8. Spin and Adding Angular Momentum

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8. Spin and Adding Angular Momentum
8A. Rotations Revisited
The Assumptions We Made

• We assumed that r? formed a basis and R(?)r?
  ?r?
• From this we deduced R(?)?(r) ?(?Tr)
• Is this how other things work?
• Consider electric field from a point particle
• Can we rotate by R(?)E(r) E(?Tr)?
• Lets try it
• This is not how electric fields rotate
• It is a vector field, we must also rotate the
  field components
• R(?)E(r) ? E(?Tr)
• Maybe we have to do somethingsimilar with ??

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Spin Matrices

• How do the D(?)s behave?
• We want to find all matrices satisfying this
  relationship
• Easy to show when ? 1, D(?) 1
• As before, Taylor expand D for small angles
• In a manner similar to before, then show

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We Already Know the Spin Matrices

• We used to have identical expressions for the
  angular momentum L
• From these we proved that L has the standard
  commutation relations
• It follows that S has exactly the same
  commutation relations
• The three Ss are generalized angular momentum
• But in this case, they really are finite
  dimensional matrices
• Logically, our wave functions would now be
  labeled
• But s is a constant, so just label them
• There are 2s 1 of them total

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Restrictions on s?

• Recall, for angular momentum, we had to restrict
  l to integers, not half-integers
• Why? Because wave functions had to be continuous
• Can we find a similar argument for spin? Consider
  s ½
• Consider a rotation by 2?
• This would imply if you rotate by 2?, the state
  vector changes by ? ? ? ? ?
• But these states are indistinguishable, so this
  is okay!
• Any value of s, integer or half-integer, is fine
• The basic building blocks of matter are all s ½
• Other particles have other spins

part. s e- ½ p ½ n0 ½
part. s ? 1 ??,? 0 0 ?s 3/2
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Basis States for Particles With Spin

• Basis states used to be labeled by r?
• But now we must label them also by
  whichcomponent we are talking about r,ms?
• Comment for spin ½, it is common to abbreviate
  the ms label
• The spin operators affect only the spin label
• Operators that concern position, like R, P, and
  L, only affect the position label
• All these position operators must commute with
  spin operators

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Sample Problem
Define J L S. Find all commutators of J, J2,
S2, and L2

• Thats 6 operators, so 6?5/2 15 possible
  commutators
• Ill just do five of them to give you the idea

• Recall, for any angular momentum-like set of
  operators, J2,J 0

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Hydrogen Revisited

• Recall our Hamiltonian
• Note that S commutes with the Hamiltonian
• We can diagonalize simultaneously H, L2, Lz, S2,
  and Sz
• It is silly to label them by s, because s ½
• Degeneracy ms takes on two values,doubling the
  degeneracy
• Do all Hamiltonians commute with spin?
• No! Magnetic interactions care about spin
• Even hydrogen has small contributions (spin-orbit
  coupling) that depend on spin

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8B. Total Angular Momentum and Addition
What Generates Rotations?

• Recall that
• Rewrite this in ket notation
• Define J
• J is what actually generates rotations
• If a problem is rotationally invariant, we would
  expect J to commute with H
• Not necessarily L or S

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What are L, S and J?

• Consider the rotation of the Earth around the
  Sun
• It has orbital angular momentumfrom its orbit
  around the Sun L
• It has spin angular momentumfrom its rotation
  around the axis S
• The total angular momentum is
• It is another set of angular momentum-like
  operators
• It will have eigenvectors j,m? with eigenvalues
• Because L and S typically dont commute with
  theHamiltonian, we might prefer to label our
  states byJ eigenvalues, which do
• To keep things as general as possible, imagine
  any two angular momentum operators adding up to
  yield a third

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Adding Angular Momentum

• Commutation relations
• We could label states by their eigenvalues under
  the following four commuting operators
• Instead, wed prefer to label them by the
  operators
• These all commute with each other
• These have the same j1 and j2 values, so well
  abbreviate them
• Two things we want to know
• Given j1 and j2, what will the states j,m? be?
• How do we convert from one basis to another,
  i.e., what is
• Clebsch-Gordan coefficients

