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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.

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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.

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


Jz is diagonal

For example:


Where |jm> is an eigenket

h-bar m is an eigenvalue

For a electron with spin up

Or spin down


These Simple Definitions have some major consequences!




raising and lowering operators
Raising and Lowering Operators

Lowering Operator

Raising Operator

proof that j is the lowering operator
Proof that Jis 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

j 2 j z 0 indicates j 2 and j z are simultaneous observables
[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!

final relation
Final Relation

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

  • 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
matrix elements of j
Matrix Elements of J

Indicates a diagonal matrix


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

a matrix approach to eigenvalues
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
Using our relations,


Pauli Spin Matrices

rotation matices
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
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


Where n-hat points along the axis of rotation

Suppose we rotated through an angle f about the z-axis

what if f 2 p
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

using the sine and cosine relation
Using the sine and cosine relation

And it should be no surprise, that a rotation of b around the y-axis is

  • 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

Wigner’s Formula (without proof)

Certain symmetry properties of d functions are useful in reducing labor and calculating rotation matrix
coupling of angular momenta
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
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 d37
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
the expansion is written as
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
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
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
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 d44
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?

step 7 keep lowering
Step 7—Keep lowering

As low as we go

an aside to simplify notation
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
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
Performing Step 8

An orthogonal vector to this could be


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
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
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
CG Helper Diagram




m1 m2

It is understood that a “C” means square root of “C” (i.e. all radicals omitted)

solution to part 1
Solution to Part 1
  • Look at 1 x 1 table
  • Find j3 = 1 and m3 = 0
  • There 3 values under these
part 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?

  • 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