This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: HW 4 Solutions Problem 1 Consider the spherically symmetric infinite well, V ( r ) = , if r < a ; ∞ , if r > a . (1) Using the variable-separable ansatz ψ = R ( r ) Y lm ( θ,φ ), the Schrodinger equation reduces to an equation involving the spherical harmonics Y lm ( θ,φ ) as well as the radial equation:- ¯ h 2 2 m d 2 u dr 2 + V ( r ) + ¯ h 2 2 m l ( l + 1) r 2 ! u = Eu, (2) where u = rR ( r ). We consider the case of l = 0 first. In this case, V = 0 inside the well which is completely allowed classically. Thus, the WKB approximation to (2) is a sum of sinusoidal terms but since u = rR ( r ) = 0 at r = 0, we only keep the sine term, and since u = 0 at r = a , we can equate Z a q 2 m ( E ) dx = nπ ¯ h (3) It is then straightforward to show that E = ( nπ ¯ h ) 2 2 ma 2 (4) This agrees with the exact result, since clearly when l = 0, the WKB solution to (2) is the exact solution for u . Let us now consider the l = 1 case. This case is more involved since we have the centrifugal potential...
View Full Document
This note was uploaded on 11/21/2010 for the course PHYSICS 137B taught by Professor Moore during the Spring '07 term at Berkeley.
- Spring '07