# Does decoherence need non-determinism?

Consider standard quantum mechanics, but forget about the collapse of the wavefunction. Instead, use decoherence through interaction with the environment to bring the evolving quantum state into an eigenstate (rspt arbitrarily close by). Question: Can this theory be fundamentally deterministic?

If one takes into account that the variables of the environment are not known, then the evolution is of course 'undetermined' in a probabilistic sense, but that isn't the question. Question is if fundamentally quantum mechanics with environmentally induced decoherence can be deterministic. Note that I'm not saying it has to be. I might be mistaken, but it seems to me decoherence could be followed by an actual non-deterministic process still, so the decoherence alone doesn't settle the question of determinism or non-determinism. Question is if one still needs a non-deterministic ingredient?

Update: Please note that I asked whether the evolution can fundamentally be deterministic, or whether it has to be non-deterministic. It is clear to me that for all practical purposes it will appear non-deterministic. Note also that my question does not refer to the prepared state after tracing out the environmental degrees of freedom, but to the full evolution of system and environment. Does one need a non-deterministic ingredient to reproduce quantum mechanics, or can it with the help of decoherence be only apparently non-deterministic yet fundamentally deterministic?

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What definition of determinism are you using? – Holowitz Jan 21 '11 at 14:45
If you know the state at time t_1, you can in principle calculate everything that's going to happen, or did happen at time t_2. – WIMP Jan 23 '11 at 10:32
You need to rephrase your question if both myself and Matt have misunderstood your intention. Are you now asking if you can deterministically collapse to a particular eigenstate? The answer to that is no, because it violates linearity. – Joe Fitzsimons Jan 23 '11 at 10:56
@ Joe: I just updated the question, hope it's clearer now? You don't need to collapse exactly to a particular eigenstate, just arbitrarily close by (as I already stated in my original question). – WIMP Jan 23 '11 at 11:15
I've posted an updated answer to answer the question as I now understand it. – Joe Fitzsimons Jan 23 '11 at 12:22

## 5 Answers

Short answer to "Can this theory be fundamentally deterministic?": No. Decoherence is the diagonalization of the density matrix in a preferred basis, with the off-diagonals vanishing at late times. Since you can get the same final diagonal matrix from several possible initial pure states of the system under consideration, there's a necessary loss of information and irreversibility. (I'm guessing this is what you meant by non-deterministic)

A bit more detail: Decoherence proceeds by the rapid establishment of entanglement-induced correlations between the system and the infinite degrees of freedom of the decohering environment. The second law prevents this process from being reversible (since S has to always increase, and S is zero for the pure state, while it is greater than zero for the decohered mixed state). If you take the second law to be fundamental, then the non-determinism here is fundamental too.

UPDATE: The updated question now refers to the full evolution of system + environment, in other words, the entire universe. Since there's nothing else for the universe to entangle with, it will remain in a pure state and evolve deterministically for ever if it always was in a pure state. I however don't know if the universe is in a pure state or a mixed state. Anyone?

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I would answer the same thing, so plus one point. ;-) – Luboš Motl Jan 21 '11 at 19:29
Another way of saying this is the process is not unitary. – Lawrence B. Crowell Jan 22 '11 at 3:34

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This isn't technically correct. Decoherence can be the result of an entirely deterministic (and unitary process), since you only care about the reduced density matrix for the system in question. Open quantum systems -do- decohere, even though the entire process is unitary. The larger wavefunction of the entire system is still pure, but the state of the local system becomes mixed. – Joe Fitzsimons Jan 22 '11 at 6:26
My understanding is that the evolution turns non-unitary once you've traced out the environmental degrees of freedom. That's not what I'm asking for. Also, time evolution doesn't need to be unitary to be deterministic. – WIMP Jan 23 '11 at 10:34

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I concur with the criticisms of this answer: you are making the mistake of only considering the subsystem in question, not the total system. The subsystem loses information, yes, but it's lost to the total system, which may well be reversible as a whole. – Greg Graviton Jan 23 '11 at 14:19
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I don't disagree with the other answers, but I want to try to use different words:

Evolution of a quantum state is deterministic in the sense that it is given by the Hamiltonian. Quantum mechanics doesn't need anything beyond unitary evolution. So in that sense, the answer is deterministic.

