finite square well normalisation

[Sarahisme]

Hi all

 

I was wondering how you go about normalising the even bound state wavefunction for the finite square well?

 

eg.

 

 

 

so i have to find F and D.

 

this what i got so far:

 

 

i can't seem to find the answer to this 'normalising' problem anywhere!

 

any ideas guys?

 

Thanks

 

Sarah

[Severian]

[math]|\phi|^2(x)[/math] is the probability of finding the particle in position x. So [math]\int dx |\phi|^2[/math] is the probability of finding the particle anywhere. What should this probability be?

[Sarahisme]

should be 1?

[5614]

Sounds right and it's what I thought... but don't take my word for it, I really don't know for sure.

[Sarahisme]

so can i go:

 

[math] \int^a_0D^2cos^2(lx)dx+\int^\infty_aF^2e^{-2kx}dx=1/2 [/math]

 

??

[swansont]

Since the square of the wave function is symmetric about x = 0, that should be fine.

[Severian]

Yes. That is what I meant.

 

You need a further condition though. This just gives you a relation between D and F. You need another relation to solve for them.

 

So, what should happen to the funtional form of [math]\psi(x)[/math] as you pass [math]x=a[/math]?

[Sarahisme]

ok

 

so could also use the continuity of psi at x = a

 

[math] Fe^{-ka} = Dcos(la) [/math]

 

?

[Sarahisme]

then i get this

 

[math] D = \sqrt{\frac{2kl}{\frac{1}{2}k(sin(2la)+la/k)+lcos^2(la)}} [/math]

 

and

 

[math] F = \sqrt{\frac{2kle^{2ka}cos^2(la)}{\frac{1}{2}k(sin(2la)+la/k)+lcos^2(la)}} [/math]

[Sarahisme]

k thanks, i got it now!

[Severian]

Incidentally, there was a slight subtlety in this problem. Normally one would expect the second derivative to be continuous also - so the gradiaent of the wavefunction would also be smooth. But in this case, the potential itself is not continuous in its gradient (it jumps from -V0 to 0 at x=a) so the gradient of psi is not continuous either.