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Consider the rotational Langevin equation in the absence of an external force: $$\frac{d \hat n(t)}{dt} =\vec{\xi}(t) \times \hat n(t)$$ where $\vec \xi(t)$ is a Gaussian white noise and $\hat n(t) \cdot \hat n(t)=1$. Take the dot product of this equation with $\hat n(0)$ to get: $$\frac{d\hat n(0)\cdot \hat n(t)}{dt}=\hat n(0)\cdot\left(\vec{\xi}(t) \times \hat n(t)\right)$$ $$=\vec{\xi}(t)\cdot\left( \hat n(t)\times \hat n(0)\right)$$ Now take the average $\langle \ldots \rangle$ over the noise: $$\frac{d\langle \hat n(0)\cdot \hat n(t) \rangle}{dt}=\langle\vec{\xi}(t)\cdot\left( \hat n(t)\times \hat n(0)\right)\rangle$$ But since $t\gt 0$ we have that $\vec{\xi}(t)$ is independent from $\hat n(0)$ as the present value of the noise cannot effect a past orientation. And also $\vec{\xi}(t)$ is independent of $\hat n(t)$ as the current orientation does not depend on the current value of the noise. Hence: $$\frac{d\langle \hat n(0)\cdot \hat n(t) \rangle}{dt}=\langle\vec{\xi}(t)\rangle \cdot \langle\left( \hat n(t)\times \hat n(0)\right)\rangle$$ but $\langle \vec \xi(t)\rangle=0$ Thus: $$\frac{d\langle \hat n(0)\cdot \hat n(t) \rangle}{dt}=0$$ $$\langle \hat n(0)\cdot \hat n(t) \rangle=\langle \hat n(0)\cdot \hat n(0) \rangle=1$$ Which actually implies that $\hat n(t)=\hat n(0)\; \forall t$ - an answer this is clearly wrong and contradicts with other calculations. That said I cannot see a floor in my reasoning. Thus my question is: What is conceptually wrong with the above calculation?

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Sorry, I just remembered I saw an explanation of something similar in "Quantum Field Theory and Critical Phenomena" by Zinn-Justin 4th ed pg63. The reason is that in general averages over the noise $\langle\ldots \rangle$ and time derivatives $\frac{d}{dt}$ do not commute i.e. in this case the equality that fails (I did it implicitly in the question) is: $$\langle \frac{d}{dt} (\hat n(t) \cdot \hat n(0))\rangle\ne\frac{d}{dt} \langle \hat n(t) \cdot \hat n(0)\rangle$$

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