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To focus this question let us consider first classical mechanics (which is time-symmetric). Given a final condition (and sufficient information) one can calculate the system conditions of an earlier time (retrodiction).

Given Quantum Mechanics (which is time-symmetric) and a final condition what is the status of retrodiction in that theory? Without choosing between them, here are three options:

(1) An earlier condition can be determined probabilistically exactly as a prediction can be.

(2) An earlier condition can be determined exactly (with enough accessible information) from a final condition.

(3) It is inappropriate to use QM for retrodictions: it is a prediction-only theory.

I have seen some thought experiments on all this in Penrose's books, but it remains inconclusive there, and standard QM texts are not interested in retrodiction.

EDIT AFTER 6 ANSWERS

Thanks for the effort on this question, which was challenging. I will continue to investigate it, but some initial thoughts on the Answers.

(a) I was expecting to receive comments on decoherence, chaos and something on Interpretations and such are amongst the answers. Perhaps the single most important sentence is Peter Shor:

The time-asymmetry in quantum mechanics beween retrodiction and prediction comes from the asymmetry between state preparation and measurement.

Lubos's introduction of Bayesian probability and response (c) is the most useful, although discussion of entropy does not seem immediately relevant. This response though suggests a different framework for retrodiction, with an apriori set of assumptions introduced for the calculation of initial state.

A complication that was not clearly enough addressed was the link with Classical Mechanics retrodiction. Statements that Quantum Retrodiction was "impossible" does not square easily with the fact that some will have corresponding classical systems easily retrodicted. Of course the fact that Quantum Prediction is probabilistic, allows Quantum Retrodiction to be probabilistic too (and thus different from classical retrodiction) was not followed up in some answers. As far as references to Chaos is concerned does not "retrodiction Chaos" result in an increase in ability to classically retrodict given that the trajectories will be converging?

On Peter Morgan's points I should say that the question is open to any interpretation of how the experimental apparatus is used - if it is relevant to giving an appropriate answer then so discuss the significance.

On Deepak's links I should note that these include references to Applications to this idea in Quantum Communication: ie what was the sent state given a received state? I think Lubos's Probability is relevant here too.

Feel free to EDIT your answers if you think of anything else!

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5 Answers 5

up vote 5 down vote accepted

Dear Roy, (3) is correct. More precisely, retrodictions have to follow completely different rules than predictions. This elementary asymmetry - representing nothing else than the ordinary "logical arrow of time" (the past is not equivalent to the future as far as the logical reasoning goes) - is confusing for a surprisingly high number of people including physicists.

However, this asymmetry between predictions and retrodictions has nothing to do with quantum mechanics per se. In classical statistical physics, one faces the very same basic problem. The asymmetry is relevant whenever there is any incomplete information in the system. The asymmetry occurs because "forgetting is an irreversible process". Equivalently, the assumptions (=past) and their logical consequences (=future) don't play a symmetric role in mathematical logic. This source of logical asymmetry is completely independent from the CPT-theorem that may guarantee a time-reversal symmetry of the fundamental laws of physics. But whenever there is anything uncertain about the initial or the final state, logic has to be used and logic has an extra asymmetry between the past and the future.

Predictions: objective numbers

In quantum mechanics, the probability of a future outcome is calculated from $|c|^2$ where $c$ is a complex probability amplitude calculated by evolving the initial wave function via Schrödinger's equation, or by an equivalent method. The probabilities for the future are completely "objective". One may repeat the same experiment with the same initial conditions many times and literally measure the right probability. And this measurable probability is calculable from the theory - quantum mechanics, in this case - too.

Retrodictions: subjective choices

However, the retrodictions are always exercises in logical inference and logical inference - and I mean Bayesian inference in particular - always depends on priors and subjective choices. There is no theoretical way to calculate "unique" probabilities of initial states from the knowledge of the final state. Also, there is no experimental procedure that would allow us to measure such retrodictions because we are not able to prepare systems in the same "final states": final states, by definition, are always prepared by the natural evolution rather than by "us". So one can't measure such retrodictions.

