# Matrix-valued stochastic DE to a vector-valued stochastic DE

Suppose we have a matrix-valued ODE: $$i\tag{1} \frac{d}{dt}U(t) = HU(t)$$ where $$U(t)$$ and $$H$$ are complex square matrices. Given a time-independent vector $$\psi$$, we can convert (1) into a vector-valued differential equation: $$i\frac{d}{dt}U(t)\psi = HU(t)\psi.$$ The above can be further rewritten as $$i \frac{d}{dt}\psi(t) = H \psi(t).$$ So, now instead of a matrix-valued ODE, we have a simpler vector-valued ODE. Some might recognize that this is the Schrodinger equation with $$\hbar=1$$.

I would like to know if the same logic applies to a stochastic matrix-valued differential equation: $$\tag{2} dX(t) = X(t)A(t)dt + X(t)\sum_j B_j dW_j,$$ where $$X(t)$$ is a matrix-valued stochastic process, $$A(t)$$ and $$B_j$$ are square complex matrices and $$dW_j$$ is a scalar increment of Wiener process.

Given a time-independent vector $$\psi$$, is it possible to bring (2) into a vector-valued stochastic differential equation?

\begin{align} \psi^*dX(t) &= \psi^*X(t)A(t)dt + \psi^*X(t)\sum_j B_j dW_j\\ d\psi(t)^* &= \psi(t)^*A(t)dt + \psi(t)^*\sum_j B_j dW_j.\\ \end{align}

Remark: $$\psi^*$$ denotes the conjugate transpose of $$\psi$$.

• In light of my comment to this similar question: what are the dimensions of $X,A,B,W$ ? Over what range runs $j$ ? Do $A$ and $B$ commute ? Feb 7 at 17:40
• It is stated in the body of the question: the matrices are all square matrices, finite-dimensional. If commutativity is not mentioned we assume that it is a general case. The sum in the index $j$ is finite. Feb 7 at 17:53
• All terms of your equation (2) are square matrices and you multiply that equation by a vector $\psi^*$ from the left. What is wrong with that ? Feb 7 at 17:57
• @KurtG. My concern is Ito's rule. If $\psi(t)$ was time-dependent, then $d\psi(t)^*$ would have to incorporate a product rule or something like that. I don't know. But since $\psi$ is time-independent, $d\psi(t)^*$ might be just what I hope it to be. Feb 7 at 18:22
• I think this is true. This question seems to be inspired by Schrödinger/Heisenberg picture from QM. They are both equivalent and in the Heisenberg picture states $\psi$ don't depend on time. Feb 7 at 18:29