PHY 491
HW Assignment #2; September 12-19, 2011
(1)
4. Using the 1st order perturbation results for 𝐸𝑚𝑣 , where mv denotes mass velocity and
(1)
𝐸𝑠𝑜 , where so denotes spin-orbit show that
(1)
(1)
(1)
(𝐸 )2
4𝑛
𝑛
𝐸𝑚𝑣 + 𝐸𝑠𝑜 = 𝐸𝑓𝑠 = 2𝑚𝑐
(3 − 𝑗+1/2),
2
where fs denontes fine structure and j is the total angular momentum, orbital plus spin.
(1)
𝐸𝑚𝑣 = −
(1)
𝐸𝑠𝑜 =
(𝐸𝑛0 )2
4𝑛
[
− 3]
2𝑚𝑐 2 𝑙 + 1/2
(𝐸𝑛0 )2 𝑗(𝑗 + 1) − 𝑙(𝑙 + 1) − 3/4
𝑛[
]
1
𝑚𝑐 2
𝑙 (𝑙 + 2) (𝑙 + 1)
(1)
If we add the above two we get 𝐸𝑓𝑠 . However for s=1/2, j=l+1/2 or j=l-1/2. This means
l=j-1/2 and l=j+1/2. Eliminate l from the above equation for each value of l. Do the
algebra and you will get the answer in terms of j.