1. FORBIDDEN TRANSITIONS IN EPR Professor P. T. Manoharan Dept. of Chemistry and RSIC I I T- Madras Chennai- 600 036 India

2. Zeeman Hamiltonian (Hydrogen atom as an example) The electron and the nucleus both interact with the steady magnetic field H 0 = g  B S z – g n  n B I z Isotropic Hyperfine Coupling is introduced into the Hamiltonian to take care of the interaction between the magnetic moments of electrons and the nucleus. H 1 = a I. S = a (I x S x + I y S y + I z S z ) where a = 8  /3 g  g n  n |  (r) | 2 Anisotropic (dipolar) part of hfcc averages out to zero since the unpaired electron is present in an s-orbital So H = g  B S z - g n  n B I z + a S. I = g  B S z - g n  n B I z + a S z I z + a [S x I x + S y I y ] Basis functions    e  n   2 = |  e  n   3 = |  e  n   4 = |  e  n 

3. Transition probability for an EPR transition is P mn =  /ħ 2 | | 2 g(  ) For an EPR transition V(t) = (g  B 1 S x ) cos  t = 2v cos  t P mn = 2  /ħ 2 g 2  2 B 1 2 | | 2 g(  ) where S x = ½ (S + + S - ) Typical matrix elements would be = = ½ = ½ i.e  m S = ± 1  m I = 0 Transition Probability P =  /2ħ 2 g 2  2 B 2 g(  ).

5. Consider now Second-order EPR levels of H-atom, inclusive of a(S x I x + S y I y ) term in the Hamiltonian Second-order ESR Spectrum/ Forbidden ESR transition Note that  e  n   e  n  e  n   e  n Strictly forbidden due to the fact  m I =  1 This transition now weakly allowed,if the oscillating field is polarized parallel to B 0 (not perpendicular as in allowed transition) due to 2 nd – order improved wave-functions. Mixing coefficient, = a/2(g  B + g n  n B) i.e., V = 2g  B 1 S z i.e.,  2 |S z |  3  =  (  e  n +  e  n ) |S z | (  e  n -  e  n )  =   e  n | S z |  e  n  - = - i.e., P = 2 2 g 2  2 B 1 2 g(  ) will be small at high field  2 negligible

7. Triplet States, S = 1 It is now possible to observe  m s =  2 transitions (in addition to the normal  m s =  1 transitions) Under Cubic field (or isotropic samples)   m s =  2 cannot be observed by a microwave field of any polarisation and is strictly forbidden Axial crystal field H =  B. g. S + D [S z 2 – 1/3 S(S + 1)] B 0 || z  “Diagonal Hamiltonian”  gives D term Behaviour Similar to Cubic field.

8. But B 0 ||  from z-axis, states gets mixed; forbidden (  m s =  2 ) transitions occur. Similarly B o || x, we get the electron spin states |  +  = a + { | 1  + | -1  } + b + | 0  |  0  = 1/  2 { |1  - | -1  } |  -  = a - { | 1  + | -1  } – b - | 0  a/b depend on relative magnitudes of g  B and D (of course, a function of  ) When B 1  B 0  0   + and  -   0 allowed. B 1 || B 0  -   + allowed since =  2 [ a - b + - a + b - ] In strong field, g  B 0 >> D, a + = a - and b - = b +, Forbidden transition intensity is zero

10. Note this is effectively the forbidden transition  m = ± 2 and the axis of quantization is determined by the applied field. Rhombic Field The term E (S x 2 – S y 2 )  E (S + 2 + S - 2 ) is now added to the spin Hamiltonian, which mixes the states | -1 > and | +1 > irrespective of the direction of B 0. Since |  + > = cos  | 1 > + sin  |-1 > |  0 > = | 0 >tan 2  = E / g  B 0 |  - > = sin  | 1 > - cos  |-1 > Here, even when B 1 ' || B 0, the transition |  - >  |  + > is allowed. Generally i.e  M s = ± 1 allowed when B’  B 0  M s = ± 2 may be allowed when B 1 ’ || B 0 If basic states are sufficiently mixed.

11. General form of transition probability (e.g) S = 3/2 in an axial or rhombic symmetry. B 0 at an arbitrary angle to the crystall axis General form of the eigen states |  n > = a n | 3/2 > + b n | ½ > + c n | ½ > + d n | - 3/2 > With n = 1, 2, 3, or 4. It is possible to have more than one “Forbidden Transitions” in terms of “  M s =  2” within the rigours of normal quantum numbers [described by the major component of a n, b n, c n, d n at a moderate external field.

14. 2D D Synder & Zager, JCP (1964) 41 1763 Triphenyl Benzene dianion(Frozen soltion)

15. free radical g yy C 10 D 8 g zz g xx

18. Cu- Cu  m s ±2  m s = ±2 Ag - Ag

19. 2D

21. I T  (h /kT) 1/[1+exp(-(D+h )/kT) + exp(-2h /kT) + exp(-(2J+1/3D+h /kT)] x (D T /h ) 2 D dip (cm -1 ) = 0.433g z 2 /r 3 System Range of T studied 2J (cm -1 ) Cu(II)/Zn(II) Ag(II)/Zn(II) 150 – 300K 180 – 300K +40.2 +62.5

22. ‘Forbidden’ Transitions in Nuclear Hyperfine Structure.

