Question

Can someone please explain the meaning of the wave function (psi) and how it applies to Schroedinger's Equation time dependent and independent equations of a wave function?

Answer 1

psi (x,y,z,t) = psi (x,y,z) e^(iwt)

Answer 2

The meaning of the wave function in plain terms is that it represents the probability density for finding an electron some distance away from the nucleus. How it relates to the time independent and dependent relations is somewhat described by the Heisenberg uncertainty principle wherein if you know the position you have no idea of its momentum and vice versa. So, the time-independent equation is sometimes referred to as the stationary equation since it relates explicitly to the position while the time dependent relation is concerned with describing the momentum. There are plenty of web resource available if you want to get a more rigorous explanation.

Answer 3

It means that each electron may be represented as a wave...and the probability of finding that particular electron at a certain point is proportional to the square of amplitude of that wave (which is a variable complex function)

Answer 4

Example?

Answer 5

Example?
Well i told you about the electron.
Thats the example...if u want the math then thats complicated

Answer 6

Wavefunction is something that gives us information about the quantum state.
psi(x) describes the quantum state of a quantum particle at a particular instant of time just like in classical mechanics x and p describes the classical state of the classical particle/body.
In case of schrodinger's equation, psi describes the quantum state of the electron.

Answer 7

But that's what I want. An example of the math and how to use Schrodinger's Equation.

Answer 8

ull notice for any wave eqn psi\[d^{2} \psi/dx^{2} = -k^{2} \psi\]

Answer 9

the most general solution to this is
\[\psi (x) = Ae^{ikx}\]

Answer 10

notice that psi is a function of x and only the space-dependent part of the eqn..u can differentiate any wave eqn and get the same differential eqn wht i wrote above

Answer 11

now we also know that
\[\psi = \psi (x) \psi (t)\]
the total wavefunction is the product of the individual space-dependent and time-dependent functions
so
\[\psi = (Ae^{ikx} ) (e^{-iwt})\]
where w is the frequency of the wavefunction
so
\[\psi = Ae^{i(kx-wt)}\]

Answer 12

Now lets see how we can find the probability density
Lets consider a FREE particle which can randomly move along the x-axis
We will consider time t=0 to make calculation simpler
The prob. density is given by the absolute value square of the wave function, since the functn itself is complex valued..

so
Probability =
\[|\psi (x,t)^{2}| = |A^{2} e^{2ikx}| = |A^{2}||e^{ikx}|^{2}\]

now remember
for any complex no z
\[\|z|^{2} = zz*\]
where z* is the complex conjugate of z
so
\[|e^{ikx}|^{2} = (e^{ikx})(e^{-ikx}) = e^{0} = 1\]

so the prob. density becomes \[A^{2}\]
which is constant
here we can see that because remember, WE CONSIDER A FREE PARTICLE, the prob. density is independent of x, ie the particle is equally likely to be at any point on the x-axis

Answer 13

now this A can be found out by a process called normalization (for a general problem)

Which follows from common sense that our particle must be SOMEWHERE along the x-axis...which means that the sum of the probabilities for finding it at all points on the x-axis must add up to 1 ..because finding it along SOME point on the axis is SURE EVENT
so
(here A^2 is the linear prob density as were looking at only one dimension)

\[\int\limits_{-\infty}^{\infty}A^{2}dx = 1\]

Answer 14

hope that satisfies you