# Can Schrödinger Equation be derived from Huygens' Principle?

Notes of Enrico Fermi start from an analogy between mechanics and optics and with 4 pages he derives the Schrödinger equation. In all my courses, I have seen as an axiom - this is how wave-particles behave. Here it is being derived from Fermat's least action principle.

• Maupertuis Principle -- $$S = \int mv\,\mathrm ds = \int p \,\mathrm dq$$

• This is integral over phase space so this is not the same as Hamilton principle $$S = \int L \, \mathrm dt$$
• Maupertuis felt space and time should be put on equal footing
• Fermat principle -- $$S = \int n \, \mathrm ds$$ principle of least time, Snell's law of refraction, etc

• this is principle of least time rather than principle of least action
• Similar to Huygens principle.    Can Quantum Mechanics be understood in terms of wave-fronts and their singularities? I would wonder what how the free particle or scattering might look in this setting.

The only other source I could find is Sir Michael Berry, who has written quite a lot. • ...and what has Huygen's principle to do with the Schrödinger equation? (The pictures you posted are completely illegible) You cannot derive quantum mechanics from classical mechanics. You may motivate it, but that is different from deriving it. Jun 1, 2016 at 20:30
• those Notes are by Enrico Fermi. hoping to gain insight to his thinking Jun 1, 2016 at 21:01
• @johnmangual What Fermi does in these notes is to retrace a simplified version of Schroedinger's own path to his celebrated equation. So you might find your answers in Schroedinger's original papers, especially the 2nd one, where he explains in detail the approach he used. For a brief overview of the background story and refs. see this answer: physics.stackexchange.com/q/242059
– udrv
Jun 1, 2016 at 23:47

Diffractive optics in the Fresnel (paraxial) approximation is exactly the same as the quantum mechanics of a single particle when thickness along the optical axis is replaced by time, refractive index is replaced by mass and the inverse angular frequency of the monochromatic light is replaced by Planck's constant. Here is a brief sketch.

## Classical Mechanics

The unity of optics and mechanics is clearest using the generators of canonical transformations (see section 2.1 of Field Theory by Pierre Ramond). The Hamiltonian for a free particle is $H=\frac{p^{2}}{2m}$. The generator for this Hamiltonian is, \begin{eqnarray*} G(q,Q)&=&\frac{m(q-Q)^{2}}{2t}\\ p&=&\frac{\partial G(q,Q)}{\partial q}\\ -P&=&\frac{\partial G(q,Q)}{\partial Q} \end{eqnarray*} By messing about with the generator we can write the canonical transformation as a matrix transformation from the initial state $(P,Q)$ to the final state $(p,q)$. \begin{equation} \left[ \begin{array}{c} p \\ q \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ \frac{t}{m} & 1 \end{array} \right] \left[ \begin{array}{c} P \\ Q \end{array} \right] \ . \end{equation} The $2\times 2$ matrix is an element of the symplectic group Sp(2,R). In other words, classical mechanics (for quadratic Hamiltonians) is synonymous with the two-dimensional defining representation of Sp(2,R) carried on the two-dimensional phase space $(p,q)$,

## Quantum Mechanics

The momentum becomes an operator $p\rightarrow -i\hbar\frac{\partial}{\partial q}$. The states become wavefunctions $\langle Q|\psi\rangle$. The canonical transformation becomes a unitary operator, \begin{equation} \langle q|U|Q\rangle \propto \exp\left(\frac{i G(q,Q)}{\hbar}\right)= \sqrt{\frac{m}{it}}\exp\left( \frac{im(q-Q)^{2}}{2t\hbar}\right) \end{equation} The amplitude $\langle q|U|Q\rangle$ is an infinite-dimensional matrix. The rows are indexed by $q$ and the columns by $Q$.The initial and final states are related by, \begin{equation} \langle q|U|\psi\rangle=\int\frac{dQ}{\sqrt{2\pi}} \langle q|U|Q\rangle\langle Q|\psi\rangle \end{equation} The integral is an infinite-dimensional matrix multiplication. Differentiating the above equation recovers Schrodinger's equation, \begin{equation} i\hbar\frac{\partial \psi}{\partial t} =\frac{1}{2m}\left(-i\hbar\frac{\partial}{\partial q}\right)^{2}\psi \end{equation} The operator $\langle q|\hat{U}|Q\rangle$ is an element of an infinite-dimensional unitary representation of the symplectic group Sp(2,R). In other words, quantum mechanics (for quadratic Hamiltonians) is synonymous with the infinite-dimensional unitary representations of Sp(2,R) carried on the space of wavefunctions $\psi(q)$.

