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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 $\psi$.

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  • $\begingroup$ 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 ? $\endgroup$
    – Kurt G.
    Feb 7, 2023 at 17:40
  • $\begingroup$ 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. $\endgroup$
    – MonteNero
    Feb 7, 2023 at 17:53
  • $\begingroup$ 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 ? $\endgroup$
    – Kurt G.
    Feb 7, 2023 at 17:57
  • $\begingroup$ @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. $\endgroup$
    – MonteNero
    Feb 7, 2023 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Feb 7, 2023 at 18:29

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