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# Quantum Mechanics - Potential Well watch

1. I have the waveforms in both regions, and the boundary conditions, and I know that the probability of both regions added together = 1.

But...I have no idea how to piece all of that together.

Thanks
2. I have the waveforms in both regions, and the boundary conditions, and I know that the probability of both regions added together = 1.

But...I have no idea how to piece all of that together.

Thanks
The attractive part of the well is the bit where 0 < x < a.

Also we know that there is 0 probability of finding the particle at x < 0, as here there is an infinite potential well.

We need to use that probability in QM is given by

where is the wavefunction and * denotes the complex conjugate.

You say you have the wavefunctions describing both regions so what I would do is one of these two things:

1) Find the probability of the particle sitting in the attractive region by the following integral

where is the wavefunction for the particle in this region (0 < x < a). The probability of it lying outside this region is then 1 - the answer from above, as the total probability adds to 1 as you say.

2) The other option is to find the probability directly, i.e. do

where this time is the wavefunction of the particle in the region x > a.

I would choose whichever gives me the easier integral to do! (At a guess it'd be 1 in this case, as infinite limits are never fun...)
3. Thanks, but I only have what the functional form of the wave functions in each region, but I do not know any of the normalisation constants, I realize that the wave functions in both regions must be continuous and smooth at x=a (the boundary). But I I dont really how to combine these together
4. Thanks, but I only have what the functional form of the wave functions in each region, but I do not know any of the normalisation constants, I realize that the wave functions in both regions must be continuous and smooth at x=a (the boundary). But I I dont really how to combine these together
Ah ok sorry didn't realise that. To find the normalisation constants you can use that that

where im calling 0<x<a region 1 and x>a region 2.

The continuous part means that

i.e. the gradients of the two wavefunctions evaluated at x=a must be equal for continuity.

Will these two equations give you enough info to find everything you need?

If not, I see that there is something about the total energy being -V/4 in the lowest energy state. This I would assume to be that in the potential well when n = 1, so the following will hold

where is the hamiltonian of the system.

If this isn't enough info post the wavefunctions you've got, as im not 100% exactly what the question has given you at the moment!
5. ok just wow..
6. ok this is what I've done so far:

So I can sort of get simultaneous equations....But it'll end up being really really disgusting
7. So I can sort of get simultaneous equations....But it'll end up being really really disgusting
I agree, making me wonder if there is a better way to do this? Im not seeing it if there is though!

One thing that may help is that

Ah wait - continuity also requires that

(using your labelling of the regions now)

because the wavefunction for x < 0 is 0, implying a gradient of 0 also.

i.e.

That certainly simplifies things a bit!
8. if B=0 then there is no wave in region 2? The gradient of the wavefunction shouldnt be zero at x=0 right? because the function is not smooth at x=0, they are only continuous....implying that:

The gradient of the wavefunction from to is discontinuous?
9. if B=0 then there is no wave in region 2? The gradient of the wavefunction shouldnt be zero at x=0 right? because the function is not smooth at x=0, they are only continuous....implying that:

The gradient of the wavefunction from \psi_1 to \psi_2 is discontinuous?
Hmmm I get what you're saying... makes sense that an infinite potential wouldn't have a smooth boundary. So if

that implies that A = 0. So the set of equations can still be simplified using this...
10. I already have A=0 lol..i might just give up on this one and let my supervisor sort things out
11. I already have A=0 lol..i might just give up on this one and let my supervisor sort things out
ah lol sorry didnt see that from your working. Yeah maybe supervisor is best on this one - sorry couldn't be of more help!
12. Thanks so much for your time though....I really appreciate it, will rep u

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