I am trying to derive the following result from this paper (Equation S3):$$e^{-ik_z\hat{z}}\hat{\rho}e^{+ik_z\hat{z}} - \hat{\rho} \approx k_z^2[\hat{z},[\hat{z}, \hat{\rho}]]$$with operators $\rho, z$.
The authors state the expansion below is sufficient to arrive at the result. The details in the paper are not relevant to understand the problem, it suffices to know that we deal with operators.
The exponential operators can be expanded to the lowest order in the one-dimensional coordinate z in the present case
I then did the expansion by using the series expansion of the exponential function ($\sum_0^{\infty}\frac{(ik)^n}{n!}z^n$) $$e^{ik_z\hat{z}} \approx \mathbb{I} + ik_z\hat{z}$$ which got me $$e^{-ik_z\hat{z}}\hat{\rho}e^{+ik_z\hat{z}} - \hat{\rho} \approx k_z^2z\rho z + i_kz[z, \rho]$$Evidently, this is not the result from the paper. I also checked the BCH-formula and found that my result would be correct if $[z, \rho]$ would commute with either $\rho$ or $z$. This can't be the case here as otherwise, the first expression would be zero.
Where is my mistake?
EDIT
I am aware that the matrix exponential only holds for square matrices, so it is questionable if I can apply the series expansion as I did above. If this is indeed the mistake, what would be the correct expansion?
If you're going up to second order in $k$, your exponential expansion also needs to be second order. This gives \begin{multline} e^{-ikz}\rho e^{ikz} = \left(\mathbb I -ikz -\frac{k^2z^2}{2} + ...\right)\rho\left(\mathbb I +ikz -\frac{k^2z^2}{2} + ...\right) \\= \rho-ikz\rho+ik\rho z + \frac{k^2}{2}\left(z^2\rho -2z\rho z + \rho z^2\right)+ ... = \rho - ik[z,\rho]+\frac{k^2}{2}[z,[z,\rho]] + ..., \end{multline} in agreement with this answer. So we should get $$ e^{-ikz}\rho e^{ikz} -\rho \approx -ik[z,\rho] + \frac{k^2}{2}[z,[z,\rho]]. $$ I'm not sure where the commutator term goes. Perhaps there's something in the paper that justifies neglecting it?