density matrix in operator form

Leave a comment

If we express the density matrix in operator form, it is more easy to use in NMR calculation. Let me show you.

for spin ½ ensemble, the density matrix has 4 matrix elements. and we are always to expand it into Pauli matirx and identity matrix, which also give us 4 coefficients.

\rho = \begin{pmatrix} \rho_+ & \rho_2 \\ \rho_1 & \rho_- \end{pmatrix}

\rho = \frac{1}{2} \left( (\rho_++\rho-)1 + (\rho_+-\rho_-) \sigma_z + (\rho_1+\rho_2) \sigma_x + (\rho_1-\rho_2) \sigma_y \right)

the Pauli matrix is just the angular momentum operator in some sort of reduced form. Thus, the magnetization or the polarization can be very easy to calculated by notice that:

\sigma_i \sigma_j = i\epsilon_{ijk} \sigma_k

and the result is the trace.

in a rotation frame, the density matrix under transform:

\rho_R = R_z(-\omega_0 t) \rho R_z(\omega_0 t)

and the Pauli matrix change under rotation is like a vector following right-hand rotation rule:

\sigma_z \rightarrow \sigma_z

\sigma_x \rightarrow \sigma_x cos(\omega_0 t) + \sigma_y sin(\omega_0 t)

\sigma_y \rightarrow -\sigma_x sin(\omega_0 t) + \sigma_y cos(\omega_0 t)

Thus, for a polarized ensemble, the off-diagonal matrix element are zero, the density matrix keeps the same under the changing of frame.

in the NMR calculation, the flipping of polarization is just applying the suitable rotation operator. for example, for a \pi_x – pulse, the density matrix becomes:

\rho_R \rightarrow R_x(-\pi)\rho_R R_x(\pi)

For the longitudinal relaxation, which is only affect the z-component.

\sigma_z \rightarrow (1-2 e^{-t/T_1}) \sigma_z

for the transverse relaxation of the FID ( free induction decay ), which is due to de-synchronization of individual spin ):

\sigma_x \rightarrow (\sigma_x cos(\omega_0 t) + \sigma_y sin(\omega_0 t))e^{-t/T_2}

\sigma_y \rightarrow (-\sigma_x sin(\omega_0 t) + \sigma_y cos(\omega_0 t))e^{-t/T_2}

since the density matrix is now written on Pauli Matrix, and the polarization can be easily to see :

P = \left< J \right> = Tr(( J_x , J_y, J_z) \rho) = (\rho_+ - \rho_- , \rho_1+\rho_2 , \rho_1 - \rho_2 )

thus, we can write it as:

\rho = f 1 + g \vec{\sigma} \cdot \vec{n}


f = \frac{1}{2} ( \rho_+ + \rho_- )

and g is the magnitude of P/4.

by seeing in this way , we can connect it the the matrix M in spin scattering.

detail treatment on Larmor Precession and Rabi Resonance

Leave a comment

a treatment on Larmor Precession and Rabi resonance

the pdf is a work on this topic. it goes through Larmor Precession and give example on spin-½ and spin-1 system.

then it introduce Density matrix and gives some example.

The Rabi resonance was treated by rotating frame method and using density matrix on discussion.

the last topic is on the relaxation.

the purpose of study it extensively, is the understanding on NMR.

the NMR signal is the transverse component of the magnetization.

on Relaxation in NMR

Leave a comment

If we only switch on the transverse magnetic field for some time \tau . after the field is off, the system will go back to the thermal equilibrium. it is due to the system is not completely isolated.

instead of consider a single spin, we have to consider the ensemble. and an ensemble is describe by the density matrix.

the reason for not consider a single spin state is, we don’t know what is going on for individual spin. in fact, in the previous section, the magnetization is a Marco effect. a single spin cannot have so many states, it can only have 2 states – up or down. if we insist the above calculation is on one spin, thus, it only give the chance for having that direction of polarization. which, is from many measurements.

so, for a single spin, the spin can only have 2 states. and if the transverse B field frequency is not equal to the Larmor frequency , and the pule is not a π-pulse, the spin has chance to go to the other state, which probability is given by a formula. and when it goes to relax back to the minimum energy state, it will emit a photon. but when it happen, we don’t know, it is a complete random process.

However, an ensemble, a collection of spins, we can have some statistic on it. for example, the relaxation time, T1 and T2.