Relation between Black-Scholes equation and quantum mechanics

I am interested in the link between the Black & Scholes equation and quantum mechanics.

I start from the Black & Scholes PDE $$ \frac{\partial C}{\partial t} = -\frac{1}{2}\sigma^2 S^2 \frac{\partial^2C}{\partial S^2} -rS\frac{\partial C}{\partial S}+rC $$ with the boundary condition $$C(T,S(T)) = \left(S(T)-K\right)_+.$$ Performing the change of variables $q=\ln(S)$ this equation rewrites $$ \frac{\partial C}{\partial t} = H_{BS}C $$ with the Black & Scholes Hamiltonian given by $$H_{BS} = -\frac{\sigma^2}{2} \frac{\partial^2}{\partial q^2}+\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r.$$

Now I compare this equation with the Schrödinger equation for the free particle of mass $m$ : $$i\hbar \frac{d\psi(t)}{dt} = H_0\psi(t),\quad \psi(0)=\psi$$ with the Hamiltonian (in the coordinate representation) given by $$H_0 = -\frac{\hbar^2}{2m} \frac{d^2}{dq^2}.$$

My problem comes from the fact that the various references I am reading for the moment explain that the two models are equivalent up to some changes of variables (namely $\hbar=1$, $m=1/\sigma^2$ and the physical time $t$ replaced by the Euclidean time $-it$). However, their justifications for the presence of the terms $$\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r$$ in the Hamiltonian seem very suspicious to me. One of them tells that these terms are "a (velocity-dependent) potential". Another one tells this term is not a problem since it can be easily removed is we choose a frame moving with the particle.

I have actually some difficulties to justify why, even with this term, we can say that the Black & Scholes system is equivalent to the one coming from the quantum free particle. I don't like this potential argument, since (for me) a potential should be a function of $q$ (so it would be ok for example for the $+r$ term) but not depending on a derivative.

Could you give me your thoughts on this problem?