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## Introduction to Quantum Theory of Angular Momentum

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**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**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**Lowering Operator Raising 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**[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 andlet mlarge=upper bound on m**mlarge cannot any larger**Final Relation**So the eigenvalue is mlarge*(mlarge +1) for any value of m**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**Matrix Elements of J**Indicates a diagonal matrix**Theorems**And we can make matrices of the eigenvalues, but these matrices are NOT diagonal**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**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**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**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**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**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