We will start the study of matrix representations of the rotation group by looking at the defining representation, and then derive the general matrix representation.
The Defining Representation of the Three-Dimensional Rotation Group
In Sect.4.1we found the condition
RãRT =1 (7.39)
for rotation matrices. This leaves the following possibilities for the matrix1 X = lnR,
XT = −X+2πin1. (7.40)
The equation
det(expX)=exp(trX), (7.41)
which follows from the existence of a Jordan canonical form (F.3) for every matrix, implies then
1See AppendixFfor the calculation of the logarithm of an invertible matrix.
7.4 Matrix Representations of the Rotation Group 137
detR =exp
trX+XT 2
=(−1)n, (7.42)
i.e detR = ±1. Pure rotations have detR =1, whereas additional inversion of an odd number of axes2yields detR= −1. We will focus on pure rotations.
The general solution of Eq. (7.40) in three dimensions and withn=0 is
X=
⎛
⎝ 0 ϕ3 −ϕ2
−ϕ3 0 ϕ1
ϕ2 −ϕ1 0
⎞
⎠=ϕiLi =ϕãL, (7.43)
where the basisLi of anti-symmetric real 3×3 matrices is
L1=
⎛
⎝0 0 0 0 0 1 0−1 0
⎞
⎠, L2=
⎛
⎝0 0−1 0 0 0 1 0 0
⎞
⎠, L3=
⎛
⎝ 0 1 0
−1 0 0 0 0 0
⎞
⎠. (7.44)
We can write these equations in short form as(Li)j k =ij k. The general orientation preserving rotation in three dimensions therefore has the form
R(ϕ)=exp(ϕãL)= ∞
n=0
ϕn
n!(ϕˆ ãL)n. (7.45) We can order the series expansion into even and odd powers ofϕ ãLusing the properties
(ϕˆãL)2= ˆϕ⊗ ˆϕT −1, (ϕˆ ãL)3= − ˆϕãL. (7.46) This yields the representation
R(ϕ)= ˆϕ⊗ ˆϕT +
1− ˆϕ⊗ ˆϕT
cosϕ+ ˆϕãLsinϕ. (7.47) Application of the matrixϕˆ ãLon a vectorr generates a vector product,
(ϕˆ ãL)ãr = − ˆϕìr, (7.48) i.e. for every vectorr, the first term in (7.47) preserves the partr= ˆϕ⊗ ˆϕTãrof the vector which is parallel to the vectorϕ, the second term multiplies the orthogonal partr⊥ =r −rby the factor cosϕ, and the third part takes the orthogonal part, rotates it byπ/2 and multiplies it by the factor sinϕ,
2Inversion of three axes is equivalent to inversion of one axis combined with a rotation.
r=R(ϕ)ãr =r+r⊥cosϕ− ˆϕìrsinϕ. (7.49) This also implies that the directionϕˆ of the vectorϕis the direction of the axis of rotation.3
Exponentiation of the linear combinations ϕ ã L of the matrices (7.44) thus generates rotations in three dimensions, and therefore these matrices are also denoted as three-dimensional representations ofgenerators of the rotation group.
They satisfy the commutation relations
[Li, Lj] = −ij kLk. (7.50) We will also use the hermitian matrices
Mi = −ihL¯ i, [Mi, Mj] =ih¯ ij kMk, [Mi,M2] =0. (7.51) It is no coincidence that the angular momentum operators
Mi =ij kxjpk (7.52)
satisfy the same commutation relations. We will see that angular momentum operators also generate rotations, and a set of operatorsMi generates rotations if and only if the operators satisfy the commutation relations (7.51). It is a consequence of the general Baker-Campbell-Hausdorff formula in AppendixEthat the combination of any two rotations to a new rotation is completely determined by the commutation relations (7.51) of the generators of rotations.