To estimate the retrodicted probabilities theoretically, one must choose competing hypotheses $H_i$ - in the case of retrodictions, they are hypotheses about the initial states. We must decide about their prior probabilities $P(H_i)$ and then we may apply the logical inference. The posterior probability of $H_i$ is this conditional probability: $$ P (H_i|F) = P(F|H_i) P(H_i) / P(F) $$ This is Bayes' formula.

Here, we have observed some fact $F$ about the final state (which may be, hypothetically, a full knowledge of the final microstate although it's unlikely). To know how this fact influences the probabilities of various initial states, we must calculate the conditional probability $P(F|H_i)$ that the property of the final state $F$ is satisfied for the initial state (assumption or condition) $H_i$. However, this conditional probability is not the same thing as $P(H_i|F)$: they are related by the Bayes formula above where $P(H_i)$ is our prior probability of the initial state $P(H_i)$ - our conclusions about the retrodictions will always depend on such priors - and $P(F)$ is a normalization factor ("marginal probability of $F$") that guarantees that $\sum_i P(H_i|F) = 1$.

Second law of thermodynamics

The logical asymmetry between predictions and retrodictions becomes arbitrarily huge quantitatively when we discuss the increase of entropy. Imagine that we organize microstates in ensembles - both for initial and final states; and this discussion works for classical as well as quantum physics. What do we mean by the probability that the initial state $I$ evolves to the final state $F$ if both symbols represent ensembles of microstates? Well, we must sum over all microstates in the final state $F$, but average over all microstates in the initial state $I$. Note that there is a big asymmetry in the treatment of the initial and final states - and it's completely logical that this asymmetry has to be present: $$ P ( F|I) = \sum_{i,j} P(F_j|I_i) P(I_i) $$ We sum over the final microstates because $P(F_1 {\rm or } F_2) = P(F_1)+P(F_2)$; "or" means to add probabilities. However, we must average over the initial states because we must keep the total probability of all mutually excluding initial states equal to one.

Note that $P(I_i)$ is the prior probability of the $i$th microstate. In normal circumstances, when all the initial states are considered equally likely - which doesn't have to be so - $P(I_i) = 1/N_{I}$ for each $i$ where $N_{I}$ is the number of the initial states in the ensemble $I$ (this number is independent of the index $i$).

So the formula for $P(F|I)$ is effectively $$ P ( F|I) = \frac{1}{N_{I}} \sum_{i,j} P(F_j|I_i) $$ Note that we only divide by the number of initial microstates but not the final microstates. And the number of the initial states may be written as $\exp(S_I)$, the exponentiated entropy of the initial state. Its appearance in the formula above - and the absence of $\exp(S_F)$ in the denominator - is the very reason why the lower-entropy states are favored as initial states but higher-entropy states are favored as final states.

On the contrary, if we studied the opposite evolution - and just to be precise, we will CPT-conjugate both initial and final state, to map them to $I', F'$ - the probability of the opposite evolution will be $$ P(I'|F') = \frac{1}{N_{F}} \sum_{i,j} P(I'_i|F'_j). $$ Now, the probability $P(I'_i|F'_j)$ may be equal to $P(F_j|I_i)$ by the CPT-theorem: they're calculated from complex amplitudes that are equal (up to the complex conjugation). But this identity only works for the individual microstates. If you have ensembles of many microstates, they're treated totally differently. In particular, the following ratio is not one: $$ \frac {P(I'|F')}{P(F|I)} = \exp(S_I-S_F) $$ I wrote the numbers of microstates as the exponentiated entropy. So the evolution from $F'$ to $I'$ isn't equally likely as the evolution from $I$ to $F$: instead, they differ by the multiplicative factor of the exponential of the entropy difference - which may be really, really huge because $S$ is of order $10^{26}$ for macroscopic objects. This entropy gets exponentiated once again to get the probability ratio!

This point is just to emphasize the people who claim that the evolution from a high-entropy initial state to a low-entropy final state is "equally likely" as the standard evolution from a low-entropy initial state to a high-entropy final state are making a mistake of a missing or incorrectly added factor of $\exp(10^{26})$ in their formulae, and it is a huge mistake, indeed. Also, there is absolutely no doubt that the inverse processes have these vastly different probabilities - and I would like to claim that I have offered the dear reader a full proof in the text above.

