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.