ∇l × , − mc rBl rAj A,B j,l

aver

where the subscript “aver” means the averaging of Eq. (H.4) and the identity A · (B × C) = (A × B) · C has been used. We have the following chain of equalities (involving4 the electronic momenta pˆ j and angular momenta LAj with respect to nucleus A, where j means electron number j ):     (0) 1  1  ie 2   (0) ˆ γA γB ψ0 |IA · = rAj × pˆ j R0 IB · rBl × pˆ l ψ0 mc i i aver A,B j,l    e 2       γA γB ψ0(0) |IA · rAj × pˆ j Rˆ 0 IB · rBl × pˆ l ψ0(0) = aver mc A,B j,l  e 2     (0) (0) γA γB ψ0 |IA · Lˆ Aj Rˆ 0 IB · Lˆ Bl ψ0 = aver mc A,B j,l   e 2      1  (0) ˆ (0) (0) (0) ψ0 |LAj,x Rˆ 0 Lˆ Bl,x ψ0 + ψ0 |Lˆ Aj,y Rˆ 0 Lˆ Bl,y ψ0 γ A γ B IA · IB mc 3 

A,B j,l

4 Let us have a closer look of the operator

in ∇j ×

 r ∇j × Aj 3 rAj

acting on a function (it is necessary to remember that ∇j

rAj rAj rAj 3 is not just acting on the components of r 3 alone, but in fact on r 3 times a wave function) f . We rAj Aj Aj

have ⎞







rAj

rAj





⎞ rAj

⎞ rAj

⎠ f = i ⎝∇j × ⎠ f + j ⎝∇j × ⎠ f + k ⎝∇j × ⎠ f= ⎝∇j × 3 3 3 3 rAj rAj rAj rAj x y z ⎛ ⎞ ∂ yAj ⎠ ∂ zAj − f + similarly with y and z = i⎝ ∂yj r 3 ∂zj r 3 Aj Aj x ⎛ ⎞ yAj zAj zAj ∂ yAj zAj yAj ∂ ⎠ f + similarly with y and z = i ⎝−3 4 + 3 +3 4 − 3 rAj rAj rAj ∂yj rAj ∂zj x ⎛ ⎛ ⎞ ⎞ zAj ∂ yAj ∂ zAj ∂ yAj ∂ ⎠ f + similarly with y and z = i ⎝ ⎠ f i⎝ 3 − 3 − 3 r ∂yj r ∂zj r 3 ∂yj r ∂zj Aj

Aj

x

Aj

Aj

  1  1  −rAj × pˆ j f = rAj × pˆ j f. +similarly with y and z = − i i

x

632 Appendix H

   (0) ˆ (0) ˆ Bl,z ψ + ψ0 |LAj,z Rˆ 0 L . 0

Thus, finally   1  e 2   (0) ˆ (0) ˆ ˆ E¯ PSO = R L γA γB IA · IB ψ0 |L ψ . Aj 0 Bl 0 3 mc A,B j,l



We have   (0) (0) E¯ SD = ψ0 |Bˆ 6 Rˆ 0 Bˆ 6 ψ0 = aver #    $ N   ˆ I · r · r s ˆ s · I j Aj A Aj j A (0) γA γB ψ0 | −3 × γel2 3 5 rAj rAj j,l=1 A,B # $

  sˆl · rBl (IB · rBl ) ˆ · I s l B (0) −3 = Rˆ 0 ψ0 3 5 rBl rBl aver #  $   N   sˆj · rAj xAj sˆj,x 1 (0) 2 γ A γ B IA · IB ψ0 | 3 − 3 γel × 5 3 r r Aj Aj j,l=1 A,B # $

  sˆl · rBl (xBl ) ˆ s l,x (0) Rˆ 0 3 − 3 ψ0 + 5 rBl rBl # $ # $

     sˆj · rAj yAj sˆl · rBl (yBl ) sˆj,y sˆl,y (0) (0) Rˆ 0 3 − 3 ψ0 + ψ0 | 3 − 3 5 5 rAj rAj rBl rBl # $ # $

      sˆj · rAj zAj sˆl · rBl (zBl ) sˆj,z ˆ s l,z (0) (0) Rˆ 0 3 − 3 ψ0 . ψ0 | 3 − 3 5 5 rAj rAj rBl rBl Therefore, #  $   N   sˆj · rAj rAj sˆj 1 (0) 2 γ A γ B IA · IB ψ 0 | 3 − 3 × E¯ SD = γel 5 3 rAj rAj j,l=1 A,B # $

  sˆl · rBl (rBl ) ˆ s l (0) Rˆ 0 3 − 3 ψ0 . 5 rBl rBl

NMR Shielding and Coupling Constants – Derivation 633 •

We have

  E¯ FC = ψ0(0) |Bˆ 7 Rˆ 0 Bˆ 7 ψ0(0) =     (0)  (0) γA γB ψ0 |δ rAj sˆj · IA Rˆ 0 δ (rBl ) sˆl · IB ψ0 γel2

aver

j,l=1 A,B

γel2

 j,l=1 A,B

=

  1  (0)   (0) ψ0 |δ rAj sˆj,x Rˆ 0 δ (rBl ) sˆl,x ψ0 + γ A γ B IA · IB 3

         ψ0(0) |δ rAj sˆj,y Rˆ 0 δ (rBl ) sˆl,y ψ0(0) + ψ0(0) |δ rAj sˆj,z Rˆ 0 δ (rBl ) sˆl,z ψ0(0) . Hence, 1 E¯ FC = 3



8π 3

2

γel2



   (0)  (0) γA γB IA · IB ψ0 |δ rAj sˆj Rˆ 0 δ (rBl ) sˆl ψ0 .

j,l=1 A,B

The results mean that the coupling constants J are such as reported on p. 324.