FORBIDDEN TRANSITIONS IN EPR

Professor P. T. ManoharanDept. 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 Sz – gn n 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 = 8/3 g gn 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 Sz - gn n B Iz + a S. I = g B Sz - gn n B Iz + a Sz Iz + a [Sx Ix + Sy Iy]

Basis functions

e n 2 = | e n 3 = | e n 4 = | e n

Transition probability for an EPR transition is

Pmn = /ħ2 | < n |V| m> |2 g()

For an EPR transition V(t) = (g B1 Sx) cos t = 2v cos t

Pmn = 2/ħ2 g2 2 B12 | < n | Sx| m > |2 g()

where Sx = ½ (S+ + S-)

Typical matrix elements would be< en | Sx | en > = < e | ½ [S+ + S-] e > < e | n> = ½ < e | S+ | e > = ½

i.e mS = ± 1 mI = 0

Transition Probability P = /2ħ2 g22B2g() .

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 en en

e n 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(gB + gn nB) i.e., V = 2g B1Sz i.e., 2|Sz|3 = (e n + e n) |Sz| (en - e n) = en | Sz | en - = -

i.e., P = 2 2g22B12g()

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. 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 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 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 – Sy

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 B0.

Since | + > = cos | 1> + sin |-1> | 0 > = | 0 > tan 2 = E / gB0

| - > = sin | 1> - cos |-1>

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.

2D

D

Synder & Zager, JCP (1964) 41 1763

Triphenyl Benzene dianion(Frozen soltion)

free radical

gyy

C10D8

gzz

gxx

Cu- Cu

ms ±2

ms= ±2

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

System Range of T studied 2J (cm-1)

Cu(II)/Zn(II)

Ag(II)/Zn(II)

150 – 300K

180 – 300K

+40.2

+62.5

‘Forbidden’ Transitions in Nuclear Hyperfine Structure.

I 1

Pure | ms, mI states

g

3A

Complete Hamiltonian

With S = ½, I = 3/2 (eg 63, 65 Cu 2+),

the eigen spinstates are |M,mWhere 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 = . g. S + S. A. I + I. P. I – n 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)]

Complex Q’ value (in 10-4cm-1)

Co(NH4)2(SO4)26H2O -0.2 0.05Na4CoPTS -0.2 0.1

Tutton Salt(theoretical estimation)

Co(BPT)

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/rL

3

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 < ¾Usually, Bmin = 2.0023/gmin [B2

0/4 –(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

(eg) Ni (S = 1)

Total transition probability for a DQ transition in a 3 level system = Pik = Pij x Pjk

If D/h increases, the resonances may be off the available magnetic field, it is necessary to work at higher fields

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 .

The angle 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.

s = 1, I = 1,

As increases, mixing of the spin states increases resulting in hyperfine (allowed) lines

H = g SZ – gnnBIz + AIz. Sz

E = gms – g. 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 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

Larger the gn better the resolution

Hence only nuclei like,1H (gn = 5.585)P9F (gn = 5.26)7Li (gn = 2.17)

(eg) 2D(gn = 0.857) gn 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 ath = g B + D (2Sz –1) gn n B + ½ (Azz 2P) (2Iz – 1) ½ Azz (2Sz – 1)

Main lines occur at h = g 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 2 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 2 / B2 reff6 ; ri = n1/6 reff

sat = satellite line main= main line

in the case of n nearby nuclei

r can be calculated by two different methods

From Intensity Isat

Imain

n= 9/8 g22 sin2i cos2 i/B2ri

i = 1From spacings

(E)2 = (gnnB)2 + (3/4 ggn n/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