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The procedure

• It is easy to figure out what the eigenvalues of
  Jz are, because
• For each basis vector j1,j2m1,m2?, there will
  beexactly one basis vector j,m? with m m1
  m2
• The ranges of m1 and m2 are known
• From this we can deduce exactly how many
  basisvectors in the new basis have a given value
  of m
• By looking at the distribution of m values,
  wecan deduce what j values must be around
• Easier illustrated by doing it than describing it

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Sample Problem
Suppose j1 2 and j2 1, and we change basis
from j1,j2m1,m2? to j,m?. (a) What values of m
will appear in j,m?, and how many times? (b)
What values of j will appear in j,m?, and how
many times?

• First, find a list of all the m1 and m2 values
  that occur
• I will do it graphically
• Now, use the formula m m1 m2 to find the
  mvalue for each of these points
• From these, deduce the m values and how
  manythere are
• I will do it graphically
• Note where the transitions are

m3
m2
m-1
m-3
m1
m0
m-2
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Sample Problem (2)
Suppose j1 2 and j2 1, and we change basis
from j1,j2m1,m2? to j,m?. (a) What values of m
will appear in j,m?, and how many times? (b)
What values of j will appear in j,m?, and how
many times?

• For any value of j, m will run from j to j
• Clearly, there is no j bigger than 3
• But since m 3 appears, there must be j 3
• This must correspond to ms from 3 to 3
• Now, there are still states with m up to 2
• It follows there must also be j 2
• This covers another set of ms from 2 to 2
• What remains has m up to 1
• It follows there must be j 1
• And thats it.
• Why did it run from j 3 to j 1?
• Because it went from j1 j2 down to j1 j2

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General Addition of Angular Momentum

• The set of all (m1,m2) pairs forms a rectangle
• The largest value of m is m j1 j2, which can
  only happen one way
• As m decreases from the max value, there is one
  more way of making each m value for each decrease
  in m until you get to j1 j2
• This implies that you get maximum jmax j1 j2
  and minimum jmin j1 j2
• So, j runs from j1 j2 to j1 j2 in steps
  of size 1

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Check Dimensions

• For fixed j1 and j2, the number of basis vectors
  j1,j2m1,m2? is
• How many basis vectors j,m? are there?
• For each value of j, there are 2j 1.
• Therefore the total is

• So dimensions work out

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Sample Problem
Suppose we have three electrons. Define the
total spin as S S1 S2 S3. What are the
possible values of the total spin s, the
corresponding eigenvalues of S2, and how many
ways can each of them be made?

• Electrons have spin s ½, so
• Combine the first two electrons
• Now add in the third
• If s12 0, this says
• If s12 1, this says
• Final answer for s
• The repetition means there are two ways to
  combine to make s ½
• For S2

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Hydrogen Re-Revisited

• Recall hydrogen states labeled by
• Because of relativistic corrections, these arent
  eigenstates
• Closer to eigenstates are basis states
• j l ? ½
• States with different mj are related by rotation
• Indeed, the value of mj will depend on choice of
  x, y, z axis
• And they are guaranteed to have the same energy
• Therefore, when labelling a state we need to
  specify n, l, j
• We label l values by letters, in a not obvious
  way
• Good to know the first four s, p, d, f
• We then denote j by a subscript, so example state
  could be 4d3/2
• Remember restrictions l lt n and j l ? ½
• Often, we dont care about j, so just label it 4d
• Remember, number of states for given n,l is

l let 0 s 1 p 2 d 3 f 4 g 5 h 6 i 7 k 8 l 9 m 10 n
11 o 12 q
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8C. Clebsch-Gordan Coefficients
How do we change bases?

