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Is the state of a particle arriving from outside an observer's light cone merely unknown, or is it quantum mechanically indeterminate?

I guess the question needs a background in the form of a gedankenexperiment, so let's assume that someplace far away and long ago, someone (person B) built a laser that emits individual vertically polarized photons spaced at uniform time intervals, then passed the laser beam through a polarizing beamsplitter tilted at 45 degrees, then directed the emerging beam in our direction. Person B also sets up a detector to detect any photons deflected by the beamsplitter, thereby obtaining a set of measurements we can represent as a random string of 1's and 0's. Person B knows that the beam on its way toward us is the complement of the beam he has detected, and can be described as the complement of the string of 1's and 0's he detected. So, from Person B's perspective the photons in the beam are sent toward us in a way that corresponds deterministically to the 1's and 0's he detected.

Because it takes time for Person B's light beam to get to us, we know that the beam was emitted at a point outside our light cone. Because the beam eventually does get to us, we know that at that point of arrival our light cone intersects with the light cone of the emission event.

IF we know (perhaps because Person B sent us a text message on his laser beam) that Person B set up his experiment the way he did, we can detect the timing with which the incoming photons arrive and thereby infer the string of 1's and 0's Person B has recorded. So far, this is essentially identical to establishing a quantum key (can't correctly say that the key is "transmitted"). Person B's random string is entangled with the random string we detect at our end.

Edit 8/11/18: Now reduce the number of photons in the beam to two. That is, we can receive only (1,1),(1,0),(0,1) or (0,0), and Person B can, correspondingly, only have (0,0),(0,1),(1,0), or (1,1) at his end. In effect, Person B is sending us the value of two qbits which can each have either the value 1 or the value 0. (I know-- some more photons would need to be sent to tell us when to start our clock, etc., but I assume we can ignore that without confusing the results.) The point is that Person B can send us a quantum object whose state he knows, and he is sending it from outside our light cone.

This seems to be a significantly different situation than the standard EPR scenario, in which travelers A and B carry entangled quantum objects to widely separated points and then perform measurements, because in the standard scenario the entangled objects and the travelers have a common origin so they are always within each others' light cones.

Here is the first part of my question, which seems easy to answer:

Does the quantum object (the (1,1),(1,0),(0,1) or (0,0) signal) arrive in a mixed state? My first inclination is to say "no", because Person B knows what it's going to be. But in an EPR experiment, the order in which the measurements occur depends on the frame of reference -- so it's improper to say that one traveler's measurement causes the other traveler's measurement to have the same value. So, in some sense, in the scenario I'm proposing, the combined states of Person B's recorded signal and our detected signal are a single quantum object. That is, from our perspective Person B's recorded signal is not determined until we detect the incoming signal (billions of years later). In other words, the signal should arrive in a mixed state.

The second part of my question is: "Does every particle arriving from outside our light cone arrive in a mixed state?". I think the answer is "yes" (complicated somewhat by possible entanglements between the arriving particles), but haven't come up with convincing arguments. The closest I've come to an argument is to point out that the state of a particle arriving from outside our light cone is inherently unknowable in advance -- which sounds a lot like the definition of a mixed state.

