I Geiger counters and measurement

[jeeves]

TL;DR Summary

Do Geiger counters cause collapse when they don't click?

I have the following elementary confusion about how measurement and collapse work in quantum mechanics.

Place an unstable particle and a Geiger counter in a sealed box. The particle can be in two states: not decayed (N) or decayed (D). Scale time so that one unit of time after the particle decays, the Geiger registers the decay product with an audible click. We neglect direct interactions between the counter and the particle itself. Start the experiment at t=0. Let the superposition of states for the particle be a(t)*N + b(t)*D after time t, where a(0) = 1 and b(0) = 0, and we have b(t) -> 1 as t -> infinity.

Suppose I wait until time t=100, and I do not hear the counter click. What is the state of the particle?

I can think of two possibilities:

1) It is in state a(100)*N + b(100)*D. This is because the particle has evolved until time t=100, and the Geiger counter has not interacted with the particle or done anything to affect its state.

2) It is in state a(1)*N + b(1)*D. At t=100, we know there has been no decay product formed until at least t=99, so at t=99 we know the particle did not decay yet. Then the state at t=99 is N, so at t=100 the particle has evolved into the superposition one second after starting at N.

Which of these is correct, and why?

 

[hutchphd]

What is the purpose of the one second delay?

 

[jeeves]

The purpose is to model the fact that the decay product takes some nonzero amount of time to reach the detector.

To make the numbers simple, I set this to be one unit of time. You could of course put everything in terms of seconds, in which case the decay product would take some extremely small number of seconds to reach the detector.

 

[hutchphd]

And this means what?

Summary: Do Geiger counters cause collapse when they don't click?

We neglect direct interactions between the counter and the particle itself.

 

[jeeves]

It means that as an approximation, I ignore any sort of entanglement between the counter and the atom itself. I believe the only relevant interactions should between the decay product and the counter (since this is how Geiger counters work – they detect the decay via the decay product). We could imagine the box is very large and the counter is very far away from the decaying atom, for example.

If you feel this assumption is problematic, please me know why.

 

[hutchphd]

So Erwin's cat is killed by the poison gas with a delay rather than by a direct radioactive event. It's just the same old question. Might as well not create new scenarios for no particular reason

 

[jeeves]

I'm sorry, but I don't follow. I am not asking about any cat, or any interpretational question. I am only asking: What is the state of the atom at t=100, given the counter has not detected decay at that time?

I believe the answer has direct experimental consequences, since the time to decay after t=100, given that no decay has been observed, will depend on which state the particle is in at t=100. One could actually run this experiment multiple times to find the empirical decay statistics and determine the answer, so it's not just a question about interpretation. It's testable in lab.

 

[Nugatory]

It means that as an approximation, I ignore any sort of entanglement between the counter and the atom itself

That assumption does indeed simplify the problem - it eliminates the possibility that the detector will ever click.

 

[jeeves]

That assumption does indeed simplify the problem - it eliminates the possibility that the detector will ever click.

OK, then let us not make this assumption. What would the answer be then?

 

[hutchphd]

How do you "know" the state of the system at t=0?

 

[jeeves]

Let's say I measure the time to the first decay after t=0. Then by definition, no relevant decay has taken place at t=0.

 

[PeterDonis]

OK, then let us not make this assumption. What would the answer be then?

The answer is that you wrote down an incorrect state for the system in your OP. The correct state will look like this:

$$
\psi = A(t) \ket{\text{atom not decayed}} \ket{\text{counter not clicked}} + B(t) \ket{\text{atom decayed}} \ket{\text{counter clicked}}
$$

In this state, the atom and the counter are entangled. And, as @Nugatory says, without this entanglement, the detector will never click.

At time $t = 0$ we can assume that $A = 1$ and $B = 0$ (because we can assume we did some kind of preparation process for the system that ended with the atom being in its non-decayed state at $t = 0$), but over time both will be nonzero functions of $t$ because there is a nonzero probability of the atom decaying and the counter clicking, so $B \neq 0$, but there is also a nonzero probability of the atom not decaying and the counter not clicking, so $A \neq 0$ as well.