• We wish to interchange bases j1,j2m1,m2? ??
  j,m?
• These are complete orthonormal basis states in
  the same vector space
• We can therefore use completeness either way
• The coefficients are called Clebsch-Gordancoeffic
  ients, or CG coefficients for short
• Our goal Show that we can find them (almost)
  uniquely
• Note that the states j1,j2m1,m2? are all
  related by J1? and J2?
• There are no arbitrary phases concerning how they
  are related
• The j,m? states with the same js and different
  ms are related by J?
• But there is no simple relation between j,m?s
  different js convention choice

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Convention Confusion

• If you ever have to look them up, be warned,
  different sources use different notations
• Recall that the other states are alsoeigenstates
  of J12 and J22
• People also get lazy and drop some commas
• In addition, the Clebsch-Gordan coefficients are
  defined only up to a phase
• Everyone agrees on phase up to sign
• As long as you use them consistently, itdoesnt
  matter which convention you use.
• They will turn out to be real, and therefore
• Because of this ambiguity, people get lazy and
  often use what is logically the wrong one

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Nonzero Clebsch-Gordan (C-G) Coefficients

• When are the coefficients meaningful and
  (probably) non-zero?
• (1) j range
• (2) m range
• (3) conservation of Jz
• Lets prove the last one using
• Act on the left with Jz and on the right with
  J1z and J2z
• Must be zero unless

j1 j2 j is an integer
j m is an integer, etc.
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Finding C-G Coefficients for m j

• Largest value for m is j, therefore
• Recall in general
• We therefore have
• Recall only if m1 m2 m ( j) are non-zero
• This relates all the non-zero terms for m j,
  all relative sizes determined
• To get overall scale, use normalization
• This determines everything up to a phase
• We arbitrarily pick

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Finding C-G Coefficients for m 1 from m

• We now have CG coefficients when m j
• I will now demonstrate that if we have them for
  m, we can get them for m 1
• First note
• Dagger this
• So
• So if we know them for m, we know them for m 1
• Since we know them for m j, we know them for m
  j 1, j 2, etc.
• Hence we have a (painful) procedure for finding
  all CG coefficients
• Sane people dont do it this way, they look them
  up or use computers

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Properties of CG-coefficients

• Adding j1 and j2 is thesame as adding j2 and j1
• Corollary if j1 j2, then the combinations of
  spins is symmetric if j1 j2 j is even,
  anti-symmetric if it is odd
• You can work your way up from m j in the same
  way we worked our way down from m j
• Adding j1 0 or j2 0 is pretty
  trivial,because these imply J1 0 or J2 0
• If you ever look things up in tables, they will
  assume j1 ? j2 gt 0, and assume you will use the
  first or third rule to get other CG coefficients
• Or you can use computer programs to get them

gt clebsch(1,1/2,1,-1/2,3/2,1/2)
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CG coefficients when j2 ½

• For j2 small, we can find simple formulas for the
  CG coefficients
• If j2 ½, then j j1 ? ½

• Example
• For one electron, J L S. Let j1 ? l, m ? mj,
  drop j2 s ½, m2 ?½ ? ?
• For adding two electron spins, drop s1 and s2,
  abbreviate mi ?½ ? ?

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Sample Problem
Hydrogen has a single electron in one of the
states n,l,m,ms? 2,1,1,? or 2,1,0,? , or
in one of the states n,l,j,mj? 2,1,3/2,1/2?
or 2,1,1/2,1/2? . In all four cases, write
explicitly the wave function

• For s ½, wave function looks like
• Spin state ms tells us which component exists
• This lets us immediately write the wave
  functionfor the first two
• For the j,mj?states we have

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Sample Problem (2)
or in one of the states n,l,j,mj?
2,1,3/2,1/2? or 2,1,3/2,1/2? . In all four
cases, write explicitly the wave function

• You can also get the CG coefficients from Maple

gt clebsch(1,1/2,1,-1/2,3/2,1/2) gt
clebsch(1,1/2,0,1/2,3/2,1/2) gt
clebsch(1,1/2,1,-1/2,1/2,1/2) gt
clebsch(1,1/2,0,1/2,1/2,1/2)
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Sample Problem
Hydrogen has a single electron in the state
n,l,j,mj? 2,1,3/2,1/2?. If one of the
following is measured, what would the outcomes
and corresponding probabilities be, and what
would the state afterwards look like E, J2, Jz,
L2, S2, Lz,Sz

