Transcript Document

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
H 0 = g b B Sz – g n b 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.
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
< en | Sx | ben > = < 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  ben
be bn  en 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| (ben -  e bn)
=  ebn | Sz | ben - 
=-
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.
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,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 I2  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)26H2O
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 cos2j - 13d rj-3 3d
Qlig =   nLi 3cos2Li – 1/rLi3 +  ZL3cos2 - 1/rL3
Q’ = (3e2 Q/84) (1 – R)  nj 3cos2j - 13d rj-33d
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 cos2i sin2i / 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 sin2i cos2 i/B2ri
i=1
(E)2 = (gnbnB)2 + (3/4 ggn b bn/r3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.
Thank You