Introduction to Quantum Theory of Angular Momentum

Angular Momentum AM begins to permeate QM when you move from 1-d to 3-d This discussion is based on postulating rules for the components of AM Discussion is independent of whether spin, orbital angular momenta, or total momentum.

Definition An angular momentum, J, is a linear operator with 3 components (Jx, Jy, Jz) whose commutation properties are defined as

Or in component form

Convention Jz is diagonal For example:

Therefore Where |jm> is an eigenket h-bar m is an eigenvalue For a electron with spin up Or spin down

Definition These Simple Definitions have some major consequences!

THM Proof: QED

Raising and Lowering Operators Raising Operator

Product of J and J+

Fallout

Proof that J is the lowering operator It is a lowering operator since it works on a state with an eigenvalue, m, and produces a new state with eigenvalue of m-1

[J2,Jz]=0 indicates J2 and Jz are simultaneous observables Since Jx and Jy are Hermitian, they must have real eigenvalues so l-m2 must be positive! l is both an upper and LOWER limit to m!

Let msmall=lower bound on m and let mlarge=upper bound on m mlarge cannot any larger

Final Relation So the eigenvalue is mlarge*(mlarge +1) for any value of m

Four Properties

Conclusions As a result of property 2), m is called the projection of j on the z-axis m is called the magnetic quantum number because of the its importance in the study of atoms in a magnetic field Result 4) applies equally integer or half-integer values of spin, or orbital angular momentum

END OF LECTURE 1

Matrix Elements of J Indicates a diagonal matrix

Theorems And we can make matrices of the eigenvalues, but these matrices are NOT diagonal

Fun with the Raising and Lowering Operators

A matrix approach to Eigenvalues If j=0, then all elements are zero! B-O-R-I-N-G! Initial m j= 1/2 final m What does J+ look like?

Using our relations, Answer: Pauli Spin Matrices

J=1, An Exercise for the Students Hint:

Rotation Matices We want to show how to rotate eigenstates of angular momentum First, let’s look at translation For a plane wave:

A translation by a distance, A, then looks like translation operator Rotations about a given axis commute, so a finite rotation is a sequence of infinitesimal rotations Now we need to define an operator for rotation that rotates by amount, q, in direction of q

So Where n-hat points along the axis of rotation Suppose we rotated through an angle f about the z-axis

Using a Taylor (actually Maclaurin) series expansion

What if f = 2p? The naïve expectation is that thru 2p and no change. This is true only if j= integer. This is called symmetric BUT for ½ integer, this is not true and is called anti-symmetric

Let j=1/2 (for convenience it could be any value of j)

Using the sine and cosine relation And it should be no surprise, that a rotation of b around the y-axis is

Consequences If one rotates around y-axis, all real numbers Whenever possible, try to rotate around z-axis since operator is a scalar If not possible, try to arrange all non-diagonal efforts on the y-axis Matrix elements of a rotation about the y-axis are referred to by

And Wigner’s Formula (without proof)

Certain symmetry properties of d functions are useful in reducing labor and calculating rotation matrix

Coupling of Angular Momenta We wish to couple J1 and J2 From Physics 320 and 321, we know But since Jz is diagonal, m3=m1+m2

Coupling cont’d The resulting eigenstate is called And is assumed to be capable of expansion of series of terms each of with is the product of 2 angular momentum eigenstates conceived of riding in 2 different vector spaces Such products are called “direct products”

Coupling cont’d The separateness of spaces is most apparent when 1 term is orbital angular momentum and the other is spin Because of the separateness of spaces, the direct product is commutative The product is sometimes written as

Proof of commutative property

The expansion is written as Is called the Clebsch-Gordan coefficient Or Wigner coefficient Or vector coupling coefficient Some make the C-G coefficient look like an inner product, thus

A simple formula for C-G coefficients Proceeds over all integer values of k Begin sum with k=0 or (j1-j2-m3) (which ever is larger) Ends with k=(j3-j1-j2) or k=j3+m3 (which ever is smaller) Always use Stirling’s formula log (n!)= n*log(n) Best approach: use a table!!!

What if I don’t have a table? And I’m afraid of the “simple” formula? Well, there is another path… a 9-step path!

9 Steps to Success Get your values of j1 and j2 Identify possible values of j3 Begin with the “stretched cases” where j1+j2=j3 and m1=j1, m2=j2 , and m3=j3, thus |j3 m3>=|j1 m1>|j2 m2> From J3=J1+J2,, it follows that the lowering operator can be written as J3=J1+J2

9 Steps to Success, cont’d Operate J3|j3 m3>=(J1+J2 )|j1 m1>|j2 m2> Use Continue to lower until j3=|j1-j2|, where m1=-j1 , m2= -j2, and m3= -j3 Construct |j3 m3 > = |j1+j2 -1 j1+j2-1> so that it is orthogonal to |j1+j2 j1+j2-1> Adopt convention of Condon and Shortley, if j1 > j2 and m1 > m2 then Cm1 m2j1 j2 j3 > 0 (or if m1 =j1 then coefficient positive!)

9 Steps to Success, cont’d Continue lowering and orthogonalizin’ until complete! Now isn’t that easier? And much simpler… You don’t believe me… I’m hurt. I know! How about an example?

A CG Example: j1 =1/2 and j2 =1/2 Step 1 Step 2 Step 3

Steps 4 and 5 and 6->

Step 7—Keep lowering As low as we go

An aside to simplify notation Now we have derived 3 symmetric states Note these are also symmetric from the standpoint that we can permute space 1 and space 2 Which is 1? Which is 2? “I am not a number; I am a free man!”

The infamous step 8 “Construct |j3 m3 > = |j1+j2 -1 j1+j2-1> so that it is orthogonal to |j1+j2 j1+j2-1>” j1+j2=1 and j1+j2-1=0 for this case so we want to construct a vector orthogonal to |1 0> The new vector will be |0 0>

Performing Step 8 An orthogonal vector to this could be or Must obey Condon and Shortley: if m1=j1,, then positive value j1=1/2 and |+> represents m= ½ , so only choice is

Step 9– The End This state is anti-symmetric and is called the “singlet” state. If we permute space 1 and space 2, we get a wave function that is the negative of the original state. These three symmetric states are called the “triplet” states. They are symmetric to any permutation of the spaces

A CG Table look up Problem Part 1— Two particles of spin 1 are at rest in a configuration where the total spin is 1 and the m-component is 0. If you measure the z-component of the second particle, what values of might you get and what is the probability of each z-component?

CG Helper Diagram m1 m2 j3 m3 C It is understood that a “C” means square root of “C” (i.e. all radicals omitted)

Solution to Part 1 Look at 1 x 1 table Find j3 = 1 and m3 = 0 There 3 values under these m1 m2 1 -1 1/2 -1/2

So the final part m2 C Prob -1 1/2 ½ 1 -1/2

Part 2 An electron is spin up in a state, y5 2 1 , where 5 is the principle quantum number, 2 is orbital angular momentum, and 1 is the z-component. If you could measure the angular momentum of the electron alone, what values of j could you get and their probabilities?

Solution Look at the 2 x ½ table since electron is spin ½ and orbital angular momentum is 2 Now find the values for m1=1 and m2=1/2 There are two values across from these: 4/5 which has j3 = 5/2 -1/5 which has j3 = 3/2 So j3=5/2 has probability of 4/5 So j3 = 3/2 has probability of 1/5