When we actually observe the counter click, as far as the basic math of QM is concerned, we replace the above state for the system with the new "collapsed" state

$$
\ket{\text{atom decayed}} \ket{\text{counter clicked}}
$$

But this doesn't have much practical effect since the atom has decayed so we can't make any further measurements on it, and the counter clicking just gets recorded in our experimental data. Whether the new state above is the "actual" state of the system or not after we observe the counter click depends which interpretation of QM you adopt, and that is out of scope for this forum (interpretation discussions belong in the interpretation subforum).

 

[Nugatory]

OK, then let us not make this assumption.

Without that assumption we must treat the problem as a single quantum system consisting of the particle and the Geiger counter (that's what entanglement means!) and given that its wave function at time $t=0$ is is $\psi_0=\psi(0)$ its wave function at time $t$ is $\psi(t)=e^{-i\hat{H}t/h}\psi_0$.
Now the answer to your original question is "neither" - that wave function cannot be written in either form that you suggest.

In practice we solve these problems using the Copenhagen interpretation or something similar, treating the Geiger counter as a classical measuring device not part of the quantum system. This approach gives us superbly accurate results for the tick rate of the Geiger counter, how long we need to wait for it to tick when we have exactly $N$ (possibly one) nuclei, and pretty much anything else that we care about. It does not, however, tell us anything about what's going on while the nucleus is interacting with the detector nor what the actual state of the total system is before the detector clicks.

 

[jeeves]

Thanks, Peter and Nugatory. That clears a lot of things up.

Let me try to ask the original question correctly. Let's follow the notation in Peter's post and let $\psi$ be the wavefunction of the entangled particle–counter system.

I perform the experiment and observe that after $t=100$ the Geiger counter has not clicked. What is the state of the atom after this observation?

Is it the state with $A=1$ and $B=0$, because I have "measured" the atom and found that it has not decayed? Or is it the superposition with coefficients $A(100)$ and $B(100)$?

I do not think this question is about interpretation, since the two possibilities should result in empirically different observed results, but please let me know if I am wrong.

 

[PeterDonis]

I perform the experiment and observe that after $t=100$ the Geiger counter has not clicked.

How can you "observe" this?

I do not think this question is about interpretation, since the two possibilities should result in empirically different observed results

Why? What would you expect the two different observed results to be?

 

[Nugatory]

What is the state of the atom after this observation?

There is no such thing as “the state of the atom” - we make a mistake as soon as we introduce the idea, the same way that we’re making a mistake when we do a division by zero and take the result at face value. The only state we have is that of the quantum system consisting of the counter and the particle, and the question is whether that state is “detector clicked and atom decayed” or “detector not clicked and atom not decayed”. These are mutually exclusive entangled states, and an observation of the detector will tell us which we have at the moment of the observation.

 

[PeterDonis]

an observation of the detector will tell us which we have at the moment of the observation.

I don't think this is correct, because "an observation of the detector", unless the OP has some very unusual (and unknown to date) experimental technique in mind, does not involve actually probing the state of the detector with any kind of interaction. If we have not observed the detector actually clicking, we have not made any measurement. In more technical language, no decoherence has occurred and we cannot say that either outcome has happened. But if we do observe the detector actually clicking, then decoherence has occurred and we can say that the "atom decayed, detector clicked" outcome has happened. So the two possibilities are not actually the same, since only one involves decoherence; but I suspect the OP is thinking of them as though they are the same and that is what is causing the OP's confusion.

 

[jeeves]

How can you "observe" this?Why? What would you expect the two different observed results to be?

I observe this by listening to the counter continuously between $t=0$ and $t=100$ and noting that there is no click during this time span. Because there was no click, and there was no decay before $t=100$.

Regarding the observed results, I admit I have no proof, but it would seem exceedingly strange to me if atoms prepared in the states with $A=1$ and $B=0$ , and $A(100)$ and $B(100)$, had the same decay statistics. Do you believe this would in fact be the case?

To steer fully clear of interpretation issues, I can ask the following related question. Repeat the experiment many thousands of times and record the time during each run where the Geiger counter first clicked. We obtain an (empirical) probability distribution $P_1$. Now, using this dataset, look at the time to first decay among the subset of atoms that first decayed after $t=100$, and subtract $100$. This gives a probability distribution $P_2$. Are $P_1$ and $P_2$ the same or different?

