6 ms·
> Rotation matrices are hermitian Is that the case? Consider |a -b| |b a| with Eigenvalues lambda = a +/- i b a = cos theta and b = sin theta gives a rota
by electricslpnsld 7y ago
> Rotation matrices are hermitian
Is that the case? Consider
|a -b|
|b a|
with Eigenvalues
lambda = a +/- i b
a = cos theta and b = sin theta gives a rotation, so the Eigenvalues are complex.
- contravariant 7y agoYeah rotation matrix are not hermitian (note that you don't need to show the eigenvalues aren't real, you just need to show it's not self-conjugate). The OP may have been confused by the fact that you can make a rotation matrix of eigenvectors of a hermitian matrix, which diagonalises the original matrix into two conjugate rotation matrices with a diagonal matrix between them.
- cshimmin 7y agoD'oh you are correct! It's a bit late here. Anyways the rest of my point stands about the author's intent. To connect that to your original comment: a rotation is not "observable", instead, it's a transformation that modifies a physical system. In quantum mechanics, physical observables (energy, momentum, position, mass, etc) are always associated with Hermitian operators that act on the Hilbert space of states for the system.