# ch17 - CHAPTER 17 1 We start with Eq(17-19 We define k as...

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

CHAPTER 17 1. We start with Eq. (17-19) . We define k as the z axis. This means that the polarization vector, which is perpendicular to k has the general form ε ( λ ) = ˆ i co s ϕ + ˆ j s in ϕ This leads to B =∇× A =- i h 2 ε 0 ϖ V k ˆ k × ( ˆ i co s ϕ + ˆ j s in ϕ ) = B 0 ( ˆ j co s ϕ - ˆ i s in ϕ ) We are now interested in M = B 0 g p - g n 2 h 2 X 0 { ( σ y ( p ) - σ y ( n ) ) co s ϕ - ( σ x ( p ) - σ x ( n ) ) s in ϕ } X 1 m The operators are of the form σ y co s ϕ - σ x s in ϕ = 0 - ico s ϕ ico s ϕ 0 - 0 s in ϕ s in ϕ 0 = 0 - ie - i ϕ ie i ϕ 0 It is simple to work out the “bra” part of the scalar product 1 2 χ + ( p ) χ - ( n ) - χ - ( p ) χ + ( n ) ( 29 0 - ie - i ϕ ie i ϕ 0 p - 0 - ie - i ϕ ie i ϕ 0 n with the help of χ + 0 - ie - i ϕ ie i ϕ 0 = 10 ( 29 0 - ie - i ϕ ie i ϕ 0 = 0 - ie - i ϕ ( 29 =- ie - i ϕ χ - and χ - 0 - ie - i ϕ ie i ϕ 0 = 01 ( 29 0 - ie - i ϕ ie i ϕ 0 = ie i ϕ 0 (29 = ie i ϕ χ + This implies that the “bra” part is

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

View Full Document
1 2 χ + ( p ) χ - ( n ) - χ - ( p ) χ + ( n ) ( 29 0 - ie - i ϕ ie i ϕ 0 p - 0 - ie - i ϕ ie i ϕ 0 n = =- 2 i ( e - i ϕ χ - ( p ) χ - ( n ) + e i ϕ χ + ( p ) χ + ( n ) ) =- 2 ie - i ϕ X 1 - 1 + e i ϕ X 1 1 ( 29 For the “ket” state we may choose X tr ip le t = α X 1 1 + β X 1 0 + γ X 1 - 1 , and then the matrix element is M =- i 2 B 0 g p - g n 2 h 2 ( e i ϕ α + e - i ϕ γ ) 2.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern