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.
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}