Deriving Bloch vector $dr$ from master equation

Question

I am trying to derive the Bloch vector $dr$ for a measurement of a observable in an 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: $$\frac{d \rho}{d t}=\mathcal{D}[X] \rho+\sqrt{\eta} \mathcal{H}[X] \rho \xi(t)$$ where \begin{align*} \mathcal{D}[X] \rho&=X \rho X^{\dagger}-\frac{1}{2}(X^{\dagger} X \rho+\rho X^{\dagger} X)\\ \mathcal{H}[X] \rho&=X \rho+\rho X-\operatorname{tr}(( X+X^{\dagger})\rho) \rho \end{align*} and $\kappa$ is the measurement strength. Using the Bloch Vector form, i.e. $\rho=\frac{1}{2}(I+x \sigma_{x}+y \sigma_{y}+z \sigma_{z})$, then, \begin{align*} \mathcal{D}[X] \rho&=2 \kappa(\sigma_{z} \rho \sigma_{z}-\rho)\\ \mathcal{H}[X] \rho&=\sqrt{2 \kappa}(\sigma_{z} \rho+\rho \sigma_{z}-2 \operatorname{tr}(\sigma_{z} \rho) \rho) \end{align*} To find $\dot x$ we compute \begin{align*} \frac{d x}{d t}&=\frac{d \operatorname{tr}(\sigma_{x} \rho)}{d t}\\ &=2 \kappa(\operatorname{tr}(\sigma_{z} \sigma_{x} \sigma_{z} \rho)-x)+\sqrt{2 \kappa \eta}(\operatorname{tr}((\sigma_{x} \sigma_{z}+\sigma_{z} \sigma_{x}) \rho)-2 x z) \xi(t)\\ &=-4 \kappa x-\sqrt{8 \kappa \eta} x z \xi(t) \end{align*} and similarly for $\dot z$ \begin{align*} \frac{d z}{d t}&=\frac{d \operatorname{tr}(\sigma_{z} \rho)}{d t}\\ &=2 \kappa(\operatorname{tr}(\sigma_{z}^{2} \rho \sigma_{z})-\operatorname{tr}(\sigma_{z} \rho))+\sqrt{2 \kappa \eta}(\operatorname{tr}(\sigma_{z}^{2} \rho+\sigma_{z} \rho \sigma_{z})-2 z \operatorname{tr}(\sigma_{z} \rho)) \xi(t)\\ &=\sqrt{8 \kappa \eta}(1-z^{2}) \xi(t) \end{align*} Now I am trying to find $\dot x,\dot z$ for the case where $X = \cos(\theta )\sigma _z+\sin(\theta )\sigma_x$ along the measurement angle $\theta$:

$$\mathcal{D}[X] \rho=X \rho X^{\dagger}-\frac{1}{2}(X^{\dagger} X \rho+\rho X^{\dagger} X)= \cos^2(\theta )\sigma _z\rho \sigma _z+\sin^2(\theta )\sigma _x\rho \sigma _x$$

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 $\frac{d \operatorname{tr}(\sigma_{x} \rho)}{d t}$ and $\frac{d \operatorname{tr}(\sigma_{z} \rho)}{d t}$ with the $\sin$ and $\cos$ terms like the above simplification.

Answer 1

While I'm of course 4 years late I'd like to answer this question for the sake of posterity. tl;dr: Your calculations for the $Z$-measurement ($X=\sqrt{2\kappa}\sigma_z$) are correct and for the general case where $X=\cos(\theta)\sigma_z+\sin(\theta)\sigma_x$ the desired expressions read \begin{align*} \mathcal D[X]\rho&=\cos^2(\theta)\sigma_z\rho\sigma_z\color{red}{+\cos(\theta)\sin(\theta)(\sigma_x\rho\sigma_z+\sigma_z\rho\sigma_x)}+\sin^2(\theta)\sigma_x\rho\sigma_x \color{red}{-\rho} \\ \mathcal H[X]\rho&=\cos(\theta)\sigma_z\rho +\sin(\theta)\sigma_x\rho +\cos(\theta)\rho\sigma_z+\sin(\theta)\rho\sigma_x - 2(\cos(\theta)z+\sin(\theta)x)\rho \end{align*} (where the expressions highlighted in red were missing from the derivation in the original question) and \begin{align*} \frac{d\,{\rm tr}(\sigma_x\rho)}{dt}&= -2\cos^2(\theta)x+2\cos(\theta)\sin(\theta)x + \sqrt\eta\xi(t)\big( 2\sin(\theta)(1-x^2)-2\cos(\theta)xz \big) \\ \frac{d\,{\rm tr}(\sigma_z\rho)}{dt}&= -2\sin^2(\theta)z+2\cos(\theta)\sin(\theta)x+\sqrt\eta\xi(t)\big( 2\cos(\theta)(1-z^2)-2\sin(\theta)xz\big)\,. \end{align*}

