Matrix exponential of infinite antisymmetric matrix with entries only next to its diagonal What is the exponential $\exp (t A)$ of the operator $A$ whose components are given by
$A_{nm} = \delta_{nm-1} \sqrt{n+1} - \delta_{nm+1}\sqrt{n}$
where the $n,m \in \mathbb{N}_0$. If we just consider indices up to 1, the answer is easy. However, it doesn't seem to be straight forward to generalize to infinite dimensions. I tried to do it with mathematica by increasing the dimension, but it didn't seem to converge.

Define the infinite matrix
$$ A_{n,m} := \delta_{n,m-1} \sqrt{n+1} - \delta_{n,m+1}\sqrt{n} \tag{1} $$
where $\,n,m\ge 0.\,$ Given a variable $\,t\,$ define another infinite matrix
$$ B := \exp(t\,A) = \sum_{n=0}^\infty A^n \frac{t^n}{n!}. \tag{2} $$
The result is that
$$ B_{n,m} = e^{-t^2/2}\,t^{m-n}\sqrt{\frac{m!}{n!}}\sum_{k=0}^n
  (-1)^k{n \choose k} \frac{t^{2k}}{(k+m-n)!} \tag{3}. $$
Note that $\, B_{n,m} = (-1)^{n+m}B_{m,n}. $
