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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: $$\mathrm{d}S_t=S_t(\mu\mathrm{d}t+\sigma\mathrm{d}W_t)\Rightarrow S(t,W_t)=S(0)e^{(\mu-\sigma^2/2)t + \sigma W_t}.$$

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  • $\begingroup$ Not sure if this really helps or if it is at all what you're asking, but $\operatorname{E}(S)=\operatorname{E}(S_0e^{(\mu-\sigma^2/2)t+\sigma W_t})=e^{\mu t}$, "thus" the drift term. Or I guess had you gone with Stratonovich as one usually does in physics, $\mathrm{d}S_t=S_t(\mu\mathrm{d}t+\sigma\circ\mathrm{d}W_t) \Rightarrow S_t=S_0e^{\mu t+\sigma W_t}$. $\endgroup$
    – alarge
    Jan 26, 2019 at 13:17

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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.(**): $$ \begin{aligned} d \psi(t) &= \psi(t + d t ) -\psi(t) \\ &=\left[U(t+d t,t)- I\right]U(t,0) \psi(0)\\ &= \left[\exp(-iH d X_t) - I \right]\psi(t) \\ &= -iH\psi(t)d X_t - \frac{1}{2}H^2\psi(t) d t. \end{aligned} $$

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