where u_s = -((g_s)(u_B))/(hbar) . S
Thus the Hamiltonian is H_0 = ((g_S)(u_B))/(hbar) . B . S = ((g_s)/2)(u_B) . B . sigma
and g_s = 2 for the electron. The eigenstates of H_0 are just the spin up and spin down states, | up > and | down >, with energies E_+ = (u_B)(B_z) and E_- = -(u_B)(B_z).
Suppose we are in the spin up state, | up >, and we add the small magnetic field (B_x)(x hat) with B_x << B_z. Take B = (B_x)(x hat) + (B_z)(z hat), and calculate the exact energies.© BrainMass Inc. brainmass.com December 20, 2018, 5:34 am ad1c9bdddf
Let's write the Hamiltonian as
H = u_b B dot sigma
Then , if we put B = B_z z-hat + B_x x-hat,
We still have a Hamilitonian of the same form as when B_x was zero, because by rotational symmetry, it doesn't matter in which direction the magnetic field happens to point in. So, if we know that the energy eigenvalues for the case B_x = 0 are given by:
E = u_b B_z and E = -u_b B_z,
Then this simply means that E = plus or minus u_b |B| with B the magnitude of the magnetic field, because we can always decide to call the direction in which the field points to be the z-direction. This does mean that the states |up> and ...
To illustrate second order perturbation theory, we consider an electron in a magnetic field, where the magnetic field has a large component in the z-direction and small component in the x-direction. The exact energy eigenvalues are easily obtained from considering the total magnetic field. We also obtain the same solution in a different way by writing down the Hamiltonian in the basis of spin up and down in the z-direction and diagonalize the Hamiltonian.
From the exact solution we can see that if we treat the magnetic field in the x-direction as a perturbation, it will make a nonzero contribution to second order in perturbation theory. We then use perturbation theory to calculate the first and second order contributions and show that the results are in agreement with the series expansion of the exact solution.