# Deriving Bloch vector $dr$ from master equation

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,

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

Then,

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

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

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.