19
$\begingroup$

I'm a student and I had to give a talk on seminar about Quantum Zeno effect and Anti-Zeno effect to my colleagues (all listeners have had a course in quantum physics, but not a heavy one with all the bra and ket stuff).

My first idea to give a simple explanation of Zeno's effect was this: Let's take a look at exponential decay where the chance for particle or state to survive some time $t$ is $P_S=e^{-t/\tau}$. If I measure it after time $\tau$ I have chance $P_S=1/e$ that it will still be intact.

If I instead allow it to do it's things only for time $\tau/N$ and then measure it, the survival chance will be $P_S=e^{-1/N}$ which approaches $1$ as $N$ increases. To achieve the same total time, I have to repeat this procedure $N$ times and the total survival probability is... $P_S=(e^{-1/N})^N=1/e$.

So it obviously doesn't work, I get no Zeno's effect in this way.

It's interesting that after I gave the talk professor rose and said "Well, this can be easily understood if we look at the exponential decay". Then he started drawing exponent and another exponent that's repeatedly interrupted and reset to initial state after small intervals. Later we agreed that this won't actually work, but the question is - why?

Why doesn't this intuitively obvious way doesn't work and what would be the correct law out of which one could see the Zeno's effect? Is there any elegant way to explain this effect without heavy math and angles of state vectors?

ADDITIONAL QUESTION (related): Is it correct to use name "Quantum Zeno Effect" for turning of polarization by series of inclined polarizers or the thing that is done in this article?

$\endgroup$
1
  • $\begingroup$ This article (arxiv.org/abs/0809.4388) also has an elegant explanation. $\endgroup$
    – emarti
    Dec 21, 2012 at 23:43

3 Answers 3

12
+50
$\begingroup$

As for me, there is no hard math, and this is maybe the best explanation:

Let $|\psi_{0}\rangle$ be the state of the system with the Hamiltonian $H$ at $t=0$ and $|\psi(t)\rangle$-the state at time $t$. The the evolution of the state can be described by a unitary operator of evolution $U(t)$, where (let $\hbar=1$) $U(t)=e^{-iHt}$:

$|\psi(t)\rangle=U(t)|\psi_{0}\rangle$

The probability that the system will be still in initial state after 1 measurement:

$P(t)=|\langle\psi_{0}|\psi(t)\rangle|^2=|\langle\psi_{0}|U(t)|\psi_{0}\rangle|^2$

If you will use $|\psi|^2=\psi\psi^{*}$, you can expand this as:

$P(t)=1-t^{2}(\langle\psi_{0}|H^{2}|\psi_{0}\rangle-\langle\psi_{0}|H|\psi_{0}\rangle^{2})+...$

If $\Delta H=\sqrt{\langle\psi_{0}|H^{2}|\psi_{0}\rangle-\langle\psi_{0}|H|\psi_{0}\rangle^{2}}$, the "survival" probability can be rewritten:

$P(t)=1-t^{2}(\Delta H)^{2}+...$

We can define Zeno time $\tau_{Z}=1/\Delta H$. Then:

$P(t)=1-\frac{t^{2}}{\tau_{Z}^{2}}+...$

For short times it can be written as:

$P(t)\approx(1-\frac{t^{2}}{\tau_{Z}^{2}})$

It was the survival probability after one measurement.

After $N$ measurements:

$P^{N}(T)=(1-\frac{T^{2}}{N^{2}\tau_{Z}^{2}})^{N}$

We can see that in the limit of continuous measurements, i.e. when $N\rightarrow \infty$, probability tends to $1$:

$$\lim_{N\to\infty}P^{N}(T)=1$$.

For your ADDITIONAL QUESTION: Yes. It seems to me that it is correct for polarization: it is a two level system. The polarization of photon is described by two basics states, which correspond to horizontal and vertical polarization:

$|H\rangle= (1,0)^{T}$,

$|V\rangle=(0,1)^{T}$

We will use the Faraday effect(see http://en.wikipedia.org/wiki/Faraday_effect):

"The Faraday effect causes a rotation of the plane of polarization which is linearly proportional to the component of the magnetic field in the direction of propagation"

So in this case the operator of evolution will have the the same matrix as rotation operator( see http://en.wikipedia.org/wiki/Rotation_operator_(vecto..).

Let the photon has horizontal polarization $|H\rangle$ in the initial time. Then at any later time $t$:

$|\psi(t)\rangle=U|H\rangle=(\cos(\beta), \sin(\beta))^{T}$, where $\beta=\beta(t)$. Then

$P_{H}(t)=|\langle H|\psi(t)\rangle|^{2}=\cos(\beta)^{2}$.