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Detailed answer

First of all I'm not sure where the measurement strength factor $\sqrt{2\kappa}$ went for the general case but if one wants to add it to $X$ the results get modified in a straightforward way: the terms prior to $\xi(t)$ (in $\dot x,\dot z$) get multiplied by $2\kappa$, and the $\xi(t)$ terms get multiplied by $\sqrt{2\kappa}$.

Now for the actual calculations: because \begin{align*} (\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)^2&=\cos^2(\theta){\bf1}+\cos(\theta)\sin(\theta)\{\sigma_z,\sigma_x\}+\sin^2(\theta){\bf1}\\ &=(\cos^2(\theta)+\sin^2(\theta)){\bf1}={\bf1} \end{align*} one finds \begin{align*} \mathcal D[&\cos(\theta)\sigma_z+\sin(\theta)\sigma_x]\rho\\ &=(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)\rho(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x) -\frac12\big\{(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)^2,\rho\big\}\\ &=\cos^2(\theta)\sigma_z\rho\sigma_z+{\cos(\theta)\sin(\theta)(\sigma_x\rho\sigma_z+\sigma_z\rho\sigma_x)}+\sin^2(\theta)\sigma_x\rho\sigma_x {-\rho} \end{align*} and \begin{align*} \mathcal H[&\cos(\theta)\sigma_z+\sin(\theta)\sigma_x]\rho\\ &=(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)\rho+\rho(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)-2{\rm tr}(\cos(\theta)\sigma_z+\sin(\theta)\sigma_x)\rho)\rho\\ &=\cos(\theta)\sigma_z\rho +\sin(\theta)\sigma_x\rho +\cos(\theta)\rho\sigma_z+\sin(\theta)\rho\sigma_x - 2(\cos(\theta)z+\sin(\theta)x)\rho\,. \end{align*} The derivatives are computated analogously, i.e. using that $\sigma_z\sigma_x\sigma_z=-\sigma_x$, $\{\sigma_x,\sigma_z\}=0$, etc. we find $\frac{d\,{\rm tr}(\sigma_x\rho)}{dt}$ to be equal to

\begin{align*} {\rm tr}\big(\sigma_x &\mathcal{D}[\cos(\theta)\sigma_z+\sin(\theta)\sigma_x] \rho\big)+\sqrt{\eta}\xi(t) {\rm tr}\big( \sigma_x\mathcal{H}[\cos(\theta)\sigma_z+\sin(\theta)\sigma_x] \rho \big)\\ &={\rm tr}\big(\cos^2(\theta)\sigma_x \sigma_z\rho\sigma_z+{\cos(\theta)\sin(\theta)(\sigma_x \sigma_x\rho\sigma_z+\sigma_x \sigma_z\rho\sigma_x)}+\sin^2(\theta)\sigma_x \sigma_x\rho\sigma_x {-\sigma_x \rho}\big)\\ &+\sqrt{\eta}\xi(t) {\rm tr}\big( \cos(\theta)\sigma_x\sigma_z\rho +\sin(\theta)\sigma_x\sigma_x\rho +\cos(\theta)\sigma_x\rho\sigma_z+\sin(\theta)\sigma_x\rho\sigma_x - 2(\cos(\theta)z+\sin(\theta)x)\sigma_x\rho )\\ &=\cos^2(\theta){\rm tr}(-\sigma_x \rho)+2{\cos(\theta)\sin(\theta){\rm tr}(\sigma_z\rho)}+\sin^2(\theta){\rm tr}(\sigma_x \rho{-\sigma_x \rho}\big)\\ &+\sqrt{\eta}\xi(t)\big[2\cos(\theta){\rm tr}(\{ \sigma_x,\sigma_z\}\rho) +2\sin(\theta){\rm tr}( \rho) - 2(\cos(\theta)z+\sin(\theta)x){\rm tr}( (\sigma_x\rho )\big]\\ &= -2\cos^2(\theta)x+2\cos(\theta)\sin(\theta)x + \sqrt\eta\xi(t)\big( 2\sin(\theta)(1-x^2)-2\cos(\theta)xz \big) \end{align*} and similarly for $\frac{d\,{\rm tr}(\sigma_z\rho)}{dt}$.