# Find wavefunction of a particle

1. Nov 4, 2005

### touqra

I am not sure how to solve this question. I can only say that, the wavelength is an integral number multiple of the circumference. Then?

A particle is confined to move in a circular tube with radius R. Work out the wavefunction of this particle.

2. Nov 4, 2005

### lightgrav

Are they talking about a torus-shaped tube (donut)? what's the other dimension?
or a circular cross-section cylinder (what length?)

3. Nov 5, 2005

### touqra

More like a string with two ends connected to each other to form a circle. Like a big letter "O" .

4. Nov 5, 2005

### siddharth

You mean a ring of radius 'r'.
Have you learnt the expression for the Hamiltonian in cylindrical co-ordinates?

5. Nov 6, 2005

### lightgrav

The only thing you can say is that there are an integer number of wavelengths.
You can write it as Aexp(imΦ) multiplied by a delta-function (r-R) .
but then you would be specifying a Φ=0 orientation,
so you should include an arbitrary phase angle in the exponent.

But this is really the same thing.

6. Nov 6, 2005

### siddharth

To find the wavefunction, you have to solve the time independent Schrodinger equation $$H \Psi = E \Psi$$. Because of the symmetry for a particle confined to move in a ring, working in the cylindrical polar co-ordinates will save you a lot of time.
So, if you write down the Hamiltonian Operator in the cylindrical polar co-ordinates(remember that the radius is constant), you will be able to work out the wavefunction of the particle.

7. Nov 7, 2005

### dextercioby

Separate variables in the time-indep SE and then, using boundary conditions (the wavefunction must be 0 on the boundar) find the solution. Hope you've seen and worked with Bessel functions before.

Daniel.

8. Nov 7, 2005

### siddharth

dextercioby, I think it's possible to solve this question without any knowledge of Bessel fuctions.
Also, won't we be using the periodic boundary condition $$\Psi(\theta) = \Psi(\theta + 2\pi)$$? How do we use the fact that the wavefunction must be 0 on the boundary?

9. Nov 9, 2005

### Dr.Brain

There is no as such some fixed wavefunction of the partyicle in a ring. Generally it is some exponential function , like for an electron in Bohr radius it is $e^(-r)$ . For deciphering the wavefunction , you also need to satisfy the boundary conditions , like how potential varies , it can assumed to be zero on the path particle moves , infinite potential generally implies the particle is forbidded to pass the barrier.

BJ