There is no such thing as “the state of the atom” - we make a mistake as soon as we introduce the idea, the same way that we’re making a mistake when we do a division by zero and take the result at face value. The only state we have is that of the quantum system consisting of the counter and the particle, and the question is whether that state is “detector clicked and atom decayed” or “detector not clicked and atom not decayed”. These are mutually exclusive entangled states, and an observation of the detector will tell us which we have at the moment of the observation.

Thank you for correcting me. I should have instead asked, what combined detector/atom state are we in after we observe at $t=100$ that the detector has not yet clicked? Do we effectively (for all practical/ observational purposes) collapse to the pure state "detector not clicked and atom not decayed"?

 

[PeterDonis]

I observe this by listening to the counter continuously between $t=0$ and $t=100$ and noting that there is no click during this time span

This does not mean what I suspect you think it means. See my post #17 just now in response to @Nugatory.

what combined detector/atom state are we in after we observe at $t=100$ that the detector has not yet clicked?

The superposed state with $A(100)$ and $B(100)$ as coefficients.

Do we effectively (for all practical/ observational purposes) collapse to the pure state "detector not clicked and atom not decayed"?

No. See above and post #17.

 

[PeterDonis]

Regarding the observed results, I admit I have no proof, but it would seem exceedingly strange to me if atoms prepared in the states with $A=1$ and $B=0$ , and $A(100)$ and $B(100)$, had the same decay statistics.

You are right, they wouldn't. But you can't create the state with $A = 1$ and $B = 0$ just by listening for the click and not hearing it. You would have to engage in some active process to prepare the system in that state (and, as you'll note from one of my earlier posts, in order to use that state as your state at $t = 0$, you must assume that some such process was done that ended right at $t = 0$ with the combined system being in that state).

 

[jeeves]

This does not mean what I suspect you think it means. See my post #17 just now in response to @Nugatory.

Thanks. You've definitely pinpointed the source of my confusion.

Could you please explain the following:

If we have not observed the detector actually clicking, we have not made any measurement. In more technical language, no decoherence has occurred and we cannot say that either outcome has happened.

Naively, it seems to me that observing the detector not clicking tells me something about the state of the atom, and hence counts as a measurement. Why is this not the case? If the answer is too complicated for a forum post, is there an accessible reference on delocalization that addresses this issue?

 

[PeterDonis]

Naively, it seems to me that observing the detector not clicking tells me something about the state of the atom, and hence counts as a measurement. Why is this not the case?

Because no decoherence occurs in this case, whereas decoherence does occur in the case where the detector does click. If you want to dig deeper into this, "decoherence" is the general term to search for; however, if you don't already have a solid understanding of the basics of QM, you should get that first.

 

[Nugatory]

I don't think this is correct, because "an observation of the detector", unless the OP has some very unusual (and unknown to date) experimental technique in mind, does not involve actually probing the state of the detector with any kind of interaction.

I was thinking in terms of such a probe…. I’m not seeing how I can say “I have observed that the detector has not clicked” without having inspected it with thermodynamically irreversible consequences (and of course the click must also have such consequences or there’s nothing to inspect for).

In practice it’s easier to Copenhagen the question and treat the detector as a classical measuring device outside the quantum system.

 

[PeterDonis]

I’m not seeing how I can say “I have observed that the detector has not clicked” without having inspected it with thermodynamically irreversible consequences

And doing that would amount to resetting the system to the $t = 0$ state, i.e., it would be the sort of "preparation process" that would produce that state. In other words, it would be something like a quantum Zeno effect experiment.

But what we are doing when we just "listen" for the detector clicking and don't hear it is not such a probe. It can't be, because if doing that was such a probe, it would have the effect described above, and we would never hear any detectors click at all, ever. Since we do hear detectors click, we can't possibly be doing the equivalent of a quantum Zeno effect experiment when we listen for detector clicks and don't hear them.

 

[jeeves]

But what we are doing when we just "listen" for the detector clicking and don't hear it is not such a probe. It can't be, because if doing that was such a probe, it would have the effect described above, and we would never hear any detectors click at all, ever. Since we do hear detectors click, we can't possibly be doing the equivalent of a quantum Zeno effect experiment when we listen for detector clicks and don't hear them.

