I am trying to derive the Bloch vector $dr$ for a measurement of a observable in any arbitrary direction $\theta$. For context this is the setup and derivation I have for continuous measurement along the $z$ axis.

The equation of continuous measurement on observable X has the following form:

$\frac{d \rho}{d t}=\mathcal{D}[X] \rho+\sqrt{\eta} \mathcal{H}[X] \rho \xi(t)$

$\mathcal{D}[X] \rho=X \rho X^{\dagger}-\frac{1}{2}\left(X^{\dagger} X \rho+\rho X^{\dagger} X\right)$

$\mathcal{H}[X] \rho=X \rho+\rho X-\left\langle X+X^{\dagger}\right\rangle \rho$

Kappa is the measurement strength.

In Bloch Vector form,

$\rho=\frac{1}{2}\left(I+x \sigma_{x}+y \sigma_{y}+z \sigma_{z}\right)$


$\mathcal{D}[X] \rho=2 \kappa\left(\sigma_{z} \rho \sigma_{z}-\rho\right)$

$\mathcal{H}[X] \rho=\sqrt{2 \kappa}\left(\sigma_{z} \rho+\rho \sigma_{z}-2 z \rho\right)$

To find $dx$

$\frac{d x}{d t}=\frac{d T r\left(\sigma_{x} \rho\right)}{d t}=2 \kappa\left(\operatorname{Tr}\left(\sigma_{z} \sigma_{x} \sigma_{z} \rho\right)-x\right)+\sqrt{2 \kappa \eta}\left(\operatorname{Tr}\left(\left(\sigma_{x} \sigma_{z}+\sigma_{z} \sigma_{x}\right) \rho\right)-2 x z\right) \xi(t)$

$=-4 \kappa x-\sqrt{8 \kappa \eta} x z \xi(t)$

$\frac{d z}{d t}=\frac{d \operatorname{Tr}\left(\sigma_{z} \rho\right)}{d t}=2 \kappa\left(\operatorname{Tr}\left(\sigma_{z}^{2} \rho \sigma_{z}\right)-\operatorname{Tr}\left(\sigma_{z} \rho\right)\right)+\sqrt{2 \kappa \eta}\left(\operatorname{Tr}\left(\sigma_{z}^{2} \rho+\sigma_{z} \rho \sigma_{z}\right)-2 z \operatorname{Tr}\left(\sigma_{z} \rho\right)\right) \xi(t)$

$=\sqrt{8 \kappa \eta}\left(1-z^{2}\right) \xi(t)$

Now I am trying to find the $\frac{d z}{d t}$ and $\frac{d x}{d t}$, in the case that the $\mathcal{D}[X]$ and $\mathcal{H}[X]$ terms for the same master equation where becomes $X = \cos(\Theta )\sigma _z+\sin(\Theta )\sigma_x$ along measurement angle $\theta$ are:

$\mathcal{D}[X] \rho=X \rho X^{\dagger}-\frac{1}{2}\left(X^{\dagger} X \rho+\rho X^{\dagger} X\right)$

$= \cos^2(\Theta )\sigma _z\rho \sigma _z+\sin^2(\Theta )\sigma _x\rho \sigma _x$

I am confused on what the simplified form of $\mathcal{H}[X]\rho$ will be as my simplification skills are not very strong

I would also need help in finding $\frac{d T r\left(\sigma_{x} \rho\right)}{d t}$ and $\frac{d T r\left(\sigma_{z} \rho\right)}{d t}$ with the $\sin$ and $\cos$ terms like the above simplification.


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