Class calendar, week 7

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Mon. 10/11:
#

Transformation	Generator
time translations	Hamiltonian, H
spatial translations	momentum, P_i
rotations	angular momentum, J^i

# Invariance under a symmetry implies conservation of the corresponding generator.
# If a rotation is represented on the state space of a quantum mechanical systems by a unitary
operator U(R), then we must have U(RR')=U(R)U(R'),
i.e. we must have a unitary representation of the rotation group.
Similarly, for translations.
# [P_i,P_j] = 0 because translations commute.
Rotations do not commute however: R_y(-a)R_x(-a)R_y(a)R_x(a)= R_z(-a^2) + O(a^3).
Writing the unitary operator as U(R) = exp(-ia_i J^i/hbar), we thus infer the commutation relations:
 

[J^i , J^j] = i hbar e^ijk J^k

# e^ijk is the alternating symbol: +1/-1 of ijk is an even/odd permutation of 123, and zero otherwise.
# For a spinless particle, J^i = L^i = e^ijk r^j p^k, i.e. the operator corresponding to the classical
definition of orbital angular momentum. We can infer this as we did for the momentum operator.
For example, for J_z,
<r,theta,phi| exp(-iaJ_z/hbar)|v> = <r,theta,phi||v>  -ia/hbar <r,theta,phi| J_z |v> + O(a^2).
Acting to the left with the rotation on the position bra on the other hand  we have
<r,theta,phi| exp(-iaJ_z/hbar)|v> = <r,theta,phi-a |v> = <r,theta,phi||v>  -a (d/d phi) <r,theta,phi |v> + O(a^2),
hence J_z = -i hbar (d/dphi) = -i hbar (x d/dy - y d/dx) = x p_y - y p_x = L_z.

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Wed. 10/13:
# Note L^i = e^ijk r^j p^k is hermitian, since r^j and p^k commute unless j=k, but e^ijk=0 if j=k.
# For a composite system, we add the rotation generators. Why? if U_A and U_B implement
the rotations on Hilbert spaces H_A and H_B, then U_A U_B does it on H_A x H_B.
But U_A U_B = (1 - ia^i J_A^i + ...)(1 - ia^i J_B^i + ...) = (1 - ia^i (J_A^i + J_B^i) + O(a^2)) ,
so the generator on the joint system is J^i = J_A^i + J_B^i.
(Strictly speaking, by this equation we mean J^i = (J_A^i x I_B) + ( I_a x J_B^i).)
The same goes for the momentum of course.
#Example: 9Be+ ion: J_total  = J_electrons + J_nucleus.
Standard notation for contributions to the angular momentum of an atom:

J_total	J_electrons	J_elec., orbital	J_elec., spin	J_nucleus
F	J	L	S	I

Thus  J_electrons = Sum_n ( L_n + S_n), where n labels the electrons.
# What Monday's argument shows is that all of these rotation generators satisfy the commutation
relations [J^i , J^j] = i hbar e^ijk J^k. If we can classify all the possible representations of this
angular momentum algebra, we will have the full set of possible unitary representations of the
rotation group that can occur in quantum mechanics.
# Since [J^i , J^j] is not zero, we can't simultaneously diagonalize all the J^i,s so pick one, J_z.
Note [J^i,J^2]=0, since J^2 is a scalar and is hence invariant under rotations. We can also just
show this by computation: [J^i,J^2]=[J^i,J^j J^j] = i hbar e^ijk (J^k J^j + J^j J^k) = 0.
[The last step follows because an antisymmetric index pair is summed with a symmetric index pair:
If A^ij = -A^ji and S^ij = S^ji, then A^ij S^ij = - A^ji S^ij = - A^ji S^ji =  - A^ij S^ij.]
Thus we have simultaneous eigenvectors of J^2 and J_z.
We showed these must fall into finite sets of the form {|jm>} with (hbar=1)
J^2| jm> = j(j+1) | jm>,     J_z|jm> = m| jm>,  j= 0,1/2,1,3/2, 2,... and m = -j, -j+1, ...,j-1,j.
For each j we get a unitary representation of the rotation group, of dimension 2j+1, called
the spin-j representation.
 

j	0	1/2	1	3/2	2
dimension: 2j+1	1	2	3	4	5
name: spin-j	singlet, scalar	spinor	vector	spin-3/2	spin-2

