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I have a radioactive particle in a box, prepared so as to initially be in a pure state

$\psi_0 =1\ \theta_U+ 0\ \theta_D$

(U is Undecayed, D is Decayed). I put a Geiger counter in the box.

Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger

$\psi_t =a\ \theta_U+ b\ \theta_D$

Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.

However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.

But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.

I want to know where the mistake lies.

In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is

$\psi_t =1\ \theta_U+ 0\ \theta_D$

which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).

How do I deal with this constant 'passive' measuring?

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    $\begingroup$ Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer. $\endgroup$ – JPattarini Sep 24 at 16:35
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    $\begingroup$ Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur. $\endgroup$ – Stéphane Rollandin Sep 24 at 17:20
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    $\begingroup$ Maybe the detector here is not actually be the thing collapsing the wavefunction of the nucleus. The detector is collapsing the wave function of the emitted gamma ray (or whatever). What actually collapses the nucleus is its interaction with whatever causes it to decay in the first place (the vacuum, stray radiation, other particles?) $\endgroup$ – KF Gauss Sep 25 at 4:31
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    $\begingroup$ The Geiger counter only detects that the radioactive atom has decayed because the atom ejects a particle that passes through the Geiger tube (and even if the ejected particle does pass through the tube, it's not 100% guaranteed that it will actually be detected). So I don't think it's fair to say that the Geiger counter is continuously measuring the state of the atom. $\endgroup$ – PM 2Ring Sep 25 at 4:42
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    $\begingroup$ Nothing profound going on here; it's just another case of wrongly treating "collapse" as a physical change/event. It's not. $\endgroup$ – R.. Sep 26 at 1:21
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Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.

Here is a physical answer to your question in the framework of QM: Look at the a position wave function of the decayed particle $\psi(x)$ (*if it exists: see bottom of post if you care). When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you get a characteristic decay time. This picture is a good intuition, but also an inexact/insufficient answer, because the notion of "reaches the detector" is only heuristic and classical. A full quantum treatment of this problem should give us more: a probability distribution in time $\rho(t)$ for when the particle is detected. I will come back to this.

So what about the Zeno effect? Based on the reasoning you gave, the chance of decaying is always zero, which is obviously a problem! Translating your question to position space $\psi(x)$, your reasoning says the wave function should be projected to $0$ in the region of the detector at every moment in time that the particle hasn't been found. And in fact you're right - doing this does cause the wave function to never arrive at the detector! (I actually just modeled this as part of my thesis). This result is inconsistent with experiment, so we can conclude: continuously-looking measurement cannot be modeled by straightforward projection inside the detector at every instant in time.

A note, in response to the comments of Mark Mitchison and JPattarini: this "constant projection" model of a continuous measurement can be rescued, by choosing a nonzero time between measurements $\Delta t \neq 0$. Such models can give reasonable results, and $\Delta t$ can be chosen based on a characteristic detector time, but in my view such models are still heuristic and a deeper, more precise explanation should be aspired to. Mark Mitchison gave helpful replies and linked sources in the comments for anyone who wants to read more on this. Another way to rescue the model is to redefine the projections to be "softer", as in the sources linked by JPattarini.

Anyway, despite the above discussion, there is still a gaping question: If continuous projection of the wave function is wrong, what is the correct way to model this experiment? As a reminder, we want to find a probability density function of time, $\rho(t)$, so that $\int_{t_a}^{t_b}\rho(t)dt$ is the probability that the particle was detected in time interval $(t_a, t_b)$. The textbook way to find a probability distribution for an observable is to use the eigenstates of the corresponding operator ($|x\rangle$ for position, $|p\rangle$ for momentum, etc) to form probability densities like $|\langle x | \psi \rangle|^2$. But there is no clear self-adjoint "time operator", so textbook quantum mechanics doesn't give an answer.

One non-textbook way to derive such a $\rho(t)$ is the "finite $\Delta t$ approach" mentioned in the note above, but besides this there are a variety of other methods which give reasonable results. The issue is, they don't all give the same results (at least not in all regimes)! The theory doesn't have a definitive answer on how to find such a $\rho(t)$ in general; this is actually an open question. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a weak point of the theory, which needs work. If you don't want to take my word for it, take a look at Gonzalo Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a completely satisfactory way. I am still learning more about these approaches, but if you are curious, the one I found most clean so far uses trajectories in Bohmian Mechanics to define when the particle arrives at the detector. That said, the measurement framework in QM in general is just imprecise, and I would be very happy if a new way of understanding measurement were found which gives a higher level of understanding of questions like this one. (yes I am aware of decoherence arguments, but even they leave questions like this unanswered, and even Wojciech Zurek, the pioneer of decoherence, does not argue that it fully solves problems with measurement)

(*note from 2nd paragraph): Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. Thus even this intuition doesn't always have mathematical backing.

