2
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Your initial state $|+,-\rangle$ is however a mix of $|1,0\rangle$ and $|0,0\rangle$. Also, you still need to solve the Schrödinger equation for the time evolution of the states. $\endgroup$ May 19, 2020 at 17:04
  • $\begingroup$ I didn't specify the details of the calculations thinking that they were too long and just said what the form of the solutions would be (I have them written them but was too lazy to actually type them here). I will edit the post to reflect this, thanks for the input nonetheless. $\endgroup$ May 19, 2020 at 17:34

1 Answer 1

1
$\begingroup$

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 $[0,T]$.

$\endgroup$
3
  • 1
    $\begingroup$ I have solved it but I didn't want to write all the calculations on this questions since it's relatively easy with respect to what I'm used at this point. I just wanted to know if the problem is exactly solvable and doesn't require any approximations (like the $t\to\pm\infty$ would have suggested), which you answered, thank you. $\endgroup$ May 19, 2020 at 17:31
  • 1
    $\begingroup$ It was not quite clear what exactly you were asking . Otherwise, as you say, it is not a very difficult problem - it can be indeed misleading when you expect more :) $\endgroup$ May 19, 2020 at 17:37
  • $\begingroup$ Sorry for not being clear and thanks again for the answer! $\endgroup$ May 19, 2020 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.