The General Matrix Representations of the Rotation Group
We wish to classify all possible representations of the commutation relations (7.51) in vector spaces. To accomplish this, it is convenient to change the basis fromMx ≡ M1andMy ≡M2to
M±=M1±iM2, Mz≡M3. (7.53)
3Please note thatr=R(ϕ)ãris just a common “vectorlike” abbreviation for the transformation equationxi = Riaxa for the Cartesian componentsxi andxa of vectors under a rotation of the basis (passive transformation) or rotation of the whole vector space (active transformation).
For the passive transformations, we found in Sect.4.1thatei =eaRaiimpliesxi =(R−1)iaxa on the vector components, see Eq. (4.14), whereas for active transformations the basis rotation ei = eaRai impliesxa = Raixi for the vector components, see Problem4.2. For the passive transformations, this implies that our rotation matricesR here correspond toR−1in Sect.4.1, whereas for active transformations it is the sameRas in Problem4.2.
7.4 Matrix Representations of the Rotation Group 139
The productM2≡MiMi is then M2=1
2(M+M−+M−M+)+Mz2=M−M++Mz2+ ¯hMz, (7.54) and we have the commutation relations in the new basis,
[Mz, M±] = ± ¯hM±, [M+, M−] =2hM¯ z. (7.55) Hermiticity4 implies that we can use a basis where Mz is diagonal with real eigenvalues,
Mz|m = ¯hm|m, m∈R. (7.56) The commutation relations then imply
M±|m = ¯hC±(m)|m±1, (7.57) C+(m−1)C−(m)=2hm¯ +C−(m+1)C+(m), (7.58) andM++=M−implies
C−(m)= m−1|M−|m =(m|M++|m−1)+=C+(m−1)+. (7.59) Substitution in Eq. (7.58) yields
|C+(m)|2= |C+(m−1)|2−2h¯2m. (7.60) Since the left-hand side cannot become negative, there must exist some maximal valueformsuch thatC+()=0,M+| =0, and we have
|C+(−1)|2=2h¯2, |C+(−2)|2=2h¯2(2−1), (7.61) and aftern−1 steps
|C+(−n)|2= |C−(−n+1)|2= ¯h2[2n−n(n−1)]. (7.62) Again, the left-hand side cannot become negative, and therefore the expression on the right-hand side must terminate for some valueN ofn,C−(−N+1) = 0, M−|−N+1 =0. This implies existence of an integerN such that 2=N−1 and
4We could do the following calculations in slightly more general form without using hermiticity, and then find hermiticity of the finite-dimensional representations along the way.
C+(−N )=C+(−(N+1)/2)=C−((1−N )/2)=C−(−)=0, (7.63) where an irrelevant possible phase factor was excluded. Therefore we have bound- aries
−= 1−N
2 ≤m≤ N−1
2 = (7.64)
andN =2+1 possible values formfor both integerand half-integer. Equation (7.62) yields with
n=N−1
2 −m (7.65)
the equation
n(N−n)= N2
4 −
m+1
2 2
= N2−1
4 −m(m+1)
=(+1)−m(m+1), (7.66)
and therefore C+(m)2= ¯h2
N2−1
4 −m(m+1)
= ¯h2[(+1)−m(m+1)], (7.67)
C−(m)2=C+(m−1)2= ¯h2[(+1)−m(m−1)]. (7.68) We have found all the hermitian matrix representations of the commutation relations (7.51). The magnetic quantum numbermcan take values−≤m≤, the number of dimensions is N = 2+1 ∈ N, and the actions of the angular momentum operators are
Mz|, m = ¯hm|, m, 2∈N0, m∈ {−,−+1, . . . , −1, }, (7.69) M+|, m = ¯h
(+1)−m(m+1)|, m+1 (7.70)
M−|, m = ¯h
(+1)−m(m−1)|, m−1 (7.71)
Mx|, m = ¯h 2
(+1)−m(m+1)|, m+1
+ ¯h 2
(+1)−m(m−1)|, m−1, (7.72)
My|, m = ¯h 2i
(+1)−m(m+1)|, m+1