You are Here: Home

Probability of the lowest eigenenergy, quantum theory

Announcements Posted on
Take our survey to be in with the chance of winning a £50 Amazon voucher or one of 5 x £10 Amazon vouchers 28-05-2016

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
Attached Thumbnails

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?

Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 18, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

Don't be a half-term hermit

How to revise this week and still have a life

Poll
Useful resources

Maths Forum posting guidelines

Not sure where to post? Read here first

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams