# Quantum Zeno Effect: What is the argument?

1. Jan 21, 2014

### James MC

Every general explanation of the quantum zeno effect I've found is (from my perspective) so full of gaps that I cannot understand the explanation. I am wondering if anyone can help me fill in the gaps here?

The most detailed explanation I've found runs something like this:

Let |ψ0> be the initial quantum state of a system at time 0 and let |ψt> be its state at some later time t.

The dynamical evolution of the system is described by a unitary operator U(t) that is a complex function of the initial system's Hamiltonian: U(t) = e-iHt. Thus: |ψt> = U(t)|ψ0>.

The "survival" probability Ps that the system will still be in the initial state at t is given by:

Ps = |<ψ0t>|2 = |<ψ0|e-iHt0>|2

So far so good. But now standard explanations assert that:

Ps = |<ψ0|e-iHt0>|2 = 1 - (ΔH)2t2
(Where (ΔH)2 = <ψ0|H20> - (<ψ0|H|ψ0>)2)

Where does that come from? Is it meant to be obvious that 1 - (ΔH)2t2 follows from the left hand side?

At any rate, we can now define the Zeno time Z = 1/ΔH so that:

Ps = 1 - $\frac{t^{2}}{Z^{2}}$

Presumably this shows that as t gets smaller the probability tends to 1 so that the faster we measure the system after time = 0 the more probable it will be found in its initial state.

Now for the final bit. If we consider N measurements then we can understand the survival probability given those N measurements as:

P$^{N}_{s}$ = (1 - $\frac{t^{2}}{N^{2}Z^{2}}$)N

...so that in the limit of continuous measurements where N → ∞ we get:

$\stackrel{Lim}{N→∞}$ P$^{N}_{s}$ = 1

I just don't see how this final bit follows. After all, if t is large then increasing N won't bring on the QZE. Surely we also need t → 0 but I don't see how the above accounts for this.

Any help would be most appreciated, thanks.

2. Jan 21, 2014

### Demystifier

A few more details can be found, e.g., in
http://lanl.arxiv.org/abs/1311.4363
Read everything from Eq. (2) to Eq. (9).
In particular, in calculating the final limit, t is neither very large nor close to 0.
I hope it helps.

3. Jan 21, 2014

### strangerep

If you wish to understand such things properly, you should also study (carefully!) Ballentine section 12.2 pp338-343.

I'm happy to (try and) fill in any details more explicitly, but only after you've read Ballentine -- so that I don't have to repeat textbook stuff here.

4. Jan 22, 2014

### wle

It's an approximation valid for small $t$ (obviously, because the expression becomes negative for $t > \Delta H$). You can obtain it by expanding the exponential as

$$e^{-iHt} = \mathbb{1} - i H t - \tfrac{1}{2} H^{2} t^{2} + \dotsb$$
and only keeping the terms up to order $t^{2}$ in the expression you get for the survival probability.

You do measure after time intervals that become arbitrarily small. Instead of doing a single measurement after a time $t$, you perform $N$ successive measurements at time intervals $\delta t = \frac{t}{N}$. That's why the $\frac{t^{2}}{Z^{2}}$ changes to $\frac{t^{2}}{N^{2} Z^{2}}$ in the expression for the survival probability.

As for the limit itself, there's more than one approach that might work. One way is to factor $1 - \frac{t^{2}}{N^{2} Z^{2}}$ as $\bigl( 1 + \frac{t}{N Z} \bigr) \bigl( 1 - \frac{t}{N Z} \bigr)$ and use that $\lim_{n \to \infty} \bigl( 1 + \frac{x}{n} \bigr)^{n} = e^{x}$.

5. Jun 21, 2015

### white elephant

I also tryed to understand this calculation and still dont get it. Can someone please help me with the next step after the expanding of the exponential?