Principles of Quantum Mechanics · Problem 14.4.3

Addition of Angular Momenta — R. Shankar

Physics — Quantum Mechanics · spin-rotating-frame

Problem

We would like to study here the evolution of a state that starts out as

\binom{1}{0}

and is subject to the

\mathbf{B}

field given in Eq. (14.4.27). This state obeys

i\hbar \frac{d}{dt}|\psi(t)\rangle = H|\psi\rangle \tag{14.4.34}

where

H = -\gamma \mathbf{S}\cdot\mathbf{B}

, and

\mathbf{B}

is time dependent. Since classical reasoning suggests that in a frame rotating at frequency

(-\omega\mathbf{k})

the Hamiltonian should be time independent and governed by

\mathbf{B}

, [Eq. (14.4.29)], consider the ket in the rotating frame,

|\psi_r(t)\rangle

, related to

|\psi(t)\rangle

by a rotation angle

\omega t

|\psi_r(t)\rangle = e^{-i\omega t S_z/\hbar}|\psi(t)\rangle \tag{14.4.35}

Combine Eqs. (14.4.34) and (14.4.35) to derive Schr\"odinger's equation for

|\psi_r(t)\rangle

in the

S_z

basis and verify that the classical expectation is borne out. Solve for

|\psi_r(t)\rangle = U_r(t)|\psi_r(0)\rangle

by computing

U_r(t)

, the propagator in the rotating frame. Rotate back to the lab and show that

|\psi(t)\rangle \xrightarrow{S_z\text{ basis}} \begin{bmatrix} \left[\cos\left(\frac{\omega_r t}{2}\right) + i\frac{\omega_0 - \omega}{\omega_r}\sin\left(\frac{\omega_r t}{2}\right)\right]e^{+i\omega t/2} \\ \frac{i\gamma B}{\omega_r}\sin\left(\frac{\omega_r t}{2}\right)e^{-i\omega t/2} \end{bmatrix} \tag{14.4.36}

Compare this to the state

|\hat{n}, +\rangle

and see what is happening to the spin for the case

\omega_0 = \omega

. Calculate

\langle\mu_z(t)\rangle

and verify that it agrees with Eq. (14.4.31).