# Zeno effect - standard derivation?

• I
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:

$|\psi_t\rangle = e^{-i H t} |\psi_0 \rangle$

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

$P_s(t) = | \langle \psi_0| e^{-i H t} |\psi_0 \rangle |^2$

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

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

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

$P_s(t) \approx 1 - ( \langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 )t^2$

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 $(\Delta H)^2$ stand for $\langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2$ to give the simpler expression:

$P_s(t) \approx 1 - (\Delta H)^2t^2$

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 $|\psi_0\rangle$ is an eigenstate of the measurement operator. In that case, we get:

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

...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!

atyy

Demystifier
Gold Member
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
$$|A|^2=1-t^2\langle H^2\rangle +t^2\langle H\rangle^2=1-t^2(\Delta H)^2$$

atyy, Agrippa, bhobba and 2 others
Demystifier
Gold Member
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
$$P_s(t)=p_s(t/n)\cdots p_s(t/n)=[p_s(t/n)]^n=[1-(\Delta H)^2(t/n)^2]^n$$

atyy, Agrippa, bhobba and 1 other person
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 ${\cal O}(t^3)$ 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?

atyy and Demystifier
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.

atyy and Agrippa
Demystifier
Nothing seems to cancel the ${\cal O}(t^3)$ term. Why, then, do we go on to drop it from subsequent expressions?