This just started to bother me after reading yet another entangled particle question, so I hate to ask one myself, but...

If we have two entangled particles and take a measurement of one, we know, with certainty, what the outcome of same measurement of the other particle will be when we measure it. But, as with all fundamental QM, we also say the second particle does not have a definite value for the measurement in advance of making it.

Of course, the particle pairs are always described by an appropriate wave function, and measurement of half the pair changes the probabilities associated with each term. But, and this is the part that bothers me, with knowledge of the first measurement, we know the outcome of the second measurement. There's nothing strange or forbidden or anything about this, no non-local affects, no FTL communication, etc.

But does it still make sense to talk about the second particle not being in a definite state prior to the measurement? Or can we consider a local measurement of one half of an entangled pair to be a measurement of the other half? The latter option feels distasteful because it implies communication between the particles when communication is possible, which is obviously wrong, but I find it hard to pretend the second particle in the pair, whose state I know by correlation, is somehow not in that state before I confirm it by measurement.

I should clarify that my uneasiness comes from a comparison with the usual "large separation" questions of entangled particles, where our ignorance of the first measurement leads us to write an "incorrect" wavefunction that includes a superposition of states. Although this doesn't affect the measurement outcome, it seems there is a definite state the second particle is in, regardless of our ignorance. But in this more commonly used scenario, it seems more fair to say the second particle really isn't in a definite state prior to measurement.

As a side note, I realize there's no contradiction in two different observers having different descriptions of the same thing from different reference frames, so I could accept that a local measurement of half the pair does count as a measurement of the other half, while a non-local measurement doesn't, but the mathematics of the situation clearly point to the existence of a definite state for the second particle, whereas we always imply it doesn't have a well-defined state (at least in the non-local case) before measurement.

At this point, my own thinking is that the second particle is in a definite state, even if that state could, in the non-local separation case, be unknown to the observer. But that answer feels too much like a hidden variables theory, which we know is wrong. Any thoughts?


3 Answers 3


According to quantum mechanics, the particles - whether it is the first particle or the second particle or any other particle in the Universe - refuse to have any well-defined state or property prior to the measurement.

The only "kind of" exception is the case - relevant for a maximally entangled scenario - in which we are interested in a property of the second particle - or any other particle - that is predicted to take a particular value with probability equal to 100 percent. Of course, if the probability is 100 percent, then you can be sure what the measured property will be, and you may assume that this value of the property exists even before the measurement.

However, for the very same particle that has some value of the quantity equal to something at 100 percent, there still inevitably exist other observables that are not known. (Just design a Hermitian observable with random off-diagonal matrix elements.)

In the basis of eigenstates of those other properties, the probability amplitudes are generic, and some of the options have probabilities that differ from 0 percent as well as 100 percent. For those quantities - and it is a majority of observables - the usual prescriptions of quantum mechanics hold: the value is not determined prior to the measurement. It is not just unknown to the physicists: it is unknown to Nature. A picture in which those observables are determined leads to wrong predictions and contradictions.

  • $\begingroup$ š All perfectly true. But my problem is not with the majority of observables that are still unknown, just with the one observable whose state is known with 100% probability. Perhaps it is merely a problem of interpretation or the sloppiness of language, but the question of whether, with regard to the one known observable, the state exists prior to the measurement, still feels up in the air to me. If you think we can declare an exception for 100% probability outcomes and want to say that the question of having a state is just language and not physically meaningful, I could accept that. $\endgroup$
    – Mitchell
    Feb 1, 2011 at 7:16
  • $\begingroup$ Dear Mitchell, it's a philosophy, but if $X=X_0$ may be predicted with certainty - if the probability is 100 percent - then assuming that $X=X_0$ was true prior to the measurement can't lead to any contradictions. On the other hand, you don't have to insist that the "reality" existed prior to the measurement. The latter attitude treats the world more uniformly because there's no real "qualitative difference" between propositions that have 100% and 99% probabilities. Whatever answer you choose for the 100%-statements is clearly just philosophy, not physics accessible to tests. $\endgroup$ Feb 1, 2011 at 8:25
  • $\begingroup$ š I'd say that's a fair assessment. I'm going with the "not a physically meaningful statement" version, as I find that answer to satisfy a lot of these scenarios. $\endgroup$
    – Mitchell
    Feb 1, 2011 at 17:00
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    $\begingroup$ Good choice, @Mitchell. This "positivist" attitude is what the founding fathers of QM emphasized. If you don't need to talk about something (a philosophical dilemma) that has no testable consequences, don't talk about it. $\endgroup$ Feb 1, 2011 at 18:41
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    $\begingroup$ I'll talk anyway, or, rather, I'll let Tomonaga talk, making your point more precise. « the statement the probability that the particle is at $x$ is not a proper statement and should be rephrased as follows.» ... [ahem, it's a phrase, not a statement...] « the probability that one obtains the experimental result $x$ when an experiment of measuring the location is carried out with the quantum-mechanical particle in this state. » In my view, Dirac took the same consistently careful attitude. $\endgroup$ Jan 17, 2012 at 20:19