S. McGrew
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  • I'm not 100% sure I follow your scenario. However, it reminds me greatly of the thinking behind the delayed choice quantum eraser, which can be set up in a way which pushes these "light cone" limits. Is this similar to the question you are exploring? If not, can you help me undersatnd why they aren't similar? – Cort Ammon Aug 11 '18 at 01:23
  • My scenario is possibly related to delayed choice quantum eraser experiments, but I think it may be different because no choices are made in my scenario. One way to look at the issue is as a variation of a Schroedinger's Cat experiment: Let's say Person B dies if he detects a (0,1), and lives if he detects a (1,0). Then when we receive and detect his signal, we know if he lived or died. But if the signal is indeterminate until we detect it, Person B's life or death is/was indeterminate (mixed) until we detect the signal. – S. McGrew Aug 11 '18 at 02:11
  • Perhaps an experiment could be designed to test the indeterminacy of the states of photons arriving from distant stars. I know the usual Bell's theorem experiment requires generating a lot of photons that are known to have precisely the same mixed state. But perhaps an experiment could make use of, e.g., superradiant emission that has been detected in the vicinity of some stars. – S. McGrew Aug 11 '18 at 02:18
  • It still feels really like the quantum erasers. Everything you say reads the way the quantum erasers are setup. The only difference is that the erasers deal with two distributions -- the bimodal distribution if the photon goes through a slit, or the interference pattern if it goes through both. In your case, you're stretching it to a binary distrition: alive or dead, but the pattern is the same. You're still seeing superpositions that feel "funny." – Cort Ammon Aug 11 '18 at 03:57
  • Maybe. The question, though is about the whether or not particles that arrive from outside the light cone - of which we inherently can have no knowledge - necessarily have mixed/indeterminate states. – S. McGrew Aug 11 '18 at 05:15
  • Maybe that's the question. How does a particle arrive from outside your light cone? The way I read it, that calls for the particle to travel faster than the speed of light. – Cort Ammon Aug 11 '18 at 05:23
  • I don't think it calls for the particle to travel faster than the speed of light. Any two events at a given moment (in the same rest frame) but separated by a long distance are in light cones that are initially separate, but that intersect at a later time. – S. McGrew Aug 11 '18 at 13:16
  • Different observers can assign different states according to the information they have available. This is perfectly normal and does not lead to any inconsistencies. From person B's point of view, the photon beam heading towards person A is in a pure state. From person A's point of view, of course the arriving photon beam is in a mixed state. With this in mind, I don't understand what you mean by "does the quantum object arrive in a mixed state"? – Mark Mitchison Aug 11 '18 at 18:09
  • I'm confused by your first sentence. The measured states are unambiguous. Let's say that Person B waits a while after sending the signal beam, then transmits a simple text message describing the photon sequence he measured. Person A receives the complementary photon sequence, then later receives B's description of the complement. Person A must find that his own sequence is the complement of the sequence person B describes. I don't think person A can "assign" a state; he can only measure it, and unless he does similar experiments many times, he can't know if the state was pure or not. – S. McGrew Aug 11 '18 at 19:49
  • I'm not sure what kind of experiment could test the purity of the signal photons' state. Even when person A receives confirmation from B, he doesn't know if the state was pure or not, because B's state is entangled with the group of signal photons he sends to A. It's not the usual kind of entanglement, but it surely must be entanglement. Rearranging the experiment so that the description arrives first would make mixed-state entanglement harder to believe, but it shouldn't matter which one of a pair of entangled objects is observed/measured first by Person A. – S. McGrew Aug 11 '18 at 20:15
  • @S.McGrew I agree. By "assigning" a state I am talking about person A assuming that the subsequent photons will obey the same statistics that they have already measured after many repeated experiments (assuming the beam of photons continues for a long time). I'm afraid I still don't really understand the question though. In this example it is clear that person A will measure a mixed state. But now remove the beamsplitter and send all photons to person A. Now they will see pure-state measurement statistics. Isn't this already a counterexample to your conjecture? – Mark Mitchison Aug 11 '18 at 23:56
  • I don't think "arrives in a mixed state" is a good way to state it. The state arrives in some (pure) state, it is our ignorance of what that state is that makes it useful for us to describe it as a mixed state. Part of an entangled system is always (describable by) a mixed state. Doesn't this answer your first question? I don't understand why you bring causation into the argument. The measurement outcomes are correlated but not causally connected (see also https://physics.stackexchange.com/a/418462/58382) – glS Aug 12 '18 at 10:59
  • Also, about your last comment, it is possible to test the purity of a state in a number of ways, essentially by testing whether it displays the effects of interference upon unitary evolution. One example that comes to mind is https://arxiv.org/abs/1112.1027 – glS Aug 12 '18 at 11:03
  • I think Bell-type experiments have mostly disproven hidden variables, and so make a clear distinction between mixed states and pure-but-unknown states. Your last link, [https://arxiv.org/abs/1112.1027], gives hope that it might be possible to design a suitable experiment using light from a distant superradiant source. Would you like to join this chat: [https://chat.stackexchange.com/rooms/81528/discussion-between-cort-ammon-and-s-mcgrew]? – S. McGrew Aug 12 '18 at 13:04
  • Here is an article about a recent experiment that's somewhat relevant, and might suggest an experiment to test this-- [https://phys.org/news/2018-08-ancient-quasars-quantum-entanglement.html]. – S. McGrew Aug 20 '18 at 13:37

2 Answers2

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To answer this, I have to be careful. We are measuring things, which may have a state described by a superposition, and then that measurement provides classical information.

