Zeno effect - standard derivation?

In summary, the conversation discusses the standard derivation of the quantum Zeno effect and how to compute the survival probability using the Born rule and the exponential function. The final expression for the survival probability is derived by considering measurements at regular intervals and assuming the initial state is an eigenstate of the measurement operator. The term of order ##t^3## is dropped as it is negligible compared to the dominant quadratic term for short times.
  • #1
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Hi, I'm trying to grasp the standard derivation of the quantum Zeno effect. But my calculus is limited. I would love some assistance!

Schroedinger evolution tells us that the state at time t, given the initial state at time t=0, is:

[itex]|\psi_t\rangle = e^{-i H t} |\psi_0 \rangle[/itex]

We can then define the survival probability [itex]P_s(t)[/itex], which is the probability that the initial state will survive or remain, at time t, using the Born rule:

[itex]P_s(t) = | \langle \psi_0| e^{-i H t} |\psi_0 \rangle |^2 [/itex]

From the definition of the exponential function, we can expand the power series:

[itex]e^{-i H t} \approx 1 - iHt - 1/2H^2t^2 + ... [/itex]

The next part is where I get stuck. It is then often said that from this we can derive:

[itex]P_s(t) \approx 1 - ( \langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 )t^2 [/itex]

But how does that follow from the previous steps? I've never seen this spelled out anywhere before. Is anyone able to fill in the gaps here?

Once we have this expression, we can let [itex](\Delta H)^2 [/itex] stand for [itex]\langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 [/itex] to give the simpler expression:

[itex]P_s(t) \approx 1 - (\Delta H)^2t^2 [/itex]

We now wish to consider the survival probability where the interval [0, t] is interrupted by n measurements at times t/n, 2t/n, ..., t. Ideally, these measurements are instananeous projections and the initial state [itex]|\psi_0\rangle[/itex] is an eigenstate of the measurement operator. In that case, we get:

[itex]P_s(t) \approx [1 - (\Delta H)^2(t/n)^2 ]^n [/itex]

...which approaches 1 as n approaches infinity.

How this last expression is derived is also not 100% clear to me. But I'm baffled by the part I get stuck by mentioned above and would love some help with that first.

Thanks!
 
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  • #2
From
$$e^{-iHt}=1-iHt-\frac{1}{2}H^2t^2$$
we first compute the transition amplitude
$$A=\langle \psi_0|e^{-iHt}|\psi_0\rangle =\langle\psi_0|\psi_0\rangle - it\langle\psi_0|H|\psi_0\rangle -\frac{1}{2}t^2 \langle\psi_0|H^2|\psi_0\rangle$$
which can be written as
$$A=1- it\langle H\rangle -\frac{1}{2}t^2 \langle H^2\rangle$$
Therefore
$$A^*=1+ it\langle H\rangle -\frac{1}{2}t^2 \langle H^2\rangle$$
so
$$|A|^2=A^*A=1+it\langle H\rangle-\frac{t^2}{2}\langle H^2\rangle-it\langle H\rangle-\frac{t^2}{2}\langle H^2\rangle
+t^2\langle H\rangle^2+{\cal O}(t^3)$$
The imaginary terms cancel, so finally
$$|A|^2=1-t^2\langle H^2\rangle +t^2\langle H\rangle^2=1-t^2(\Delta H)^2$$
 
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  • #3
The survival probability after a short time ##t## without measurement is
$$p_s(t)=1-(\Delta H)^2t^2$$
Hence the survival probability after time ##t/n## without measurement is ##p_s(t/n)##.

Now assume that after time ##t/n## there is a measurement which reveals that the state has survived. Immediately before the measurement the state is ##|\psi(t/n)\rangle =e^{-iHt/n} |\psi_0\rangle##, but immediately after the measurement the state is ##|\psi(t/n)\rangle =|\psi_0\rangle##. At time ##t/n## the state "collapsed" from ##e^{-iHt/n} |\psi_0\rangle## to ##|\psi_0\rangle## due to the measurement. At this time, the evolution is not described by the Hamiltonian ##H##. The Hamiltonian evolution is valid only between the measurements, while the measurement itself sets the initial condition.

Now suppose that measurements are performed at times ##t/n##, ##2t/n##, ... ,##nt/n=t##. What is the probability of survival after the ##n##'th measurement? It is the probability that it survived at each time of measurement, which is the product of individual survival probabilities. Between any two measurements there is a time ##t/n##, so each individual survival probability is ##p_s(t/n)##. Therefore the total survival probability is
$$P_s(t)=p_s(t/n)\cdots p_s(t/n)=[p_s(t/n)]^n=[1-(\Delta H)^2(t/n)^2]^n$$
 
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  • #4
Thanks. I can see why they call you Demystifier, that explains a lot!
The only thing I didn't quite get is this. In going through the A*A calculation, I found that:
$${\cal O}(t^3) = (-\frac{1}{2}t^2 \langle H^2\rangle) \times (-\frac{1}{2}t^2 \langle H^2\rangle)$$
Nothing seems to cancel the [itex] {\cal O}(t^3) [/itex] term. Why, then, do we go on to drop it from subsequent expressions?
I assume it has something to do with its physical interpretation, perhaps it somehow does not represent the short time period we are considering?
 
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  • #5
It is a common strategy in physics. As we look into very short times t << 1, higher powers of t will give much smaller corrections and are thus neglected. The dominating contributions close to the limit t -> 0 come from the lowest powers of t, which is in this case the quadratic term.

Also look at the consistency: If we decided for some reason, that we wanted to take all terms up to t^4 into account for example, we also should expand the exponential up to this order to really get all contributions for the final expression.
 
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  • #6
Agrippa said:
Nothing seems to cancel the [itex] {\cal O}(t^3) [/itex] term. Why, then, do we go on to drop it from subsequent expressions?
Because we keep only the lowest non-trivial contribution. Further contributions are smaller than that, so are less interesting. Besides, if one wants to keep terms of the order of ##t^3##, one must keep them from the beginning in the expansion of ##e^{-iHt}##.
 
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  • #7
Got it. Many thanks!
 

1. What is the Zeno effect?

The Zeno effect, also known as the "watched pot" effect, is a phenomenon in quantum mechanics where a system's evolution is slowed down or halted by frequent measurements or observations.

2. How is the Zeno effect demonstrated?

The Zeno effect can be demonstrated through thought experiments or laboratory experiments using quantum systems such as photons or atoms.

3. What is the standard derivation of the Zeno effect?

The standard derivation of the Zeno effect involves applying the principles of quantum mechanics, specifically the Schrödinger equation, to a system undergoing frequent measurements. This results in a reduction of the system's wave function, leading to a slower evolution or even a halted evolution.

4. What is the significance of the Zeno effect in quantum mechanics?

The Zeno effect highlights the role of observation and measurement in quantum systems, and challenges our understanding of the concept of time and the behavior of particles at the quantum level. It also has practical applications in quantum computing and quantum information processing.

5. Can the Zeno effect be observed in macroscopic systems?

No, the Zeno effect is a phenomenon that occurs at the quantum level and is not observable in macroscopic systems. It is a result of the probabilistic nature of quantum mechanics and does not apply to classical physics.

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