Their mistake may also be phrased as the incorrect assumption that conditional probabilities $P(A|B)$ and $P(B|A)$ are the same thing: their mistake is this elementary, indeed. These two conditional probabilities are not the same thing and the validity of the CPT-theorem in a physical theory can't change the fact that these two conditional probabilities are still very different numbers, regardless of the propositions hiding behind the symbols $A,B$.

Just to emphasize how shocking it is for me to see that those elementary issues about the distinction of past and future are so impenetrable for so many people in 2011, watch Richard Feynman's The Messenger Lecture number 5, "The Distinction of Past and Future" (Internet Explorer needed):

http://research.microsoft.com/apps/tools/tuva/index.html

The very first sentence - the introduction to this very topic is - "It's obvious to everybody that the phenomena in the world are self-evidently irreversible." Feynman proceeds to explain how the second law of thermodynamics and other aspects of the irreversibility follow even from the T-symmetric dynamical laws because of simple rules of mathematical logic. So whoever doesn't understand that the past and future play different roles in physics really misunderstands the first sentence in this whole topic - and in some proper sense, even the very title of it ("The Distinction of Past and Future").

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" there is no experimental procedure that would allow us to measure such retrodictions because we are not able to prepare systems in the same "final states": final states, by definition, are always prepared by the natural evolution rather than by "us". So one can't measure such retrodictions." - Not correct; one can always repeat the experiment many times and restrict the analysis to experiment instances where the final state matches our requirement within a given error criteria. Although this alone might not change the final conclusion though –  lurscher Mar 15 '11 at 18:44
    
However, the proportion of the repetitions in which the initial state of the experiment will satisfy a property will depend not only on the final states in the experiments but also on their initial states, and by very definition, you don't know which initial states you should use to end up with the final state. What you write makes no sense whatsoever. –  Luboš Motl Mar 15 '11 at 19:05
    
Just imagine any single example to see that what you write is just pure nonsense. For example, take a healthy man and try to calculate the probability that he suffered from cancer but was cured from it. What do you think is the probability? How do you make your procedure? For example, you may do repeated experiments with men, and storing them in Fukushima reactor, to make sure they will catch cancer. Some of them will be cured, and the probability you were asked to measure will turn out to be 100% because all your healthy men will be cancer survivors. –  Luboš Motl Mar 15 '11 at 19:08
    
However, one may organize the measurement to get any other number between 0 and 100 percent, too. Prepare lots of healthy men, repeatedly, and they will stay healthy at the end, and 100% of the men who are healthy at the end have always been healthy, so the cancer survivors will be 0%. Clearly, I can get any other number in between, too. It's the very fact that the final state is determined - at least probabilistically (because the future evolves from the past!) - by the initial state, but the converse is simply untrue (because the past is not evolving from the future!). –  Luboš Motl Mar 15 '11 at 19:10
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I didn't say it's a "metaphysical" argument. I wrote it is a strictly mathematical argument. And I only wrote that just like metaphysics, mathematics doesn't depend on dynamical details of any physical theory. Instead of your statement that it is a "metaphysics that shouldn't be used in arguments", I wrote - and I still write - that it is "mathematics that is absolutely essential in any rational argument in science or elsewhere". If you want to do physics without mathematical logic, you will never be able to understand anything about the real world. –  Luboš Motl Mar 15 '11 at 19:34
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Doesn't this depend on which interpretation of quantum mechanics you subscribe to? The many-worlds interpretation and the pilot-wave interpretation would both seem to allow for completely accurate retrodiction, given the final state (which the Heisenberg uncertainty principle makes it impossible to actually discover in the many-worlds interpretation, and the inobservability of the pilot wave makes it impossible to discover in the Bohmian interpretation). The Copenhagen interpretation would say that retrodiction is only possible probabilistically (and this opens a can of worms, both classically and quantum mechanically, because putting a Bayesian probability distribution on the prior states of a system needs what might be seemingly irrelevant knowledge about the world -- see Lubos's answer).

