This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 11 1. The first order contribution is E n (1) = λ 〈 n  x 2  n 〉 = λ h 2 m ϖ 2 〈 n  ( A + A + )( A + A + )  n 〉 To calculate the matrix element 〈 n  A 2 + AA + + A + A + ( A + ) 2  n 〉 we note that A +  n 〉 = n + 1  n + 1 〉 ; 〈 n  A = n + 1 〈 n + 1  so that (1) the first and last terms give zero, and the second and third terms yield ( n + 1) + ( n – 1)=2 n . Thus the first order shift is E n (1) = λ h m ϖ n The second order calculation is quite complicated. What is involved is the calculation of E n (2) = λ 2 h 2 m ϖ 2 〈 n  ( A + A + ) 2  m 〉〈 m  ( A + A + ) 2  n 〉 h ϖ ( n m ) m ≠ n ∑ This is manageable but quite messy. The suggestion is to write H = p 2 2 m + 1 2 m ϖ 2 x 2 + λ x 2 This is just a simple harmonic oscillator with frequency ϖ * = ϖ 2 + 2 λ / m = ϖ + λ ϖ m 1 2 λ 2 ϖ 3 m 2 + ... Whose spectrum is E n = h ϖ * ( n + 1 2 ) = h ϖ ( n + 1 2 ) + λ h ϖ m ( n + 1 2 ) λ 2 h 2 ϖ 3 m 2 ( n + 1 2 ) + ... The extra factor of 1/2 that goes with each n is the zeropoint energy. We are only interested in the change in energy of a given state  n > and thus subtract the zeropoint energy to each order of λ . Note that the first order λ calculation is correct. 2. The eigenfunction of the rotator are the spherical harmonics. The first order energy shift for l = 1 states is given by ∆ E = 〈 1, m  E cos θ 1, m 〉 = E d φ sin θ d θ cos θ  Y 1. m  2 π ∫ 2 π ∫ For m = ±1, this becomes 2 π E sin θ d θ cos θ 3 8 π π ∫ sin 2 θ = 3 E 4 duu (1 u 2 ) = 1 1 ∫ The integral for m = 0 is also zero. This result should have been anticipated. The eigenstates of L 2 are also eigenstates of parity. The perturbation cos θ is odd under the reflection r  r and therefore the expectation value of an odd operator will always be zero. Since the perturbation represents the interaction with an electric field, our result states that a symmetric rotator does not have a permanent electric dipole moment. The second order shift is more complicated. What needs to be evaluated is ∆ E (2) = E 2  〈 1, m  cos θ  L , M 〉  2 E 1 E L L , M ( L ≠ 1) ∑ with E L = h 2 2 I L ( L + 1) . The calculation is simplified by the fact that only L = 0 and L = 2 terms contribute. This can easily be seen from the table of spherical harmonics. For L =1 we saw that the matrix element vanishes. For the higher values we see that cos θ Y 1, ± 1 ∝ Y 2, ± 1 and cos θ Y 1,0 ∝ aY 2,0 + bY 0,0 . The orthogonality of the spherical harmonics for different values of L takes care of the matter. Note that because of the φ integration, for m = ±1 only the L = 2 ,M = ± 1 term contributes, while for the m = 0 term, there will be contributions from L = 0 and L = 2, M = 0. Some simple integrations lead to ∆ E m =± 1 (2) =  2 IE 2 h 2 1 15 ; ∆ E m = (2) =  2 IE 2 h 2 1 60 3. To lowest order in V the shift is given by...
View
Full
Document
This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.
 Spring '05
 mokhtari

Click to edit the document details