Hermitian and unitary operators are closely related. Both are diagonalizable with an orthonormal basis of eigenvectors. The only difference is that the eigenvalues of a Hermitian operator are real, while those of a unitary operator lie on the unit circle. In other words, if $H$ is Hermitian, we can write $$H = P D P^{-1}$$ where $P$ is a unitary matrix and $D$ is real, and if $T$ is unitary we can write $T = PDP^{-1}$ where $P$ is unitary and $$ D = \begin{pmatrix} e^{i \theta_1} & 0 & \cdots & 0 \cr 0 & e^{i \theta_2}& \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & \cdots & e^{i\theta_n} \end{pmatrix} $$ In particular, if $H$ is a real symmetric matrix, then the eigenvectors are real, and the matrix $P$ is orthogonal, and can be chosen to be a rotation. That is, the eigenvectors of $H$ are obtained by just rotating the standard coordinate axes.

If we take a Hermitian matrix $H$ with eigenvalues $\lambda_1, \ldots, \lambda_n$, then $e^{iH}$ is a matrix with the same (orthonormal!) eigenvectors as $H$, and with eigenvalues $e^{i\lambda_1}, \ldots, e^{i\lambda_n}$. In other words the exponential of $i$ times a Hermitian matrix is a unitary matrix. Likewise, a unitary matrix with eigenvalues $\{e^{i\theta_j}\}$ is the exponential of $i$ times a Hermitian matrix with eigenvalues $\{\theta_j\}$. In the first video, we explore these relations between Hermitian and unitary matrices.

Let $SO(3)$ denote the set of all $3 \times 3$ orthogonal matrices with determinant $+1$, and let $\mathfrak{so}(3)$ be the set of anti-symmetric real $3 \times 3$ matrices. In the second video we show that $SO(3)$ is a group, and that every element of $SO(3)$ is the exponential of an element of $\mathfrak{so}(3)$. $\mathfrak{so}(3)$ is called the Lie algebra of the Lie group $SO(3)$. Both $SO(3)$ and $\mathfrak{so}(3)$ come up a lot in physics, especially quantum mechanics.