# I Question about the quantum Zeno effect again

1. Jun 20, 2017

### Trollfaz

When measuring the decay of a radioactive atom with a geiger counter, we collapses the superposition of the decay and undecayed state. This will affect the rate of decay of the atom.
However the geiger counter does not actively interact with the atom. It does not shoot any particle at the atom, it only absorbs the particle emitted from the decay. And this means it does nothing to the atom.
So is the rate of decay of the atom altered?

2. Jun 20, 2017

### Paul Colby

It's fair to say physical theories describe nature they don't affect nature.

3. Jun 20, 2017

### Trollfaz

Or is that only certain type of measurements can affect the speed of decay?

4. Jun 20, 2017

### Paul Colby

Well, in this case you have a hard time explaining an effect because the level of interaction between measurement and experiment seems small and insufficient to affect the decay rate. This is a correct assessment. You're confused/mystified because the language you use to describe quantum theory is poor and seems to imply effects which aren't there. I would suspect this as an artifact of the language not the theory and reject it based on my initial comment.

5. Jun 20, 2017

### Jilang

Perhaps it is not sufficiently local?

6. Jun 20, 2017

### Staff: Mentor

Correct.

The statement quoted above should make it obvious that the answer is "no".

Which, btw, means this is not an example of the quantum Zeno effect, as your thread title implies. The quantum Zeno effect requires an interaction with the system to which the effect applies.

7. Jun 20, 2017

### Trollfaz

Thanks

8. Jun 21, 2017

### thephystudent

From a somewhat more technical perspective, is this the same as saying that Zeno effect only applies to projective measurements and not to weak measurements?

9. Jun 21, 2017

### vanhees71

The quantum Zeno effect is basically just the change of the lifetime of an unstable state due to interactions.

10. Jun 24, 2017

### stevendaryl

Staff Emeritus
The heuristic derivation of the quantum zeno effect doesn't seem to involve interactions in the normal sense. It involves measurement of the state, but it's not clear that all measurements amount to an interaction. If the decay of an unstable particle produces a photon, for instance, then failing to detect a photon counts as an indirect measurement of the state: it's still in the undecayed state. But it's hard to see how failing to detect something counts as an interaction.

The origin of the Zeno effect is the difference in short-time behavior of a decaying quantum system and a decaying classical system.

A classical decay would have a characteristic decay constant $\tau$ and the probability of the particle being undecayed after a time $t$ would be given by: $P(t) = e^{-\frac{t}{\tau}}$. The short-term behavior, when $\delta t \ll \tau$ is $P(\delta t) \approx 1 - \frac{\delta t}{\tau}$. If you measured the particle every $\delta t$ seconds, then the probability of being undecayed after the $N$th measurement would be given by:

$(1 - \frac{\delta t}{\tau})^N \approx e^{-\frac{N \delta t}{\tau}} = e^{-\frac{t}{\tau}}$ (letting $t = N \delta t$)

So frequent measurements don't make any difference. But the short-term behavior of a quantum system is different. The probability that the system is still in the initial state $|\psi\rangle$ after $\delta t$ seconds is given by: $P(\delta t) \approx 1 - \frac{\delta t^2}{T^2}$, where $T$ is the short-term Zeno constant given by: $\frac{1}{T^2} = \langle H^2 \rangle - \langle H \rangle^2$. (where $H$ is the Hamiltonian, and $\langle \rangle$ means taking expectation values in state $|\psi\rangle$. The short-term behavior is quadratic in $\delta t$, instead of linear. This makes all the difference in the world, because if you perform measurements $N$ times, you get:

$(1 - \frac{\delta t^2}{T^2})^N \approx 1 - \frac{N \delta t^2}{T^2} = 1 - \frac{t^2}{N T^2}$ (letting $t = N \delta t$)

which goes to 1 as $N \rightarrow \infty$, keeping $t$ fixed.

The nature of the measurement interaction doesn't come into play except that we have to use the heuristic that if you don't detect a decay, then that means that the system is still in state $|\psi\rangle$.

Last edited: Jun 24, 2017
11. Jun 24, 2017

### stevendaryl

Staff Emeritus
There is a discussion of the Quantum Zeno Effect and interaction-free measurements here:

12. Jun 24, 2017

### thephystudent

What confuses me is the compatibility between on the one hand this quantum zeno effect (measuring affects time-evolution), and on the other hand, the quantum jump method https://en.wikipedia.org/wiki/Quantum_jump_method, which assumes the evolution of the master equation of the system (regardless of measurements being performed on its environment) is equivalent to an average over evolutions of the system given stochastic measurement records on its environment.

13. Jun 24, 2017

### stevendaryl

Staff Emeritus
I don't know about the jump method, but I would guess that the "jumps" correspond to interactions with the environment, which are like measurements. The quantum Zeno effect only happens if the measurements take place before there is significant chance of interacting with the environment.

14. Jun 24, 2017

### thephystudent

That was the reason why I asked the question before: performing a (possibly projective) measurement on an environment that is entangled with the system means also performing a weaker POVM measurement on the system itself. In this case, the atom would be the system and the fieldmode where particles decay into would be the environment. The geiger counter measures this fieldmode (e.g. the electron-field for beta-decay) in order to tell something about the system.

