Is the circuit unitary in Nielsen's method for calculating complexity?

Question

I was trying to learn how to calculate circuit complexity (1707.08570) when I chanced upon a seemingly confusing concept. The "Nielsen method" involves looking at a unitary transformation $U$ that takes a reference state $\Psi^R$ to a target state $\Psi^T$: $$|\Psi^T(t)\rangle=U(t)|\Psi^R(0)\rangle$$ The idea is to rewrite $U$ as an optimized circuit of fundamental quantum gates, and finding the cost associated with this circuit. Examples of these gates include : \begin{aligned} H \psi\left(x_{1}, x_{2}\right) &=e^{i \epsilon p_{0} x_{0}} \psi\left(x_{1}, x_{2}\right) & & \text { (global) phase change } \\ J_{1} \psi\left(x_{1}, x_{2}\right) &=\psi\left(x_{1}+\epsilon x_{0}, x_{2}\right) & & \text { shift } x_{1} \text { by constant } \epsilon x_{0} \\ K_{1} \psi\left(x_{1}, x_{2}\right) &=e^{i \epsilon p_{0} x_{1}} \psi\left(x_{1}, x_{2}\right) & & \text { shift } p_{1} \text { by constant } \epsilon p_{0} \\ Q_{21} \psi\left(x_{1}, x_{2}\right) &=\psi\left(x_{1}+\epsilon x_{2}, x_{2}\right) & & \text { shift } x_{1} \text { by } \epsilon x_{2} \quad \text { (entangling gate) } \\ Q_{11} \psi\left(x_{1}, x_{2}\right) &=e^{\epsilon / 2} \psi\left(e^{\epsilon} x_{1}, x_{2}\right) & & \text { scale } x_{1} \rightarrow e^{\epsilon} x_{1} \quad \text { (scaling gate) } \end{aligned} where $\epsilon$ is an infinitesimal parameter. In (1810.02734), this procedure leads to the following form of $U$ (Eq 3.38): $$U(\tau)=\exp \left(\sum_{k=0}^{N-1} \alpha^{k}(\tau) M_{k}^{\text {diag }}\right)$$ where $\{\alpha^k\}$ are complex (and not infinitesimal) and $\{M_k\}$ are diagonal matrices with an identity in the $k^{th}$ position (rest are zero). How can I show that $U$ is indeed unitary as per the original assumption?

Edit : There was a dubious error in the question as was pointed out by @Jahan Claes, regarding unitarity of the gate $Q_{aa}$. I have edited the question accordingly.

Answer 1

Let's compute $Q_{aa}^\dagger Q_{aa}$. For the rest of this post, I'll drop the $a$ subscript for convenience. Remember that $[x,p]=i$, so that we have $px=xp-i$.

We have: \begin{align} Q^\dagger Q & = e^{\epsilon}(e^{i\epsilon xp})^\dagger e^{i\epsilon xp}\\ & = e^{\epsilon}e^{-i\epsilon px}e^{i\epsilon xp}\\ &= e^\epsilon e^{-i\epsilon(xp-i)}e^{i\epsilon xp}\\ &= e^\epsilon e^{-\epsilon}e^{-i\epsilon xp}e^{i\epsilon xp}\\ &=1 \end{align}

So $Q$ is indeed unitary!