Central potential in quantum mechanics

Bavon
Messages
5
Reaction score
0
Hi all,
I'm new to his forum. I'm having trouble with the following question on central potentials. It's an exercise from (Bransden and Joachain, Introduction to quantum mechanics).

Homework Statement


Suppose V(r) is a central potential, expand around r=0 as V(r)=r^p(b_0+b_1r+\ldots). When p=-2 and b_0&lt;0[\tex], show that physically acceptable solutions exist only when b_0&amp;gt;-\frac{\hbar}{8\mu}<br /> <br /> <br /> <h2>Homework Equations</h2><br /> R(r) is the radial component of the wave function<br /> u(r)=r^{-1}R(r)=r^s\sum{c_kr^k} is a solution of<br /> -\frac{\hbar^2}{2\mu}\frac{d^2u(r)}{dr^2}+V_{eff}u(r)=Eu(r)<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> The effective potential is r^{-2}(b_0+\frac{l(l+2)\hbar^2}{2\mu}+b_1r+\ldots)<br /> <br /> When p&gt;-2, the case that is discussed in the textbook, they compare the lowest order terms in the radial Schroedinger equation. For p=-2, I get:<br /> <br /> The lowest order term of \frac{d^2u}{dr^2}=s(s-1)r^{s-2}\sum{c_kr^k}<br /> The lowest order term of V_{eff}(r)u(r)=(b_0+\frac{l(l+2)\hbar^2}{2\mu})r^{s-2}\sum{c_kr^k}<br /> <br /> The difference is apparently that b_0 appears in the lowest order terms.<br /> <br /> Now I need some constraint that excludes non-physical solutions. For p&gt;-2, that is u(0)=0. But for b_0&amp;lt;0 that can&#039;t be used. Because the potential is attracting, I think the probability of finding a particle at the origin should be positive. The problem is I can&#039;t think of any constraint that should be imposed, that could limit the allowed values of b_0.<br /> <br /> Any hints would be greatly appreciated.
 
Physics news on Phys.org
I solved this myself. It was so easy after all!
 
Dear Bavon
Would u please help me to solve this problem too?
thanks
I am looking forward for ur reply. It is an emergency situation!
 
Last edited:
If I decyphered my notes from almost 3 years ago correctly, I constructed an equation in s, and then expressed that it should have real solutions.
 
How can I undrestand that an equation have real solutions?
thanks
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top