What is the relation between the Black Scholes equation and the Schrodinger equation?

1) The Schrodinger Equation (Skip to section 2 if you are not into physics)

Let me derive step by step the Schrodinger equation. Schrodinger based most of his work on the special relativity theory of Albert Einstein, and the Maxwell’s Equations. In order to derive the equation we must revise some electromagnetics and relativity.

i) The four Maxwell Equations explain how electricity and magnetism behave. These equations are:

$\nabla\cdot E=\frac{\rho}{\epsilon_0}$ (The electric field that leaves a volume is proportional to the density of the charge inside)

$\nabla \cdot B=0$ (There are no magnetic monopoles)

$\nabla \times E=-\frac{\partial B}{\partial t}$ (The voltage accumulated around a closed circuit is proportional to the time rate of change of the magnetic flux it encloses.)

$\nabla \times B=\mu_0 (J+\epsilon_0 \frac{\partial E}{\partial t})$ (Electric currents are proportional to the magnetic field circulating about the are they pierce)

Now, let us see how these equations behave in the vacuum of space, where there is no charge permeability or diffusion medium.

$\nabla\cdot E=0$

$\nabla \cdot B=0$

$\nabla \times E=-\frac{\partial B}{\partial t}$

$\nabla \times B=\mu_0 \epsilon_0 \frac{\partial E}{\partial t}$

Applying the curl operator $\nabla \times$ to the third equation:

$\nabla \times (\nabla \times E)=-\frac{\partial \nabla \times B}{\partial t}=-\mu_0 \epsilon_0 \frac{\partial ^2}{\partial t^2}E$

Using  the following theorem:

$\nabla \times \nabla \times E =-\nabla^2 E+\nabla \cdot \nabla \cdot E$

Since $\nabla \cdot E=0$

$\nabla^2 E=\mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}E$

This is a three dimensional wave equation, assuming the wave only moves in one dimension gives us:

$\frac{\partial^2 E}{\partial x^2}=\mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}E$

Wave equations usually can be written in the form

$\frac{\partial^2 E}{\partial x^2}-\frac{1}{v^2} \frac{\partial^2 E}{\partial t^2}=0$

Where $v=\frac{1}{\sqrt{\mu_0 \epsilon_0}}=c$ is the speed of light in the vacuum.

Let us move now to some of the work of Albert Einstein, and derive the famous $E=mc^2$ formula formally.

The starting point is to usethe Lorentz transformation to understand the relativistic mass of an object. This means, the mass of the object when its speed approaches the speed of light. The relation between the relativistic mass $m$ and its rest or invariant mass $m_0$ is given by:

$m=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m_0=\gamma m_0$ where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is sometimes called the Lorentz factor. Let’s compute now the total energy of an object with rest mass $m_0$ and travelling at a speed $v$:

The kinetic energy of an object can be seen as the integral of the force with respect to distance:

$K =\int_0^s F ds$

Using Newton’s second law of motion $F=\frac{d mv}{dt}$

$K =\int_0^s \frac{d mv}{dt}ds$

Solving the integral

$K=\int_0^{mv}vd(mv)=\int_0^{mv}vd(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}v)$

Integrating by parts

$K=\frac{m_0 v^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0 \int_0^{v}\frac{v dv}{\sqrt{1-\frac{v^2}{c^2}}}$

$K=\frac{m_0 v^2}{\sqrt{1-\frac{v^2}{c^2}}}+[m_0c^2\frac{m_0 c}{\sqrt{1-\frac{v^2}{c^2}}}]_0^v$

$K=\frac{m_0 c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2$

$K=mc^2-m_0c^2$

Since the total energy of the object is its rest and kinetic energy, we have that total energy is equal to:

$R+K=E=m_0c^2+mc^2-m_0c^2=mc^2$

Now, before continuing we need a relation between energy and momentum. Momentum squared equals:

