Some problems still need clarification.  I will update them once we ask professor Cory.

Dynamics


$s(t)=e^{-t/T_{2}}\int P(r)e^{-i\int^{t}_{0}\omega(r,t')dt'}dr$


ω(r,t') = resonant frequency
P(r) = probability distribution


Periodic

Frequency that an arbitrary location will see

$\omega(t) = \gamma r \frac{\partial B_{z}}{\partial x} cos(\omega _{s} t + \phi)$


$exp(i\int^{t}_{0}\omega(t')dt'=exp(i[\gamma \frac{\partial B_{z}/\partial x}{\omega_{s}}r sin(\omega_{s}t+\phi])$


$exp^{iRsin\alpha}=\sum J_{k}(R)e^{ik\alpha}$

for one location in the sample

Static Spectrum

Problem 1


Nuclear Spin


$H_{z}=\omega _{0}I_{z}$


$H_{cs}=-\omega _{0}\sigma I_{z}$


PAS (Principle axis system) = coordinate system that leave the molecule in diagonal ??

ω in transverse plane (slow) can be suppressed if rotation around z-axis is fast


$\sigma _{z} \sigma _{z}'$

= secular part of the chemical shift, lead to small rotation in x-y direction

Problem 2


$\sigma = \sigma_{iso} + (\frac{\sigma}{2})(3 cos^{2}\theta -1)- \frac{\delta^{eta}}{4}sin^{2}\theta(e^{i2\phi}+e^{-i2\phi})$


$\sigma_{iso}=(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})/3$


$\delta=\frac{2}{3}\sigma_{zz}-\frac{1}{3}(\sigma_{xx}+\sigma_{yy})$


$\eta=3(\sigma_{yy}-\sigma_{xx})/2(\sigma_{zz}-\sigma_{xx}-\sigma_{yy})$



$< \sigma > = \sigma _{iso}$


It average out any non-isometric parts, so we have a homogeneous sample. So the result does not depend on the orientation of the sample.

When η = 0 -> < 3cos(θ)^2 -1 > = 0, average over sphere


Decoherence

Bloc = field that a test spin would see (every spin averagely see the same distribution of B)

average vector still pointing along y => |Bloc> of time or ensemble = 0

Problem 3


Carl-Purcell Sequence

Problem 4


Chemical Exchange

let

Problem 5


Slow Exchange

choose Δ ≥ τ exchange, Δ << T1, Δ > T2

Problem 6


$e^{i\omega_{A}t_{1}}e^{i\omega_{A}t_{2}} , e^{i\omega_{A}t_{1}}e^{i\omega_{B}t_{2}} , e^{i\omega_{B}t_{1}}e^{i\omega_{A}t_{2}} , e^{i\omega_{B}t_{1}}e^{i\omega_{B}t_{2}}$


then do phase cycle and collect data set


$cos(\omega_{A/D}T_{1})e^{i\omega_{A/D}t_{2}} , sin(\omega_{A/D}T_{1})e^{i\omega_{A/D}t_{2}}$


Then we get pure absorptive line-shape