However, decoherence means that eventually a quantum state may evolve into a superposition of very nearly orthogonal states, which for large enough systems will resemble to arbitrarily high precision the answer you might get from assuming that there is a nondeterministic, nonunitary process of "wavefunction collapse."

For all practical purposes, since we are large classical observers, we can observe only one such nearly-orthogonal combination, and the interference with other possible outcomes will be unmeasurably small. So, in this sense the outcome is "nondeterministic."

If this seems counterintuitive to you, consider something analogous about classical statistical mechanics and thermodynamics. If I start with a collection of gas molecules all bunched up in one corner of the room, it is a very atypical (low-entropy) state under any natural coarse-graining of phase space. Now, by entirely reversible interactions, it can become a typical (high-entropy) state with molecules scattered all over the room. This process appears to have lost information, in the sense that I would have to do many, many difficult measurements to ascertain that a short time before this was a very special state indeed. But really, the underlying physics is deterministic, so in principle the final state remembers where it came from, although for all practical purposes if I tried to evolve it backwards I would never discover the right answer. (To be clear, I'm not claiming a very sharp analogy here. But I'm saying that the notion that microscopically deterministic evolution can be consistent with apparent or "for all practical purposes" loss of determinism in real observations is something that might be more intuitive in this context.)

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Good answer! – Joe Fitzsimons Jan 22 '11 at 6:28
It doesn't seem counterintuitive to me, but that wasn't my question. I'm not asking "for all practical purposes." I know that for all practical purposes it will appear non-deterministic in the sense that we won't be able to predict what's going to happen -- too many variables in the environment. I'm asking if the time evolution of the system is fundamentally 'in principle' deterministic. That of course can only be before you've averaged over the variables of the environment. – WIMP Jan 23 '11 at 10:41

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Then I would say yes, it is "in principle" deterministic; the Hamiltonian specifies how everything evolves. Most of my answer was just trying to explain how to reconcile this with the observation that quantum effects look nondeterministic. – Matt Reece Jan 23 '11 at 15:41
Thanks. Yes, that was also my understanding. – WIMP Jan 24 '11 at 7:40
+1 for thermodynamics mention, I think some day decoherence and 2nd law will shake hands – HDE Jan 19 '12 at 0:16

UPDATE: It seems that we have not been answering the question WIMP intended. Here is an updated answer to deal with what I now understand to be the question: Given any unknown quantum state $|\psi\rangle$, can there be any deterministic process which will make it collapse onto a particular state $|\phi \rangle$, if $|\phi \rangle\langle \psi| \neq 0$?

The answer to this question is no, because it violates the linearity of quantum mechanics, allowing us to distinguish between non-orthogonal states. This is trivial, because states orthogonal to $|\phi \rangle$ will have zero probability of collapsing onto it. This may not seem like a big deal, but it turns out that linearity is fundemantal to quantum mechanics on many levels. If we remove this constraint, then entanglement can be used to signal, and hence create problems with causality. No signalling seems one of the most fundamental features of physics, showing up in many independent theories (electromag, quantum mechanics, relativity, etc.).

To see how this can be done, consider an entangled state $\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$. This is the anti-symmetric state: for any basis $\sigma$ a measurement resulting in outcome $m$ will leave the other qubit in the opposite eigenstate of $\sigma$. Thus, if you could deterministically collapse onto the state $|0\rangle$ then you can be sure the your half of the EPR pair was not left in state $|1\rangle$ after the measurement on the other half. So, for Alice to communicate with Bob, she need only choose to measure in the $X$ or $Z$ basis. Measuring in $X$ will mean Bob receives the output $|0\rangle$ with probability 1, where as measuring in $Z$ will return result $|1\rangle$ with probability $\frac{1}{2}$. Although this is probabilistic, you can repeat the process arbitrarily many times to get exponentially close to perfect communication. This instantaneous communication breaks causality.

If you allow all states to collapse to the target state, then the only solution is a channel which swaps the state with another ancilla system. Systems which can perform such deterministic collapse can always be used to signal, as well as allowing all sorts of additional weirdness like efficient solutions to PSPACE-complete problems in computation and time travel. As a result, this is totally impossible within the current framework of physical theories, and there are very substantial reasons to believe that it is a feature of any physical theory that is valid in our world.

The answer is no, if by deterministic you mean possessing a local hidden variable interpretation. This follows directly from the observed violations of Bell's inequality, which ever interpretation of quantum mechanics you choose (what you are referring to is known as the Everett interpretation of quantum mechanics).