15. Jun 25, 2017

### Trollfaz

stevendaryl, i dont see how a geiger counter really prevents an atom from decaying as we detect radiation with it all the time.

16. Jun 25, 2017

### stevendaryl

Staff Emeritus
Obviously, a geiger counter doesn't prevent an atom from decaying. I was just pointing out the argument for the quantum zeno effect, and it's not immediately why it wouldn't apply to the case of a geiger counter. So I'm really asking what are the conditions under which the quantum zeno effect applies, and why doesn't it apply in the case of a geiger counter.

One step in the argument is the assumption that every decay is detectable, so the lack of detection implies no decay, which implies that the particle is still in the undecayed state. I don't know whether geiger counters are accurate enough to detect every decay.

Another step in the argument is that there is a unique undecayed state, so knowing that the particle is undecayed uniquely determines what state it's in. That might be a false assumption, as well.

17. Jun 25, 2017

### vanhees71

Of course, if you want to affect something you have to interact with it. The influence of a Geiger counter on an unstable nucleus, however, can be neglected since it rather will interact with the decay product (He nuclei, electrons, photons for $\alpha$-, $\beta$, and $\gamma$ radiation respectively). It also doesn't help to simply put a Geiger counter somewhere to tell that the nucleus is not yet decayed. For that purpose you must put it there, i.e., to make sure that it is still not decayed. Then you put $t=0$ in the decay law. The "survival probability" after a time $t$ then is $P(t)=\exp(-\lambda t)$.

Also the exponential decay-law is an approximation. From unitarity of the time evolution of an unstable state you can show that it cannot be exact (see, e.g., Sakurai, Moder Quantum Mechanics).

18. Jun 25, 2017

### stevendaryl

Staff Emeritus
That's not clear. The quantum zeno effect is about measurement, not interaction. Normally, there has to be an interaction in order to have a measurement, but in certain circumstances the lack of an interaction tells you something about the state of the system, so the lack of an interaction can count as a measurement. If you arrange things so that if an electron is spin-up, then a bell will sound, then the fact that the bell does not sound implies that it is spin-down. If you arrange things so that when a particle decays, a geiger counter clicks, then the lack of click implies that the particle has not decayed.

Of course, you can't have a measurement unless there is the possibility of an interaction.

That's the whole basis of the quantum zeno effect, as I said in an earlier post. For quantum decay, the short-term behavior is not like the short-term behavior of an exponential; instead of the Taylor expansion being linear in t, it's quadratic. That's the reason that repeated measurements can delay the decay. That would not be true if it were exponential.

19. Jun 25, 2017

### stevendaryl

Staff Emeritus
Here's an informal summary of the situation with the quantum zeno effect.

Suppose we do the following:
1. At time $t_1$, we prepare a system in state $|\psi\rangle$
2. We let it evolve according to the Hamiltonian $H$ until time $t_2$, and then perform a measurement of some operator $A$. We get eigenvalue $\lambda$.
3. We let it evolve further until time $t_3$, and perform a second measurement (possibly of a different operator, but let's assume it's the same operator).
We want to figure out the probabilities for the second measurement by using the Born rule. To actually compute these from first principles would be enormously complex, because step 2 involves the system interacting with a measuring device, which is a macroscopic system. So instead we use the following "rule of thumb":
• Assume that immediately after measuring $A$, the system is in the state $|\lambda\rangle$, which is an eigenstate of $A$ with eigenvalue $\lambda$ (for simplicity, let's assume that there is only one eigenstate with that eigenvalue).
Then we can compute probabilities for the second measurement using the Born rule with the state $|\psi(t_3)\rangle = e^{-i H (t_3 - t_2)} |\lambda \rangle$

Described this way, the "collapse of the wave function" is not a matter of interpretation, it's a rule of thumb for computing the results of sequential measurements. Presumably, this rule of thumb could be justified theoretically, but for a rule of thumb, it's enough that it has empirical support.

This rule of thumb implies the quantum zeno effect. If the time between measurements of $A$ is very short, it's very likely that the second measurement will give the same result as the first measurement. If you perform measurements often enough, you can force the system to remain in the eigenstate $|\lambda\rangle$ longer than it would if there were no measurements.

Now, the question arises: What kind of measurement is involved at steps 2 and 3? Is it possible to measure $A$ without interacting with the system?

You can't say that it's impossible to measure a property of a system without interacting with it. The most striking counterexample is EPR: You have twin particles that have correlated spins (or correlated polarizations). Then you can measure the spin/correlation of one particle by measuring the spin/correlation of the other particle.

20. Jun 25, 2017

### stevendaryl

Staff Emeritus
I want to point out that using entanglement as a way of doing interaction-free measurement is a one-shot deal. After you've performed a measurement of one of the particles, then afterward, the other particle is effectively unentangled, so you can't perform subsequent measurements remotely. So EPR-type entanglement can't be used for the quantum zeno effect, which requires repeated measurements of the same operator.