Trace Differentiation with Pauli operators, finding $\frac{d x}{d t}$ and $\frac{d z}{d t}$ from the 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:
$\mathcal{D}[X] \rho=X \rho X^{\dagger}-\frac{1}{2}\left(X^{\dagger} X \rho+\rho X^{\dagger} X\right)$
$=\cos ^{2}(\Theta) \sigma_{z} \rho \sigma_{z}+\sin ^{2}(\Theta) \sigma_{x} \rho \sigma_{x}+\cos (\Theta) \sin (\Theta) \sigma_{z} \rho \sigma_{x}+\cos (\Theta) \sin (\Theta) \sigma_{x} \rho \sigma_{z}$
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.
 A: To answer your first question: since $\mathcal H[X]$ is a linear with respect to $X$, we have
$$
\mathcal H[\cos(\Theta)\sigma_z + \sin(\Theta)\sigma_x] = 
\cos (\Theta) \mathcal H[\sigma_z] + \sin (\Theta) \mathcal H[\sigma_x].
$$ 
Assuming your computation for $\mathcal H[X] \rho$ is correct, taking $X = \cos(\Theta)\sigma_z + \sin(\Theta)\sigma_x$ should give us
$$
\mathcal H[X]\rho = 
\cos(\Theta)\sqrt{2}\left(\sigma_{z} \rho+\rho \sigma_{z}-2 z \rho\right) + 
\sin(\Theta)\sqrt{2}\left(\sigma_{x} \rho+\rho \sigma_{x}-2 x \rho\right).
$$

We should find that
$$
\frac{dx}{dt} = \left[-4\cos^2 (\Theta)\, x-\sqrt{8 \eta} \cos (\Theta) \,x z\, \xi(t) + \right] + 
 \left[
\sqrt{8 \eta}\sin (\Theta)\left(1-x^{2}\right) \xi(t)
\right]
$$
and 
$$
\frac{dz}{dt} = 
 \left[ \cos \Theta\sqrt{8 \eta}\left(1-z^{2}\right) \xi(t) \right]
+ \left[-4\sin^2 \Theta z-\sqrt{8 \eta}\sin \Theta\, x z \,\xi(t)\right].
$$

From scratch:
\begin{align}
\frac{dx}{dt} &= \frac{d}{dt}\operatorname{Tr}(\sigma_x \rho) 
= \operatorname{Tr}\left(\sigma_x \frac{d\rho}{dt}\right) 
= 
\operatorname{Tr}\left(\sigma_x 
\mathcal{D}[X] \rho+\sqrt{\eta} \sigma_x\mathcal{H}[X] \rho \xi(t)\right)
\\ & = \operatorname{Tr}\left(\sigma_x 
\mathcal{D}[X] \rho\right)+ 
\sqrt{\eta}\operatorname{Tr}\left( \sigma_x\mathcal{H}[X] \rho \xi(t)\right)
\end{align}
...
\begin{align}
\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 + \cos \Theta \sin \Theta [\sigma_z \rho \sigma_x + \sigma_x \rho \sigma_z] - \rho
\end{align}
...
\begin{align}
\operatorname{Tr}\left(\sigma_x 
\mathcal{D}[X] \rho\right) &= 
\operatorname{Tr}\left(
\cos^2 \Theta \,\sigma_x\sigma_z \rho \sigma_z + \sin^2 \Theta\,\sigma_x^2 \rho \sigma_x + \cos \Theta \sin \Theta [\sigma_x\sigma_z \rho \sigma_x + \sigma_x^2 \rho \sigma_z] - \sigma_x\rho
\right)\\
&= \operatorname{Tr}\left(
\cos^2 \Theta \,[\sigma_z\sigma_x\sigma_z] \rho + \sin^2 \Theta\,\rho \sigma_x + \cos \Theta \sin \Theta [\sigma_x^2\sigma_z \rho + \sigma_z\sigma_x^2 \rho] - \sigma_x\rho
\right)
\\ &=
\operatorname{Tr}\left(
-\cos^2 \Theta \,\sigma_x \rho + \sin^2 \Theta\,\rho \sigma_x + \cos \Theta \sin \Theta [\sigma_z \rho + \sigma_z \rho] - \sigma_x\rho
\right)
\\ &=
(\sin^2\Theta - \cos^2\Theta - 1)\operatorname{Tr}(\sigma_x \rho) + 2 \cos \Theta \sin \Theta\operatorname{Tr}(\sigma_z \rho)
\\ & = 
(\sin^2\Theta - \cos^2\Theta - 1)x + 2 \cos \Theta \sin \Theta z
\end{align}
... Noting that $X = X^\dagger$,
\begin{align}
\mathcal H[X] \rho &= 
X \rho+\rho X-\left\langle X+X^{\dagger}\right\rangle \rho = 
X \rho+\rho X- 2\left\langle X\right\rangle \rho
\\ & = 
\cos \Theta[\sigma_z \rho + \rho\sigma_z] + \sin \Theta[\sigma_x \rho + \rho \sigma_x] - 2(\cos \Theta \,z + \sin \Theta \,x) \rho 
\end{align}
....
\begin{align}
\operatorname{Tr}(\sigma_x \mathcal H[X] \rho) &= 
\operatorname{Tr}(
\cos \Theta[\sigma_x\sigma_z \rho + \sigma_x\rho\sigma_z] + \sin \Theta[\sigma_x^2 \rho + \sigma_x\rho \sigma_x] - 2(\cos \Theta \,z + \sin \Theta \,x) \rho 
)
\\ & = 
\operatorname{Tr}(
\cos \Theta[(\sigma_x\sigma_z + \sigma_z \sigma_x)\rho] + \sin \Theta[\rho + \sigma_x^2\rho] - 2(\cos \Theta \,z + \sin \Theta \,x) \rho 
)
\\ & = 
\operatorname{Tr}(
\cos \Theta[0\cdot\rho] + \sin \Theta[2\rho] - 2(\cos \Theta \,z + \sin \Theta \,x) \rho 
)
\\ & = 
2(\sin \Theta - \cos \Theta \,z - \sin \Theta x)\operatorname{Tr}(\rho) = 
2(\sin \Theta - \cos \Theta \,z - \sin \Theta x).
\end{align}
....
\begin{align}
\frac{d x}{dt} &= 
(\sin^2\Theta - \cos^2\Theta - 1)x + 2 \cos \Theta \sin \Theta z + 
2\sqrt{\eta}\xi(t)(\sin \Theta - \cos \Theta \,z - \sin \Theta x)
\\ & = 
2 \sqrt{\eta}\xi(t) \sin \Theta + 
(\sin^2\Theta - \cos^2\Theta - 1 - 2 \sqrt{\eta}\xi(t) \sin \Theta)x
+ 2(\cos \Theta \sin \Theta - \sqrt{\eta}\xi(t) \cos \Theta)z
\end{align}