The time-asymmetry in quantum mechanics beween retrodiction and prediction comes from the asymmetry between state preparation and measurement. It is theoretically possible to prepare a quantum system in an arbitrary state (and experiments can often come very close to doing so) but Heisenberg's uncertainty principle makes it impossible to accurately measure what state a quantum system is in.

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This is just completely silly, Peter. Retrodictions - and whether they can be quantitative and objective - surely don't depend on interpretations of quantum mechanics. If this were the case, one could also claim that e.g. dinosaurs have never lived on Earth according to Shor's interpretation of quantum mechanics, or whatever of this kind. –  Luboš Motl Mar 15 '11 at 18:23
    
Dear Peter, concerning your updated version: it's not just Copenhagen interpretation that makes retrodictions only possible probabilistically (more precisely, dependent on subjective priors). Just take any question that could be retrodicted and calculate the probability that the answer is Yes using any interpretation of QM you may think of. If the probability is required to differ from 0% and 100%, you will fail. Just like it's not possible to retrodict with certainty, because events' irreversibility, it's also not possible to retrodict probabilities with values that are sharply well-defined. –  Luboš Motl Mar 15 '11 at 19:02
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I think what you are looking for is work done by Yakir Aharonov and Lev Vaidman (nice name ;) ) on something called the "two-time" formulation of QM which is incorporates time-symmetry from the get-go and treats prediction and retrodiction (equivalently forward and backward evolution) on an equal footing. I haven't read enough to take a stand but given the credential's of one of the authors (Aharonov as in Aharonov-Bohm effect) I'll take it as an article of faith that there is some intellectual merit therein. Of course the old argument regarding the correlation betwen old-age and "crazy" ideas will likely be raised. I try not to pay heed to such generalizations. Some references are:

Reznik and Aharonov, On a Time Symmetric Formulation of Quantum Mechanics, PRA, 1995

Pregnell & Pegg, Retrodictive quantum optical state engineering, J. Mod. Opt., 2004

Vaidman, Backward Evolving Quantum States, J. Phys. A, 2006

As for the other answers I see reason in @Peter Shor's and @sb1's answer. @sb1's answer would seem to rule out such a picture, or at least any practically useful application of the same. Presumably the "two-time" formulation avoids @sb1's argument in some way. Again I have yet to read the papers and so cannot sensibly comment on this yet.

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I suppose it is illogical to assume that the same individual has down-voted all (competing) answers to this question, but if not, then I guess it would also be illogical to expect any sort of justification for their down-vote to my answer! Here's hoping otherwise (fingers crossed ;) –  user346 Mar 15 '11 at 20:55
    
@Deepak: I never have read those papers so like you I also can't take any stance on them. But the references have certainly given an opportunity to learn something new. Therefore +1 from me. –  user1355 Mar 16 '11 at 3:50
    
+1 from me too. It is a nice post, and this way I am sticking it to the man. I had a vague memory of this but couldn't find it, so I am glad you posted it. –  MBN Mar 16 '11 at 4:21
    
@MBN, Sticking it to the man? Good one :D –  user346 Mar 16 '11 at 12:00
    
Just to avoid confusion, by that I meant beating the system. The voting/reputation system, which in my eyes is quite silly. –  MBN Mar 16 '11 at 15:15
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Your question would look different if you consider what a discussion of experimental confirmation of the retrodicted "earlier conditions" would look like. Sure, we can compute what the quantum or classical state must have been at some earlier time, given (1) the quantum or classical state now; and (2) whatever particular dynamics we have good reason (on the basis of past experience) to believe the system to be well-described by; but this is of limited interest unless we experimentally confirm that the quantum or classical state was in fact what our computation tells us. For the most part, we are just more interested in what the state in the future will be because of the state now than in what the state in the past must have been to cause the state now.

One possible experimental confirmation of a retrodicted state R computed on the basis of a state X would be to look for systems that we measure to be in a similar state to R, and have reason to believe to be subject to the same dynamics, and then verify that those systems in fact evolve into a state similar to X. This kind of discussion, however, seems some distance from the concerns in your question.

