solving "Cascaded measurements are single measurements"

Question

This is an exercise from p. 86 Quantum Computation and Quantum Information by Michael A. Nielsen and Isaac L. Chuang.

Here I think I have to show that $p\left(l\right)p\left(m\right)=p\left(lm\right)$, meaning the probability of getting $l$ and then getting m is equal to the probability of getting $lm$. But I can't prove it... The definition of $p\left(m\right)$ (same for $p\left(l\right)$ of course) is like below.

Could anyone help me how to prove $p\left(l\right)p\left(m\right)=p\left(lm\right)$?

Answer 1

Just insert the outer product $\left|\psi\right\rangle\left\langle\psi\right|$ in between operators.

$p\left(lm\right) = \left\langle\psi\right|M_{m}^{\dagger}L_{l}^{\dagger}L_{l}M_{m}\left|\psi\right\rangle $

then insert $\left|\psi\right\rangle\left\langle\psi\right|$ between $M_{m}^{\dagger}$ and $L_{l}^{\dagger} $

and again insert $\left|\psi\right\rangle\left\langle\psi\right|$ this time between $L_{l}$ and $M_{m} $

$p\left(lm\right) = \left\langle\psi\right|M_{m}^{\dagger}\left| \psi \right\rangle \left\langle\psi\right|L_{l}^{\dagger}L_{l}\left|\psi\right\rangle\left\langle\psi\right|M_{m}\left|\psi\right\rangle $

then bring the central value $p\left(l\right)$ out front

$p\left(lm\right) = p\left (l\right) \left\langle\psi\right|M_{m}^{\dagger}\left| \psi \right\rangle \left\langle\psi\right|M_{m}\left|\psi\right\rangle $

and collapse/remove the remaining outer product ... you're left with $p\left(l\right) p\left(m\right)$

Insertions and removals of $\left|\psi\right\rangle\left\langle\psi\right|$s are akin to multiplying and dividing by unity (1)

Answer 2

Using ${L_l}$ to measure $|\varphi\rangle$, the probability of resulting $l$ is $\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle$, and this will make the state collapse to $\frac {L_l|\varphi\rangle}{\sqrt {\langle{\varphi} |{L_l}^{\dagger} L_l|{\varphi}\rangle}}$.

Using ${M_m}$ to measure $\frac {L_l|\varphi\rangle}{\sqrt {\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle}}$, the probability of resulting $m$ is $\frac {\langle{\varphi} |M_m^{\dagger} L_l^{\dagger}L_l M_m|{\varphi}\rangle}{\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle}$, and this will make the state collapse to $\frac {L_l M_m|{\varphi}\rangle}{\langle{\varphi} |M_m^{\dagger} L_l^{\dagger}L_lM_m|{\varphi}\rangle}$.

Easy to verify this consecutive process is equivalent to $M_mL_l$