Bell's inequality works as follows: Given to possible local measurement operators ($A_i$ and $B_i$) at each of two localions $i \in \{1,2\}$, what is the maximum value of the expectation value of $\langle A_1 B_1 + A_1 B_2 + A_2 B_1 - A_2 B_2\rangle$. What Bell showed was that this can take on a value of at most 2 for any local hidden variables theory. However quantum mechanics allows it to take on values up to $2\sqrt{2}$, and many experiments have recorded violations of this inequality, showing values in the range $2 < v \leq 2\sqrt{2}$. This essentially rules out a local hidden variable model.

If, however, you mean can the unitary interaction of two particles give rise to decoherence, then the answer is yes, as follows: Imagine two particles initially in the state $1/\sqrt{2}(|0\rangle + |1\rangle)$. Now imagine they interact via an Ising interaction. After a certain time, they will be in the joint state $1/2(|00\rangle - |10\rangle - |01\rangle + |11\rangle)$. This is still a pure state, and so no decoherence has occurred. However, imagine one of these particles moves off far away (into the environment). If we only have access to one of these particles, then its reduced density matrix will be $1/2(|0\rangle \langle 0|+|1\rangle \langle 1|)$, which is simply a classical random distribution over the two orthogonal states, the same as would occur do to a collapse of the wavefunction.

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Thanks for the reply. Even though you've correctly interpreted the meaning of deterministic, you haven't answered my question. You're saying the theory can't be deterministic because that would be in conflict with experimental tests on Bell's inequality. That's not correct. The theory could also be non-local instead, you just need to violate one of the assumptions to prove the theorem. – WIMP Jan 23 '11 at 10:38

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@WIMP: The first line of my answer reads: "if by deterministic you mean possessing a local hidden variable interpretation". Certainly global hidden variables can be made to work, but then they always can. – Joe Fitzsimons Jan 23 '11 at 10:42
Yes, I know. But that non-local hidden variable theories are not excluded by Bell's theorem isn't an answer to my question. If you want to pursue that line of thought, the question would then be whether decoherence, rspt the entanglement with the environment is a sort of non-locality that spoils the assumptions to Bell's inequality and thus, isn't excluded by experiment. Also, I was wondering about the possibility of deterministic evolution from a theoretical rather than an experimental point of view. – WIMP Jan 23 '11 at 10:53
@WIMP: Can you please revise the question to make it clear exactly what you are asking? It's currently not clear either from the question or subsequent comments. – Joe Fitzsimons Jan 23 '11 at 10:59
@ Joe: Thanks for the update. Two things: First, you don't need to collapse exactly into |0>, you just need to get close enough so it would 'for all practical purposes' appear to be an eigenstate, see question & update. Second, that the evolution is fundamentally deterministic doesn't mean that you could in practice deterministically collapse. (Besides this, I didn't ask if it leads to instantaneous messaging as you seem to think. Also, instantaneous messaging doesn't necessarily cause problems with causality, but that's a different point.) – WIMP Jan 24 '11 at 9:13
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Not sure if you've ever heard of the De Broglie-Bohm pilot wave interpretation of QM, but it is a fundamentally deterministic interpretation.

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Yes, I've heard of it. But my question wasn't if there is any deterministic interpretation of QM, but if the standard interpretation with decoherence instead of collapse can be deterministic. – WIMP Jan 24 '11 at 7:45

My answer is NO. There is a problem in many QM considerations that we have an isolated system obeying (deterministic!) Schrödinger equation that is a subject of mystical "measurements" that introduce non-deterministic behavior. But one can't make a measurement and leave system isolated (indeed saying that system is isolated is always an approximation in QM, and much heavier than in CM). In fact measurement is an act of introducing interactions with measuring apparatus, measurer, coffee measurer drinks, ... -- so the measurement result can be in theory calculated, but would involve inaccessible and enormous amounts of information. This makes it practically non-deterministic, but fundamentally it is no better than classical chaos.

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Actually, you are arguing that the answer is yes, rather than no. I haven't asked whether it is practically non-deterministic, but whether it can be fundamentally deterministic though appear non-deterministic (much like chaos indeed). – WIMP Jan 24 '11 at 7:48
My answer to "Does decoherence need non-determinism?" is No (-; – mbq♦ Jan 24 '11 at 10:19
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