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Quantum Zeno Effect: What is the argument?

  1. Jan 21, 2014 #1
    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 - [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.
  2. jcsd
  3. Jan 21, 2014 #2


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    A few more details can be found, e.g., in
    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.
  4. Jan 21, 2014 #3


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    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. :biggrin:
  5. Jan 22, 2014 #4


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

    [tex]e^{-iHt} = \mathbb{1} - i H t - \tfrac{1}{2} H^{2} t^{2} + \dotsb[/tex]
    and only keeping the terms up to order [itex]t^{2}[/itex] 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 [itex]t[/itex], you perform [itex]N[/itex] successive measurements at time intervals [itex]\delta t = \frac{t}{N}[/itex]. That's why the [itex]\frac{t^{2}}{Z^{2}}[/itex] changes to [itex]\frac{t^{2}}{N^{2} Z^{2}}[/itex] 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 [itex]1 - \frac{t^{2}}{N^{2} Z^{2}}[/itex] as [itex]\bigl( 1 + \frac{t}{N Z} \bigr) \bigl( 1 - \frac{t}{N Z} \bigr)[/itex] and use that [itex]\lim_{n \to \infty} \bigl( 1 + \frac{x}{n} \bigr)^{n} = e^{x}[/itex].
  6. Jun 21, 2015 #5
    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?
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