FORBIDDEN TRANSITIONS IN EPR Professor P. T. Manoharan Dept. of Chemistry and RSIC I I T- Madras Chennai- 600 036 India Zeeman Hamiltonian (Hydrogen atom as an example) The electron and the nucleus both interact with the steady magnetic field H0 = g b B Sz – gn bn B Iz Isotropic Hyperfine Coupling is introduced into the Hamiltonian to take care of the interaction between the magnetic moments of electrons and the nucleus. H1 = a I. S = a (Ix Sx + Iy Sy + Iz Sz) where a = 8p/3 g b gn bn | (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 B Sz - gn bn B Iz + a S. I = g b B Sz - gn bn B Iz + a Sz Iz + a [Sx Ix + Sy Iy] Basis functions 1 = | e n 2 = | e bn 3 = | be n 4 = | be bn Transition probability for an EPR transition is Pmn = p/ħ2 | < n |V| m> |2 g() For an EPR transition V(t) = (g b B1 Sx) cos t = 2v cos t Pmn = 2p/ħ2 g2 b2 B12 | < n | Sx| m > |2 g() where Sx = ½ (S+ + S-) Typical matrix elements would be < en | Sx | ben > = < e | ½ [S+ + S-] be > < e | n> = ½ < e | S+ | be > = ½ i.e mS = ± 1 mI = 0 Transition Probability P = p/2ħ2 g2b2B2g() . Consider now Second-order EPR levels of H-atom, inclusive of a(SxIx+ SyIy) term in the Hamiltonian Second-order ESR Spectrum/ Forbidden ESR transition Note that ebn ben be bn en Strictly forbidden due to the fact mI = 1 This transition now weakly allowed,if the oscillating field is polarized parallel to B0 (not perpendicular as in allowed transition) due to 2nd – order improved wave-functions. Mixing coefficient, = a/2(gbB + gn bnB) i.e., V = 2g bB1Sz i.e., 2|Sz|3 = (e bn + be n) |Sz| (ben - e bn) = ebn | Sz | ben - =- i.e., P = 2 2g2b2B12g() will be small at high field 2 negligible Triplet States, S = 1 It is now possible to observe ms = 2 transitions (in addition to the normal ms = 1 transitions) Under Cubic field (or isotropic samples) ms = 2 cannot be observed by a microwave field of any polarisation and is strictly forbidden Axial crystal field H = b B. g. S + D [Sz2 – 1/3 S(S + 1)] B0 || z “Diagonal Hamiltonian” gives D term Behaviour Similar to Cubic field. But B0 || from z-axis, states gets mixed; forbidden (ms = 2 ) transitions occur. Similarly Bo || 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 B and D (of course, a function of ) When B1 B0 0 + and - 0 allowed. B1 || B0 - + allowed since < - | Sx | + > = 2 [ a- b+ - a+ b- ] In strong field, g b B0 >> D, a+ = a- and b- = b+, Forbidden transition intensity is zero 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 (Sx2 – Sy2) E (S+2 + S-2) is now added to the spin Hamiltonian, which mixes the states | -1> and | +1> irrespective of the direction of B0. Since | + > = cos | 1> + sin |-1> |0>=|0> | - > = sin | 1> - cos |-1> tan 2 = E / gbB0 Here, even when B1' || B0, the transition | - > | + > is allowed. Generally i.e Ms = ± 1 allowed when B’ B0 Ms = ± 2 may be allowed when B1’ || B0 If basic states are sufficiently mixed. General form of transition probability (e.g) S = 3/2 in an axial or rhombic symmetry. B0 at an arbitrary angle to the crystall axis General form of the eigen states | n > = an | 3/2 > + bn | ½ > + cn | ½ > + dn | -3/2> With n = 1, 2, 3, or 4. It is possible to have more than one “Forbidden Transitions” in terms of “Ms = 2” within the rigours of normal quantum numbers [described by the major component of an, bn, cn, dn at a moderate external field. Synder & Zager, JCP (1964) 41 1763 D Triphenyl Benzene dianion(Frozen soltion) 2D free radical gyy C10D8 gxx gzz ms ±2 ms= ±2 Cu- Cu Ag - Ag 2D IT (h/kT) 1/[1+exp(-(D+h)/kT) + exp(-2h/kT) + exp(-(2J+1/3D+h/kT)] x (DT/h)2 Ddip(cm-1) = 0.433gz2/r3 2J (cm-1) System Range of T studied Cu(II)/Zn(II) 150 – 300K +40.2 Ag(II)/Zn(II) 180 – 300K +62.5 ‘Forbidden’ Transitions in Nuclear Hyperfine Structure. I1 Pure | ms, mI states g 3A 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 = ai|M, 3/2 + bi|M, 1/2 + ci|M, -1/2 + di|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 mI= 1 as against the EPR allowed mI= 0. Similarly the quadrupolar operators I2 mix the functions to cause “Forbidden transition with” mI= 2 H = bB. g. S + S. A. I + I. P. I – bn B. gn . I. The quadrupole part of the Hamiltonian can then be expressed as H EQ =[ e2q Q/ 4I (2I –1)] [3Iz2 – 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 = (Vxx – Vyy)/Vzz If the electric field gradient has axial symmetry, then = 0, Since Vxx = Vyy, and HEQ Becomes. HEQ = e2qQ/4I(2I – 1) [3Iz2 - I(I + 1)] Q’ value (in 10-4cm-1) Complex Co(NH4)2(SO4)26H2O Na4CoPTS Tutton Salt (theoretical estimation) Co(BPT) -0.2 0.05 -0.2 0.1 0.8 +1.60 0.05 Q’ = 3e2qQ/84 and Q’’ = (e2qQ/84) (/2) = (Q’/3) (/2) From Q’ = 1.6 x 10-4cm-1 and Q’’ = 0.1 x 10-4cm-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) qval + e(1-) qlig where (1 – R) and (1 - ) are the Sternheimer antishielding factors. Qval = nj 3 cos2j - 13d rj-3 3d Qlig = nLi 3cos2Li – 1/rLi3 + ZL3cos2 - 1/rL3 Q’ = (3e2 Q/84) (1 – R) nj 3cos2j - 13d rj-33d Double Quantum Transition and Regular Half-field Transition Involves ms = 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 ms = 1 transition) Occur (in S = 1) systems when D/h < ¾ Usually, Bmin = 2.0023/gmin [B20/4 –(D’)2 / 3 – (E’)2]12 BDq = 2.0023/gav [(B02 – (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 If D/h increases, the resonances may be off the available magnetic field, it is necessary to work at higher fields (eg) Ni (S = 1) Total transition probability for a DQ transition in a 3 level system = Pik = Pij x Pjk 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 Be+ and Be- which in turn will orient with respect to the applied field. The resulting field will be along, say Bx and By, at an angle b. The angle b will be almost close to zero in the case of very weakly interacting nucleus, i.e B > Be and will be large in strongly interacting system ie. Be > B. It produces spin flip transition with simultaneous flip of electron and nuclear spins. Ms = 1, MI = 1, As b increases, mixing of the spin states increases resulting in hyperfine (allowed) lines H = g b B SZ – gn bnBIz + AIz. Sz E = gbBms – gb. B mI + ams mI 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 gn bn 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 Larger the gn better the resolution Hence only nuclei like, 1H (gn = 5.585) P9F (g = 5.26) n 7Li (g = 2.17) n (eg) 2D(gn = 0.857) gn bn 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 B + D (2Sz –1) gn bn B + ½ (Azz 2P) (2Iz – 1) ½ Azz (2Sz – 1) Main lines occur at h = g b B + D (2Sz –1) + AzzIz Depending on the number of interacting neighbours, complexities increase. Intensities of spin-flip Satellites n Isat / Imain = 9/8 g2 b2 cos2i sin2i / ri6 B2 i=1 for n different nuclei For an effective single proton in a randomly oriented system Isat / Imain = 3/20 g2 b2 / B2 reff6 ; ri = n1/6 reff sat = satellite line in the case of n nearby nuclei main= main line r can be calculated by two different methods From Intensity From spacings Isat Imain n = 9/8 g2b2 sin2i cos2 i/B2ri i=1 (E)2 = (gnbnB)2 + (3/4 ggn b bn/r3E)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. Thank You