{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

angular momentum operator

# angular momentum operator - Angular Momentum Operator I...

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

Angular Momentum Operator I. GENERATOR OF ROTATIONS In 3D real space, a vector is represented by a column matrix V = ( V x V y V z ) T . After a rotation, it is changed to another column matrix V 0 = ( V 0 x V 0 y V 0 z ) T , where V 0 x V 0 y V 0 z = R V x V y V z where R is a 3 × 3 matrix representing the rotation. R must be orthogonal, i.e., R T R = 1. The group of all orthogonal matrices in 3D is O (3). Orthogonal matrices have determinant equal to ± 1. The group of all rotation matrices is SO (3) which is a subgroup of O (3) and determinant equal to 1. Rotation matrix must possess an eigenvalue 1, the associated eigenvector points toward the axis of rotation. Consider an infinitesimal rotation in real space along the direction of a unit vector ˆ n by a small angle δφ . After rotation, we have r r + δ r = r + δφ ˆ n × r , ψ ( r ) ψ 0 ( r ) where ψ is a function of ( r ). Obviously, we shall have ψ 0 ( r + δ r ) = ψ ( r ), or ψ 0 ( r ) = ψ ( r - δ r ) (1 - δ r · ∇ ) ψ ( r ) = (1 - δφ · r × ∇ ) ψ ( r ) = ˆ 1 - i ˆ J · ˆ n ~ δφ ! ψ ( r ) where we have defined an angular momentum operator ˆ J ≡ - i ~ r × ∇ and we shall call ˆ D ˆ n ( δφ ) 1 - i ˆ J · ˆ n ~ δφ an infinitesimal rotation operator. Once we have established the infinitesimal rotation, we can carry out a rotation by any finite angle φ by compounding an infinite number of infinitesimal rotations with δφ = lim N →∞ φ N The finite rotation ˆ D ˆ n ( φ ) is then given by ˆ D ˆ n ( φ ) = lim N →∞ ˆ D ˆ n ( δφ ) · N = e - i ˆ J · ˆ n φ/ ~ Therefore, angular momentum is the generator of rotation. It is easy to see that rotations do not commute, even for infinitesimal ones. Consider an infinitesimal rotation in real space along z -axis by an angle ε . The matrix describing this rotation is R ε z = cos ε - sin ε 0 sin ε cos ε 0 0 0 1 1 - ε 2 / 2 - ε 0 ε 1 - ε 2 / 2 0 0 0 1

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

View Full Document
2 where we have kept terms up to second order in ε . By a cyclic substitution, we have R ε x = 1 0 0 0 cos ε - sin ε 0 sin ε cos ε 1 0 0 0 1 - ε 2 / 2 - ε 0 ε 1 - ε 2 / 2 R ε y = cos ε 0 sin ε 0 1 0 - sin ε 0 cos ε 1 - ε 2 / 2 0 ε 0 1 0 - ε 0 1 - ε 2 / 2 from which we have R ε x R ε y - R ε y R ε x = 0 - ε 2 0 ε 2 0 0 0 0 0 = R ε 2 z - 1 (1) Quantum mechanically, this means cartesian components of angular momentum operators do not commute. In fact, we have [ ˆ J i , ˆ J j ] = i ~ ² ijk ˆ J k However, [ ˆ J i , ˆ J 2 ] = 0 In terms of the Euler angles, an arbitray rotation can be characterized by three angles: ˆ D ( α, β, γ ) = ˆ D z ( α ) ˆ D y ( β ) ˆ D z ( γ ) II. SPIN-1/2 PARTICLE For spin-1/2 particle, the rotation operators are represented by 2 × 2 matrices which form the group SU (2) (i.e., unimodular unitary 2D matrices). A 2 π rotation along an arbitray axis changes an arbitrary spin state to itself with a minus sign: ˆ D ˆ n (2 π ) | ψ i = -| ψ i An arbitrary rotation can be represented by ˆ D ˆ n ( φ ) = e - i ˆ S · ˆ n φ/ ~ = e - i σ · ˆ n φ/ 2 = cos φ 2 - i ( σ · ˆ n ) sin φ 2 In terms of Euler angles, we have under the S z -representation ˆ D (
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