Matrix exponential of a subdiagonal matrix

Question

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.

Answer 1

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 $i\ge j$.

Answer 2

An $n\times n$ subdiagonal matrix $A$ has all elements of $A^k$ zero except along the diagonal that is $k$ below the main diagonal. This means we have $A^n=0$, and the finite sum:

$$e^{\alpha A}=\sum_{k=0}^{n-1}\frac{\alpha^k}{k!}A^k$$

The entires along the $k$th subdiagonal of $A^k$ are given by the product of $k$ consecutive terms in $A$'s subdiagonal. That is, for your case,

$$A^k=\begin{bmatrix}0&\cdots&\cdots&\cdots&0\\\vdots&\ddots&\ddots&\ddots&\vdots\\0&\ddots&\ddots&\ddots&\vdots\\\sqrt{1\times\dots\times k}&0&\ddots&\ddots&\vdots\\0&\sqrt{2\times\dots\times(k+1)}&\ddots&\ddots&\vdots\\\vdots&0&\ddots&\ddots&\vdots\\\vdots&\ddots&\ddots&\ddots&0\\0&\cdots&\cdots&0&\sqrt{(n-k)\times\dots\times(n-1)}\end{bmatrix}$$

Which gives a direct closed form for your exponential (the $k$th subdiagonal of $e^{\alpha A}$ comes from $A^k$).

Answer 3

An example will help to understand the structure of the exponential of such a category of matrices (a particular case of nilpotent matrix : $(a^{\dagger})^n=0$ for some $n$, here $n$ because we are dealing with with $6\times 6$ matrices).

Let us make a colored parallel between

1) the matrix exponential using definition :

$$\exp(a^{\dagger})= \color{cyan}{I}+ \color{blue}{a^{\dagger}}+ \color{magenta}{\tfrac12 (a^{\dagger})^2}+\color{green}{\tfrac16 (a^{\dagger})^3}+\color{red}{\tfrac{1}{24} (a^{\dagger})^4}+\tfrac{1}{120} (a^{\dagger})^5$$

(we stop here because $(a^{\dagger})^6=0$, giving $(a^{\dagger})^n=0$ for any $n \ge 6$).

2) and the different subdiagonals of the result :

$$\exp(a^{\dagger})=\begin{pmatrix} \color{cyan}{1}& 0& 0& 0& 0& 0 \\ \color{blue}{1}& \color{cyan}{1}& 0& 0& 0& 0 \\ \color{magenta}{\sqrt{2}/2}& \color{blue}{\sqrt{2}}& \color{cyan}{1}& 0& 0& 0 \\ \color{green}{\sqrt{6}/6}& \color{magenta}{\sqrt{6}/2}& \color{blue}{\sqrt{3}}& \color{cyan}{1}& 0& 0 \\ \color{red}{\sqrt{6}/12}& \color{green}{\sqrt{6}/3}& \color{magenta}{\sqrt{3}}& \color{blue}{2}& \color{cyan}{1}& 0 \\ \sqrt{30}/60& \color{red}{\sqrt{30}/12}& \color{green}{\sqrt{15}/3}& \color{magenta}{\sqrt{5}}& \color{blue}{\sqrt{5}}& \color{cyan}{1} \end{pmatrix}$$

exhibiting the fact that the powers of $a^{\dagger}$ make progressively "receding" contributions, each one on a specific subdiagonal, till a certain rank beyond which there is no longer any contribution.

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Edit : in fact, the preceding result can be cast into a more general one :

If $M$ has values $M_{k,k+1}=a_k$ and $0$ otherwise, its exponential is :

$$\begin{pmatrix} \color{cyan}{1}& 0& 0& 0& 0& 0 \\ \color{blue}{a_1}& \color{cyan}{1}& 0& 0& 0& 0 \\ \color{magenta}{\dfrac12 a_1a_2}& \color{blue}{a_2}& \color{cyan}{1}& 0& 0& 0 \\ \color{green}{\dfrac16 a_1a_2a_3}& \color{magenta}{\dfrac12 a_2a_3}& \color{blue}{a_3}& \color{cyan}{1}& 0& 0 \\ \color{red}{\dfrac{1}{24} a_1a_2a_3a_4}& \color{green}{\dfrac16 a_2a_3a_4}& \color{magenta}{\dfrac12 a_3a_4}& \color{blue}{a_4}& \color{cyan}{1}& 0 \\ \dfrac{1}{120} a_1a_2a_3a_4a_5 & \color{red}{\dfrac{1}{24} a_2a_3a_4a_5}& \color{green}{\dfrac16 a_3a_4a_5}& \color{magenta}{\dfrac12 a_4a_5}& \color{blue}{a_5}& \color{cyan}{1} \end{pmatrix}$$

giving at once the formula for the general entry of the exponential.