• For the first five choices, our state is an
  eigenstate of the operator
• The eigenstate will be unchanged by this
  measurement
• For the last two, we write it in terms
  ofeigenstates of Lz or Sz
• Then we have
• State afterwards is
• Or we have
• State afterwards is

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8D. Scalar, Vector, Tensor
Definition and Commutation with J

• A scalar operator S is anything that is unchanged
  under rotation
• Examples
• Scalar operators commute with the generator of
  rotations J
• Vector operators V are operators that rotate like
  a vector
• Examples
• They have commutation relations with J given by
• A rank 2 tensor Tij under rotation rotates as
• Can show that
• Rank k tensor has k indicesand commutation
  relations
• Scalar rank 0 tensor, Vector rank 1 tensor
• Rank 2 tensor is sometimes just called a tensor

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How to Make a Tensor From Vectors

• If V and W are any two vector operators,then we
  can define a rank-2 tensor operator
• One can similarly define higher rank tensor
  operators
• This tensor has nine independent components
• But it has pieces that arent very rank-2
  tensor-like
• Dot product V?W is a scalar operator
• Cross product V?W is a vector operator
• The remaining five pieces are the truly rank-2
  part
• We want figure out how to extract the various
  pieces

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Spherical Tensors

• We start with a vector operator V
• Define the three operators Vq by
• You can then show the following
• Proof by homework problem
• Compare this with
• Another way to write it
• Generalize this formula
• Define a spherical tensor of rank k as 2k 1
  operators
• It must have commutation relations
• Trivial example A scalar is a spherical tensor
  of rank 0

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Combining Spherical Tensors (1)
Theorem Let V and W be spherical tensors of rank
k1 and k2 respectively. Then we can build a new
spherical tensor T of rank k defined by

• Those matrix elements are CG coefficients
• Proof

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Combining Spherical Tensors (2)

• We have complete set of states k1,k2q1,q2?
• Now insert complete set of states k,q?
• J doesnt change the k value, so k k
• So we have proven it

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How it Comes Out

• This sum only makes sense if CG coefficients are
  non-zero
• Only non-zero terms are when q1 q2 q
• So its really just a single sum
• By combining two vectors, we can get k 0, 1, 2
• k 0 Scalar (dot product)
• k 1 Vector (cross product)
• k 2 Truly rank 2 tensor part
• We can then combine rank 2 tensors with more
  vectors to make rank 3 spherical tensors

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Sample Problem
If we combine two copies of the position operator
R, what are the resulting components of the
rank-2 spherical tensor Tq(2)?
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8E. The Wigner-Eckart Theorem
Why it should work

• Suppose we have an atom or other rotationally
  invariant system
• Eigenstates should be eigenstates of J2, Jz,
  probably other stuff
• It is common to need matrix elementsof operators
  between these states
• We know how the ket and bra rotate
• If we also know how the operator in the middle
  rotates, we should be able to find relations
  between these various quantities
• Suppose the operator is a spherical tensor,or
  combinations thereof
• Then we know how T rotates, and we should be able
  to find relations
• This helps us because
• If the calculation is hard, we do it a few times
  and deduce the rest
• If the calculation is impossible, we measure it a
  few times and deduce the rest

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Similarities With CG coefficients (1)

• I want to compare the matrix element above to the
  CG coefficient above
• Recall relations for Jz
• Use commutationrelation
• Let Jz act on the braor the ket on the left
• Hence matrix elements are zero unless
• Compare to the CG coefficient above
• This vanishes unless

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Similarities With CG coefficients (2)

• Recall relations for J?
• For m j,
• Implies
• Our commutation relations tell us
• Compare to the CG coefficients

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Similarities With CG coefficients (3)

• We have
• Equivalent to
• Our commutation relations tell us

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These Matrix Elements are CG Coefficients

• We have three relations that are identical for
  these two expressions
• These expressions were all that were used to find
  the CG coefficients
• Plus, we had a normalization condition
• Hence, these two expressions are identical
• Up to normalization

