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Sphereically Symmetric potential well

  • Thread starter gronke
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  • #1
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Homework Statement


A particle that movies in three dimensions is trapped in a deep spherically symmetric potential V(r):

V(r) = 0 at r < r[itex]_{}0[/itex]
--> ∞ at r ≥ r[itex]_{}0[/itex]

where r[itex]_{}0[/itex] is a positive constant. The ground state wave function is spherically symmetric, so the radial wave function u(r) satisfies the one-dimensional Schroedinger energy eigenvalue equation (6.17) with the boundary condition u(0) = 0 (eq. 6.18).

Using the known boundary conditions on the radial wave function u(r) at r = 0 and r = r[itex]_{}0[/itex], find the ground state energy of the particle in this potential well.


Homework Equations


6.17
-[itex]\hbar^{2}/2m[/itex] [itex]d^{2}/dr^{2}[/itex]u(r) + V(r)u(r) = Eu(r)

6.18
u(r) = rψ(r)

The Attempt at a Solution


I know that there is no potential V(r) to deal with, since it is zero inside the well and infinite outside.

I also know that ψ is zero at r[itex]_{}0[/itex] and finite at zero.

So, that leaves me with finding a solution to

-[itex]\hbar^{2}/2m[/itex] [itex]d^{2}/dr^{2}[/itex]u(r) = Eu(r)

which I am a bit confounded with. Even then, how do I go from a solution to that to finding the ground state energy?

Thanks.
 

Answers and Replies

  • #2
368
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This is a differential equation that should hopefully be familiar to you from your basic DiffEq class. It may help to eliminate some of the constants to clean it up a bit:

[tex]\frac{\partial^2}{\partial r^2} u(r) = -A^2 u(r)[/tex]

Where [itex]A = \sqrt{2mE/\hbar^2}[/itex]. See if you can work out the basic type of function that will satisfy this equation (hint: what function yields minus itself when differentiated twice?)
 

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