Question

The behavior of a spin- particle in a uniform magnetic field in the z-direction, $vec{B}=B_0hat{z}$ , with the Hamiltonian

You found that the expectation value of the spin vector undergoes Larmor precession about the z axis. In this sense, we can view it as an analogue to a rotating coin, choosing the $hat{S_x}$ eigenstate with eigenvalue to represent heads and the eigenstate with eigenvalue to represent tails. Under time-evolution in the magnetic field, these eigenstates will “rotate” between each other.

(a) Suppose at $t=0$ , we measure $hat{S_x}$ to be . Find the probability that a measurement of $hat{S_x}$ will again yield at time . (Where applicable, write your answers for all parts of this problem using the Larmor frequency, .)

(b) For small , what is the probability, to leading order in , that the measurement will instead yield ? What is meant by “small” here (i.e., small compared to what)?

Hint: you may find it useful to recall that the Taylor expansion of to leading order in x is given by:

Answer 1

H = -gamma B_0 s^z = -gamma B_0 h/2 (1 0 0 - 1) the Eigen states of H are the same these of S_2 Y_+ with energy E_+ = -(gamma B_0 h)/2 X_- with energy E_- = + (gamma B_0 h)/2 Now, time dependent Schr�dinger I h delta X/delta t = H x X (1) = a X_+ e^- I E + t/h + b X_- e^- I E - t/h = (a e^I gamma B_0 t/2 b e^- I gamma_0 t/2) now, for initial condition, X (0) = (a b) Where a^2_1 + b^2_1 = 1 so, we are taking a = Cos (alpha/2) b = sin (alpha/2) x(t) = (Cos alpha/2 e^i gamma B_0 t/2 Sin alpha/2 e^- I gamma B_0 t/2) Now, if we measure s_x at time t, C_+ (x) = X_+ (x) + X