After $N$ measurements we will have the probabiliity that photon still will have horizontal polarization:

$P_{H}(T)=\cos(\beta)^{2N}$.

so if you will use Taylor series, you will have

$$\lim_{N\to\infty}P_{H}(T)=1$$.

$\endgroup$
4
  • $\begingroup$ And when and how can we see exponential decay for your system? $\endgroup$
    – Džuris
    Dec 31, 2012 at 10:50
  • $\begingroup$ I believe Faraday rotation is more properly termed the "Anti-Quantum Zeno Effect" $\endgroup$
    – Mikhail
    Sep 29, 2014 at 15:19
  • $\begingroup$ This Zeno time is the caracteristic time of the time evolution of that state. $\endgroup$
    – Nogueira
    May 13, 2015 at 19:20
  • $\begingroup$ This is some sort of a proof that quantum states represent only a state of knowledge. The evolution of a quantum state take place after the preparation. If we keeping measuring the system, then the evolution of the state never take place. This is a beautiful way for nature protect herself on definite trajectories. Nature is shy! $\endgroup$
    – Nogueira
    Jul 4, 2015 at 6:23
5
$\begingroup$

Consider a system that has the two eigenfunctions: $\psi_a(x)$ and $\psi_b(x)$. It's not hard to build a Hamiltonian for which the time evolution of the wavefunction would be: $$ \psi(x,t) = \frac{1}{2} \left[ e^{\frac{2 \pi i t}{\tau}} \left( \psi_a(x) + \psi_b(x) \right) +e^{\frac{\pi i t}{\tau}} \left( \psi_a(x) - \psi_b(x) \right) \right]. $$ This way the wavefunction satisfies: $\psi(x,0)=\psi_a(x)$ and $\psi(x,\tau)=\psi_b(x)$.
The probability of measuring the system in the state $\psi_a(x)$ is: $$ P_a(t) = \left| \left\langle \psi_a(x) , \psi(x,t) \right\rangle\right|^2 = \frac{1}{2} \left[ 1 + \cos{ \left( \frac{\pi t}{\tau} \right) } \right]. $$ Obviously $P_a(0) = 1$ and $P_a(\tau) = 0$, because the system starts at state $a$ and then decays from it (but not exponentially!).

Now if we measure the system $N$ times at $\tau/N$ time intervals, using the same calculation in your question we get: $$ P_a^N(\tau/N) = \frac{1}{2^N} \left[ 1 + \cos{ \left( \frac{\pi}{N} \right) } \right]^N \xrightarrow[N \rightarrow \infty]{} 1 $$ Therefore we have prevented the system from decaying just by repeatedly measuring it, and this is the Quantum Zeno effect.

$\endgroup$
1
  • $\begingroup$ See this answer for some clarification about when we can still expect exponential decay. $\endgroup$
    – Joe
    Dec 20, 2012 at 13:19
3
$\begingroup$

Joe's answer gives a nice clean example of where the quantum Zeno effect does work. But to answer your very good question of why it doesn't work for exponential decay:

The exponentially decaying probability of being in the initial state is an example of a memoryless distribution, which you should see in any basic statistics course. In fact, the exponential PDF is the unique continuous memoryless distribution. What this means in quantum language is the probability per unit time of flipping to the other state is independent of how much time has elapsed. Thus "resetting the clock" by making intermittent observations does nothing to change the probability of changing states.

Your question is great because it illustrates a limitation of the quantum Zeno effect. You cannot easily apply it to anything with an exponential probability distribution, which for instance describes a free particle (the classic example being a neutron) that decays into others.

$\endgroup$
3
  • $\begingroup$ Your answer rightfully addresses the issue of exponential decay, which is indeed missing from my answer. I'm still a little confused though (and maybe the OP is too) as to how come we get exponential decays in quantum mechanics in the first place, in opposed to the result of the calculation I showed in my answer. This answer makes an attempt to explain this issue, but it's still not entirely clear to me. It would be great if you would have anything to add on that. $\endgroup$
    – Joe
    Dec 21, 2012 at 10:16
  • $\begingroup$ @Joe I'm admittedly a bit rusty with the finer points of QM - haven't needed it in years - but I'll think on it nonetheless. $\endgroup$
    – user10851
    Dec 21, 2012 at 10:29
  • $\begingroup$ @Joe This reference gives a treatment (section 3) which shows how, long term, you get the exponential behaviour and short term the quadratic (and hence zeno-friendly) behaviour. I don't know how to explain it in simple terms as the OP requested though. $\endgroup$
    – twistor59
    Dec 26, 2012 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.