Simple variational setups yielding Newton’s second law and Schrödinger’s equation

It is a delightful fact that one can get both the fundamental equation of classical mechanics (Newton’s Second Law) and the fundamental equation of quantum mechanics (Schrödinger’s equation) by solving very simple variational problems based on the familiar conservation of mechanical energy equation

$K + U = E$

In the present note I want to briefly set out the relevant calculations in a unified way emphasising the common underlying structure provided by the conservation of mechanical energy and the Calculus of Variations. The kinetic energy $K$ will be taken to be

$K = \frac{1}{2}m \dot{x}^2 = \frac{p^2}{2m}$

where $\dot{x} = \frac{\mathrm{d}x}{\mathrm{d}t}$ is the particle’s velocity, $p = m\dot{x}$ is its momentum, and $m$ is its mass. The potential energy $U$ will be regarded as some function of $x$ only.

To obtain Newton’s Second Law we find the stationary path followed by the particle with respect to the functional

$S[x]= \int_{t_1}^{t_2} L(t, x, \dot{x}) dt = \int_{t_1}^{t_2} (K - U) dt$

The function $L(t, x, \dot{x}) = K - U$ is usually termed the Lagrangian in classical mechanics. The functional $S[x]$ is usually called the action. The Euler-Lagrange equation for this Calculus of Variations problem is

$\frac{\partial L}{\partial x} - \frac{\mathrm{d}}{\mathrm{d}t}\big(\frac{\partial L}{\partial \dot{x}}\big) = 0$

and this is Newton’s Second Law in disguise! We have

$\frac{\partial L}{\partial x} = -\frac{\mathrm{d}U}{\mathrm{d}x} \equiv F$

$\frac{\partial L}{\partial \dot{x}} = m\dot{x} \equiv p$

and

$\frac{\mathrm{d}}{\mathrm{d}t} \big(\frac{\partial L}{\partial \dot{x}}\big) = \frac{\mathrm{d}p}{\mathrm{d}t} = m\ddot{x} \equiv ma$

so substituting these into the Euler-Lagrange equation we get Newton’s Second Law, $F = ma$ .

To obtain Schrödinger’s equation we introduce a function

$\psi(x) = exp\big(\frac{1}{\hbar}\int p\mathrm{d}x\big)$

where $p = m \dot{x}$ is again the momentum of the particle and $\hbar$ is the reduced Planck’s constant from quantum mechanics. (Note that $\int p dx$ has units of $\text{length}^2 \ \text{mass} \ \text{time}^{-1}$ so we need to remove these by dividing by $\hbar$ which has the same units, as the function $\psi(x)$ in quantum mechanics is dimensionless). We then have

$\text{ln} \psi = \frac{1}{\hbar}\int p\mathrm{d}x$

and differentiating both sides gives

$\frac{\psi^{\prime}}{\psi} = \frac{1}{\hbar} p$

$p^2 = \hbar^2 \big(\frac{\psi^{\prime}}{\psi}\big)^2$

Therefore we can write the kinetic energy as

$K = \frac{\hbar^2}{2m}\big(\frac{\psi^{\prime}}{\psi}\big)^2$

and putting this into the conservation of mechanical energy equation gives

$\frac{\hbar^2}{2m} \big(\frac{\psi^{\prime}}{\psi}\big)^2 + U = E$

$\iff$

$\frac{\hbar^2}{2m} (\psi^{\prime})^2 + (U - E) \psi^2 = 0$

We now find the stationary path followed by the particle with respect to the functional

$T[\psi] = \int_{-\infty}^{\infty} M(x, \psi, \psi^{\prime}) \mathrm{d}x = \int_{-\infty}^{\infty} \big(\frac{\hbar^2}{2m} (\psi^{\prime})^2 + (U - E) \psi^2\big)\mathrm{d}x$

The Euler-Lagrange equation for this Calculus of Variations problem is

$\frac{\partial M}{\partial \psi} - \frac{\mathrm{d}}{\mathrm{d}x}\big(\frac{\partial M}{\partial \psi^{\prime}}\big) = 0$

and this is Schrödinger’s equation in disguise! We have

$\frac{\partial M}{\partial \psi} = 2(U - E)\psi$

$\frac{\partial M}{\partial \psi^{\prime}} = \frac{\hbar^2}{m} \psi^{\prime}$

and

$\frac{\mathrm{d}}{\mathrm{d}x} \big(\frac{\partial M}{\partial \psi^{\prime}}\big) = \frac{\hbar^2}{m} \psi^{\prime \prime}$

so substituting these into the Euler-Lagrange equation we get

$2(U - E) \psi - \frac{\hbar^2}{m} \psi^{\prime \prime} = 0$

$\iff$

$-\frac{\hbar^2}{2m} \frac{\mathrm{d}^2 \psi}{\mathrm{d} x^2} + U \psi = E \psi$

and this is the (time-independent) Schrödinger equation for a particle of mass $m$ with fixed total energy $E$ in a potential $U(x)$ on the line $-\infty < x < \infty$ .