I am having great difficulty checking a step involving operations with 2×2 matrix exponentials.
The expression I would like to simplify is
$$\lim _{x\to \infty} e^{-iHt-Vt}$$
where, for some $\epsilon, x, y \in \mathbb{R}$, we define
$$H=\begin{bmatrix}
0 & \epsilon \\
\epsilon & 0
\end{bmatrix} \text{ }\text{ }\text{ }\text{ }\text{ }
V=\begin{bmatrix}
x & 0 \\
0 & y
\end{bmatrix}$$
the undecipherable line says "we easily find that this expression equals
$$\lim _{x\to \infty} e^{-iHt-Vt}=e^{-yt}\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}$$
…". After more than an hour of work, I am not convinced it is so easy. Any assistance or ideas would be of great value.
The necessary information is all on this post, but here is an image of the source if it is helpful to anyone. That image is a standalone section. I was able to prove every step until this one (used BCH formula for A6).
Edit: Here is a previous approach which seems to have gotten to the wrong conclusion (maybe limit composition is illegal in this situation?). The last steps uses the same property as in Omnomnomnom's answer for exponentiating projectors.
$$\lim _{x\to \infty} e^{-iHt-Vt}$$
$$=\lim_{x\to \infty} (e^{-itH/x-t P_- -ytP_+ /x})^x$$
$$\to \lim_{x\to \infty} (e^{-t P_-})^x$$
$$=\lim_{x\to \infty} (I+(e^{-tx}-1)P_-)$$
$$\to P_+ \neq e^{-yt} P_+$$
where $P_-, P_+$ are diag(1,0) and diag(0,1) respectively.
Best Answer
It would be equivalent to show that $$ \lim _{x\to \infty} e^{yt}e^{-iHt-Vt}=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. $$ With that said, we have $$ \exp(yt)\exp(-iHt-Vt) = \exp(Iyt)\exp(-iHt-Vt) = \exp(Iyt - iHt - Vt) =\\ \exp\left[t\pmatrix{y - x & -i\epsilon\\ -i\epsilon & 0} \right]. $$ In other words: it suffices to show that for $k,z \in \Bbb R$, we have $$ \lim_{z \to \infty} \exp \pmatrix{-z&-ik\\-ik&0} = \pmatrix{0&0\\0&1}. $$ Take $Q(z,k) = \pmatrix{-1&-\frac{ik}{z}\\ - \frac{ik}z & 0}$. We find that $Q$ has eigenvalues $$ \lambda_\pm = \frac{-1 \pm \sqrt{1 - 4 \epsilon^2/z^2}}{2}. $$ Note that for large $z$, $\lambda_+ \approx 0$ and $\lambda_- \approx -1$. We can write $$ Q(z,k) = \lambda_+ P_+ + \lambda_-P_-, $$ where $P_\pm$ denote the projections onto the eigenspaces corresponding to $\lambda_{\pm}$. By the continuity of eigenvalues/eigenvectors, we have $$ \lambda_-(z) \to -1,\quad z\lambda_+(z) \to 0, \quad P_-(z) = \pmatrix{1&0\\0&0}, \quad P_+(z) = \pmatrix{0&0\\0&1}. $$
Note that $P_{\pm}$ commute. So, we have $$ \lim_{z \to \infty} \exp \pmatrix{-z&-ik\\-ik&0} = \lim_{z \to \infty} \exp(z\lambda_+ P_+ + -z\lambda_-P_-) = \\ \lim_{z \to \infty} \exp(z\lambda_+ P_+) \exp(-z\lambda_-P_-) = \\ \lim_{z \to \infty} \exp(z\lambda_+ P_+) \cdot \lim_{z \to \infty} \exp(z\lambda_-P_-) $$ Now, it suffices to note that for any projection $P$, we have $\exp(tP) = I + (\exp(t) - 1)P$. Thus, $$ \begin{align} \lim_{z \to \infty} \exp(z\lambda_+ P_+) &= \lim_{z \to \infty} I + (\exp(z \lambda_+) - 1)P_+ \\ &= \lim_{z \to \infty} I + (\exp(z \lambda_+) - 1)\pmatrix{0&0\\0&1} = I\\ \lim_{z \to \infty} \exp(z\lambda_- P_-) &= \lim_{z \to \infty} I + (\exp(z \lambda_-) - 1)P_-\\ &= \lim_{z \to \infty} I + (\exp(z \lambda_-) - 1)\pmatrix{1&0\\0&0} = \pmatrix{0&0\\0&1}. \end{align} $$ The conclusion follows.
An abbreviation of the above derivation: note that
\begin{align} \lim_{x \to -\infty}\exp\left[t\pmatrix{y - x & -i\epsilon\\ -i\epsilon & 0} \right] &= \lim_{x \to \infty}\exp\left[x t\pmatrix{\frac{y - x}x & \frac{-i\epsilon}x\\ \frac{-i\epsilon}x & 0} \right] \\ & = \lim_{x \to \infty}\exp\left[x t\lim_{x \to \infty}\pmatrix{\frac{y - x}x & \frac{-i\epsilon}x\\ \frac{-i\epsilon}x & 0} \right] \\ & = \lim_{x \to \infty}\exp\left[x t\lim_{x \to \infty}\pmatrix{-1&0\\0&0} \right] = \pmatrix{0&0\\0&1}, \end{align} which was what we wanted.