# Quantum Zeno Effect: What is the argument? (Part II)

1. Mar 31, 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. My understanding has improved due to responses to a previous post on this topic as well as to a detailed but still gappy paper that was recommended to me, but I'm still confused. Please help!

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

Time evolution is given by a unitary operator U(t) = e-iHt where H is the system's Hamiltonian, such that |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

Since:
$$e^{-x} = \sum_{N=0}^{\infty} \frac {(-x)^N}{N!}$$
We have:

e-iHt = 1 - iHt - 1/2(H2t2) + ...

Since the values not explicitly written become negligibly small we now have:

Ps ≈ | 1 - <ψ0|iHt|ψ0> - <ψ0|1/2(H2t2)|ψ0> |2

We may remove the Born rule (the mod square) since |1|2=1 and rearrange the equation:

Ps ≈ 1 - it<ψ0|H|ψ0> - t2/2(<ψ0|(H2)|ψ0>)

This brings us up to equation (6) in the detailed but still gappy paper. But that paper then becomes incomprehensible to me...

At this stage standard presentations introduce the following term:

(ΔH)2 = <ψ0|H20> - (<ψ0|H|ψ0>)2

I do not understand the significance of this expression. Whatever it is it purportedly allows us to deduce:

Ps ≈ 1 - (ΔH)2t2

Perhaps I'm just being dumb but I do not see how this follows from the previous two expressions. If anyone could fill this gap in for me that would be wonderful!

Anyway now we introduce N measurements such that as N tends to infinity the time between measurements tends to zero:

Ps ≈ [1 - (ΔH)2(t/N)2]N

I get why we raise to the power of N: because raising by e.g 2 represents applying the Born rule twice i.e. measuring twice. I sort of get why we also divide by N, since that's now the time it takes for a measurement to occur. But why apply division by N to THAT t? Well it's the only one left after (ΔH)2 wiped the other t away. But I don't understand where that t went.

In the detailed but still gappy paper the author rearranges this expression:

Ps ≈ 1 - N(ΔH)2(t/N)2

...presumably to make it more clear as to why the expression tends to 1 - 0 as N tends to ∞. But why should it tend to 0? Well, we have division by infinity (by N). Unfortunately we also have the numerator multiplied by N. So depending on what (ΔH)2 is we may end up with 1 - 1 = 0! So (ΔH)2 is clearly causing me some bother.

I hope you can help! Thanks.

2. Apr 1, 2014

### Jilang

Well, you have an N in the numerator, but an N squared in the denominator.

3. Apr 1, 2014

### Demystifier

Your second equation above is wrong. You should remind yourself about some basics of complex algebra.
In particular, if z=x+iy iz a complex number, then
|z|^2=zz*=(x+iy)(x-iy)=x^2-y^2

The author of the mentioned paper does not write these details explicitly because he obviously assumes that the reader should already know that.

4. Apr 1, 2014

### Demystifier

No we don't. As Jilang noted, we have the numerator multiplied by N^2.

5. Apr 1, 2014

### Demystifier

Here is how you can get the second equation from the first. In your first equation you can rename t as t', so you get
Ps ≈ 1 - (ΔH)2t' 2
This is valid for any small value of t', so you can take a special case
t'=t/N
Inserting this into my first expression, you get
Ps ≈ [1 - (ΔH)2(t/N)2]
The power of N in your second quoted expression is obtained as you explained by yourself, by applying N measurements.

Does it help?

Last edited: Apr 1, 2014
6. Apr 1, 2014

### James MC

Great! Thank you.

So if I've understood, then I should return to the expression right before things went wrong:

Ps ≈ | 1 - <ψ0|iHt|ψ0> - <ψ0|1/2(H2t2)|ψ0> |2

This can be transformed into:

Ps ≈ | 1 - t2/2(<ψ0|(H2)|ψ0> - it<ψ0|H|ψ0>) |2

We can recognise this as taking the form:

Ps ≈ | 1 - x - iy |2

where x-iy = z = a complex number! And since we know that "if z=x+iy is a complex number, then |z|^2=zz*=(x+iy)(x-iy)=x^2-y^2" then we can say:

Ps ≈ 1 - x2 - y2

We now substitute the values back in for x and y:

Ps ≈ 1 - (t2/2(<ψ0|H20>)2 - (t<ψ0|H|ψ0>)2

Ps ≈ 1 - 2(t2/2(<ψ0|H20>) - t20|H|ψ0>2

Ps ≈ 1 - t20|H20> - t20|H|ψ0>2

Ps ≈ 1 - t2(<ψ0|H20> - <ψ0|H|ψ0>2)

And NOW it is clear what the motivation for (ΔH)2 is. It's purely to abbreviate the above expression a bit! In particular, if we let:

(ΔH)2 = <ψ0|H20> - <ψ0|H|ψ0>2

Then:

Ps ≈ 1 - t2(ΔH)2

The wording is a bit odd e.g. why not just use ΔH or just Δ, but nevermind. I'll look at your other post more carefully in the morning, but I'm glad that the part that was worrying me the most has been sorted. Cheers!

7. Apr 1, 2014

### Demystifier

In general, ΔA is a standard notation for the amount of quantum uncertainty of the quantity A. Quite generally, the uncertainty of A in a quantum state |ψ> is given by
ΔA^2 = <ψ|A^2|ψ> - <ψ|A|ψ>^2

8. Apr 1, 2014

### James MC

Yes, I understand this, thanks. The one thing I don't understand is that in equation (9) of the original paper the author, unlike us, uses equality instead of ≈. Is that a typo? The author gives:

(9) Ps = [1 - (ΔH)2(t/N)2]N ≈ 1 - N(ΔH)2(t/N)2

What's confusing is that the first should be ≈ if it derives from (7). The second can't be equality since in general xN ≠ Nx but it's also unclear why even ≈ holds.

9. Apr 2, 2014

### Demystifier

Yes.

Because for small x
(1-x)^N≈1-Nx
You can show it by yourself by using binomial theorem
http://en.wikipedia.org/wiki/Binomial_theorem
and neglecting higher powers of x.

10. Apr 2, 2014

### James MC

Great. Well I think I understand the whole thing now. Thanks for your help!