 
The method we used to determine the representations was like that of the ladder operators used in the harmonic oscillator case:
Define J+\- = J_x +\- i J_y. Then from the commutation relations we have
[J_z, J_+\- ] = +\-  J_+\-,
J_+ J_- = J^2 - J_z^2 + J_z,   (*)
J_- J_+ = J^2 - J_z^2 - J_z.   (**)
Now suppose |a,m> is an eigenstate of J^2 and J_z with eigenvalues a and m respectively. Then
J^2 J_+\- |a,m> = a J_+\- |a,m>, and
J_z J_+\- |a,m> =( [J_z, J_+\- ] + J_+\- J_z) |a,m> = (m+1) J_+\- |a,m>,
so J_+\- |a,m> = c |a,m+\-1>, i.e. J_+\- are raising and lowering operators for the eigenvalue of J_z.
So we seem to get an infinite ladder:
  ... |a,m-2>>, |a,m-1>, |a,m>, |a,m+1>, |a,m+2>, ...
But J^2 = J_x^2 + J_y^2 + J_z^2, so, taking expectation values, we see that the square of the
J_z eigenvalue must be less than or equal to the J^2 eigenvalue, i.e. m^2 <= a. In order not to violate
this bound we must reach a top value m_t  where raising gives zero, J_+ |a,m_t> = 0,
and a bottom value m_b where lowering gives zero, J_-  |a,m_b> = 0.
Acting on the bottom and top states with (*) and (**) respectively we learn that
a = m_b^2 - m_b  and  a = m_t^2 + m_t, which implies that m_b = - m_t.
Since we get from m_b to m_t by adding some integer n, it must be that
m_t - m_b = 2 m_t = n, or m_t = n/2.
Instead of labeling the representation by the eigenvalue a of J^2 it is conventional to label
by the maximum eigenvalue of J_z, i.e. m_t, which is denoted by the letter j, the possible values
of which are the half-integers.

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Fri. 10/15:
# If we rotate a state |jm> the eigenvalue of J^2 does not change, since [J^i,J^2]=0. When the generators
of rotations act on the state |jm>, J_z gives back  a multiple of  |jm>, while J_x and J_y, being combinations
of J_+ and J_-, give back combinations of |j,m+1> and |j,m-1>. Hence a general rotation just mixes the states
|jm> for a fixed j: exp(-ia^iJ^i/hbar) |jm> = Sum_m' c_m' |jm'> for some c_m's.
There is no subset of m's that only mixes into itself, so the representation is said to be irreducible.
# The irreducible unitary representations of the rotation group are all finite dimensional. How does the
infinite dimensional Hilbert space for a particle decompose into irreducible representations?
Write | r > = |r theta phi> = |r> |theta phi>. The radial ket is invariant under rotations, while the angular
one rotates. The angular Hilbert space is still infinite dimensional. It decomposes into angular momentum eigenkets |lm>, with angular components <theta phi|lm> = Y_lm(theta,phi), the usual spherical harmonics.
All angular functions can be expanded in spherical harmonics.
The Y_lm with a fixed value of l mix into each other under rotations.
Thus each representation j = l = 0,1,2,3,... occurs infinitely many times, once for each radius r.
# Why do only integer j's occur in the position space of a particle? Because a rotation  through 2pi must be the
identity in space, since it is (for instance) just rotating the vectors  r  labeling the position eigenkets  | r >, and
under a 2pi rotation r goes into itself.
Look at the general case for rotations about the z-axis: exp(-i2pi J_z/hbar) |jm> = exp(-im2pi).
For integer j, all the m's are integers, so we always get 1.
For half-integer j, all the m's are half-integers, so we get -1.
Thus the Hilbert space of position eigenkets must decompose into integer spin representations.
# Why aren't the half-integer spin reps excluded? If we really have a representation,  U(R1R2)=U(R1)U(R2),
then U(R)=U(IR)=U(I)U(R), so U(I)=I. For  half-integer spins, U(I)=-I, so we don't have a representation!
On the other hand, we should never have asked for a representation. In quantum mechanics, states are not
vectors in Hilbert space, they are rays, i.e. vectors up to a phase. Thus, in realizing the action of rotations on the Hilbert space, we need only require that U(R1R2) = exp(i f (R1,R2)) U(R1) U(R2),
for some phase f (R1,R2).
That is, a "representation up to a phase", or "ray representation", or "projective representation".
For a good discussion of this see Chapter 2 of vol. I of The Quantum Theory of Fields, S. Weinberg.
# And the half-integer spin reps do occur. First discovered in the 1920's, from spectroscopic evidence
(doublet lines), and the Stern-Gerlach experiment. Pauli found a two-dimensional representation of the
angular momentum commutation relations, J^i = hbar/2 sigma^i, with sigma^i the Pauli matrices, which I can't
write decently in html. Spin-1/2 is a non-classical, two-valued angular momentum.
# We don't need Pauli to work out the matrices though---for each j, we get (2j+1)x(2j+1) matrices representing
the J's in the basis {|jm>}:
<jm' | J_z |jm> = m hbar delta_m',m,
<jm' | J_x |jm> =  hbar (C^+_jm/2  delta_m',m+1 +  C^-_jm/2  delta_m',m-1),
<jm' | J_y |jm> =  hbar (C^+_jm/2i  delta_m',m+1 - C^-_jm/2 i delta_m',m-1),
where the C's are defined by J_+/- |jm> = C^+/-_jm |jm> = Sqrt[j(j+1) - m(m +/- 1)] |jm>.
In the case j=1/2, we get 2x2 matrices, and recover Pauli's matrices.
# The Pauli matrics provide a basis for the trace-free hermitian 2x2 matrices, from the angular momentum
commutation relations they inherit [sigma^i,sigma^j] = 2i e^ijk sigma^k, and they satisfy
 

