determined by the matrix $M_k(p)$ can be chosen arbitrary as well such that we can use it to enforce $\partial_k h_\Phi(p)=0$, i.e. $$h_\Phi(p,\vec q, \vec q^\prime)=V(\vec q-\vec q^\prime) \quad \text{for all } k.$$ In summary, we use $$\begin{aligned}
\nonumber
\partial_k Q^{-1} &=& -\partial_k A_k^{-1},\\
\nonumber
\partial_k M_k &=& (\partial_k A_k^{-1})^\dagger(P_\Phi+V)^\dagger,\\
\partial_k M_k^\dagger &=& (P_\Phi+V) (\partial_k A_k^{-1}).
\label{eq:dkQdkM}\end{aligned}$$ For $P_\Phi$ this leads to the flow equation $$\begin{aligned}
\nonumber
\partial_k P_\Phi &=& - V (\partial_k A_k^{-1}) V + P_\Phi (\partial_k A_k^{-1}) P_\Phi\\
\nonumber
&& - (P_\Phi + V)(\partial_k A_k^{-1}) P_\Phi - P_\Phi (\partial_k A_k^{-1})(P_\Phi+V)\\
&=& - (P_\Phi+V)(\partial_k A_k^{-1})(P_\Phi+V)\end{aligned}$$ which is solved by $$P_\Phi = A_k - V.$$ Using the definition of $A_k$ in Eq. , we see that, as expected, the zero crossings of $P_\Phi$ at $k=0$ correspond to the solution of Schrödingers equation for the two particle problem. Note that we can now also determine via Eqs. and the $k$-dependent matrix $Q$ in the Hubbard-Stratonovich transformation .

To summarize, we found the solution to the flow equation by first adjusting the scale-dependent Hubbard-Stratonovich transformation $\partial_k Q^{-1}$ such that no additional term $\lambda_\psi$ is generated by the flow. All contributions to an interaction of this form are dynamically expressed by bound state 