I've seen it stated that it "can easily be seen" that if $\psi(r,t)$ is a solution of the Schrodinger equation : $ih \dfrac{\partial \psi(r,t)}{\partial t} = H \psi(r,t)$, then $\psi^*(r,-t)$ is also a solution.
But how can this be shown in detail? In particular, what is the partial derivative with respect to time of $\psi^*(r,-t)$? If I just apply the chain rule to $\psi^*(r,-t)\equiv conjugate(\psi(r,-t))$, then I get $\dfrac{\partial conjugate(\psi(r,-t))}{\partial t} = \dfrac{\partial conjugate}{\partial \psi} * \dfrac{\partial \psi(r,-t)}{\partial t}$. But what's the derivative of the "conjugate" function?
Given
$$ i \hbar \frac{\partial \psi(x,t)}{\partial t} = H \psi(x,t), $$
whence
$$ \left( i \hbar \frac{\partial \psi(x,t)}{\partial t} \right)^* = \Big( H \psi(x,t) \Big)^*, $$
so
$$ i^* \hbar \frac{\partial \psi^*(x,t)}{\partial t} = H \psi^*(x,t). $$
As $i^* = -i$, we can write
$$ - i \hbar \frac{\partial \psi^*(x,t)}{\partial t} = H \psi^*(x,t), $$
thus
$$ - i \hbar \frac{\partial \psi^*(x,-t)}{\partial [-t]} = H \psi^*(x,-t), $$
yielding
$$ i \hbar \frac{\partial \psi^*(x,-t)}{\partial t} = H \psi^*(x,-t), $$
so IF $\phi(x,t)$ is a solution, then $\phi^*(x,-t)$ is also a solution.