p137bp9sol - Physics 137B, Fall 2007, Moore Problem Set 9...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 137B, Fall 2007, Moore Problem Set 9 Solutions 1. Let a denote the hydrogen 1s state and let b denote the the 2p state with magnetic quantum number m . In the notation | n`m i , we have | a i = | 100 i , and | b i = | 21 m i . The transition rate for spontaneous emission, calculated via the statistical argument due to Einstein, is given by Eq. (11.76): W s ab = ω 3 ba 3 πc 3 ~ ± 0 | D ba | 2 . (1) Here ω ba = E 2 - E 1 ~ = - 3 4 E 1 ~ = 3 μα 2 c 2 8 ~ (2) and D ba = h b | - e r | a i = - e h 21 m | r | 100 i . For a fixed value of the magnetic quantum number m , the dipole moment D ba can be calculated most easily by expressing the vector r in terms of so-called spherical tensor operators . To introduce this notion, first define three new basis vectors e +1 = - 1 2 x - i ˆ y ) , e - 1 = 1 2 x + i ˆ y ) , e 0 = ˆ z . (The fact that we are using complex vectors to span real 3-dimensional space looks fishy, but it turns out to be harmless. Note that these vectors are orthogonal in the complex sense, e * m · e m 0 = δ m,m 0 .) Next we define new coordinates r m e * m · r for m = 0 , ± 1. Explicitly r +1 = - 1 2 ( x + iy ) , r - 1 = 1 2 ( x - iy ) , r 0 = z . 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Using the orthogonality property mentioned above, we can write | D ba | 2 = e 2 ( |h b | r +1 | ψ a i| 2 + |h ψ b | r - 1 | a i| 2 + |h ψ b | r 0 | ψ a i| 2 ) = e 2 X m 0 |h 21 m | r m 0 | 100 i| 2 . The motivation for these definitions is that they allow us to express r m simply in terms of the spherical harmonic functions that we all know and love. Looking at Table 6.1, for instance, we see that with these definitions r m = r 4 π 3 rY 1 m ( θ, φ ) , m = 0 , ± 1 . Now the hydrogen atom wavefunctions are ψ n`m ( r, θ, φ ) = R n` Y `m ( θ, φ ), so h 21 m | r m 0 | 100 i = Z d 3 r ψ * 21 m ( r ) r 4 π 3 rY 1 m 0 ( θ, φ ) ψ 100 ( r ) = r 4 π 3 Z 0 r 3 R 21 ( r ) R 10 ( r ) dr Z Y 1 m ( θ, φ ) * Y 1 m 0 ( θ, φ ) Y 00 ( θ, φ ) sin θ dθ dφ = 1 3 δ m,m 0 Z 0 r 3 R 21 ( r ) R 10 ( r ) dr , since Y 00 = 1 4 π and the spherical harmonics have the orthogonality property Z Y `m ( θ, φ ) * Y ` 0
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

p137bp9sol - Physics 137B, Fall 2007, Moore Problem Set 9...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online