23. I  1

24. Pure | m s, m I  states g 3A

25. Complete Hamiltonian With S = ½, I = 3/2 (eg 63, 65 Cu 2+ ), the eigen spinstates are |M,m  Where M =  1/2, m =  3/2,  ½ A sample function would be like  i |M,m  = a i |M, 3/2  + b i |M, 1/2  + c i |M, -1/2  + d i |M, -3/2  with M =  1/2 I   mixes by raising or lowering the m to m  1 causing a “Forbidden Nuclear Hyperfine Line” with  m I =  1 as against the EPR allowed  m I = 0. Similarly the quadrupolar operators I  2  mix the functions to cause “Forbidden transition with”  m I =  2

28. H = . g. S + S. A. I + I. P. I –  n B. g n. I. The quadrupole part of the Hamiltonian can then be expressed as H EQ =[ e 2 q Q/ 4I (2I –1)] [3I z 2 – I (I + 1) + ½  (I + 2 + I - 2 )] where eq is the electric field gradient at the nucleus eQ is the nuclear quadrupole moment  is the asymmetry parameter  = (V xx – V yy )/V zz If the electric field gradient has axial symmetry, then  = 0, Since V xx = V yy, and H EQ Becomes. H EQ = e 2 qQ/4I(2I – 1) [3I z 2 - I(I + 1)]

36. Complex Q’ value (in 10 -4 cm -1 ) Co(NH 4 ) 2 (SO 4 ) 2  6H 2 O -0.2  0.05 Na 4 CoPTS -0.2  0.1 Tutton Salt (theoretical estimation) Co(BPT) 0.8 +1.60  0.05 Q’ = 3e 2 qQ/84 and Q’’ = (e 2 qQ/84) (  /2) = (Q’/3) (  /2) From Q’ = 1.6 x 10 -4 cm -1 and Q’’ = 0.1 x 10 -4 cm -1, Hence asymmetry parameter,  = 0.375 The field gradient at the nucleus can be separated into valence and lattice contributions as eq = e(1 – R) q val + e(1-   ) q lig where (1 – R) and (1 -   ) are the Sternheimer antishielding factors. Q val =  n j  3 cos 2  j - 1  3d  r j -3  3d

37. Q lig =   n Li  3cos 2  Li – 1/r Li 3  +  Z L  3cos 2  - 1/r L 3  Q’ = (3e 2 Q/84) (1 – R)  n j  3cos 2  j - 1  3d  r j -3  3d

39. Double Quantum Transition and Regular Half-field Transition Involves  m s =  2 but it arise from the rapid consecutive absorption of two single Quanta i.e final state is reached via transition to an intermediate state. Requirement: Energy separations between the adjacent levels be equal Occurs at magnetic fields comparable to those for g = 2 However, half-field transitions (since they occur at a half field of the main  m s =  1 transition) Occur (in S = 1) systems when D/h < ¾ Usually, B min = 2.0023/g min [B 2 0 /4 –(D’) 2 / 3 – (E’) 2 ] 12 B Dq = 2.0023/g av [(B 0 2 – (D’) 2 / 3 – (E’) 2 ] 1/2 If D/h increases, the resonances may be off the available magnetic field, it is necessary to work at higher fields

40. (eg) Ni (S = 1) Total transition probability for a DQ transition in a 3 level system = P ik = P ij x P jk If D/h increases, the resonances may be off the available magnetic field, it is necessary to work at higher fields

42. Spin Flip Transitions Any nucleus in the environment of a unpaired spin will feel a magnectic field Be coming from the isotropic and dipolar fields. Depending on the orientation of the electron, the fields felt by the nucleus can be termed B e + and B e - which in turn will orient with respect to the applied field. The resulting field will be along, say B x and B y, at an angle . The angle  will be almost close to zero in the case of very weakly interacting nucleus, i.e B > B e and will be large in strongly interacting system ie. B e > B.

45. It produces spin flip transition with simultaneous flip of electron and nuclear spins.  s =  1,  I =  1, As  increases, mixing of the spin states increases resulting in hyperfine (allowed) lines H = g  S Z – g n  n BIz + AI z. S z E = g  m s – g . B m I + am s m I In otherwords, the second term is overtaken by the third term. The spin flip lines occur on either side of the EPR lines with an energy separation of g n  n B hence Better resolution at higher frequencies(say Q - band) But Higher Intensitiesat lower fields. (say s - band) Medium freq i,.e X-band most appropriate

46. Larger the g n better the resolution Hence only nuclei like, 1 H (g n = 5.585) P9 F (g n = 5.26) 7 Li (g n = 2.17) (eg) 2 D(g n = 0.857) g n  n B ~ 2.9G at X band and with in the linewidth and its spin – flips cannot be observed Systems with S>1/2 and I>1/2 give rise to zero field splitting due to electrons and quadrupole coupling constants due to interaction between the electric field gradient and quadrupole moment of the nucleus. Satellite lines occur at h = g  B + D (2S z –1)  g n  n B + ½ (A zz  2P) (2I z – 1)  ½ A zz (2S z – 1) Main lines occur at h = g  B + D (2S z –1) + A zz I z Depending on the number of interacting neighbours, complexities increase.

47. Intensities of spin-flip Satellites n I sat / I main = 9/8  g 2  2 cos 2  i sin 2  i / r i 6 B 2 i = 1 for n different nuclei For an effective single proton in a randomly oriented system I sat / I main = 3/20 g 2  2 / B 2 r eff 6 ; r i = n 1/6 r eff sat = satellite line main= main line in the case of n nearby nuclei

48. r can be calculated by two different methods From Intensity I sat I main n = 9/8  g 2  2 sin 2  i cos 2  i/B 2 r i i = 1 From spacings (  E) 2 = (g n  n B) 2 + (3/4 gg n   n /r 3  E ) 2  E is the average distance to all of the nearest matrix nuclei interaction with the electron spins. Possibility of Errors (i)At x-band, the limit of weak mixing and high Field approximation assumed in this equation does not hold good and may introduce an error. (ii) Insufficient resolution at x-band frequency (iii) Contribution to the satallite intensity from the hyperfine coupling.

59. Thank You