Measuring  T1 in NMR, we apply follow  pulse sequence:

\pi_x \longrightarrow \tau \longrightarrow (\pi/2)_x

according to previous post on density matrix in operator form, we can evaluate the polarization. suppose the magnetization is pointing with the external B-field.

\rho_R = \rho = k\sigma_z

\downarrow \pi_x

\rho_R = - k \sigma_z

\downarrow \tau

\rho_R = -k \left( 1- 2 e^{-\tau/T_1} \right) \sigma_z

\downarrow (\pi/2)_x

\rho_R = k \left( 1- 2 e^{-\tau/T_1} \right) \sigma_y

in lab frame:

\rho = k \left( 1-2 e^{-\tau/T_1} \right) (\sigma_y cos(\omega_0 t) - \sigma_x sin(\omega_0 t) )

if the NMR coil is placed alone with x-axis. the magnetization is proportional to :

\left<\sigma_x\right> = 2 k\left( 1- 2 e^{-\tau/T_1} \right) sin(\omega_0 t)

the amplitude of the magnetization is only a function of \tau . bu measuring the amplitude with different \tau , we can determine the T_1 .

To measure the T2, we use follow pulse sequence:

(\pi/2)_x \longrightarrow \tau \longrightarrow (\pi)_y \longrightarrow \tau

again we use the same initial state, but this time, we are going watch it as lab frame.

\rho = k\sigma_z

\downarrow (\pi/2)_x

\rho = - k \sigma_y

\downarrow \tau

\rho = -k (\sigma_y cos( \omega_0 \tau ) - \sigma_x sin(\omega_0 \tau) ) e^{- \tau/T_2}

\downarrow \pi_y

\rho = -k (\sigma_y cos( \omega_0 \tau ) + \sigma_x sin(\omega_0 \tau) ) e^{- \tau/T_2}

\downarrow \tau

\rho = - k \sigma_ye^{- 2 \tau/T_2}

at  the last step, the free-induction decay is not decay but a revert process of decay. we see that the state back to its beginning state! thus, this method also called the spin echo. this can be see in pictorial  way. after the (\pi/2)_x pulse, the spin go to – y axis. now, due to the incoherence of each spin, some spin are faster and some are slower. after a time \tau , the \pi_y pulse flips all the spin by 180 degree. now, the “slower” spin become ahead of the “faster” spin. After the same time interval, the “faster” spin will catch up the “slower” spin and all the spin becomes coherence at that moment again! thus, the amplitude of the signal will become large again. However, since the spins are not at same Larmor frequency, some will flip more than 180 degrees while some flip less, thus, the decay are still there.

Note:

the FID ( free induction decay) rate is not exactly equal to T2, since there are many other way to make the spin polarization decay. the FID decay rate T_2^* should be :

\frac{1}{T_2 ^*} = \frac{1}{T_2} + \frac{1}{T_{others}}

and the spin echo method can eliminate the others.

Advertisements