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R.G. Griffin Winter School 2020 Page 1 Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in a electron nuclear DNP experiment, Overhauser is (gS/gI)~660. 1.1 Energy levels and transitions of a model four level system Consider the Hamiltonian with electron and nuclear Zeeman terms and a hyperfine term of magnitude A. The hyperfine term couples two spin-1/2 particles I=1/2 and S=1/2. We assume that and therefore with mS=±1/2 and mI=±1/2 we obtain resonance frequencies and the transitions given below. NMR transtions EPR transitions In the absence of applied alternating fields the populations are in thermal equilbrium where is the transition rate from state à and ’s are the populations of the two states. This can be rewritten in terms of Boltzmann factors as H = γ S !B 0 S Z γ I !B 0 I Z + A ! I i ! S γ S !B 0 > A and γ S >> γ I H = γ S !B 0 S Z γ I !B 0 I Z + AI Z S Z ω S = γ S B 0 + A ! m I and ω I = γ I B 0 A ! m S W εη , ε ' η' P εη = W ε ' η', εη P ε ' η' W εη , ε ' η' εη ε ' η' P ij
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Page 1: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 1

Notes on Dynamic Nuclear Polarization

1. Overhauser Effect The following simple derivation shows that the enhancement in a electron nuclear DNP

experiment, Overhauser is (gS/gI)~660.

1.1 Energy levels and transitions of a model four level system Consider the Hamiltonian with electron and nuclear Zeeman terms and a hyperfine term of magnitude A.

The hyperfine term couples two spin-1/2 particles I=1/2 and S=1/2. We assume that

and therefore

with mS=±1/2 and mI=±1/2 we obtain resonance frequencies and the transitions given below.

NMR transtions EPR transitions

In the absence of applied alternating fields the populations are in thermal equilbrium

where is the transition rate from state à and ’s are the populations of

the two states. This can be rewritten in terms of Boltzmann factors as

H = γ S!B0SZ − γ I!B0IZ + A!I i!S

γ S!B0 > A and γ S >> γ I

H = γ S!B0SZ − γ I!B0IZ + AIZSZ

ωS = γ SB0 +A!mI and ω I = γ IB0 −

A!mS

Wεη,ε 'η'Pεη =Wε 'η',εηPε 'η'

Wεη ,ε 'η' εη ε 'η' Pi j

Page 2: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 2

1.2 Overhauser effect (A. Overhauser, 1953 and Carver and Slichter, 1953) Now consider the same system subject to irradiation of one of the electron transitions, we, as illustrated below.

Under irradiation we can write rate equations for the populations

(1)

(2)

(3)

(4)

The probabilities sum to unity . We solve the equations in the steady state when the

driving field is strong and saturates the transition. Therefore,

and (4) yields

Wεη,ε 'η'Wε 'η',εη

=Pε 'η'Pεη

= e Eεη−Eεη⎛⎝⎜

⎞⎠⎟ / kT

dP1dt = P2W21−P1W12 + (P1−P2 )ωe

dP2dt = P1W12 −P2W21+P3W32 −P2W23 + (P1−P2 )ωe

dP3dt = P2W23 −P3W32 +P4W43 −P3W34

dP4dt = P3W34 −P4W43

Pi1

4

∑ =1

ωe >>W12,W21 1 → 2P1=P2

Page 3: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 3

which is thermal equilibrium for and . Equation (3) yields

Since and and and are in equilibrium. Therefore, and are in equilibrium. We define

Therefore,

and

.

The expectation value is

We evaluate this in the high temperature limit

and

.

Using

Therefore the Overhauser polarization is

P3 = P4W43W34

P3 = P2W23W32

Pj = PieEi−Ej⎛⎝⎜

⎞⎠⎟ / kT = piBi j

P1 = P2 P3 = P2B23 P4 = P2B24

P1 = P2 =1

2+B23 +B24P3 =

B232+B23 +B24

P4 =B24

2+B23 +B24

IZ = Pi i IZ1

4

∑ i = 12 P1+P2 −P3 −P4( )

IZ = 122−B23 −B242+B23 +B24

⎣⎢⎢

⎦⎥⎥

Bi j =1+Ei −Ej( )kT

IZ OV= 12

E3 −E2( )+ E4 −E2( )4kT

⎝⎜⎜

⎠⎟⎟

E3 −E2( ) = γ S!B0 +γ I!B0 E4 −E2( ) = γ I!B0 + A2

Page 4: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 4

.

In the absence of a saturating microwave field

Using

we find

and therefore,

If both EPR transitions are irradiated

which is the desired result.

IZ OV= 12

γ S!B0 + 2γ I!B0 +A2

4kT

⎜⎜⎜

⎟⎟⎟≈ 12

γ S!B04kT

⎝⎜

⎠⎟

IZ Boltz= 12

1+B12 −B23 −B241+B21+B23 +B24

⎣⎢⎢

⎦⎥⎥= 12

E2 −E1( )+ E3 −E2( )+ E4 −E2( )4kT

⎝⎜⎜

⎠⎟⎟

E2 −E1( ) = −γ S!B0 −A2

IZ Boltz= 12

γ I!B02kT

⎝⎜

⎠⎟

IZ OV

IZ Boltz

=γ S2γ I

IZ OV

IZ Boltz

=γ Sγ I

Page 5: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 5

2. Solid effect -- DNP with forbidden transitions (Jefferies and Abragam)

Similar arguments can be extended to the Solid Effect that is based on forbidden transitions.

The solid effect utilizes zero and double quantum transitions that require operators of the

form

However, in the DNP experiment we excite these nomally DQ and ZQ transitions with a microwave field. This is represented be . In a separate note we show how this is accomplished. Assume that we have admixtures of states which permit excitation of these transitions. We saturate the transition with microwaves . Thus, . Proceeding as before

+ −S+I− +S−I+ − + ≠ 0 + +S+I+ +S−I− − − ≠ 0

P2 = P1B12 P3 = P4B43

P1 = P4 =1

2+B12 +B43P2 =

B122+B12 +B43

P3 =B43

2+B12 +B43

IZ = 12B12 −B432+B12 +B43

⎣⎢⎢

⎦⎥⎥

IZ SE= 12

E1−E2( )− E4 −E3( )4kT

⎝⎜⎜

⎠⎟⎟

Page 6: 3 Notes on Dynamic Nuclear Polarization · 2020. 1. 6. · Notes on Dynamic Nuclear Polarization 1. Overhauser Effect The following simple derivation shows that the enhancement in

R.G. Griffin Winter School 2020 Page 6

Which is the desired result for the solid effect

E1 =12γ S!B0 −

12γ I!B0 −

A4 E2 = − 12γ S!B0 −

12γ I!B0 +

A4

E1−E2( ) = γ S!B0

E3 =12γ S!B0 +

12γ I!B0 +

A4 E4 = − 12γ S!B0 +

12γ I!B0 +

A4

E4 −E3( ) = −γ S!B0

IZ SE= 12

γ S!B02kT

⎝⎜

⎠⎟

IZ SE

IZ Boltz

=γ Sγ I


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