This [question was originally posted on MSE](https://math.stackexchange.com/questions/2659397/matrix-exponential-containing-a-thermal-state), and I'm cross posting it here. 


Define an infinite matrix $$ M = 
\begin{bmatrix}
0 & -1 & 0 & 0 & \cdots \\
1 & 0 & -2 & 0 & \cdots \\
0 & 2 & 0 & -3 & \cdots \\
0 & 0 & 3 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$


Numerically, I've found that the first column of $\exp(M)$ is given by $\alpha(1,e^{-\lambda},e^{-2\lambda},e^{-3\lambda},\dots)^T$, where $\lambda \approx 0.27$ and $\alpha = \sqrt{1-e^{-2\lambda}} \approx 0.65$. 



> 
> **Question:** How can we prove analytically that the first column of $\exp(M)$ has the stated form?
> 
> 
> 




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**Context:** This question comes from a quantum mechanical model of two harmonic oscillators coupled by a Hamiltonian of the form $\hat{H} \propto \hat{a}\_1^\dagger \hat{a}\_2^\dagger - \hat{a}\_1 \hat{a}\_2$, where $\hat{a}\_i$ is the lowering operator for oscillator $i$. The matrix $M$ is essentially this Hamiltonian on the subspace spanned by $\{\left|nn\right>\}$, and so $\exp(M)$ is a time evolution operator. The exponentially decaying coefficients in the first column indicates that either of the individual harmonic oscillators (after tracing out the other one) is in a thermal state.


I've tried several tricks with the [BCH formula](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula), but was discouraged by the fact that the commutators are not very cooperative. 



Matrix exponential, containing a thermal state
It is actually easier to compute the first column of $e^{tM}$ for each $t \in \mathbb{R}$ rather than computing only the first column of $e^M$.


Indeed, let $u(t) = e^{tM}e\_1$ for each $t \in \mathbb{R}$, where $e\_1$ denotes the first canonical unit vector. Then $u(t)$ is the first column of $e^{tM}$.


**Claim:** For every $k \in \mathbb{N} := \{1,2,\dots\}$ and every $t \in \mathbb{R}$ we have
\begin{align\*}
 u\_k(t) = \frac{1}{\cosh(t)} \tanh(t)^{k-1}. \tag{\*}
\end{align\*}
**Proof.** Let $v\_k(t)$ denote the right handside of $(\*)$ for each $k$ and each $t$. A short computation shows that $\dot v\_k(t) = (k-1)v\_{k-1}(t) - k v\_{k+1}(t) = (Mv(t))\_k$ for each $k \in \mathbb{N}$ and each $t \in \mathbb{R}$ (where we assign an arbitary value to $v\_0(t)$ to make sure that the equality also makes sense for $k = 1$). Hence, $v(t)$ solves the differential equation $\dot v(t) = Mv(t)$. Moreover, we have $v(0) = e\_1$, so indeed $v(t) = e^{tM}e\_1 = u(t)$ for all $t \in \mathbb{R}$.


**Remarks:**


(1) In the above "proof" I was a bit sloppy about properties of "unique solutions of differential equations". More precisely: (i) I only computed the derivative of $v(t)$ componentwise, which does not show that the mapping $\mathbb{R} \ni t \mapsto v(t) \in \ell^2$ is differentiable with respect to the norm in $\ell^2$. (ii) The "matrix" $M$ is not a continuous linear mapping, but an unbounded linear operator on $\ell^2$, so to make precise assertions about "unique solvability" of initial value problems, one needs to employ the theory of $C\_0$-semigroups. 


However, given the context of the question, I don't think that this is the most relevant point here, so I guess there's no need to go into detail here.


(2) Concerning the question how to find the solution presented in $(\*)$: The numerical computation of the OP suggests that $u(t)$ is of the form $u\_k(t) = g(t) f(t)^{k-1}$ for two functions $f$ and $g$. The differential equation for $u(t)$ yields the differential equation
\begin{align\*}
 \dot f(t) = 1 - \frac{k}{k-1} f(t)^2 - \frac{1}{k-1} \frac{\dot g(t)}{g(t)} f(t) \tag{A}
\end{align\*}
for $k \ge 2$ and the differential equation
\begin{align\*}
 -f(t) = \frac{\dot g(t)}{g(t)} \tag{B}
\end{align\*}
for $k = 1$ (provided that $g(t)$ is non-zero). Plugging $(B)$ into $(A)$, we obtain $\dot f(t) = 1 - f(t)^2$, which is solved by $\tanh(t)$. Then we can plug this into $(B)$ again and compute $g(t) = \frac{1}{\cosh(t)}$.


(3) By setting $t = 1$ we obtain $u\_k(1) = \frac{1}{\cosh(1)} \tanh(1)^{k-1}$, which is exactly the solution suggested by the OP's numerical experiment (where $\alpha = \frac{1}{\cosh(1)} \approx 0.6481$ and $\lambda = -\ln(\tanh(1)) \approx 0.2723$).


