Unitary operators

We have $\braket{\hat Uu}{\hat Uv}=\braket{u}{\hat U^\dagger\hat Uv}$. If $\hat U^\dagger\hat U=\mathbf{1}$, this is $\braket uv$. Conversely, if the inner product is preserved, $\braket{u}{(\hat U^\dagger\hat U-\mathbf{1})v}=0$ for all $\ket u$, hence $(\hat U^\dagger\hat U-\mathbf{1})\ket v=0$ for every $\ket v$, and $\hat U^\dagger\hat U=\mathbf{1}$. Taking $\ket u=\ket v$ shows that the norm is preserved. Finally, the $j$th column of the matrix of $\hat U$ in an orthonormal basis $(\ket{e_j})$ represents the vector $\hat U\ket{e_j}$, and $\braket{\hat Ue_j}{\hat Ue_k}=\braket{e_j}{e_k}=\delta_{jk}$: the columns are orthonormal. In finite dimension, $\hat U^\dagger\hat U=\mathbf{1}$ implies that $\hat U$ is invertible with inverse $\hat U^\dagger$, and therefore also $\hat U\hat U^\dagger=\mathbf{1}$.

If $\hat U\ket v=\lambda\ket v$ with $\ket v\neq0$, preservation of the norm gives $\norm v=\norm{\hat Uv}=|\lambda|\norm v$, hence $|\lambda|=1$: we may write $\lambda=e^{i\theta}$. If $\hat U\ket u=\lambda\ket u$ and $\hat U\ket v=\mu\ket v$, then $\braket uv=\braket{\hat Uu}{\hat Uv}=\lambda^*\mu\braket uv$. Since $|\lambda|=1$, $\lambda^*=1/\lambda$, and $\lambda^*\mu=\mu/\lambda\neq1$ if $\lambda\neq\mu$: hence $\braket uv=0$. Finally, $\det(U^\dagger U)=\det(U)^*\det(U)=|\det U|^2=\det\mathbf{1}=1$.

The columns $\frac{1}{\sqrt2}(1,i)$ and $\frac{1}{\sqrt2}(i,1)$ are normalised, and their inner product is $\frac12\bigl(1\cdot i+(-i)\cdot1\bigr)=0$: they form an orthonormal basis, and $U$ is unitary. Since $U$ has the form $\alpha\mathbf{1}+\beta\sigma_x$, its eigenvectors are those of $\sigma_x$, namely $\frac{1}{\sqrt2}(1,1)$ and $\frac{1}{\sqrt2}(1,-1)$. We find

$$U\begin{pmatrix}1\\1\end{pmatrix}=\frac{1+i}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}, \qquad U\begin{pmatrix}1\\-1\end{pmatrix}=\frac{1-i}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}.$$

The eigenvalues are $e^{i\pi/4}$ and $e^{-i\pi/4}$, of modulus $1$, and the eigenvectors are orthogonal. The determinant is $\frac12(1-i^2)=1$, the product of the two eigenvalues.

Since $\sigma_x^2=\mathbf{1}$, the even powers of $\sigma_x$ equal $\mathbf{1}$ and the odd powers equal $\sigma_x$. Separating the even and odd terms of the exponential series gives $e^{i\theta\sigma_x}=\cos\theta\,\mathbf{1}+i\sin\theta\,\sigma_x$. For $\theta=\pi/4$, $\cos\theta=\sin\theta=\frac{1}{\sqrt2}$, and we recover $U$. More generally, if $\hat H$ is self-adjoint, it can be diagonalised in an orthonormal basis, $\hat H=\sum_k\lambda_k\ket{k}\bra k$ with real $\lambda_k$, and $e^{i\hat H}=\sum_ke^{i\lambda_k}\ket k\bra k$. Its adjoint is $\sum_ke^{-i\lambda_k}\ket k\bra k$, and their product is $\sum_k\ket k\bra k=\mathbf{1}$. This is why the evolution operator $e^{-i\hat Ht/\hbar}$ is unitary.