# Time travel and nuclear decay

Reading a previous closed question an interesting variation has come to my mind.

Suppose that time travel to the past was possible:

• I wait for an atom to decay and measure the time, $t_{1a}$
• I travel back in time at $t_0<t_{1a}$
• I wait for the same atom to decay and measure the time, $t_{1b}$

Let's think about the two values, $t_{1a}$ and $t_{1b}$.

If they coincide, then from my point of view the measured $t_1$ time would not be governed by chance (the second time I would know $t_1$ a priori). It would therefore prove some form of existence of hidden variables. Would this violate any known laws of quantum mechanics? Would this prove the existence of hidden variables (e.g. Bohm's interpretation)?

If they are different, then what to make of $t_{1a}$ and $t_{1b}$? Which would be the correct value? To me this doesn't make any sense, but maybe it could be compatible with the multiverse interpretation (I don't know how though).

In other words, does this gedankenexperiment:

1. ...help us select viable QM interpretations and exclude others? Which?
2. ...lead us to conclude that backwards information time travel is incompatible with QM?
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Of course this Gedankenexperiment applies to any QM problem, so feel free to use a different experiment if it makes more sense. –  Sklivvz Jan 3 '11 at 1:06
It is an interesting point, but it is not answerable. No one can say if it's true or false. You could pose a question like: "Assuming it's true, could it still be compatible with orthodox QM?". That would be answerable. –  Malabarba Jan 3 '11 at 1:58
What exactly is not answerable? If the values are the same, then there is a process by which the particular outcome of the experiment is defined a priori. It is therefore a question on whether this implies hidden variables (and if it violates Bell's theorem). If not, maybe because of some weird multiverse interpretation detail, then it's a question about that QM interpretation. In any case I don't see why it's not answerable. –  Sklivvz Jan 3 '11 at 2:05
@Sklivvz: The part that wasn't answerable was 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 –  Malabarba Jan 3 '11 at 2:27
All these hit and run down votes are very annoying. –  Sklivvz Jan 4 '11 at 22:22

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.

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Very interesting: can you provide some links for the interested visitors? –  Sklivvz Jan 6 '11 at 13:46

Because we can't actually go back in time to test the idea behind this question, what you're asking in some sense boils down to a question of what theory you really believe governs the universe.

According to the mathematics of QM, if you run experiments backwards in time (mathematically, of course), you only recover your starting state if you do not disturb the experimental system (with a measurement, for example) before running it backwards. So, implicit in the question is a desire to take a measurement and then undo the measurement by traveling back in time. This is impossible in standard QM because it would allow violation of uncertainty and all sorts of other fundamental underpinnings of the theory. So, the question fundamentally can't be asked in a meaningful way. This is the view I would tend to take.

If, on the other hand, you believe in, say, the many-worlds interpretation of QM, then traveling back in time probably corresponds to taking a different path down the infinitely bifurcating tree of futures, meaning the question can be asked and the answer will be that your second time measurement may be different from the first. Both are "correct" in that one takes place in one section of the multiverse while the other takes place in a different section.

If you believe in a hidden variables interpretation, that implies (in the sense that if you hold that view, you probably don't believe in the probabilistic nature of QM) that the decay will happen at a set time. In that case, you'd get the same measurement both times.

So, to answer your numbered questions with this subset of interpretations:

1. The real experiment would help us select between some interpretations of QM. However, the thought experiment does not, as the result depends upon what assumptions you use, which reduces, in this case, to which interpretation you believe in.

2. The second paragraph of my answer implies not that backwards information travel is prohibited, but that information about the state of the system is outdated because the system has been altered. In fact, I would personally argue that, based on the reversibility of the mathematics, backwards information travel has occurred in the form of the measurement altering the system. Of course, that's a clear violation of causality, that the effect of the measurement is felt before the measurement is done, but obviously, in this posited situation, causality violation should be expected.

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It depends on what type of universe you think we live in. There are at least two schools of thought here. One says that when you go back in time the universe splits into two realities. The original, and the alternate universe. So in the alternate universe anything can happen, and the atom has two distinct decay times, one for each universe.

The second school of thought says that all time travel into the past has already happened (that is, you can't change the past, because your travel to the past has already been taken into account in our timeline). So in that case, you go back, wait around, and get to watch yourself take the first measurement. It is impossible for you to interact with yourself, because then you would have remembered that you saw your self from the future in the first place.

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## protected by Qmechanic♦Mar 17 '13 at 21:45

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