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## Homework Statement

Two identical spin-1/2 particles interact with Hamiltonian

**H0**=ω0

**S1.S2**where ω0>0. A time dependent perturbation is applied,

**H'**=ω1 (S1z-S2z) θ(t) Exp[-t/τ], where ω1>0 and ω1<<ω0. What are the probabilities that a system starting in the ground state will be excited into each of the three excited states to first order?

## Homework Equations

P[initial state -> final state] = Norm[ Integrate[ < final l H'[t'] l initial > Exp[ I (Ef - Ei) t'], {t',0,t}]]^2

## The Attempt at a Solution

Sorry if the statement is unclear, but I just wanted to make sure my thinking on this is correct. We have two identical fermions interacting through a spin-spin interaction, so the first thing to do in my head is to use the coupled basis states representing the particles. We only have spin interactions so spatial states aren't needed. In a non-identical particle situation, we would have four possible states:

l 1 1 > = l + +>

l 1 0 > = 1/Sqrt[2] (l + - > + l - + >)

l 0 0 > = 1/Sqrt[2] (l + - > - l - + >)

l 1 -1 > = l - - >

But since we have two identical spin-1/2 particles, only the anti-symmetric state is allowed, so our initial state must be the l 0 0 > state.

Now here is where I get confused. Our perturbation Hamiltonian is still only dependent on spin interactions, so our final state should still be one of the four kets listed above. If this is the case I don't see why all probabilities aren't zero (besides the l0 0> -> l0 0> probability) since we can't find fermions in a symmetric state. For all I know this is a trick question and I could be thinking correctly about it, but I feel like there should be some sort of computation that goes along with this.