$p^2=(mv)^2=m^2 v^2=m_0^2 \frac{v^2}{1-\frac{v^2}{c^2}}$

Which means that $\frac{v^2}{c^2}=1-\frac{m_0^2 v^2}{p^2}$

Replacing in our $E=mc^2$ formula leads us to:

$E=m_0 c^2 \sqrt{1+(\frac{p}{m_0 c})^2}$

$E^2=m_0^2 c^4 +\frac{m_0^2 c^4 p^2 }{m_0^2 c^2}=m_0^2 c^4 +p^2 c^2$

This is one of the most bad ass equations in physics. Look that the term $E^2$ actually has a positive and a negative root. This means that the equation also holds for particles with negative energies. Starting with this observation Paul Dirac predicted the existence of antiparticles decades before their discovery.

Now that we have derived both an equation for electromagnetic waves, the formula for the total energy of a particle, and the relationship between energy and momentum, we can  continue with the Schrodinger equation.

Recall the wave equation

$\frac{\partial^2 E}{\partial x^2}-\frac{1}{c^2} \frac{\partial^2 E}{\partial t^2}=0$

The general solution of this equation has the following form:

$E(x,t)=E_0 e^{i(kx-\omega t)}$

Where $k=\frac{2\pi}{\lambda}$ is the spatial frequency and $\omega=2\pi \upsilon$ is the temporal frequency of the  wave. Replacing this equation into the differential equation gives us:

$(\frac{\partial^2}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})E_0 e^{i(kx-\omega t)}=0$

$[-k^2+\frac{1}{c^2} \frac{\partial \omega^2}{\partial t^2}]E_0e^{i(kx-\omega t)}=0$

This means that:

$-k^2+\frac{\omega^2}{c^2}=0$

$k=\frac{\omega}{c}$ which relates both the spatial and temporal frequencies of the wave and the speed of light.

An important result from Max Planck was that the energy and the frequency of photons are related. Specially

$\mathbb{E}=\hbar \omega$ is the energy of the particle and $p=\hbar k$ the momentum, where $\hbar$ is the normalized Planck’s constant. Here I will use $\mathbb{E}$ to denote the energy of the particle and $E$ the electromagnetic field. Substituting into the equations

$-\frac{1}{\hbar^2}(p^2-\frac{\mathbb{E}^2}{c^2})E_0e^{\frac{i}{\hbar}(px-\mathbb{E}t)}=0$

Which leads to $\mathbb{E}^2=\rho^2 c^2$ which is offcourse the relativistic total energy $\mathbb{E}^2=m^2 c^4+\rho^2 c^2$ for massless particles like the photon.

Let’s try now to understand how particles with mass (e.g. electrons, positrons, neutrinos) move with out working with electromagnetic waves any more. Instead let’s just define a wave function $\Psi(x,t)$ that will lead us to the following differential equation:

$-\frac{1}{\hbar^2}(p^2 -\frac{\mathbb{E}^2}{c^2}+m^2 c^2)\Psi e^{\frac{i}{\hbar}(px-\mathbb{E}t)}=0$

With general wave equation

$\Psi(x,t)=\Psi_0e^{\frac{i}{\hbar}(px-\mathbb{E}t)}$

Normalized to unit probability

$\int \Psi^*\Psi dx=1$

The equation satisfying these properties is known as the Klein-Gordon equation (extending it to more than one dimensions)

$\nabla^2 \psi-\frac{m^2 c^2}{\hbar^2}\Psi=\frac{1}{c^2} \frac{\partial^2 \Psi}{\partial t^2}$

This a relativistic equation, Schrodinger’s equation is not. In order to make it non relativistic let’s approximate $\mathbb{E}^2$ as follows:

$\mathbb{E}=mc^2 \sqrt{1+\frac{p^2}{m^2 c^2}}$

$\mathbb{E}\approx mc^2 (1+\frac{1}{2}\frac{p^2}{m^2 c^2})$

$\mathbb{E} \approx mc^2 +\frac{p^2}{2m}$

Recall that the last term is nothing but the Kinetic energy of the particle. We can rewrite the wave equation as:

