7—Operators and Matrices 154the other. This is no different from the way you look at ordinary real valued functions. The exponential
and the logarithm are inverse to each other because*
ln(ex) =x for allx.
For the rotation operator, Eq. (7.10), the inverse is obviously going to be rotation by the same angle
in the opposite direction.
RαR−α=I
Because the matrix components of these operators mirror the original operators, this equation must
also hold for the corresponding components, as in Eqs. (7.27) and (7.29). Setβ=−αin (7.29) and
you get the identity matrix.
In an equation such as Eq. (7.7), or its component version Eqs. (7.8) or (7.9), if you want to
solve for the vector~u, you are asking for the inverse of the functionf.
~u=f(~v) implies ~v=f−^1 (~u)
The translation of these equations into components is Eq. (7.9)
(u 1
u 2
)
=
(
f 11 f 12
f 21 f 22
)(
v 1
v 2
)
which implies1
f 11 f 22 −f 12 f 21
(
f 22 −f 12
−f 21 f 11
)(
u 1
u 2
)
=
(
v 1
v 2
)
(7.31)
The verification that these are the components of the inverse is no more than simply multiplying the
two matrices and seeing that you get the identity matrix.
7.6 Rotations, 3-d
In three dimensions there are of course more basis vectors to rotate. Start by rotating vectors about
the axes and it is nothing more than the two-dimensional problem of Eq. (7.10) done three times. You
do have to be careful about signs, but not much more — as long as you draw careful pictures!
x
y
z
x
y
z
x
y
z
α
β
γ
The basis vectors are drawn in the three pictures:~e 1 =ˆx, ~e 2 =yˆ, ~e 3 =zˆ.
In the first sketch, rotate vectors by the angleαabout thex-axis. In the second case, rotate by
the angleβabout they-axis, and in the third case, rotate by the angleγabout thez-axis. In the first
case, the~e 1 is left alone. The~e 2 picks up a little positive~e 3 , and the~e 3 picks up a little negative~e 2.
Rα~e 1
(
~e 1
)
=~e 1 , Rα~e 1
(
~e 2
)
=~e 2 cosα+~e 3 sinα, Rα~e 1
(
~e 3
)
=~e 3 cosα−~e 2 sinα (7.32)
* The reverse,elnxworks just for positivex, unless you recall that the logarithm of a negative
number is complex. Then it works there too. This sort of question doesn’t occur with finite dimensional
matrices.