Rotations using matrix exponentials
Posted by peeterjoot on July 27, 2010
In  it is noted in problem 1.3 that any Unitary operator can be expressed in exponential form
where is Hermitian. This is a powerful result hiding away in this problem. I haven’t actually managed to prove this yet to my satisfaction, but working through some examples is highly worthwhile. In particular it is interesting to compute the matrix for a rotation matrix. One finds that the matrix for such a rotation operator is in fact one of the Pauli spin matrices, and I found it interesting that this falls out so naturally. Additionally, it is rather slick that one is able to so concisely express the rotation in exponential form, something that is natural and powerful in complex variable algebra, and also possible using Geometric Algebra using exponentials of bivectors. Here we can do it after all with nothing more than the plain old matrix algebra that everybody is already comfortable with.
The logarithm of the Unitary matrix.
By inspection we can invert 1.1 for , by taking the logarithm
The problem becomes one of evaluating the logarithm, or even giving meaning to it. I’ll assume that the functions of matrices that we are interested in are all polynomial in powers of the matrix, as in
and that such series are convergent. Then using a spectral decomposition, possible since Unitary matrices are normal, we can write for diagonal
Provided the logarithm has a convergent power series representation for , we then have for our Hermitian matrix
Evaluate this logarithm for an plane rotation.
Given the rotation matrix
We find that the eigenvalues are , with eigenvectors proportional to respectively. Our decomposition for is then given by
Taking logs we have
With the Pauli matrix
we then have for an plane rotation matrix just:
Immediately, since , this also provides us with a trigonometric expansion
By inspection one can see that this takes us full circle back to the original matrix form 2.7 of the rotation. The exponential form of
2.12 has a beauty that is however far superior to the plain old trigonometric matrix that we are comfortable with. All without any geometric algebra or bivector exponentials.
Three dimensional exponential rotation matrices.
By inspection, we can augment our matrix for a three dimensional rotation in the plane, or a rotation, or a rotation. Those are, respectively
Each of these matrices can be related to each other by similarity transformation using the permutation matrices
Exponential matrix form for a Lorentz boost.
The next obvious thing to try with this matrix representation is a Lorentz boost.
where , and .
This matrix has a spectral decomposition given by
Taking logs and computing we have
Again we have one of the Pauli spin matrices. This time it is
So we can write our Lorentz boost 2.17 as just
By inspection again, we can come full circle by inspection from this last hyperbolic representation back to the original explicit matrix representation. Quite nifty!
It occurred to me after the fact that the Lorentz boost is not Unitary. The fact that the eigenvalues are not a purely complex phase term, like those of the rotation is actually a good hint that looking at how to characterize the eigenvalues of a unitary matrix can be used to show that the matrix is Hermitian.
 BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.