I agree, but I still don't understand the general principle at work here. Is there some rule that would have allowed us to deduce that listening to the detector won't reset the state continuously without actually doing the experiment first? Or, more to the point, is there a rule that would allow us to distinguish between mere "observations" (whatever you want to call listening to the detector, that does not reset the state) and measurements that affect the state in experiments that we have not yet performed, or are not feasible to perform?

 

[PeterDonis]

I still don't understand the general principle at work here.

The general principle at work is decoherence. That is not something we're going to be able to explain in detail in a single thread. You will need to spend some time learning about it (and, as I said, learning a solid understanding of basic QM first, if you don't already have that).

 

[PeterDonis]

The general principle at work is decoherence.

Note, btw, that decoherence is not something that the original discoverers of QM knew about. That's why the concept of "measurement" is so vague in all of the early writings about QM, and why, if you don't understand decoherence theory, it's easy to go astray trying to think about what does and does not count as a "measurement". It wasn't until about 5 decades after QM was first developed, in the 1970s, that decoherence was discovered, and it took a few decades after that for decoherence theory to be developed in detail (and it's still a work in progress). Not all QM textbooks even now have a good treatment of decoherence.

 

[jeeves]

Fair enough. Is there a decoherence reference you could recommend? Or a place in the literature where this kind of example is discussed in detail? Or even a (grad-level) textbook with a good decoherence discussion.

 

[PeterDonis]

For textbooks, IIRC Ballentine has a decent discussion of decoherence. As for papers, I would look on arxiv.org for papers by Zurek on decoherence. I believe there are some fairly recent ones that give an overview of decoherence theory in the light of all the work that has been done over the past few decades.

 

[vanhees71]

A very nice book is

E. Joos et al, Decoherence and the Appearance of a Classical World in Quantum Theory, Springer (2010)

 

[gentzen]

Summary: Do Geiger counters cause collapse when they don't click?

Even if your interpretation would claim that they do, this would not lead any problems (like the quantum zeno effect). Radioactive decay is described by the exponential distribution, which is memoryless. This does not contradict radioactive decay being a quantum mechanical process:

In the early 20th century, radioactive materials were known to have characteristic exponential decay rates, or half-lives. At the same time, radiation emissions were known to have certain characteristic energies. By 1928, Gamow in Göttingen had solved the theory of the alpha decay of a nucleus via tunnelling, with mathematical help from Nikolai Kochin...

... In quantum mechanics, however, there is a probability the particle can "tunnel through" the wall of the potential well and escape. Gamow solved a model potential for the nucleus and derived from first principles a relationship between the half-life of the alpha-decay event process and the energy of the emission, which had been previously discovered empirically and was known as the Geiger–Nuttall law.

 

[vanhees71]

The exponential-decay law is an approximation, called the Wigner-Weisskopf approximation. Quantum mechanically it cannot be exact. See, e.g.,

J.J. Sakurai, Modern Quantum Mechanics, extended edition, Addison-Wesley (1994)

 

[jeeves]

Thanks, Peter and vanhees, for the references. I was not able to locate a section on decoherence in Ballentine's book (it does not appear in the index or table of contents). I read Zureck's arxiv paper 2107.03378, since this was the only survey of his that I could find, but it was more of a remembrance than an expository article. I have yet to read the Schlosshauer book.

One reference I found helpful was "The Metaphysics of Decoherence" by Vassallo and Ramano. My understanding now is that decoherence explains why we see mixed states and not superpositions in real life. That is, using the example of that paper, decoherence theory explains why we see that Schrodinger's cat is alive or dead, but we never see a superposition of an alive and dead cat. This is great, however it seems not directly relevant to my question of what constitutes a measurement, since opening the box to view the cat is obviously a "measurement."

In particular, there is a difference with the Geiger counter case from my the first post in this thread that I do not know how to accommodate. Namely, the cat is put in the box, some time elapses, and then the box is opened. Hence, there is a single observation taken at a fixed point in time. While listening to the Geiger counter, it seems that I am taking a continuous series of observations of the atom. How precisely does decoherence theory show that these observations do not act like Copenhagen-esque collapsing measurements?

