Relation between Black-Scholes equation and quantum mechanics

Question

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?

Answer 1

The Black-Scholes equation is given by,

$$\frac{\partial V}{\partial t}+ \frac{1}{2}\sigma^2S^2 \frac{\partial^2 V}{\partial S^2} + rs \frac{\partial V}{\partial S}-rV=0$$

If we write the equation in the form $\partial_tV=\dots$ and complete the square, we obtain,

$$\frac{\partial V}{\partial t}= -\frac{1}{2}\sigma^2 \left[ S\frac{\partial }{\partial S}-\frac{1}{2}\left( 1-\frac{2r}{\sigma^2} \right)\right]^2V+\frac{1}{8}\sigma^2 \left( 1+\frac{2r}{\sigma^2}\right)^2V$$

Set the momentum operator $p=-i\sigma(\partial_x -(1-2r/\sigma^ 2)/2)$ with $x=\log S$ and,

$$U=\frac{1}{8}\sigma^2\left( 1+\frac{2r}{\sigma^2}\right)^2$$

allows us to express the equation in the form of a Schrödinger equation, namely,

$$\partial_t V =\left[ \frac{1}{2}p^2 + U\right]V=HV$$

The equation is a Wick rotated expression for a charged particle in an electromagnetic field, with $V$ now having the interpretation of a gauge field. Notice the commutator,

$$[x,p]=i\sigma$$

implies the volatility $\sigma$ has the interpretation of the reduced Planck's constant, $\hbar$. However, in more realistic scenarios, often $S$ is a stochastic process $S_t$, as stock price fluctuates almost randomly. Hence the Black-Scholes becomes a stochastic differential equation, and then this duality to the Schrödinger equation sort of breaks down because of course $\psi$ is a regular function, not dependent on a stochastic process, even though what it describes may be random.