Question about an Exercise with Time-Dependent Hamiltonian

Question

I've been recently been assigned this exercise:

Consider two spin 1/2 particles which are coupled through a time dependent interaction: $$ H(t) = a(t) s_1 \cdot s_2 $$ where $a(t)$ is a function which is constant in the interval $[0,T]$ and zero elsewhere. The system is in the state $|+,->$ for $t \to -\infty$.

The exercise then asks various questions about the probability of finding the state in another state for $t\to +\infty$.

It appears as if this exercise is extremely easy as it can be solved exactly for any state in the $|S,M_S>$ base of eigenstates of $S^2, S_z$. More precisely, in this base the Time-dependent Schrodinger Equation becomes a system of four decoupled first-order linear differential equations in the coefficients of $|\psi(t)>$ in this base, since $H(t)$ is diagonal in this base:

$$ H(t) = a(t) \bigg[ \frac{S^2}{2} - \frac{S_1^2}{2} - \frac{S_2^2}{2} \bigg] = \frac{a(t)\hbar^2}{2} [S^2 - 3/2] = \frac{a(t)\hbar^2}{2} \begin{pmatrix} 1/2 &&&& \\ & 1/2 &&&\\ && 1/2 &&\\ &&& -3/2 \end{pmatrix} $$ where the states are ordered as such: $|1,1>,|1,-1>,|1,0>,|0,0>$. It's pretty easy from here since $a(t)$ is either constant or zero, which means the coefficients evolve with an imaginary exponential in the interval $[0,T]$ and stay constant elsewhere. More precisely:

$$ |\psi(t)>\;\; = \begin{pmatrix} b_1(t) \\ b_2(t) \\ b_3(t) \\ b_4(t) \end{pmatrix} \Rightarrow i\hbar\frac{d}{dt}|\psi(t)> = H(t)|\psi(t)>\; \Rightarrow \begin{cases} i\hbar \frac{db_j}{dt}(t) = \frac{a(t)\hbar^2}{4}b_j(t) & j=1,2,3\\ i\hbar \frac{db_j}{dt}(t) = -3\frac{a(t)\hbar^2}{4}b_4(t) \end{cases}$$ and finally:

$$ b_j(t) = \begin{cases} b_j(0) & t<0\\ b_j(0)e^{-i a_0\hbar t/4} & 0\leq t\leq T\\ b_j(0)e^{-i a_0\hbar T/4} & t>T\\ \end{cases} \quad j=1,2,3 \qquad b_4(t) = \begin{cases} b_4(0) & t<0\\ b_4(0)e^{i 3a_0\hbar t/4} & 0\leq t\leq T\\ b_4(0)e^{i 3a_0\hbar T/4} & t>T\\ \end{cases} $$

and in the case $|\psi(-\infty)> = |\psi(0)> = |+,->$ the initial conditions give $b_1(0) = b_2(0) = 0$ and $b_3(0) = b_4(0) = 1/\sqrt{2}$, which means it will oscillate between different linear combinations of $|1,0>,|0,0>$ or equivalently $|+,->,|-,+>$.

Then the probability of finding the system in a given state is a simple scalar product between 4-dimensional vectors.

Is this the case or am I forgetting something? It's been the easiest homework assigned so far and I find it very strange. Also, why would the exercise specify the state of the system for $t\to\pm\infty$ since the system only evolves between $t = 0$ and $t = T$, should it not be for $t<0$ and $t>T$ instead? What I mean is it seems like unnecessary detail, since the state changes only in a finite amount of time.

EDIT: Inserted the calculations necessary to find the time evolution of the system in any given initial state.

Answer 1

You are right that at $t=0$ the system is still in state $|+,-\rangle$, since $a(t)=0$ for $t<0$ and this state is an eigenstate of the Hamiltonian. However, for $t>0$ it is not an eigenstate anymore, but a superposition of the two eigenstates $|1,0\rangle$ and $|0,0\rangle$. Each of these evolves with different time exponent, so the result at $t=T$ may be rather different from what you had at $t=0$. Note that the problem is solvable, even if $a(t)$ is an arbitrary function in $[0,T]$.