Can you give me an example using that equasion, I am not good with understanding tex formatting. And can you please explaine what all those variables are?
In Quantum mechanics (QM) all you can know about a partice is it's
state. Where in Classical mechanics you're solving the equation of motion (Newtons F=ma eg) to find the position of a particle as a function of time, in QM you solve (in the nonrelativistic case) the Schrödinger equation. This equation:
i\hbar\frac{\partial \Psi}{\partial t}=\hat{H}\Psi
gives you, when you know the phsyical circumstances (H), the state \hat{\psi}. This state does not directly give you the position of a particle as a function of time. You can only use it to find a 'probabality density' for the particle's position, from which an 'expectation value' (mean) for the particles position can be found. In QM this is all we
can know about a particles position, the probablility that upon measurement you find it here or there!
Now for a molecule this Schrödinger equation can only be a analytically solved for a hydrogen atom (and some similar atoms), a system of a proton and an electron. The relevant part of the state you need to say something about the electrons position or velocity is called the wavefunction. It's 'square' (actually its modulus) gives you the aforementioned probability distribution. The result in the form of a wavefuncion you can find in almost any textbook on QM. It is denoted by \psi_{n,l,m} and is actually a whole collection of wavefunction depending on what numbers of n, l and m you are interested in. They represent the different orbitals, or excited states, of the hydrogen molecule. Note that n is the same variable as in Bohrs theorem of the hydrogen atom and is called the 'principle quantum number'.
As an example I will try to answer your question for the ground state (n=1, l=m=0) of a hydrogen atom. The wavefunction as a function of the radial coordinate r is
\psi_{1,0,0}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a} with a=\frac{4 \pi \epsilon_0 \hbar^2}{me^2}=0,529E-10 m (Bohr radius in meters)
As said before the 'square' gives you the probability density:
|\psi_{1,0,0}|^2=\frac{1}{\pi a^3}e^{-2r/a}
And tells you the elektron has the greatest probability to be found at r=o wch is at the nucleus! Furtermore the wavefunction is spherically symmetric. But all this is not really relevant for finding out the velocity, so let's move on.
As said before you cannot find the position of the electron, simply because i has no position. You
can find the expectation value for the position though. Suppose you know a particle can be found with probability at x=1, probability (1/4) at x=2 and probability (1/4) x=4 you can find the expectation value by calculating the following sum:
<x>=\sum_i P(x_i)x_i = (1/2)*1+(1/4)*2+(1/4)*4=2.
But now we don't deal with a discrete spectrum of probabilities ((1/2), (1/4) and (1/4)) but a continuum of probabilities. So instead of summing over the probabilities you will have to integrate:
<x>=\int |\psi_{1,0,0}|^2 x dx
Where the probabilities P are replaced by the probability density |\psi_{1,0,0}|^2. This integral has to be evaluated over all space.
Also the electron has no definite speed, but only a probability to be measured that speed, or that speed. So I will have to dissapoint you, again all we can find out about the velocity is it's expectation value <v>. Now:
<v>=\frac{d<x>}{dt}=\int \frac{\partial |\psi_{1,0,0}|^2}{\partial t} x dx = \frac{-i \hbar}{m} \int \psi_{1,0,0}^* \frac{\partial \psi_{1,0,0}}{\partial x} dx
Where the last step follows after some juggling with the integral and using the Scrödinger equation. [I don't know have far your mathematics skills reach but |\psi_{1,0,0}|^2 = \psi_{1,0,0}^* \psi_{1,0,0} is called the 'modulus' and a star denotes 'complex conjugation'. In case your wavefunction is real and this means just respectively squaring and doing nothing.]
This is (for the groundstate of hydrogen) the result dextercioby and gokul pointed out and hopefully you now know more or less know where it came from.
But hopefully u're aware of the tedious proof that