## Ray Optics

Fermat's principle of least time in ray optics plays the r\^{o}le of the principle of least action in classical mechanics.

Let $s$ be the distance along a ray. The time to move by distance $ds$ is $dt=nds/c$ where $c$ is the speed of light in a vacuum and $n$ is the refractive index of the material. The total time on the path is, \begin{equation} S=\int dt=\int \frac{nds}{c}=\int \frac{n\sqrt{dq^{2}+dz^{2}}}{c} =\int\frac{ndz}{c}\sqrt{1+\dot{q}^{2}}\ . \end{equation} where $\dot{q}=dq/dz$ and the $z$ coordinate in ray optics plays the r\^{o}le of time in classical mechanics. The Lagrangian in ray optics is therefore, \begin{equation} L=\frac{n}{c}\sqrt{1+\dot{q}^{2}} \ . \end{equation} The canonically conjugate momentum is, \begin{equation} p=\frac{\partial L}{\partial \dot{q}} =\frac{n\dot{q}}{c\sqrt{1+\dot{q}^{2}}} =\frac{ndq}{c\sqrt{dq^{2}+dz^{2}}} =\frac{n}{c}\frac{dq}{ds}\ . \end{equation} The ray optics Hamiltonian is, \begin{equation} H=p\dot{q}-L=-\sqrt{\left(\frac{n}{c}\right)^{2}-p^{2}} \simeq\frac{p^{2}c}{2n}-\frac{n}{c} \end{equation} and the last result is for small momentum. Ray optics is the same as classical mechanics with $\frac{n}{c}$ in place of mass $m$. The $2\times 2$ matrices of the defining representation of Sp(2,R) are the ray transfer matrices in ray optics.

## Diffractive Optics

The momentum becomes an operator $p \propto -i\frac{\partial}{\partial q}$. The constant of proportionality cannot be Planck's constant $\hbar$ because the dimensions are wrong. In ray optics, the momentum has dimensions of second/metre and so the constant has to have dimensions of second. The physical quantity with dimensions of second is the inverse frequency of the light. A factor of $2\pi$ appears just as in Planck's constant $\hbar$ and so the constant with dimensions of second is, in fact, the inverse angular frequency $\omega^{-1}$ of the light. The momentum operator is now, \begin{equation} p= -i\omega^{-1}\frac{\partial}{\partial q} \end{equation} and everything in the quantum mechanics of a particle goes over to diffractive optics by replacing $\hbar\rightarrow \omega^{-1}$. For example, the plane wave in quantum mechanics is, \begin{equation} \psi(q)=\exp\left(\frac{ipq}{\hbar}\right) \end{equation} and so the plane wave in diffractive optics is, \begin{equation} \psi(q)=\exp\left(\frac{ipq}{\omega^{-1}}\right) =\exp\left(\frac{in\omega}{c}\frac{dq}{ds}q\right) =\exp\left(\frac{i2\pi n\sin(\theta)q}{\lambda}\right) \end{equation} using the definition of the ray optics momentum.

Quantum mechanics is not so strange because it is the same theory as diffractive optics in the Fresnel (paraxial or small momentum) approximation. I first learnt this from reading the introductory chapter of "Symplectic techniques in physics" by Victor Guillemin and Shlomo Sternberg.

## Huygen's Principle and Quantum Mechanics

The amplitude $\langle q|U|Q\rangle$ for a free particle given on the RHS of the first equation of the section on quantum mechanics is a cylindrical wave centred at coordinate $Q$. The integral in the second equation is then a sum over cylindrical waves and this is Huygen's Principle. So, Schrodinger's equation can be derived from Huygen's principle.

Even mathematically, Schrödinger's equation cannot be derived from the principle of least action because it only depends on the first derivatives of time, $\psi' = \partial \psi / \partial t$. This proves that $\psi'$ would have to appear in the action, but then $\psi''$ would unavoidably appear in the Euler-Lagrange equations, too, unless the action had just some terms of the form $\psi' \psi$. But $\psi'\psi = (\psi^2)'/2$ is a total derivative, so it doesn't contribute to the equations of motion. (The complex conjugation makes the right derivation more messy but the conclusion is ultimately the same because the action has to be made real etc.)