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    $\begingroup$ It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at. $\endgroup$ – Juan Perez Sep 24 at 21:39
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    $\begingroup$ Speaking of which - modelling continuous measurement as infinitely many measurements in sequence would allow you to measure time with infinite precision, which would require infinite energy via the uncertainty principle. Perhaps we could model continuous measurement as a sequence of anytime-within-interval measurements where the interval length is determined by the energy of the measurement process. $\endgroup$ – John Dvorak Sep 25 at 5:22
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    $\begingroup$ All of this has been well known since the 90s and the corresponding theory is routinely tested in quantum optics laboratories (for example). It is extremely misleading to claim that there is somehow something mysterious or unknown about the description of continuous measurements. It's textbook material, see for example Wiseman & Milburn's book or this review, or just Google to find many more references. $\endgroup$ – Mark Mitchison Sep 25 at 10:06
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    $\begingroup$ OP's question asked about a model in which projection happens at every moment in time, and it is true that in such a model the particle is never detected. @MarkMitchison (and probably also the link from JPattarini ) is talking about a different (more fruitful) model for continuous measurement which chooses a time between measurements $\Delta t$. This model can yield reasonable results, though it depends on a parameter $\Delta t$ which doesn't have physical grounds (why would nature measure in particularly that interval?). I'm going to edit the answer to include that now. $\endgroup$ – doublefelix Sep 25 at 11:32
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    $\begingroup$ 1. The last few lines of OP's question pretty clearly suggest a continuously measuring device $\Delta t=0$, otherwise mentioning the zeno paradox would be irrelevant, & there would be no confusion. 2. While I think the "finite $\Delta t$ approach" is one of the better ones, in my subjective view the choice of $\Delta t$ requires some loose interpretation: choosing $\Delta t$= "minimal time in between two successive counts" is not clearly the same thing as "time in between successive measurements when no particles are found". $\endgroup$ – doublefelix Sep 25 at 12:28
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No, the detector is not always collapsing the state.

When the particle is in an undecayed state its wave function is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.

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    $\begingroup$ So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier. $\endgroup$ – Juan Perez Sep 24 at 21:42
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My take on this is that in the original thought experiment, you don't get to monitor the detector. When the detector detects, it kills the cat. But it doesn't tell you then. You only find out when you open the box.

If it tells you immediately, then you know immediately. And then there's the question whether the detector detects 100%.

If the Geiger counter detects 100%, then you could have 100 Geiger counters or 10000, and they would all detect the particle decaying. If they were all the same distance, they should all detect it at the same time. (Assuming the particle was not moving relative to them. Otherwise relativity might give them different times which would be 100% predictable.

I think it's more plausible that each detector detects a different photon. And the first single detector might easily miss a particular gamma ray photon.

So if there is only one radioactive particle, then if the geoger counter does detect it, then you know it's been detected and you know pretty much when. But if it hasn't detected it yet, there's an increasing chance with time that the particle has decayed and the geiger counter did not detect it and will never detect it.

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  • $\begingroup$ Deserves more upvotes. The state of the cat is purely a visual representation of the state of the atom, as we can't see whether an atom has decayed or not visually. It also serves as a memorable (if possibly a bit horrific) identifier for the thought experiment. - the whole point is to make it approachable to a layman. $\endgroup$ – Baldrickk Sep 25 at 16:23
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    $\begingroup$ @Baldrickk - that may be the purpose of the cat now when the Copenhagen interpretation is widely accepted, but Schoedinger introduced the cat in an argument against that interpretation, exactly because he thought the idea of a cat being in a superposition of alive and dead to be ridiculous. $\endgroup$ – Paul Sinclair Sep 25 at 17:58
  • $\begingroup$ With physics, if you have two different interpretations, but for every experiment you could possibly do they would both give the exact same result, then they are not really two different theories. They are the same theory. Talking about the difference is not physics but metaphysics or paraphysics or something. It isn't ridiculous if it corresponds with reality as well as we can measure it. $\endgroup$ – J Thomas Sep 26 at 23:17
  • $\begingroup$ How would 10000 Geiger counters detect the particle emitted by a single decaying atom? I suppose a very energetic gamma photon could pass through multiple detectors, but that's not very likely for a beta particle, and almost impossible for an alpha. $\endgroup$ – PM 2Ring Sep 29 at 8:45
  • $\begingroup$ I find it exhausting to make up explanations for things that don't happen. If the classical idea was right and the gamma ray traveled in a spherical shell, intensity decreasing with the inverse square law, then each detector would have the same chance of firing. If instead the photon travels in precisely one direction, not spreading at all, then the chance that an particular detector will be hit will fall by the inverse square law. is one of those right? I don't know. Probably not. But if there's room for 100 non-overlapping detectors, then wouldn't the chance for any one be less than 1%? $\endgroup$ – J Thomas Sep 30 at 2:45
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Your statements treat the quantum mechanical distribution as physical, whereas it is a mathematical function fitting the boundary condition of your experiment, i.e. it is the mathematical function describing a particle's probability of decay.

