# Quantum Zeno Effect [closed]

This is a homework exercise that is very useful to think about:

A quantum system of two levels is described by two unitary vectors on $\mathscr{H}=\mathbb{C}^2$. The initial state of the system is

$$\mid\psi (0) \rangle =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}.\tag{1}$$ The hamiltonian of the system is $H=\varepsilon \sigma_z$, where $\varepsilon>0$ is a constant with energy units and $\sigma_z$ the Pauli matrix $$\sigma_z \equiv \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\tag{2}$$ I know, resolving the Schrödinger equation, which is the wave function $\mid\psi (t)\rangle$ in a time $t$: $$\mid\psi(t)\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \text{ e}^{-\text{i} \varepsilon t/\hbar}+\frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \text{ e}^{\text{i} \varepsilon t/\hbar}\equiv \frac{1}{\sqrt{2}} \mid\uparrow\rangle \text{ e}^{-\text{i} \varepsilon t/\hbar}+\frac{1}{\sqrt{2}} \mid \downarrow \rangle \text{ e}^{\text{i} \varepsilon t/\hbar}.\tag{3}$$ If we measure the energy we can obtain $E_\uparrow = \varepsilon$ with a probability of $$p_\uparrow[1] = \dfrac{1}{2}.\tag{4}$$ and $E_\downarrow =-\varepsilon$ with a probability of $$p_\downarrow [1] =\dfrac{1}{2}\tag{5}$$ where the braket $[\cdot ]$ means the times the particle has been measured. Now we measure the observable $\sigma_x$ in regular intervals of time $t=\Delta t, 2 \Delta t, ..., N\Delta t$ ¿Where is the probability of obtaining the result $+1$ on the $N$ measurements? Study the probability of $+1$ when $\Delta t \rightarrow 0$ and $N \rightarrow \infty$ very large being the total time of the measurement finite $\tau\equiv N\Delta t$. Draw the probability function of $\tau$ for $\Delta t =1$ fs $=10^{-15}$ s and $\varepsilon=10^{-3}$ eV and discuss the physic relevance of the previous limit.

My answer is $$p_\uparrow [N] = 1-\dfrac{1}{2^N}\tag{6}$$ (binomial distribution ¿and the limit is Poisson distribution?)

and the reduction of state vector $\textbf{R}$ is permanent, so $$p_\uparrow [\infty]=p_\uparrow [1]=\dfrac{1}{2}.\tag{7}$$

El camino a la realidad, Roger Penrose, p.714, Debate (2006)

What is the correct answer? Am I wrong?

## closed as off-topic by Bosoneando, DanielSank, user36790, John Rennie, Jon CusterOct 5 '16 at 18:24

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• And your question is? – Bosoneando Oct 5 '16 at 16:30
• What is the correct answer? Am I wrong? – Clare Francis Oct 5 '16 at 16:39
• 1. Next time, include your actual question in your post. 2. Check-my-work questions are frowned upon in this site (see the site policies ) – Bosoneando Oct 5 '16 at 16:48