Matrix exponential

We have $A^k=SDS^{-1}SDS^{-1}\cdots SDS^{-1}=SD^kS^{-1}$, since the intermediate factors $S^{-1}S$ cancel. By linearity, $e^A=S\bigl(\sum_kD^k/k!\bigr)S^{-1}=Se^DS^{-1}$, and the exponential of a diagonal matrix is the diagonal matrix of the exponentials. Thus $\det e^A=\det e^D=\prod_ke^{\lambda_k}=e^{\sum_k\lambda_k}=e^{\tr A}$, since the trace is the sum of the eigenvalues. This result remains true for any matrix, by density of the diagonalisable matrices.

We have $N^2=0$, so all powers $N^k$ with $k\geq2$ vanish, and the series terminates:

$$\begin{aligned} e^{tN}=\mathbf{1}+tN=\begin{pmatrix}1&t\\0&1\end{pmatrix}. \end{aligned}$$

The matrix $N$ is not diagonalisable, and the method of the previous question does not apply; but the direct calculation is immediate here. We check that $\det e^{tN}=1=e^{\tr(tN)}$.

Since $A^2=\mathbf{1}$, we have $A^{2k}=\mathbf{1}$ and $A^{2k+1}=A$. Separating the even and odd terms in the series,

$$e^{i\theta A}=\sum_k\frac{(i\theta)^{2k}}{(2k)!}\mathbf{1}+\sum_k\frac{(i\theta)^{2k+1}}{(2k+1)!}A=\cos\theta\,\mathbf{1}+i\sin\theta\,A,$$

since $i^{2k}=(-1)^k$ and $i^{2k+1}=i(-1)^k$. Without the factor $i$, we similarly obtain the series for $\cosh$ and $\sinh$. For $A=\sigma_x$,

$$\begin{aligned} e^{i\theta\sigma_x}=\begin{pmatrix}\cos\theta&i\sin\theta\\ i\sin\theta&\cos\theta\end{pmatrix},\qquad e^{\sigma_x}=\begin{pmatrix}\cosh1&\sinh1\\ \sinh1&\cosh1\end{pmatrix}. \end{aligned}$$

The first matrix is unitary; the second is not.

From the second question, $e^N=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, and similarly $e^{N^{\mathsf T}}=\begin{pmatrix}1&0\\1&1\end{pmatrix}$. Their product is

$$\begin{aligned} e^Ne^{N^{\mathsf T}}=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}2&1\\1&1\end{pmatrix}. \end{aligned}$$

Now $N+N^{\mathsf T}=\sigma_x$, and from the previous question $e^{\sigma_x}$ has entries $\cosh1\simeq1{,}543$ and $\sinh1\simeq1{,}175$. The two matrices are different: in general, $e^Ae^B\neq e^{A+B}$. The reason is that $N$ and $N^{\mathsf T}$ do not commute. If $AB=BA$, we may expand $(A+B)^k$ using the binomial formula as for numbers, and then show that $e^{A+B}=e^Ae^B$. This observation is essential in quantum mechanics: the exponential of a sum of noncommuting operators, such as $e^{-i(\hat P^2/2m+V(\hat X))t/\hbar}$, does not factor into a product of exponentials.

Diagonalise $\hat H$ in an orthonormal basis: $\hat H=VDV^\dagger$, with $V$ unitary and $D$ real diagonal. Then $U(t)=Ve^{-iDt}V^\dagger$, and $U(t)^\dagger=Ve^{iDt}V^\dagger$, since the complex conjugates of the diagonal entries $e^{-i\lambda_kt}$ are $e^{i\lambda_kt}$. Thus $U(t)^\dagger U(t)=Ve^{iDt}e^{-iDt}V^\dagger=\mathbf{1}$. Moreover, $-i\hat Ht$ and $-i\hat Hs$ commute, since they are multiples of the same matrix; from the previous question, $U(t)U(s)=e^{-i\hat H(t+s)}=U(t+s)$. This is the group law for the time-evolution operator.