sigma^i sigma^j = delta^ij + i e^ijk sigma^k

(This can be verified directly from the matrices, but it also follows directly from the general properties of
the J's in the j=1/2 case. For example,  0 = (S_+)^2 = S_x^2  - S_y^2 + i (S_x S_y + S_y S_x).
The hermitian and anti-hermitian parts of this must vanish separately, hence S_x and S_y anti-commute.)
Thus (a.sigma)(b.sigma) = a.b + i (axb).sigma, and in particular (a.sigma)^2 = a^2.
# Magnetic moment M = gamma J, gamma = gyromagnetic ratio. This M is a vector operator.
The "magnetic moment" also refers to a number  mu, the maximum eigenvalue of M:
mu = j gamma hbar, where j is the spin.
# Interaction energy in a magnetic field: U = - M.B.
# In a situation where the only relevant term in the Hamiltonian is this, and B is time independent,
the evolution operator is
U(t) = exp(-iHt/hbar) = exp(i (gamma t/hbar) B.J),
which is a rotation through the angle  -gamma |B| t/hbar about the axis along B.
Thus the angular momentum just precesses about B.
We can see it also in the Heisenberg picture:
dJ^i/dt = (1/ihbar) [J^i, -gamma B^j J^j] = -gamma e^ijk B^j J^k = -gamma (B x J )^i,
which describes the same precession.
# Stern-Gerlach experiment: pass silver atoms through an inhomogeneous B-field.
The classical force is grad(M.B).
If B is in the z-direction and is changing only in the z-direction the force is gamma J_z dB_z/dz.
In quantum mechanics, J_z takes on quantized values, so the force is quantized,
so the spin is deflected in one of  2j+1 ways, depending on the eigenvalue of J_z.
Orbital magnetic moments can only give odd numbers of beams in such an experiment,
but for the silver atoms exactly two beams were observed, implying that the silver atom has spin-1/2.
This is the spin of the valence electron.
# To understand better what is going on in such an experiment, let's consider the Heisenberg equations of motion for the spin and momentum operators. The spin satisfies the precession eqn given above, provided it is all internal spin and not orbital angular momentum (orbital would not commute with a spatially dependent B-field, so there would be an additional term in the eqn). Using the notation J^i = S^i, the momentum satisfies
dp^i/dt = (1/ihbar)[p^i, -gamma B^j S^j] = gamma S^j B^j,i,
where we used [p^i,B^j] = -ihbar B^j,i, and B^j,i means partial_i B^j.
Now div B = 0, so if B has only a z-component it must be constant.
So let's take B = B_0 z^ + b(z z^ - x x^) for example.
The force comes only from the b-part: dp_z/dt = b gamma S_z, and dp_x/dt = -b gamma S_x.
If B_0 is large enough, the  spin is precessing rapidly enough about the z-axis that the x-force averages to zero, and just puts a wiggle on the path, which is deflected in the z-direction.

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