There's a prescription by Deutsch for the quantum mechanics of closed timelike curves. It works on the level of density states, instead of Hilbert space states. Given his prescription, he showed that a fixed point solution always exists no matter what the initial conditions are. However, this solution isn't unique in general. Also, pure states can be converted into mixed states. Not only that, the solution isn't even linear in the initial density state. It has been shown by Aaronson and Watrous that this prescription allows time travelling computers to solve PSPACE-complete problems.
There's another prescription based upon post-selection, as expounded by Seth Lloyd where you would observe the same decay time. This prescription has the drawback that not every initial condition admits a solution. It has been shown that time-travelling computers can only solve PP-complete problems.
Both prescriptions violate unitarity. In fact, the density matrix evolves nonlinearly in both prescriptions. This nonlinearity has the potential to wreak havoc upon quantum mechanics.
The difference between both prescriptions can be seen most clearly with the grandfather paradox. To simplify matters, consider a qubit which can only take on two states, $| 0 \rangle$ and $| 1 \rangle$. Let's also suppose this qubit is sent around on a closed timelike curve and loops back onto itself. During a cycle around the loop, the qubit is flipped. Now classically, there clearly isn't any consistent solution. However, according to Deutsch's prescription, the mixed density state $\begin{pmatrix} \frac{1}{2} & a\\ a & \frac{1}{2} \end{pmatrix}$ where $a$ is a real number between $-\frac{1}{2}$ and $\frac{1}{2}$ is a fixed-point solution. He chooses to interpret this using the many worlds interpretation as follows; say the qubit starts off with a value of $0$ in one world. After a loop, it ends up with a value of $1$ in a different parallel universe.
According to Lloyd's prescription, on the other hand, there is no solution at all!
However, for the example you presented, both prescriptions will give the same answer, namely the time traveller will observe the same decay time both times around. This is because the nuclear decay is not an integral part of the closed timelike loop. To see this, suppose we have the original prepared unstable particle, plus some experimental apparatus with a clock and a clock pointer which will be set to the time of the nuclear decay. After waiting for some time much longer than the half-life, the clock pointer will end up in the state $\sqrt{k}\int^\infty_0 dt\, e^{-kt/2} | t \rangle$. The whole point is, it doesn't make any difference if the time traveller doesn't have direct access to the unstable particle, but only to the clock pointer, and it doesn't make any difference either if the clock pointer is prepared outside the time machine, i.e. is part of the initial conditions.
Of course, it might turn out the correct prescription is none of the above. Or it might also turn out that closed timelike curves are absolutely forbidden in a complete theory of quantum gravity. Yet another possibility might be closed timelike curves are allowed, but a complete theory of quantum gravity somehow manages to preserve unitarity, just as it presumably preserves unitarity in evaporating black holes.
will $t_b=t_a$?because nobody can state the outcome of that experiment beyond pure speculation. Your last edit fixes it. Assuming an outcome and asking what are the consequences of that specific outcome is much more answerable in a logical manner. +1 – Bruce Connor Jan 3 '11 at 2:27