(The Dirac equation or any equation for fermions is a loophole – it may be first-order and derived from the least action. That's because the extra fermionic signs and/or signs from the Dirac matrices prevent you from writing $\psi'\psi = (\psi^2)'/2$ – the Dirac Lagrangian is not a total derivative.)

So the equation for the wave function in quantum mechanics does not result from the least action. Instead, one may say that the Heisenberg equations for the operators in quantum mechanics (in the Heisenberg picture that may be shown to be physically equivalent to the Schrödinger's picture) are basically the same equations as the classical equations, and those can be derived from the least action.

Can the least action be used directly in quantum mechanics, without the reference to the classical equations? Well, with a modification. The right way to use the action $S$ in quantum mechanics is called the Feynman's path integral. The quantum mechanical system actually "probes" all possible trajectories and histories at the same moment and $\exp(iS/\hbar)$ is the integrand that contributes to the probability amplitude of some evolution.

This complex exponential is basically a "random, quickly oscillating phase" and almost all of them cancel in the limit where $\hbar$ is small relatively to the parameters of the problem. It's primarily the trajectories near $\delta S = 0$, an extremized action, that are exceptions. Because $S$ is rather stable there, the amplitudes constructively interfere. That's an explanation why near the classical limit, the classical trajectories dominate the evolution in Feynman's approach to quantum mechanics.

• Lubos, you might want to know that Shrodinger equation actually can be derived from lagrangian formulation. Try Lagrangian density $\mathcal L = \frac{i}{2} \left( \psi ^* \partial_t \psi - (\partial_t \psi ^*) \psi \right) - \vec{\nabla} \psi^* \cdot \vec{\nabla} \psi$. This shows that complex numbers are actually enough to introduce first powers of first order derivatives in Lagrangian without making these boundary terms. Jun 3, 2016 at 9:01
• It is interesting to note that this lagrangian differs only by boundary term from another one (which is no longer real): $\mathcal L' = \psi ^* (i \partial_t + \vec{\nabla}^2 ) \psi$. Another nice observation is that conserved probability current in one particle Shrodinger QM is directly related to global change of phase symmetry of the real Lagrangian $\mathcal L$. Jun 3, 2016 at 9:06
• Oops, that's interesting, it works. It's still a misguided way to approach the equation, I think, but mathematically, it does what we need. Jun 3, 2016 at 10:06

The Huygen's principle is that a wave amplitude $A(t_0)$, usually a plane wave, is modified into a spherical wave with an amplitude $$A(t,r) = \frac{A(t_0)e^{ikr + i\delta}e^{-i\omega t}}{r}.$$ The radial distance at $t_0$ is $r \simeq \lambda$ and the phase $\delta$ is set so that $k\lambda = \delta$, which is set to $2\pi$. Now consider a Taylor expansion of this wave front with respect to $r$, and \begin{align} A(t,r) &= A(t_0) + \frac{\partial A(t,r)}{\partial r}\Big|_{r = \lambda}r + \frac{1}{2}\frac{\partial^2 A(t,r)}{\partial^2 r}\Big|_{r = \lambda}x^2 + \dots\\ & = A(t_0) + \frac{e^{ikr}(ikr - 1)}{r^2}\big|_{r = \lambda}\delta r - \frac{1}{2}\frac{e^{ikr}(k^2r^2 + 2ikr – 2)}{r^3}|_{r=\lambda}\delta r^2\end{align}

We set $r = \lambda$ and we find some subtractions so that $$A(t, r) \simeq A(t_0) - \frac{k^2}{\lambda}e^{ikr}\delta r^2$$ To treat the time dependency explicitly we use $A(t,r) = e^{-i\omega t}A(r)$ and the expansion on the left hand side is then justification $$A(t, r) = A(r)\Big|_{t = t_0} + i\frac{\partial A(t,r)}{\partial t}|_{t = t_0}\delta t + \dots,$$ where the $i\frac{\partial}{\partial t}$ is used to get a real valued term linear in $\delta t$. We now set all the variations according to the wave length at set $k = p/\hbar$ so that $$i\hbar\frac{\partial}{\partial t}e^{ikr - i\omega t} = \frac{1}{2}p^2e^{ikr – i\omega t}$$ This is pretty close to being a Schroedinger equation for a free particle.

This is close, but I have a mass term missing. I should have $1/2m$ instead of $1/2$, which is a manifestation of deriving this from a result involving EM waves that have no mass. This is an interesting question and it made me do some quick calculations.

• This is the eikonal approximation or some other fancy PDE concept. I trust Fermi's intuition, even though strange. Jun 1, 2016 at 21:30