I think the answer to your question is that the experiment is flawed because the 2 photon system can transmit more than 2 symbols. It can transmit (0, 0) (0, 1) (1, 0) and (1, 1). However, to cover my bases, I've got a bit more written down here to see if things can be straightened out. I don't think any actual quantum weirdness enters the story. Nothing is in a meaningful mixed state until we add some additional constraints to your problem which weren't there in your original wording.

The light cone details turn out to not be important for understanding this case. Light cones only become important when you start to get into systems which appear to transmit classical information faster than the speed of light. Nothing in your system transmits such information.

You mention a random string being entangled. I think that may be causing confusion. A string is classical information. It is not entangled. Particles can be entangled. The photons which encode the string may get entangled.

So, from a QM perspective, Person B's machine emits two photons. Each of these reach the beam splitter, and each becomes an entangled pair of photons, such that if the photon is on one path it is not on the other. Person B observes which path the photon took, and acts accordingly. The other half of each entangled pair travels to Person A. Person A also observes which path the photon took, and acts accordingly. The laws of QM indicate that these two observations should be consistent. Either both people agree that the photon took the path to Person A, or both people agree that the photon took the path to Person B. Fortunately for us, this is also consistent with or normal understanding of physics, so this doesn't create any weird behaviors which need QM to explain.

What makes this a little confusing is that you ask if a signal arrives in a mixed state. To the best of my knowledge, when people start using the word "signal," they are referring to something classical. That classical signal is not in a mixed state. We can say that the two photons are in a superposition of states until they are observed.

Now, where I think this breaks down is that you state that the photon-pair can only be in one of two states (0, 1) or (1, 0). Actually, there are two other states which can occur if the photons are independent: (0, 0) and (1, 1). This is a natural probabalistic outcome.

So how could we ensure that the pair is never in those (0, 0) or (1, 1) states? One approach is a purely classical one: we simply reject pairs like that when we see them, and both Person A and Person B continue the process until we see (0, 1) or (1, 0). Such an approach does not need any quantum mechanics. (As an aside, this is actually a procedure is a classic trick for unbiasing a random number source which is accredited to von Neumann).

A second approach is to entangle the photons before they hit the beam splitter such that they must take opposite paths. This is easy to do with electrons and spins, so a lot of experiments use them to do this. Now both parties recieve an entangled pair of photons instead of receving two independent photons. Now we will be able to observe a "mixed state." Of course we can use classical measurements to see this probalistically (like we always can with mixed states), or we can do quantum measurements (such as weak measurement) which yield fun behaviors.

Now, fun things happen if B does not act immediately. What if B doesn't observe this entangled pair right away. What if B finds a way to hold onto them until after A observes it classically, but before A can use a classical channel to provide B with the results of that observation? What does A see? The answer is that A sees something consistent with B's future decision to observe the particles or not observe them.

This situation is the delayed choice quantum eraser. It's behaviors are well understood, and really nonintuitive unless you are comfortable with QM.

