TITLE: Matrix exponential for non-commutative operator entries of matrix
QUESTION [5 upvotes]: I would like to find the matrix exponential $e^{iHt}$ of the Hermitian matrix $H$ where
$$
H=\begin{pmatrix}
\delta& \sqrt{2}a & 0\\
\sqrt{2}a^\dagger &0& \sqrt{2}a\\
0 &\sqrt{2}a^\dagger&-\delta
\end{pmatrix}
$$
$$[a,a^\dagger]=a a^\dagger-a^\dagger a=1$$
$$[a,\delta]=[a^\dagger,\delta]=0.$$
The dagger $\dagger$ denotes complex conjugate. My thought is to make use of the definition of matrix exponential $e^{X}=\sum_{k=0}^{\infty} \frac{1}{k !} X^{k}$ and calculate $H^2$ and $H^3$ to find some pattern to exploit. For example, in the special case where $\delta=0$, we can find a diagonal matrix $D$
$$
D=\begin{pmatrix}
4a^\dagger a+6 &0&0\\
0&4a^\dagger a+2&0\\
0&0&4a^\dagger a-2
\end{pmatrix}
$$
such that $H^{2n+1}=D^{n}H$ and $H^{2n+2}=D^{n}H^2$, with which we can obtain a neat result of $e^{i H t}$. However, it seems that no such diagonal matrix exists for $\delta \neq 0$ case. Are there any other ways to calculating matrix exponential like this with non-commuting entries? Any input is much appreciated!

REPLY [2 votes]: A tip of the hat to @Semiclassical 's insightful suggestion. Now that he has unravelled the group structure of H through the symmetry operator N, you may classify its eigenstates of eigenvalue n and specify them further to find eigenstates of the hamiltonian.
Recall $a^\dagger |n\rangle=\sqrt{n+1}|n+1\rangle$ and $a|n\rangle= \sqrt{n}|n-1\rangle$, so impose $H V(n)=\epsilon(n) ~ V(n)$ to fix the labile coefficients b and c for the unnormalized N eigenstate,
$$
V(n)\equiv \begin{pmatrix} \sqrt{2} b(n) |n-1\rangle \\ |n\rangle \\\sqrt{2} c(n) | n+1\rangle    \end{pmatrix}.
$$
The Hamiltonian eigenvalue condition then restricts
$$
\delta b +\sqrt{n}=\epsilon b \\
2\sqrt{n} b + 2\sqrt{n+1} c = \epsilon\\
\sqrt{n+1}-\delta c =\epsilon c, \leadsto \\
b=\frac{\sqrt n}{\epsilon -\delta}, ~~ c=\frac{\sqrt {n+1}}{\epsilon +\delta} ,\\
\epsilon (\epsilon^2 -\delta^2 -4n-2)+2\delta =0.
$$
The (readily depressed!) cubic equation may be solved for $\epsilon$, but I won't go into it... For vanishing δ, you have
$$
\epsilon =\pm \sqrt{4n+2},\\
b(n) = 1/\sqrt{4+2/n}, ~~~ c(n)= 1/\sqrt{4-2/(n+1)},
$$
so that
$V(n,t) = \exp \left( \pm it \sqrt{4n+2}\right )~~ V(n,0) $, etc. Even if $D-H^2\neq 0$, $V(n)$ is in its kernel.
I have not fully explored the non-vanishing δ solutions of the (depressed!) cubic.
If you wished to find the operator exponentiation (but why?) ,
you could write its spectral resolution in terms of these states $V(n)$, after you normalize them.