 

[DrChinese]

In particular, there is difference with the Geiger counter case my the first post in this thread that I do not know how to accommodate. Namely, with the cat, the cat is put in the box, some time elapses, and then the box is opened. Hence, there is a single observation taken at a fixed point in time. While listening to the Geiger counter, it seems that I am taking a continuous series of observations of the atom. How precisely does decoherence theory show that these observations do not act like Copenhagen-esque collapsing measurements?

Many things (other than radioactive decay) have a probability of occurring per unit of time. Example: when an electron drops to a lower orbital and emits a photon. I would not normally call a detection "non-event" to be equivalent to a "continuous series of observations" of the particle in question. (There might be a few cases where it is difficult to suitably define a "non-event" or a "continuous series of observations".)

Also, in case this was not already clear: There is no known difference in the state of a radioactive particle at T=0 and T=100 in the sense that it is no more (or less) likely to decay at T=100 than at any other time. That likelihood remains constant.

 

[WernerQH]

1) It is in state a(100)*N + b(100)*D. This is because the particle has evolved until time t=100, and the Geiger counter has not interacted with the particle or done anything to affect its state.

2) It is in state a(1)*N + b(1)*D. At t=100, we know there has been no decay product formed until at least t=99, so at t=99 we know the particle did not decay yet. Then the state at t=99 is N, so at t=100 the particle has evolved into the superposition one second after starting at N.

Which of these is correct, and why?

The short answer is neither. You've probably been led to believe that the wave function accurately represents an individual system. But it is really only a statistical description. Already Schrödinger struggled to make sense of the wave function for non-stationary states.

But you hit an important point relating to the time scales of the problem. The lifetime of a neutron, for example, is several minutes, whereas the actual decay occurs in less than a microsecond. (That's an upper limit on the time it takes the created anti-neutrino to leave the laboratory.) As has been explained already by gentzen and DrChinese, a "newly prepared" neutron is not discernible in any way from a neutron that has "aged" in the apparatus for 100 seconds. For the experimenter they will be in the same state (if it hasn't decayed in the meantime). An individual neutron decays at a rather well defined (measurable!) but random time. Only the average number of neutrons varies in a deterministic and continuous way. Continuous and deterministic evolution according to the time-dependent Schrödinger equation simply doesn't square with the discontinuous and random character of the events that we observe in the real world! The wave function is but a piece in a bigger mathematical apparatus, and there's more to quantum theory than Schrödinger's equation.

 

[jeeves]

The general principle at work is decoherence. That is not something we're going to be able to explain in detail in a single thread. You will need to spend some time learning about it (and, as I said, learning a solid understanding of basic QM first, if you don't already have that).

I have read more, and am still a bit confused.

Consider the Renninger negative-result experiment.

I place a single unstable atom, say Carbon-14, at the center of a sphere consisting of two hemispherical detectors.

First, remove one of the hemispheres and start the experiment. Suppose that the remaining detector does not signal a detection after a long period of time, long enough that for all practical purposes, we are certain the atom has decayed. Then the non-detection of the decay product on the remaining detector is logically equivalent to knowing that the particle escaped out the other hemisphere (a detection on the removed detector), and hence locates the trajectory of the particle in a subset of the original possible trajectories. In Copenhagen language, we have a partial collapse of the wave function.

Next, restore the second detector to complete the sphere, and begin the experiment anew. Suppose after some period of time neither detector registers decay. If I understand correctly, you claimed earlier that this non-detection does not qualify as a measurement, and does not collapse the wave function.

How can it be that non-detection partially collapses the wave function in the first scenario but not the second?

 

[PeroK]

How can it be that non-detection partially collapses the wave function in the first scenario but not the second?

Because they are different experiments. The implications of non detection depend on how closely you have been monitoring something.

That applies classically as well as quantum mechanically.

 

[jeeves]

Because they are different experiments. The implications of non detection depend on how closely you have been monitoring something.

What precisely is the mechanism that causes collapse in the first experiment but not the second? The detectors themselves operate the same way in both setups.

 

[PeroK]

What precisely is the mechanism that causes collapse in the first experiment but not the second? The detectors themselves operate the same way in both setups.

There's no mechanism. Wave function collapse isn't a physical thing. It's a consequence of knowledge about a system.

 

[PeroK]

If your friend is in his house and you are watching the front door, then after a certain time there is a probability he has left by the back door. Whereas, if you are monitoring all possible exits you know he's still in the house.

