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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 to prove analytically that the first column of $\exp(M)$ has the stated form?

[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 precisely this Hamiltonian on the subspace spanned by $\{\left|nn\right>\}$. I've tried using a BCH approach, but was discouraged by the fact that the commutators are not very cooperative. The exponential coefficients in the first column indicates (roughly) a thermal state.]

Edit: I cross-posted this on mathoverflow, and got a beautiful solution by Jochen Glueck. The exact value of $\lambda$ is $-\ln(\tanh(1))$.

Yly
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  • Trying this numerical for the $3 \times 3$ version, I do not get the same result – Ben Grossmann Feb 21 '18 at 00:16
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    That being said, it might be helpful to consider the first column of $\exp(Mt)$ (function of $t$ for $t \geq 0$) as the unique solution to the linear differential equation $$ \frac {d \mathbf x}{dt} = M \mathbf x, \qquad \mathbf x(0) = (1,0,\dots,0)^T $$ – Ben Grossmann Feb 21 '18 at 00:17
  • (see the $3 \times 3$ computation on WA here) – Ben Grossmann Feb 21 '18 at 00:17
  • Oh, I missed that this was an infinite matrix earlier... sorry – Ben Grossmann Feb 21 '18 at 01:21
  • @Omnomnomnom I added the word "infinite" in the description to hopefully avert further confusion. – Yly Feb 21 '18 at 04:44
  • @Omnomnomnom For the numerical stuff, I of course had to truncate the matrix. Roughly the first half of the first column for a truncated matrix has the stated form. So if you want to see this behavior for 10 elements, you should truncate to at least a 20 by 20 matrix. – Yly Feb 21 '18 at 04:49
  • I tried to find the spectrum of $H$ since if it was discrete (with analytic eigenvectors) one could write $H=U^\dagger DU$ with $U$ unitary, $D=\operatorname{diag}(\lambda_1,\lambda_2,\ldots)$ containing the eigenvalues of $H$. That would mean $\exp(H)=U^\dagger\operatorname{diag}(\exp(\lambda_1),\exp(\lambda_2),\ldots)U$. Sadly the spectrum of $H$ does not behave as nicely as the usual Hamiltonian of the harm. oscillator $H'=\operatorname{diag}(\frac12,\frac32,\ldots)$ as so far I only could show that $\sigma_p(H)\subseteq(-1,1)$ so I don't think my approach will be of too much use here. – Frederik vom Ende Feb 21 '18 at 10:24
  • @Yly Good to know – Ben Grossmann Feb 21 '18 at 12:18
  • Cross posting this on MO: https://mathoverflow.net/questions/293637/matrix-exponential-containing-a-thermal-state – Yly Feb 22 '18 at 18:37

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