Scrodinger equation and Quantum mechanic stuff required

[KKtm]

Apply Schrodinger’s Wave equation to calculate energies of particle in a box .
kindly refer me to some web material and send some stuff regarding it , also if some one has the PDF files related to Quantum mechanics with solved examples... would be immense help.
Regards

 

[Mindscrape]

Are you using a book to learn quantum mechanics?

You should know that
\hat{H}\Psi = E \Psi

So all you have to do is solve the PDE for a particle in a box. Is your box infinite, or finite? If it is finite you would have to solve the problem numerically, which is more work, but still not too difficult.

 

[arunma]

So this is actually a really easy problem. All you do is take the Schrodinger Equation:

-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}+V(x)\psi(x) = E\psi(x)

Now you plug in the potential. This can be centered at 0, but it's usually more easily centered at L/2, where L is the well width. The potential is,

V(x) = 0 when 0<x<L, and

V(x) = \infty when x<0 or x>L

So inside the well you've just got:

-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2} = E\psi(x)

Or,

\dfrac{d^2\psi}{dx^2}+ \dfrac{2mE}{\hbar^2}\psi(x)=0

What does this look like? Hint: it's exactly the same mathematical form as the simple harmonic oscillator from classical mechanics (but it doesn't actually have anything to do with oscillators physics-wise). You should be able to solve the equation using the ansatz,

\psi(x) = Asin(kx) + Bcos(kx)

The value of k is found from the boundary conditions. Remember that the potential outside the well is zero. Thus the wavefunction has to either be 0 or infinity at those places. Common sense dictates it'll be zero, so the wave has to "fit" inside the box. Thus the wavenumber will have to be such that the wave always goes to zero at the boundaries. This eliminates all cosine solutions, since cos(0)=1. You can figure out what k is. From that you can solve for the energy. And if you really want, you can figure out what A is by finding the complex modulus squared of psi and treating it as a probability density.

Let me know what you come up with!