Your statements that classical and quantum mechanics are time-symmetric is true enough of those theoretical frameworks in themselves, but in practical use such models of experimental apparatus rarely include all degrees of freedom that affect the experiment. Even at only moderate levels of accuracy, models usually have to include the effects of, for example, heat flows into and out of the experimental apparatus.

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For classical chaotic and for most quantum systems, it is impossible to measure enough information to retrodict their behavior accurately. –  Peter Shor Mar 15 '11 at 18:38
    
@Peter Shor Good point for actual experimental apparatus. Your comment touches on what I think is a problem in the Question, the lack of a distinction between models and experimental apparatus. For a classical chaotic model, we can predict its evolution to the ultimate future, we can just feed in an initial value like $\sqrt{2}$; for real systems, we may not be able to measure parameters of a dynamical ansatz or initial conditions accurately enough even for milliseconds into the future, even on the almost certainly false assumption that our dynamical ansatz is perfectly correct. –  Peter Morgan Mar 15 '11 at 20:35
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The only (potential, i'm speculating here) issue i see in retrodiction is that you would have to isolate how the system (i.e: the observer and measuring aparatus together) couples to the retrodicted state all the time up to the final measurement which is behaving as your initial condition.

Probably if you can guarantee no coupling for a time $\Delta t$, there is still the problem that the state will still be coupled to the system for earlier times, and hence making it not entirely independent.

In the case of prediction, its is implicitly assumed that your coupling to a future state is zero. Weirdly enough, it doesn't seem you can do the same for postdiction. Definitely a interesting question to ponder

EDIT:

Lets elaborate a bit more how an experiment of retrodiction could be done;

lets take system A to be observed and measure the state at time $t_{0}$, store the measurement result somewhere where it would not decohere with the measuring apparatus. I don't have any evidence to assert that this can always be done consistently, but given the results of the delayed quantum erasing experiment, there is a good chance there is nothing in principle avoiding one to do this.

After the measurement is done, if decoherence has happened, the system might be in a (Collapsed) state and if you were to measure the state further, you would find it in a pure-diagonal density matrix (a fully classical state with classical probabilities) so at $t_{1}$ you would observe a certain final state $|\Psi_{final}(t_{1})>$ and the probability of it would be a classical weighted probabilities (of the initial state mixed state classical probabilities after the measurement) times the probability of each individual initial state in the mixture naturally evolving and finally collapsing into $|\Psi_{final}(t_{1})>$

So the net probability of this state should be (in the case of decoherence):

$$ \sum_{ |\Psi_{final}(t_{0}) > } (P(|\Psi_{final}(t_{0}) >) x <\Psi_{final}(t_{1})| e^{-i(t_{1} - t_{0}) H } |\Psi_{final}(t_{0}) > )^2 $$

in the case of no decoherence, i'm not sure what should happen. I will think on this an update the answer later.

However, whatever is the resulting probability equation (and how it depends on the fact of measuring or not the delayed measurement state), the experimental procedure to follow (which doesn't depend on the final math that much) would be:

experiment template:

  • save the $t_{0}$ state somewhere that will not decohere with your apparatus or anything else. This is probably the hardest part to be achieved, if at all possible

  • let the system evolve to $t_{1}$ and measure the final state

now you execute the experiment template many times, and you restrict your dataset to experiments where final state are your $|\Psi_{final}(t_{1})>$. In this way you have properly and consistently defined the problem of retrodicting the state evolution based on a final state

Now that the experimental procedure is properly defined, the question is:

assumming no decoherence happens in the delayed measurement, would the subset of experiments where final state = $|\Psi_{final}(t_{1})>$, have probabilities amplitudes at $t_{0}$ that are consistent with a propagator of the form $ e^{-i(t_{0} - t_{1}) H }$?

I think the question is yes, but only if you guarantee that the available physical information of the state at $t_{0}$ is fully encoded in the delayed measurement storage and nowhere else, otherwise you are back in the decoherence scenario

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