# Why and how is nondegenerate perturbation theory used for time evolution under $\vec{L}.\vec{S}$ coupling?

Let us say that we start with an electron which is in a spin up state and has a spatial wave-function of the form $xf(r)$. Then one turns on a perturbation of the form $\frac{U(r)\vec{L}\cdot\vec{S}}{\hbar ^2}$.

• Now one wants to know what is the probability that the state that one started out with will be still found in the same state after a time $t$.

A few steps towards that which I can imagine,

• write the coupling $\vec{L}\cdot\vec{S}$ as $\frac{1}{2}(J^2 - L^2 - S^2)$

• Using the identification of $Y_{1\pm 1} = \lvert1\pm1\rangle$ and one can rewrite the initial wave-function as $xf(r)\otimes \frac{1}{2} = \Bigl[0.5\sqrt{\frac{8\pi}{3}}rf(r)\Bigr]\bigl\{ \lvert 1-1\rangle\otimes \frac{1}{2} - \lvert 11\rangle\otimes \frac{1}{2}\bigr\}$

• one can probably guess that there are $2$ more states at the same $l=1$ value which are of the similar form but are a sum of the $m = \pm1$ states and another one is just the $m=0$ state. (..I would like to know if there is a systematic way to get these $3$ states using the action of some symmetry operator which rotates between these three states..can't immediately see what that operator is beyond the fact that this is just the $3$-vector representation of $SO(3)$..)

• In first order perturbation theory one would take the expectation value of these three states above in the perturbing potential of $\frac{U(r)\vec{L}\cdot\vec{S}}{\hbar ^2}$. Here when for any of the above three states one takes the expectation value in the $J^2$ operator one has to rewrite these states above in the $J$ basis like, $\lvert 1-1\rangle\otimes \frac{1}{2} = \frac{1}{\sqrt{3}}\lvert \frac{3}{2} -\frac{1}{2}\rangle_J - \sqrt{\frac{2}{3}}\lvert\frac{1}{2} -\frac{1}{2}\rangle_J$ etc. (..but for taking the expectation value in the $L^2$ and the $S^2$ operator one can continue with the writing as the tensor product of the $l=1$ and $s=1/2$ states.

One can find the 3 expectation values of the perturbing $\vec{L}\cdot\vec{S}$ potential in the initially 3 degenerate states and these expectation values are the first order corrections to those energies that split that degeneracy.

• But I would think that to answer the question of probaility as asked earlier, one needs to find the (perturbative) eigenstates of the perturbed potential and only then one can answer the time-evolution question. And this is what is not being clear to me as to how one can get that..

Is there a regime in the perturbation theory where one can continue to think of the 3 initially degenerate states as being eigenstates of the new 3 split energy levels?

• Very honestly I would have thought that one should have been using degenerate perturbation theory and calculating that determinant to determine the new energy levels. But conventionally why is the above non-degenerate perturbation theory done for $\vec{L}\cdot\vec{S}$ coupling?
• This might be too simplified if an answer: You can calculate the energy shift of spin-orbit coupling, and by the Uncertainty Principle (in its limit) we get the time shift $T$ for this to take place. Assume uniform probability, then it's $1-t/T$ to stay put? – Chris Gerig Aug 5 '12 at 23:22
• @ChrisGerig I am looking for a more full fledged quantum calculation! – user6818 Aug 14 '12 at 20:11
• Very honestly I would have thought that one should have been using degenerate perturbation theory and calculating that determinant to determine the new energy levels. But conventionally why is the above non-degenerate perturbation theory done for L⃗ ⋅S⃗ coupling? I have the same question, why don't we use degenerate perturbation, as the energy levels before the spin-orbit interaction are degenerated.. – user11478 Aug 18 '12 at 5:10

You've avoided having to use degenerate PT in the first step, by writing $L.S$ in terms of $J^2$, $L^2$ and $S^2$.
Instead of using eigenstates of the operators $L_z$ and $S_z$, which do not commute with the perturbed Hamiltonian, we use eigenstates of $J^2$ and $J_z$, which do commute with $H$, because the total angular momentum $J$ is classically conserved.
This means that there exists a set of simultaneous eigenstates of $J^2$, $J_z$ and $H$ - we have solved the problem. This technique works generally (see 'Introduction to Quantum Mechanics' by David Griffiths for a full proof and discussion), and it's a good way of avoiding the hard work of degenerate perturbation theory.
• Then to work out the probability of being in one of the perturbed states, given that it's in an unperturbed one, you just need to work out the overlap $\langle\psi'|\psi\rangle$ – Benjamin Hodgson Aug 18 '12 at 20:47