An electron is in a strong, uniform, constant magnetic field with magnitude B_0 aligned in the +x direction. The electron is initially in the state |+> with x component of spin equal to hbar/2. A weak, uniform, constant magnetic field of B_1 (where B_1 << B_0) in the +z direction is turned on at t=0 and turned off at t=t_0. Let P(i -> f) be the probability that the electron is in the state |-> with x component of spin equal to -hbar/2 at a later time t_f > t_0. Show that
P(i -> f) = (B_1/B_0)^2 sin^2(u_B B_0 t_0 /hbar)© BrainMass Inc. brainmass.com October 10, 2019, 1:25 am ad1c9bdddf
The Hamiltonian is:
H = u_b B dot sigma
We can write this as:
H = u_b B_0 sigma_x + u_b F(t) B_1 sigma_z
where F(t) equals 1 for 0<=t<=t_0 and zero otherwise.
Let us denote the eigenstate corresponding to spin in the +x direction by |+> and in the minus x-direction by |->, the energy eigenvalues are E_1 = u_b B_0 and E_2 = - u_b B_0, respectively. The probability that the spin will ...
We explain how using time dependent first order perturbation theory, the transition probability for the spin flip can be computed.