$\Psi(x,t)=e^{\frac{i}{\hbar}(mc^2 t)}\Psi_0e^{\frac{i}{\hbar}(px - Kt)}$

The term $\Psi_0e^{\frac{i}{\hbar}(px - Kt)}$  does not oscillate as fast as the first one since the speed of the particle is not close to $c$. We can call this term $\phi$ and see that:

$\Psi(x,t)=e^{\frac{i}{\hbar}(mc^2 t)} \phi$

The first and second derivatives with respect to time are:

$\frac{\partial \Psi}{\partial t}=-\frac{i}{\hbar}mc^2e^{-\frac{i}{\hbar}(mc^2 t)}\phi+e^{-\frac{i}{\hbar}(mc^2 t)}\frac{\partial \phi}{\partial t}$

$\frac{\partial^2 \Psi}{\partial t^2}=(-\frac{m^2 c^4}{\hbar^2}e^{-\frac{i}{\hbar}(mc^2 t)}\phi-\frac{2i}{\hbar}mc^2 e^{-\frac{i}{\hbar}(mc^2 t)}\frac{\partial \phi}{\partial t})+e^{-\frac{i}{\hbar}(mc^2 t)}\frac{\partial^2 \phi}{\partial t^2}$

The first term in brackets is large and the last term is small. We keep the large terms and discard the small one. The Klein Gordon equation is now:

$e^{-\frac{i}{\hbar}(mc^2 t)}[\frac{\partial^2}{\partial x^2}+\frac{2im}{\hbar}\frac{\partial}{\partial t}]\phi=0$

Which leads to the well known (in this case one-dimensional) Schrodinger equation

$-\frac{\hbar^2}{2m}\frac{\partial^2\phi}{\partial x^2}=i\hbar \frac{\partial \phi}{\partial t}$

Which can be written also as:

$\frac{\partial}{\partial t}V=HV$

Where $H$ is the Hamiltonian operator.

2) The Black Scholes Equation

The Black Scholes Partial Differential Equation (BSPDE) is an equation that explains how the value of any derivative must behave under no arbitrage opportunities. In their “seminal” paper (The Pricing of Options and Corporate Liabilities (1973) Journal of Political Economy) Black and Scholes derive the following equation for a derivative with value $V(S,t)$ that depends on a underlying asset $S$ and time $t$.

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

Im going to skip the derivation of this equation since it is widely known. However let’s see what is its relation with the Schrodinger’s equation. Organizing the time derivative on one side:

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

Now, we can complete the square on the right hand side of the equation

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

Now let’s make some transformations. Define $p=-i \sigma [\frac{\partial}{\partial x}-\frac{1}{2}(1-\frac{2r}{\sigma^2})]$ where $x=log(S)$ and $U=\frac{\sigma^2}{8}(1+\frac{2r}{\sigma^2})^2$ the Black Scholes equation can be written as:

$\frac{\partial}{\partial t}V=\frac{p^2}{2}V+UV=HV$

The equation is a particular case of the Schrodinger equation with Hamiltonian equal to $\frac{p^2}{2}+U$. Analyzing in more detail the equation we see that the term $\sigma^2$ works as a Planck’s constant relating frequency with energy. Also the Black Scholes equation is like a Schrodinger equation for imaginary time since every solution lies on the Real Line.

We can understand prices as particles in which their position can not be known in advance. Instead, we assign to them a probability function that give us the probability of finding the particle in some interval tomorrow.

Some references:

http://arxiv.org/pdf/physics/0610121.pdf

https://phorgyphynance.wordpress.com/2008/06/03/black-scholes-and-schrodinger/

http://physics.stackexchange.com/questions/60242/relation-between-black-scholes-equation-and-quantum-mechanics