Probabilities are the same in classical mechanics, in economics in gambling, in population interactions. Take the probability of throwing a dice and coming up with six. For a true dice (not weighted) it is 1/6 of the time no matter whether you throw the dice or not , if you throw it you have a probability of 1/6. If a gambler has weighted the dice, maybe the probability curve is weighted towards 6, so it could be you have 1/3 probability to get a 6 with a weighted die.

You have a particle that can decay while sitting alone. The probability of its decay is given by $Ψ^*Ψ$ , by the solution of a mathematical quantum mechanical differential equation ( or maybe lattice QCD, which uses the solutions). Whether a geiger counter was there or not, one can calculate how many nuclei will have decayed given the probability distribution for the nucleus (a function of time in this case) and the time past.

The geiger counter is incidental , a second interaction with a $Ψ^*Ψ$ locally that has ideally 100% probablity to interact when a charged particle hits it. A tool for recording a decay. (as your eyes do not affect the probability of the dice coming up 6).

The states you write down are not quantum mechanical states. They may be logical mnemonics, but they do not have to obey quantum mechanical equations or postulates , they are not a $Ψ^*Ψ$ .

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    $\begingroup$ "You have a particle that can decay while sitting alone" The probability of its decay is, just as you say, given by Born rule. But Born rule is only used when a measurement is performed. What is the measurement here, since the particle is alone? $\endgroup$ – Stéphane Rollandin Sep 25 at 15:26
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    $\begingroup$ @StéphaneRollandin Probabilities are accumulated statistically measurements and are predicted by the solution of quantum mechanical equations. in the case of quantum systems. Decaying particles are quantum systems. One instance of a proabbility distribytion does not make a distribution.. Th measurement is a cumulative acquisition of data, that is what particle experiments are about, in order to test the quantum predictions. $\endgroup$ – anna v Sep 25 at 15:35
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    $\begingroup$ By itself, the system will evolve according to Schroedinger's equation, into a superposition of Decayed and Undecayed states. It won't change its state into Decayed unless it interacts with something (and the counter is the only other thing in the experiment). Or at least that's how I thought QM modeled. $\endgroup$ – Juan Perez Sep 26 at 17:38
  • $\begingroup$ @JuanPerez Quantum mechanics models pure states. Schrodinger's equation holds for pure states , a particle in a potential well. The system "decaying nucleus + Geiger counter" cannot be in a pure state .( like the one you wrote) The density matrix formalism is used if one wants to theoretically describe everything by quantum mechanics, which is certainly not your description. en.wikipedia.org/wiki/Density_matrix $\endgroup$ – anna v Sep 27 at 4:16
  • $\begingroup$ @JuanPerez It is a miss conception ( the famous Schrodinger cat) to think that an observer is needed for every quantum mechanical probability to manifest. If you have a box of uranium, and nobody observes it (and nothing) it will decay according to the decay law , inexorably, no matter how much time one measures it later. That is atoms decaying, that can be calculated. $\endgroup$ – anna v Sep 27 at 4:24
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Consider the Many-Worlds approach.

You have a wavefunction (an immensely complicated one, of course). Your amplitude for having heard a click steadily grows in magnitude.

No paradox if you look at it like this.

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    $\begingroup$ While I too find the many-worlds formulation best for thinking about this, the quantum Zeno effect persists in the many-worlds formulation as well, so I don't think this really answers the question. In other words, if the setup exhibits quantum zeno, your amplitude for hearing a click would not steadily grow in magnitude. $\endgroup$ – aquirdturtle Oct 11 at 22:27
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I think that “listening” even in the case of silence is already the measurement. You can only hope to hear something when there is a medium (air) that will carry the sound waves. This medium causes a continuous interaction between you and the Geiger counter. Only without the medium there is no interaction but then you can also not tell that the Geiger counter kept silence.

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    $\begingroup$ I think the box is only a good box if it does not transmit the sound of the Geiger counter outwards. So the Geiger counter also will be in mixed state until you open the box. $\endgroup$ – Adder Sep 25 at 13:27
  • $\begingroup$ @Adder You are right. But OP said that the missing click indicates that the particle did not decay. Thus, there is a contradiction: the wave function is still in a mixed state but we know for sure that the particle is not decayed. I simply wanted to resolve the contradiction: the wave function is not in a mixed state because the existence of the medium provides a continuous measurement thus a continuous collapse of the wave function (in the Copenhagen interpretation) $\endgroup$ – Hartmut Braun Sep 27 at 7:00

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