Given a subdiagonal matrix (every element is zero except the elements directly below the main diagonal), is there an analytic form to calculate the elements of the matrix exponential?
In particular, I want an analytic expression of the elements $m_{ij}$ of $M = e^{\alpha\hat{a}^\dagger}$ where $\alpha$ is an arbitrary complex number, and $\hat{a}^\dagger$ is the $N\times N$-matrix
$$
a^{\dagger} = \begin{pmatrix}
0 & 0 & 0 & \dots & 0 \\
\sqrt{1} & 0 & 0 & \dots & 0\\
0 & \sqrt{2} & 0 & \dots & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
0 & 0 & \dots & \sqrt{N-1} & 0 \\
\end{pmatrix},
$$
which is also known as the creation operator.
Alternatively, if there are analytic expressions for $P$ and $\lambda$ in the eigen-decomposition $\alpha\hat{a}^{\dagger} = P \lambda P^{-1}$, then that approach might suffice.
Best Answer
Let $D=\operatorname{diag}(\sqrt{0!},\sqrt{1!},\ldots,\sqrt{n!})$ and let $J$ be the lower triangular nilpotent Jordan block of size $n$ (i.e. the entries on the first subdiagonal of $J$ are ones and all other entries are zero). Then $a^\dagger=DJD^{-1}$. Therefore $$ M:=e^{\alpha a^\dagger}=De^{\alpha J}D^{-1}=D\left(\sum_{k=0}^{n-1}\frac{\alpha^k}{k!}J^k\right)D^{-1}. $$ Since $(J^k)_{ij}=1$ when $i-j=k$ and zero elsewhere, $M$ is a lower triangular matrix with $$ m_{ij}=\frac{\alpha^{i-j}}{(i-j)!}\frac{d_{ii}}{d_{jj}}=\frac{\alpha^{i-j}}{(i-j)!}\sqrt{\frac{i!}{j!}} $$ when $i\ge j$.