Before measurement, we have a bipartite entanglement between the two particles. After measurement, we have a multipartite entanglement between the two particles, the measuring apparatus, and the environment, which includes the observers. It's only with respect to an Everett branch of the wavefunction that we have definite values.


One aspect that has not been covered by the two answers is more fundamental, I think. Since the particles are entangled, they are no longer separate systems: it makes no sense to say the first particle has a quantum state at all, the axioms of QM do not allow us to say that because the first particle is no longer a system... IMHO it would be clearer if we didn't even call it « the first particle » anymore: that kind of language is only valid for isolated systems or separable states. In particular, there isn't « one term » that corresponds to one particle: $$\vert up \rangle \otimes \vert down\rangle - \vert down\rangle \otimes \vert up \rangle $$ has two summands, but if those are the terms you mean, neither one has any close connection with one rather than the other particle, and each summand has two factors, but that makes four factors altogether, again none of which have any physical significance by themselves for the actual « first particle » merged in this combined system.

Being scrupulously careful like this avoids the paradoxes and questions automatically, since they cannot even be phrased in QM if one is careful enough.

Using this consideration allows us to answer the OP

« At this point, my own thinking is that the second particle is in a definite state, even if
that state could, in the non-local separation case, be unknown to the observer. But that answer
feels too much like a hidden variables theory, which we know is wrong. Any thoughts?»

rather differently and, I think, more fundamentally, more radically. The phrase « the second particle » makes no sense so the question whether or not it is in a definite quantum state, with its own quantum-mechanical wave function, cannot even be asked, it is nonsensical.

The way to talk carefully about the physical set-up is that a combined system of
entangled-pair-of-particles has a certain spread in position space, and a measurement which is localised to one location is getting ready to measure the combined system. We have arrange the production and dynamics of the production of this pair-system so that if it had not been an entangled pair, the first particle would now be located in the location of that measurement apparatus. And, similarly, the second particle would now be located where we have also prepared another measurement apparatus. Then we switch the apparati on. The apparati are designed and prepared so that if the pair had not been entangled, they would be ready to measure the spin of a single particle. ETc. etc. This is phrased extra carefully in case it is an entangled electron--proton system, for example.

But the usual situation is entanglement of two identical particles, and this introduces even another reason why it makes no sense to distinguish the particles by calling them « first, second » and in other experiments, one lets quite a few entangled pairs evolve to places where you haven't even put any detectors, you haven't even arranged that you are sure the first particle would have gone to where you put the detector, you just wait until the detector fires at which point you know, for various reasons, the other detector should also. Still, the general principle that it is now a combined system and there are no particles anymore (i.e., after the entanglement and before a measurement disturbs the entanglement and re-establishes separability---now not every measurement will do that but we are particularly interested in one that does) holds good and makes all seeming paradoxical questions nonsensical, but for a different reason than Prof. Motl brought up.

That is, it is also the first axiom of QM which forbids these phrases, not merely the fifth, the one Prof. Motl brings up, which does have some exceptions, as he admits.


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