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# Probability of the lowest eigenenergy, quantum theory Watch

1. So I have my solutions of part b, but I'm not sure how to go about c.

What eigenfunction am I using, exactly?

For the region I have the eigenfunction . The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

Edit: Oh of course I'm using and
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2. (Original post by wanderlust.xx)

So I have my solutions of part b, but I'm not sure how to go about c.

What eigenfunction am I using, exactly?

For the region I have the eigenfunction . The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

Edit: Oh of course I'm using and
Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f
3. (Original post by ben-smith)
Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f
Oh yeah.

So after normalising I got .

Thus

Hence

Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

Thanks for the help.
4. (Original post by wanderlust.xx)
Oh yeah.

So after normalising I got .

Thus

Hence

Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

Thanks for the help.
Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.
5. (Original post by ben-smith)
Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.
The problem now is that normalising becomes a pain in the arse, since we'd have
6. Okay, here's another idea. Since the potential is symmetric, I can split our solutions into even and odd states, yielding

Even:

Odd:

Hence our lowest eigenenergies for Even and Odd parity states respectively will be

but now clearly we have , so we use the even eigenfunction

in

Does this work? If not, why not?

Updated: April 18, 2012
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