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The Wigner-Eckart Theorem

• What can the proportionality constant depend on?
• Not m, m, nor q
• It can depend on ?, ?, j, j, and of course T
• The Wigner-Eckart Theorem
• The square root in the denominator is a choice,
  neither right nor wrong
• That other thing is called a reduced matrix
  element
• You dont calculate it (directly)
• You may be able to calculate left side for one
  value of m, m, q
• Or you may be able to measure left side for one
  value of m, m, q
• Then you deduce the reduced matrix element from
  this equation
• Then you can use it for all the other values of
  m, m, q

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Why Is the Wigner-Eckart Theorem Useful?

• The number of matrix elements is
• For example, if j 3, j 2, k 1, this is 105
  different matrix elements
• Calculating them computationally may be difficult
  or impossible
• Measuring them may be a great deal of work
• By doing one (difficult) computation or one
  (difficult) measurement you can deduce a lot of
  others
• Comment why is the factor of 2j 1 there?
• If T0(k) is Hermitian, then you can show

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Sample Problem
The magnetic dipole transition of hydrogen
causing the 21 cm line is governed by the matrix
element , where F is the total angular
momentum quantum number and mF is the
corresponding z-component, and S and L are spin
and orbital angular quantum operators for the
electron. Deduce as much as you can about these
matrix elements for mF 1, 0, or 1.

• We have no idea what most of this means, but its
  clear
• F and mF are angular quantum number, effectively,
  j ?? F and m ?? mF
• S and L are vector operators
• Call reduced matrix element A
• Non-vanishing only if q mF 0
• Get the CG coefficients from program
• All other matrix elements vanish

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Sample Problem
Deduce as much as you can about these matrix
elements for mF 1, 0, or 1.

• For mF 1, we also have
• Vx and Vy two equations, two unknowns
• We therefore have
• Similarly

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8F. Integrals of Spherical Harmonics
Products of Spherical Harmonics

• Consider the product of any two spherical
  harmonics
• By completeness, this can be written as a sum of
  spherical harmonics
• The coefficients clm can be found
  usingorthogonality
• Think of the expression
• as an operator acting on a wave function
• It is not hard to see that this operator is a
  spherical tensor operator
• Think of clm then as a matrix element
• By the Wigner-Eckart theorem
• All that remains is to findthe reduced matrix
  elements

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Working on the Reduced Matrix Element

• Substitute the top equation in the bottom
• Multiply this expression by and sum over
  m1, m2
• Rename l as l

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Finishing the Computation

• Must be true at all angles
• Evaluate at ? 0
• Formula for the Ys at ? 0 is simple
• Now we solve for thereduced matrix element
• We therefore have

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When doesnt it vanish?

• We want to know when this is non-zero, or likely
  to be non-zero
• We need
• We need
• Under parity, each of the spherical harmonics
  transforms to
• So the whole integral satisfies
• We need

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Sample Problem
Atoms usually decay spontaneously by the electric
dipole process, in which case the rate is
determined by the matrix element , where?I
and ?F are the initial and final states. For
hydrogen in each of the following states, which
states might be the final n and l quantum states
if the initial state is 4s, 4p, 4d, 4f?

• The initial state has n 4 and l 0, 1, 2, or
  3
• Final state has unknown n and l, but
• Must have n lt 4 because energy goes down
• Must have l lt n
• The position operators can be written in terms of
  l 1 spherical harmonics
• So we have
• To not vanish, we need
• For 4s, l 0
• Must have l lt n lt 4

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Sample Problem (2)
Atoms usually decay spontaneously by the electric
dipole process, in which case the rate is
determined by the matrix element , where?I
and ?F are the initial and final states. For
hydrogen in each of the following states, which
states might be the final n and l quantum states
if the initial state is 4s, 4p, 4d, 4f?

• For 4p l 1, so
• Must have l lt n lt 4
• For 4d l 2, so
• Must have l lt n lt 4
• For 4f
• Must have l lt n lt 4