What happens if you modulate a Hamiltonian with white noise? Consider the Hamiltonian $f(t)H,$ where $H$ is time-independent and $f(t)$ is classical white noise. Then I would write a Schrodinger equation
$$\mathrm{d}\psi=-iH\psi\ \mathrm{d}W_t,$$
where $W_t$ is a Wiener process. But using Ito calculus, the solution is un-normalized$^{[a]}$:
$$\psi(t,W_t)=e^{+H^2 t/2}e^{-iHW_t}\psi(0) \, .$$
What happened here? I could modify the Schrodinger equation to get rid of the $e^{H^2 t/2}$ term, but I don't know how to justify that.

$[a]$: You can get this solution as a special case of the example on Wikipedia: $$\mathrm{d}S_t=S_t(\mu\mathrm{d}t+\sigma\mathrm{d}W_t)\Rightarrow S(t,W_t)=S(0)e^{(\mu-\sigma^2/2)t + \sigma W_t}.$$
 A: The Schrödinger equation you wrote conserves the averaged probability density if it is seen in the Stratonovich form. Now, its solution, which is normalized if $||\psi(0)||^2 = 1,$ would be $$\psi(t) = U(t)\psi(0) = \exp(-iX_tH)\psi(0) \quad (*), $$
where $X_t = \int_0^t \xi(s)d s$ is the Brownian motion in $\mathbb{R}$ and the integral is in the Stratonovich form, i.e., whenever evaluating, say, expectation values and delta functions ($\Theta(t)$) pop-up, you must take $\Theta(0) = 1/2$ --Conversely, in the Ito form, $\Theta(0) = 0$.
However, if you want to write your equation in the Ito form, the correct equation would be $$ d \psi(t) = -iH\psi(t)d W_t - \frac{1}{2}H^2\psi(t)d t. \quad (**) $$
One can demonstrate this equation either by just transforming Eq. (*) to the Ito form (see page 11 in https://arxiv.org/abs/quant-ph/9702030), or by more "physical" arguments (see, e.g., page 45 of van Kampen's Stochastic processes in physics and chemistry). Namely, starting with a stochastic Schrödinger equation with two generators: a unitary, fluctuating one $-iHd X_t$;  and a dissipative one of the form $-V d t$, i.e., $$ d \psi(t) = -iH \psi d X_t - V\psi(t)d t. $$
After imposing the conservation of averaged probability density in infinitesimal times, i.e., $\mathbb{E}\left[||\psi(t + d t)||^2\right] = \mathbb{E}\left[||\psi(t)||^2\right]$, one finds that $U = \frac{1}{2}H^2$.
To conclude, Eq. (*) is normalized and, when taken in the Ito form, solves Eq.(**):
$$
\begin{aligned}
d \psi(t) &= \psi(t + d t ) -\psi(t) \\
&=\left[U(t+d t,t)- I\right]U(t,0) \psi(0)\\
&=  \left[\exp(-iH d X_t) - I \right]\psi(t) \\
&= -iH\psi(t)d X_t - \frac{1}{2}H^2\psi(t) d t.
\end{aligned}
$$
