QM M00 1
From QED
Consider a spin- particle constrained to move on a 1D line with a harmonic oscillator potential and a magnetic field so that the Hamiltonian is:
The first energy level is not degenerate but all the other levels are doubly degenerate.
Now add a small magnetic field in the direction with a magnitude proportional to x . The Hamiltonian is:
Calculate the energy difference in the levels to lowest order.
Define the original the Hamiltonian:
And add the perturbation:
We define the basis of eigenstates of H0 to be , where nk is the quantum number for the kinetic energy, and defines either the + or - spin state. The energy from the state is .
Since we know:
so that the properties of Sx are:
We can also decompose the x operator:
where a and are the annhiliation and creation operators. The energy perturbation to first order is given by:
Where we must diagonalize the states:
So the change in energy is given by:
Now it simply remains to calculate the values of :
So that for :
And for :
So that the total energy difference is just given by:
We must treat the ground state separately, since it is nondegenerate. We must take it out to second order:
Finding the nonzero values of this, we get:
as the only one. Thus:
At last we get:
where the pair for n > 0 indicates the splitting of a degenerate pair.
This page was recovered in October 2009 from the Plasmagicians page on Prelim_M00_QM1 dated 22:28, 23 December 2005.