- #1
Pete5876
- 7
- 0
Misplaced Homework Thread
Show that this dephasing operation is Markovian
$$E_t(\rho)=\left(\frac{1+e^{-t}}{2}\right)\rho + \left(\frac{1-e^{-t}}{2}\right)Z\rho Z$$
The operation is supposed to be Markovian when
$$E_g E_h = E_{g+h}$$
So this is what I get when I apply this multiplication
$$
E_g E_h=\frac{1+e^{-g}}{2}\frac{1+e^{-h}}{2} \rho^2 + \frac{1-e^{-g}}{2} \frac{1-e^{h}}{2}Z\rho ZZ\rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} = \rho Z \rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} Z \rho Z \rho
$$
We can lose ZZ because
$$ZZ=I$$
but I can't see what else can be done. ρ after all only commutes with Z if it's diagonal but we can't assume that for ρ in general?
$$E_t(\rho)=\left(\frac{1+e^{-t}}{2}\right)\rho + \left(\frac{1-e^{-t}}{2}\right)Z\rho Z$$
The operation is supposed to be Markovian when
$$E_g E_h = E_{g+h}$$
So this is what I get when I apply this multiplication
$$
E_g E_h=\frac{1+e^{-g}}{2}\frac{1+e^{-h}}{2} \rho^2 + \frac{1-e^{-g}}{2} \frac{1-e^{h}}{2}Z\rho ZZ\rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} = \rho Z \rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} Z \rho Z \rho
$$
We can lose ZZ because
$$ZZ=I$$
but I can't see what else can be done. ρ after all only commutes with Z if it's diagonal but we can't assume that for ρ in general?
