- #1
James MC
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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 = |<ψ0|ψt>|2 = |<ψ0|e-iHt|ψ0>|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|H2|ψ0> - (<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.
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 = |<ψ0|ψt>|2 = |<ψ0|e-iHt|ψ0>|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|H2|ψ0> - (<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.