[NOTES/QM-11006] Optics Mechanics Analogy Road to Wave Mechanics

The motion of waves and particles, both, can be described by variationl principles. For waves we have the Fermat's principle. The motion of particles is described my Maupertuis Principle. In the table given below, the frameworks of the variational principles for a particle and for a wave are summarized.$$\begin{array}{ll} \mbox{Point Particle}\hspace{1 cm}&\mbox{Waves}\\ \mbox{Mass Point}&\mbox{Wave packet}\\ \mbox{Trajectory}&\mbox{Ray}\\ \mbox{Velocity,$\vartheta$}&\mbox{Group Velocity,$\tilde{\vartheta}$}\\ & \mbox{Phase Velocity, (v)}\\ \mbox{Potential Energy $\vartheta(x)$}&\mbox{Refractive Index},\\ &\mbox{Function of position $n(x)$}\\ \mbox{Energy E}&\mbox{Frequency $\nu$}\\ &\mbox{Dispersive medium, $n(,\nu,x),v(\nu,x)$}\\ \mbox{Manupertuis Principle:}&\mbox{Fermat Priniciple:}\\ \end{array} $$

\begin{equation*} \int{\sqrt{E-\vartheta(x)}\,dx}= \text{min}\hspace{1cm} \int{\frac{ds}{v(\nu,x)}}=\text{min} \end{equation*}

Remark
Note that,for a particle, $\int{\sqrt{E-V(x)}\,dx}=$min, means $\int{pdx}=$ {min} which, for \(E=\)const, implies $\int(p dx-H dt)=\text{min}.$\\ Particles have dual nature, therefore consistency requires that both Maupertuis and Fermat principles should give the same answer. Therefore, we must have \begin{equation} \frac{1}{v(\nu, x)}=f(\nu)\sqrt{E-V(x)} \end{equation} Velocity of a point mass is \begin{equation} v=\frac{p}{m}=\sqrt{\frac{2}{m}\Big(E-V(x)\Big)}. \end{equation} On the other hand, the group velocity, for waves, is given by \begin{equation} v_{g}=\frac{d\omega}{dk}=1/\Big(\frac{dk}{d\omega}\Big) \end{equation} Note that $v=\nu\lambda, \omega=2\pi\nu$ and $k=2\pi/\lambda,$ \begin{equation} v_{g}=1\Big/\frac{d}{d\nu}(1/\lambda)=1\Big/\frac{d}{d\nu}\Big(\frac{\nu}{v(\nu) } \Big) \end{equation} Using (1) the group velocity $\widetilde{\vartheta}$ of waves is given by \begin{equation} \frac{1}{\widetilde{\vartheta}}=\frac{d}{d\nu}\Big(\frac{\nu}{v(\nu,x)}\Big) = \frac{d}{d\nu}\Big(\nu f(\nu)\sqrt{E(\nu)-V(x)}\Big). \end{equation} Velocity of a mass point $v$ corresponds to the group velocity $v_{g}$ of wave packet ($=\widetilde{\vartheta}$). Hence from (2) and (5) we obtain \begin{equation} \sqrt{\frac{m}{2}}\frac{1}{\sqrt{E-V(x)}}=\frac{d}{d\nu}\Big(\nu f(\nu)\sqrt{E(\nu)-V(x)}\Big) \end{equation} \begin{equation} =\frac{d}{d\nu}(\nu f(\nu))\sqrt{E(\nu)-V(x)}+\frac{\nu f(\nu)}{2}\frac{dE(\nu)}{d\nu}\frac{1}{\sqrt{E(\nu)-V(x)}} \end{equation} This equation will be correct for all $x$ and all $\vartheta(x)$ if \begin{equation} \frac{d}{d\nu}(\nu f(\nu))=0~ \text{or}~ \nu f(\nu)=\text{const,K.} \end{equation} and the coefficient of $\frac{1}{\sqrt{E-V(x)}}$ on both sides are equal: \begin{equation} \sqrt{\frac{m}{2}}=\frac{\nu f(\nu)}{2}\frac{dE(\nu)}{d\nu}=\frac{K}{2}\frac{dE(\nu)}{d\nu} \end{equation} \begin{equation} \Rightarrow\frac{dE(\nu)}{d\nu}=\text{const},h\Rightarrow E=h\nu+\text{const.,} \end{equation} Setting this last constant to zero gives \begin{equation} \boxed{E=h\nu} \end{equation} and Eq.(10) determines the constant $K=\frac{\sqrt(2m)}{h}.$ Also $\nu f(\nu)=K$ along with Eq.(1)implies \begin{equation} \frac{1}{v(\nu,x)}=\frac{K}{\nu}\sqrt{E(\nu)-V(x)}=\frac{\sqrt{2m}}{h\nu}\sqrt{ E(\nu)-V(x)} \end{equation} or \begin{equation} \frac{h}{\sqrt{2m(E(\nu)-V(x))}}=\frac{v}{\nu}=\lambda \end{equation} This gives \begin{equation} \boxed{\lambda=\frac{h}{p}} \end{equation}

Derivation of Schr\"{o}dinger Equation: 
For monochromatic waves \begin{equation} \nabla^{2}\psi-\frac{1}{v^2}\frac{d^2\psi}{dt^2}=0 \end{equation} We set $\psi(x,t)=u(x)e^{-i\omega t}$, assuming $\omega$ to be fixed, using Eq. (13) we get \begin{equation} \nabla^2 u+\frac{\omega^2}{v^2}u=0 \end{equation} \begin{equation} \Rightarrow \nabla^2u+\frac{4\pi^2\nu^2}{h^2\nu^2}(2m(E(\nu)-V(x))u=0 \end{equation} \begin{equation} \Rightarrow \nabla^2u+\frac{2m}{\hbar^2}(E(\nu)-V(x))u=0 \end{equation} For states with fixed energy, we have \begin{equation} E\psi=h\omega\psi\sim i\hbar\frac{\partial\psi}{\partial t}, \end{equation} giving the time dependent Schr\"{o}dinger equation \begin{equation} ih\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\psi+V(x)\psi \end{equation} Clap!Clap!

Remarks: 
The highlight of this route is the derivation of de Broglie relation $\lambda=h/p$ and the Einstein relation $E=h\nu$ for point particles by demanding the action principles for matter and waves give the same result.\\ I thank Bindu Bambah for providing Fermi's Chicago University lecture notes to me. I wish Fermi's Lecture Notes were available to the present generation of students.