Cort Ammon
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  • Thanks for your careful answer. However, in the experiment I described, the laser source emits single photons, not pairs of photons. After the beamsplitter, each photon either goes to Person B's detector or travels a long distance to Person A's detector. So, there are no entangled pairs of photons. – S. McGrew Aug 11 '18 at 17:34
  • I don't see a problem with describing a signal as having a mixed quantum state. That's just shorthand for saying that the physical objects that convey the signal (in this case, the photons received by Person A) have mixed quantum states. The essence of the Schroedinger's Cat paradox is that the cat - a classical object - must exist in a mixed state unless some extra (unproven and unnecessary) rule is invoked. In the past few years, experiments have shown that a mechanical resonator - a classical object - can be put into a mixed state: https://arxiv.org/pdf/1703.02548.pdf. – S. McGrew Aug 11 '18 at 17:43
  • @s. Mcgrew if the laser is outputting uncorrelated photons, the situation is well modeled as a purely classical system. No qm is needed. – Cort Ammon Aug 11 '18 at 20:13
  • Aren't the photons from the laser just about as fully correlated as they can possibly be? They're all identical, with identical (vertical) polarization, emitted one at a time at precisely regular intervals. Where the QM randomness comes in is at the tilted beamsplitter which intercepts the laser beam. – S. McGrew Aug 11 '18 at 20:22
  • They are coherent, but not entangled. Their interactions with the beam splitter is independent. If the first photon goes to A, the second one is not obliged to go to B – Cort Ammon Aug 11 '18 at 20:43
  • I guess I'm not communicating clearly. B fires two photons, one at a time. There are four possible results: B and A detect (1,1)&(0,0), (1,0)&(0,1), (0,1)& (1,0), or (0,0)&((1,1) respectively. That is, A always detects the complement of what B detects. But what B detects is indeterminate until he detects it (because of QM). The two photons (first photon, second photon) are not correlated. What is correlated is the sequence of detections by A and B. If B' detects a photon corresponding to the laser's first photon emission, A will not detect it - and vice versa. – S. McGrew Aug 11 '18 at 22:49
  • Ah-- I made an error in framing the question. I'll fix it now, and maybe that will help. – S. McGrew Aug 11 '18 at 22:53
  • I would say that the sequence of detections by B is a quantum object and that the sequence of detections by A is as well. A definition of entanglement (from wikipedia): Quantum entanglement is a physical phenomenon which occurs when pairs or groups of particles are generated, interact, or share spatial proximity in ways such that the quantum state of each particle cannot be described independently of the state of the other(s), even when the particles are separated by a large distance—instead, a quantum state must be described for the system as a whole. – S. McGrew Aug 11 '18 at 23:10
  • Of course B is convinced that the photons he's sending on to A have fully determined polarizations. But because B sends the photons from outside A's light cone, B appears to be in a Schroedinger box from A's perspective. When A detects the photons, it's like peeking into the box to observe B's state. Another way to formulate the question is: Are B and his apparatus in a mixed state until A detects the photons? – S. McGrew Aug 11 '18 at 23:19
  • @S.McGrew As a related question, a coin can be flipped head or tails. While the coin is in the air, is the coin in a mixed state? Is the coin 50% heads and 50% tails? – Cort Ammon Aug 11 '18 at 23:55
  • Let us continue this discussion in chat. – Cort Ammon Aug 12 '18 at 00:00
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In this question I asked if a particle arriving from outside our light cone has a totally undetermined state. I've come to the conclusion that the question was poorly posed.

The problem is that, even if the answer were "yes", it is impossible to know by measurement what the full quantum description of an arriving particle is. We can't even know if the particle arrives, much less its type or momentum or spin, without measuring it; and measurement "fixes" its state. If the particle's wave packet is totally undetermined before measurement, then the identity of the particle and its other descriptors exist only in a probabilistic sense prior to measurement.

Some of the experiments I've been able to imagine that might answer the question require a stream of identical particles to arrive from outside the light cone: if they all turn out to be electrons, or all spin-up, then we can conclude that their quantum properties are at least partially determined before they arrive. However, this kind of experiment would require prior knowledge (from outside the light cone) that the arriving particles really are identical.

Another class of experiments that might answer the question would simply look for patterns or correlations among measurements of vast numbers of non-identical particles arriving from outside the light cone – and in fact we're doing that type of experiment all the time: all optical and radio astronomy is measurement of non-identical particles arriving from outside the light cone. And, we do observe correlations among those measurements: photons are highly correlated with respect to direction of arrival, time of arrival, frequency, and even polarization. Simply put, telescopes can form images of distant stars and galaxies, which is only possible because of those correlations.

That said, the original question should be re-phrased: "Are there correlations between quantum particles arriving from outside the light cone?". The answer to that question is an obvious "yes". Those correlations reveal – or constitute- the structure of the universe that exists independently of our perspective.

S. McGrew
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