- #1
Agrippa
- 78
- 10
Hi, I'm trying to grasp the standard derivation of the quantum Zeno effect. But my calculus is limited. I would love some assistance!
Schroedinger evolution tells us that the state at time t, given the initial state at time t=0, is:
[itex]|\psi_t\rangle = e^{-i H t} |\psi_0 \rangle[/itex]
We can then define the survival probability [itex]P_s(t)[/itex], which is the probability that the initial state will survive or remain, at time t, using the Born rule:
[itex]P_s(t) = | \langle \psi_0| e^{-i H t} |\psi_0 \rangle |^2 [/itex]
From the definition of the exponential function, we can expand the power series:
[itex]e^{-i H t} \approx 1 - iHt - 1/2H^2t^2 + ... [/itex]
The next part is where I get stuck. It is then often said that from this we can derive:
[itex]P_s(t) \approx 1 - ( \langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 )t^2 [/itex]
But how does that follow from the previous steps? I've never seen this spelled out anywhere before. Is anyone able to fill in the gaps here?
Once we have this expression, we can let [itex](\Delta H)^2 [/itex] stand for [itex]\langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 [/itex] to give the simpler expression:
[itex]P_s(t) \approx 1 - (\Delta H)^2t^2 [/itex]
We now wish to consider the survival probability where the interval [0, t] is interrupted by n measurements at times t/n, 2t/n, ..., t. Ideally, these measurements are instananeous projections and the initial state [itex]|\psi_0\rangle[/itex] is an eigenstate of the measurement operator. In that case, we get:
[itex]P_s(t) \approx [1 - (\Delta H)^2(t/n)^2 ]^n [/itex]
...which approaches 1 as n approaches infinity.
How this last expression is derived is also not 100% clear to me. But I'm baffled by the part I get stuck by mentioned above and would love some help with that first.
Thanks!
Schroedinger evolution tells us that the state at time t, given the initial state at time t=0, is:
[itex]|\psi_t\rangle = e^{-i H t} |\psi_0 \rangle[/itex]
We can then define the survival probability [itex]P_s(t)[/itex], which is the probability that the initial state will survive or remain, at time t, using the Born rule:
[itex]P_s(t) = | \langle \psi_0| e^{-i H t} |\psi_0 \rangle |^2 [/itex]
From the definition of the exponential function, we can expand the power series:
[itex]e^{-i H t} \approx 1 - iHt - 1/2H^2t^2 + ... [/itex]
The next part is where I get stuck. It is then often said that from this we can derive:
[itex]P_s(t) \approx 1 - ( \langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 )t^2 [/itex]
But how does that follow from the previous steps? I've never seen this spelled out anywhere before. Is anyone able to fill in the gaps here?
Once we have this expression, we can let [itex](\Delta H)^2 [/itex] stand for [itex]\langle \psi_0| H^2 |\psi_0 \rangle^2 - (\langle \psi_0| H |\psi_0 \rangle )^2 [/itex] to give the simpler expression:
[itex]P_s(t) \approx 1 - (\Delta H)^2t^2 [/itex]
We now wish to consider the survival probability where the interval [0, t] is interrupted by n measurements at times t/n, 2t/n, ..., t. Ideally, these measurements are instananeous projections and the initial state [itex]|\psi_0\rangle[/itex] is an eigenstate of the measurement operator. In that case, we get:
[itex]P_s(t) \approx [1 - (\Delta H)^2(t/n)^2 ]^n [/itex]
...which approaches 1 as n approaches infinity.
How this last expression is derived is also not 100% clear to me. But I'm baffled by the part I get stuck by mentioned above and would love some help with that first.
Thanks!