# Help! particel in spherical delta potential well

1. Apr 24, 2007

### valtorEN

1. The problem statement, all variables and given/known data

A particle with mass M is moving in a spherical delta-potential well, V(r)=-Vo*delta(r-a); Vo>0, a>0
Find the minimum Vo value so that there is at least 1 bounded eigenstate for the particle.

2. Relevant equations

looking through my quantum book (griffiths of course), i found a similar problem (4.9) i assume i put the potential into the 3d schrodinger eqn and solve from there

3. The attempt at a solution

solve Schrödinger's eq in 3d
i assume l=0
i "plug" -Vo*delta(r-a) into the radial equation, -h^2/2m*d^2/dr^2+[-Vo*delta(r-a)]u=Eu

after plug n chug, i get
d^2u/dr^2=-k^2u k=SQRT(2m(E+Vo)/h^2)u

so is the solution just sines and cosines? i am confused
what is the minimal value for a bound state(E<0)?

any and all feedback is much appreciated
cheers
nate

2. Apr 24, 2007

### valtorEN

ok i will solve myself as u all r morons
i will post solution here

3. Apr 24, 2007

### Dick

Thanks for your txt insult. I'm sure you'll be well received next time you post. FYI I have been thinking about your problem. Your radial part of the schrodinger equation didn't look right and I wanted to wait till I had access to materials and was away from work before I answered. It's only a second derivative, not some other stuff I would expect in a 3d laplacian. In general you should be aware that the delta function is zero except at r=a. So the solutions away from r=a are the same as with no potential, hence some form of damped exponential which I don't feel motivated to look up right now, u moron. And the presence of the delta function is generally a clue to look for a discontinuity in the first derivative. In a way in which I'm sure you can find yourself. Don't bother to post. I doubt anyone is interested.

4. Apr 25, 2007

### valtorEN

hey dick at least u replied to my post!
thank u! i apologize but i am frustrated with this problem
i found a solution on the internet in a published paper that showed the 1D and 3D delta potential "attractive" well, but i am reluctant not to derive it myself
blah blah...

5. Apr 25, 2007

### Dick

That's ok, but sometimes a question simply doesn't find someone who has the expertise to handle it. You want to splice two solutions corresponding to the same negative value of E together at r=a in such a way the there is a jump in the derivative at r=a corresponding to the size of V0. What is the form of solutions of SE corresponding to -E?

Last edited: Apr 25, 2007
6. Apr 26, 2007

### valtorEN

the solutions are of the form Ae^ikr+Be^ikr if E<Vo

7. Apr 26, 2007

### Dick

Even if you weren't in three dimensions with spherical symmetry that should still be A*e^(-kr)*B*e^(kr) since it's a bound state (negative energy solution). But you are in 3d and so it's even more wrong. You looked up a solution for a 3d spherical 'square well' potential, right? Review it. You need the laplacian in spherical coordinates and the solutions are spherical bessel and hankel functions.

8. Apr 26, 2007

### valtorEN

ah, i see, i see the solutions as
u(r)=A*r*jl(kr)+B*r*nl(kr)
where jl(kr) is the spherical Bessel function of order l
and nl(kr) is the spherical Neumann function or order l
is this correct?

9. Apr 26, 2007

### Dick

I think you are getting there. My book calls it Hankel instead of Neumann, but I presume they are the same thing. You can set l=0 since you are looking for a lowest energy bound state. Now in the square well case you match the derivatives at the boundary. In this case the delta function tells you the derivatives don't match - but it does let you determine the difference of the derivatives as a function of V0.

10. Apr 27, 2007

### valtorEN

in my book they say that B= zero, sol u(r)=A*sin(kr)

now do i take that solution and use boundary conditions to solve for the minimum Vo

uin(a)=uout(a) and u'in(a)=u'out(a)?

do i use the radial equation to solve for Vo?

11. Apr 27, 2007

### Dick

Ok, but what's the exterior solution like? As I keep SHOUTING u'in(a)=u'out(a) is WRONG for this case. uin(a)=uout(a) is fine. Once you know u'in(a) you can figure out u'out(a) because their difference is controlled by the size of the delta function.

12. Apr 27, 2007

### valtorEN

the exterior has no potential, so i assume that V=0, so there is no wavefunction that exists outside the well.

13. Apr 27, 2007

### Dick

No potential doesn't mean no wavefunction. I've already agreed with you that uin(a)=uout(a). If that common value is non-zero its going to be difficult to claim there is no exterior wavefunction. The exterior solutions look exactly like they do for the square well. The only difference is the matching conditions between interior and exterior.

14. Apr 27, 2007

### valtorEN

so unlike the inf square well, the delta well does NOT meet the u'in(a)≠u'out(a)
boundary condition?

15. Apr 27, 2007

### valtorEN

so for r≥a the solutions are called modified spherical Bessel Functions (lol)

kl(kr) k^2=(2m|E|)/hbar^2

for boundary conditions (l=0)

r=a

(djl(Kr)/dr)/jl(Kr)|r=a- = (dkl(kr)/dr)/kl(kr)|r=a+

now since l=0, and j0(x)=sin(x)/x and k0(x)=e^-x/x (these are the solutions

that join together at the r=a condition right?!)

Kcot(Ka)=-k (l=0)

is this correct?

i found in a published article but did not derive as i do not have the math

16. Apr 27, 2007

### Dick

You have the outer solution right. But the inner solution also needs to be a negative energy V=0 solution. That makes it j0(ikr) not j0(kr). That comes out easy on the eye if you write it in exponentials, (e^(kr)-e^(-kr))/(2kr). So now you've got A*uin and B*uout to match. A*uin(ka)=B*uout(ka) and B*uout'(ka)-A*uin'(ka)=-2mV0*A*uin(ka)/(hbar^2). The latter is the term coming from the delta function. Use the first equation to eliminate A or B in the second one. Now you will notice if V0=0 there are no solutions! You can't match. What's the minimum V0 that will give a solution? I solved for V0 and took the limit as k->0 to find out. That's a LOT of hints.

17. Apr 28, 2007

### valtorEN

u'in(a)≠u'out(a)

uout(a)=C*e^-(l*a) (simple enough)

since l=0, the Neumann function nl(ka)=-B*a*(cos(a)/a)

uin(a)=-B(cos(a)) ?

u'in(a)B*sin(a)

u'out(a)=-l*a*C*e^(-l*a)

now i use u'in(a)≠u'out(a) to solve for Vo?

can i get away with setting A=0 and D=0 like in the finite spherical well (griffiths problem 4.9)?

18. Apr 28, 2007

### Dick

You didn't read my last post, did you? I actually TOLD you what uin was.