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Time dependent perturbation: electron spin problem

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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)

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Solution Preview

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 ...

Solution Summary

We explain how using time dependent first order perturbation theory, the transition probability for the spin flip can be computed.

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Quantum Mechanics: Calculate the probability that at a later time tf the electron spin has flipped to the -z direction.

An electron is in a strong, static, homogenous magnetic field with magnitude B0 in the z direction. At time t=0, the spin of the electron is in the +z direction. At t=0 a weak, homogenous magnetic field with magnitude B1 (where B1 << B0) is turned on. At t=0 this field is pointing in the x direction, but it rotates counterclockwise in the x-z plane with angular frequency &#969;, so that at any later time t this field is at an angle &#969;t relative to the x-axis.

Calculate the probability that at a later time tf the electron spin has flipped to the -z direction. Do not assume anything about the particular value of &#969;.

Please see the attached PDF file for question along with a diagram.

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