There's no mechanism there, only inference from knowledge about the system and its possible evolution. In this case probabilistic calculations based on your friend's likely movements.

 

[PeterDonis]

What precisely is the mechanism that causes collapse in the first experiment but not the second?

What the mechanism is, or even whether there is any mechanism at all involved, depends on which interpretation of QM you adopt. Discussion of QM interpretations is out of scope for this forum; it belongs in the interpretations subforum. All the basic math of QM can do is make predictions about the probabilities of various possible experimental results; it cannot tell you "what really happens".

 

[Nugatory]

Namely, the cat is put in the box, some time elapses, and then the box is opened. Hence, there is a single observation taken at a fixed point in time.

Opening the box is pretty much irrelevant to the quantum mechanical evolution of the system. The macroscopic elements of the quantum system in the box (detector and trigger mechanism, vial of poison, living breathing warm wiggly cat, countless air molecules drifting around inside the box, ...) have enough degrees of freedom that the quantum system consisting of the unstable nucleus entangled with all this macroscopic stuff almost immediately decoheres into the mixed "either we have a dead cat in the box or a live cat in the box" state. This happens long before and whether or not we even open the box and look to see which we have. Generally any thermodynamically irreversible interaction leading to decoherence counts as a "observation", and a system as complex as Schrodinger's cat in a box can be considered to be continuously observing itself.

It is worth noting that Schrodinger did not accept the idea that the cat was in a superposition of alive and dead until opening the box collapsed the wave function. The point of his thought experiment was that something had to be wrong with the then-current understanding of QM because it suggested that opening the box mattered.

 

[jeeves]

Opening the box is pretty much irrelevant to the quantum mechanical evolution of the system. The macroscopic elements of the quantum system in the box (detector and trigger mechanism, vial of poison, living breathing warm wiggly cat, countless air molecules drifting around inside the box, ...) have enough degrees of freedom that the quantum system consisting of the unstable nucleus entangled with all this macroscopic stuff almost immediately decoheres into the mixed "either we have a dead cat in the box or a live cat in the box" state. This happens long before and whether or not we even open the box and look to see which we have. Generally any thermodynamically irreversible interaction leading to decoherence counts as a "observation", and a system as complex as Schrodinger's cat in a box can be considered to be continuously observing itself.

Thanks. I guess I am confusing myself by thinking of "measurement" as some magic wand. It is as you and Peter said: the system rapidly decoheres into a mixed state which is a combination of "cat dies and particle has decayed" or "cat is alive and particle hasn't decayed." And if I observe the system at time $t$, the probability that the cat is alive can be computed from the coefficient $A(t)$ of the "alive and no decay" part of the mixed state (using Peter's notation).

I think I am fine with this. Certainly this gives us a way to compute the empirical frequency of living cats if we perform the experiment repeatedly.

What I still don't understand is how decoherence explains what happens when there are multiple observations. For example, suppose I want to answer the question: "If I observe that the cat is alive at time $t=1$, what is the probability the cat is still alive at time $t=2$?"

To answer this question, it suffices to know the state of the cat at $t=1$, in particular $A(t)$ and $B(t)$. Then I just run the Schrodinger evolution and get a mixed state at $t=2$ that gives me the answer. So my question becomes: if I observe the cat is alive at $t=1$, what is the state of the cat after the observation?

My inclination is to say that if I see the cat is alive, I'm observing a pure state, so I'm in the state with $A(t)=1$.

But this is obviously wrong because then we could get a quantum Zeno effect by observing the cat many times in succession.

So what is the state after observation and why? Can decoherence help explain this too?

 

[PeterDonis]

what is the state after observation and why?

The "observation" you describe--"observing" that the cat is alive--does not tell you anything useful. What you would need to do is "observe" the radioactive atom whose decay will trigger the process that kills the cat, and see "how close it is to decaying". But, as others have already pointed out in this thread, there is no such thing; the probability of the atom decaying per unit time is constant (at least if we ignore the small corrections to the exponential approximation). So there is no way to "observe" anything that can give you more information than the exponential decay law about what will happen to the cat in the future.

