Does the wave function of a particle completely describe the state of the particle?

In classical mechanics, if you know the position and momentum of a particle at time $t$ and the Hamiltonian, you can predict the particle's position and momentum at any time.

In quantum mechanics, if you know the wave function of a particle at time $t$ and the Hamiltonian, can you predict the wave function at any time? Or do you also need to know $d\Psi/dt$ at time $t$?

My understanding is that, from the Schrödinger equation, you only need to know $\Psi$ at time $t$ and $H$ to predict $\Psi$ at any time. That is, you don't need to know $d\Psi/dt$, because it can be obtained from calculation.

The wavefunction $\Psi(r,t)$ is a function of position and time.
So in summary, if you know the wavefunction $\Psi(r,t)$, you already know the time dependence. You don't need additional information or calculation.
• Thank you. Just to clarify, if I know the values of $\Psi(r, t)$ for all $r$ and $t = t_0$, then I know $\Psi(r, t)$, for all $r$ and all $t$, right? – user45996 May 7 '14 at 1:55
• Yes, since the Schrödinger equation is linear, the Hamiltonian is Hermitian and can be diagonalized, and the time evolution of an eigenstate of the Hamiltonian is trivial. You could also sledgehammer it with the fundamental theorem of ODEs but that gives a weaker conclusion, it only guarantees existence and uniqueness in an interval around $t_0$. – Robin Ekman May 7 '14 at 2:04
• Just a "small" remark: all this is true unless a measurement was made between $t_0$ and $t$. – Wildcat May 7 '14 at 11:32