- #1
James MC
- 174
- 0
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 = |<ψ0|ψt>|2 = |<ψ0|e-iHt|ψ0>|2
So far so good. But now standard explanations assert that:
Ps = |<ψ0|e-iHt|ψ0>|2 = 1 - (ΔH)2t2
(Where (ΔH)2 = <ψ0|H2|ψ0> - (<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 - [itex]\frac{t^{2}}{Z^{2}}[/itex]
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[itex]^{N}_{s}[/itex] = (1 - [itex]\frac{t^{2}}{N^{2}Z^{2}}[/itex])N
...so that in the limit of continuous measurements where N → ∞ we get:
[itex]\stackrel{Lim}{N→∞}[/itex] P[itex]^{N}_{s}[/itex] = 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.
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 = |<ψ0|ψt>|2 = |<ψ0|e-iHt|ψ0>|2
So far so good. But now standard explanations assert that:
Ps = |<ψ0|e-iHt|ψ0>|2 = 1 - (ΔH)2t2
(Where (ΔH)2 = <ψ0|H2|ψ0> - (<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 - [itex]\frac{t^{2}}{Z^{2}}[/itex]
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[itex]^{N}_{s}[/itex] = (1 - [itex]\frac{t^{2}}{N^{2}Z^{2}}[/itex])N
...so that in the limit of continuous measurements where N → ∞ we get:
[itex]\stackrel{Lim}{N→∞}[/itex] P[itex]^{N}_{s}[/itex] = 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.