Infinite well with one wall

[DevonV]

Homework Statement

An electron is trapped in a 1D potential described by:

V(x) = 0 if x < R₀
V(x) = infinity if x > R₀

Electron is in lowest energy state, and experiment shows that:
(\Delta)x = sqrt(<x²> - <x>²) = 0.181 x 10^-10

Show that <x> = 0.5R₀

Homework Equations

The Attempt at a Solution

I started by treating it like an infinite well, with:
(\psi)(x) = Asin(kx) + Bcos(kx)

and applying the boundary conditions as usual, however in this case there is only one (at R₀):

(\psi)(R₀) = Asin(kR₀) + Bcos(kR₀) = 0

Normally the BC at x = 0 eliminates the Bcos(kx), but that doesn't exist in this case.

Any guidance would be greatly appreciated!

(PS. sorry for bad formatting, latex was being extremely uncooperative)

 

[fzero]

Normally the BC at x = 0 eliminates the Bcos(kx), but that doesn't exist in this case.

You can use your boundary condition to solve for B. Then you can rescale A and use a trig identity to simplify the wavefunction.

 

[DevonV]

Thanks for the reply!

I solved for B, yielding -Atan(kRo), and subbing that back into the BC equation gave me the condition:

Asin(kR₀) -Atan(kR₀) = 0

which i think is only true for:

k = n(\pi)R₀

However in order to normalize this and find A, I need to integrate from -infinity to R₀ right? Which isn't going to work out well for a trig function, so I must have done something wrong.

Anyone give me another nudge in the right direction?

Thanks!

 

[fzero]

Thanks for the reply!

I solved for B, yielding -Atan(kRo), and subbing that back into the BC equation gave me the condition:

Asin(kR₀) -Atan(kR₀) = 0

which i think is only true for:

k = n(\pi)R₀

If you substitute back into the boundary condition, you should find

Asin(kR₀) -Atan(kR₀) cos(kR0) =0

which is automatically true, since you're just going in circles.

However in order to normalize this and find A, I need to integrate from -infinity to R₀ right? Which isn't going to work out well for a trig function, so I must have done something wrong.

Anyone give me another nudge in the right direction?

Thanks!

Are you sure it's just a one-sided barrier? If it is, there's a continuous spectrum and the question doesn't make any sense. There is no lowest-energy state and $$\langle x\rangle$$ is not defined. $$\langle x\rangle = R_0/2$$ would be correct if there's another barrier at $$x=0.$$

 

[DevonV]

Thanks again for the help!

Using a BC of (\psi)(0) = 0 the question works out much better. I found one mistake in the question, and I'm guessing another was made and that it was intended to be a 1D version of a radial problem, given the notation of R₀ which was only used for those types of questions, and the implied BC at x = 0.

Thanks again!