If you could somehow "observe" the atom continuously and verify that it continued to be in the undecayed state, you would be running a quantum Zeno effect experiment on the atom; but that is not the situation you have been describing. Certainly "observing" that the cat is alive is not such an experiment.

Can decoherence help explain this too?

Decoherence can explain why there is no interference between the "atom undecayed, cat alive" and "atom decayed, cat dead" states. But it cannot, by itself, explain the "collapse" of the system to one state or the other as a result of some "observation". Basic QM does not even attempt to explain that; it just tells you what to do mathematically when you know a particular result has been recorded for some observation. It does not tell you "what really happens" or "what the actual state of the system is" or anything like that. Particular QM interpretations do make such claims, but, as has already been noted, discussion of interpretations is off limits for this forum.

 

[jeeves]

The "observation" you describe--"observing" that the cat is alive--does not tell you anything useful. What you would need to do is "observe" the radioactive atom whose decay will trigger the process that kills the cat, and see "how close it is to decaying". But, as others have already pointed out in this thread, there is no such thing; the probability of the atom decaying per unit time is constant (at least if we ignore the small corrections to the exponential approximation). So there is no way to "observe" anything that can give you more information than the exponential decay law about what will happen to the cat in the future.

If you could somehow "observe" the atom continuously and verify that it continued to be in the undecayed state, you would be running a quantum Zeno effect experiment on the atom; but that is not the situation you have been describing. Certainly "observing" that the cat is alive is not such an experiment.

I agree that if we are in the regime where the decay is almost exactly exponential, the distribution is (almost) memoryless and there is not much to be said.

So let's not assume that. Suppose we are in the small time regime where the decay dramatically deviates from exponential.

In this regime, does observing the cat tell me something useful? That is, is the conditional distribution for survival conditioned on the cat surviving until $t=1$ the same as the unconditional distribution for survival, like in the memoryless case, or is it different? I would suppose it is different, because after all, we do seem to learn something about the atom after observing the living cat (it has not decayed), and because the distribution is not memoryless this affects the (conditional) frequency of decay in the future.

So in the non-exponential regime, what is the state of the cat after I observe that it is alive at $t=1$, in terms of the $A(t)$ and $B(t)$ notation from earlier?

 

[PeroK]

I agree that if we are in the regime where the decay is almost exactly exponential ...

Let me ask you a question. I would like you to understand the nature of classical probabilities before tackling the complex probability amplitudes of QM. Which share some of the properties of classical probabilities.

We have a pack of cards and you draw a card (face down). What is the probability that it is some particular card like the six of clubs?

Now, we start looking at the other cards in the pack one by one. As each card is found not to be the six of clubs, does the probability that your card is the six of clubs change?

Do you know how to analyse a problem like that?

 

[jeeves]

Yes, I know how to analyze such problems.

 

[PeroK]

Yes, I am aware of how to analyze such problems.

Okay. Re QM. If, instead, we have 52 slits in a barrier and we carry out an experiment, then we get a 52-slit interference pattern. But, if we monitor one of the slits and run the experiment, then we get either a detection event at that slit or a 51-slit interference pattern. And, if we monitor $n$ slits, then we get either a detection event at one of the slits or a 52-$n$ slit interference pattern.

The calculation in QM is different in that each slit corresponds not to a probability but a probability amplitude. It doesn't work to say that the probability that the particle went through slit 1 is 1/52 (or 1/51 or whatever). Then there would be no intereference, but simply the sum of 52 single-slit patterns Instead, there is a complex probability amplitude associated with the path through each slit. We then combine these amplitudes (which, being complex, can cancel each other out) and we get probabilitistic quantum interference (of a particle with itself).

Okay so far?

 

[jeeves]

Okay so far?

Yes, this is fine.

 

[PeroK]

Yes, this is fine.

The same logic applies to time evolution of a state. For a radioactive decay, there is a probability amplitude for the state of decay after some time $t$ and a probablity amplitude for the state of non-decay after some time $t$. Usually the state of non-decay remains physically identical - in the sense that decay over the next time $t$ remains equally probable. In this case a definite measurement that results in "no decay" does not physically change the system.

If instead you postulate that the probability amplitude for decay changes over time (perhaps it gets less and less likely over time). Then a definite measurement of non-decay is meaningful, as you know that the state has evolved into a more stable state.

Does that make sense?