Quantum Field Theory for the Gifted Amateur Exercise 1.5 Surfing the Quantum Waves

Welcome back, fellow aficionados of quantum mechanics. The journey through Quantum Field Theory for the Gifted Amateur has led us to Exercise 1.5. Here we look at a three-dimensional elastic medium and show, from the action principle, that its scalar displacement field obeys the wave equation.

Table of Contents

Setting the Stage: Elastic Media and Energies

Consider an elastic medium described by a scalar field $\psi(\mathbf{x}, t)$. This medium is characterized by two crucial energies: the potential energy $V$ and the kinetic energy $T$:

\[V = \frac{\mathcal{T}}{2} \int d^3x\, (\nabla \psi)^2\]

\[T = \frac{\rho}{2} \int d^3x\, \left( \frac{\partial \psi}{\partial t} \right)^2\]

Here $\mathcal{T}$ is the elastic stiffness parameter and $\rho$ is the mass density. The potential energy measures the cost of spatial deformation, while the kinetic energy measures the cost of time variation.

The Lagrangian

The Lagrangian $L$, a cornerstone concept in classical mechanics, is the difference between kinetic and potential energies:

\[L = T – V\]

For this field, that means

\[L = \int d^3x\, \left[ \frac{\rho}{2} \left(\frac{\partial \psi}{\partial t}\right)^2 – \frac{\mathcal{T}}{2} (\nabla \psi)^2 \right].\]

The Action

The action $A$ is defined as the integral of the Lagrangian over time:

\[A = \int (T – V)\, dt\]

or, equivalently,

\[A = \int dt \int d^3x\, \left[ \frac{\rho}{2} \left(\frac{\partial \psi}{\partial t}\right)^2 – \frac{\mathcal{T}}{2} (\nabla \psi)^2 \right].\]

The classical field configuration is the one for which the action is stationary under small variations of $\psi$.

The Equation of Motion

Functional Variation

We aim to find the equation of motion by taking the functional derivative $\delta A / \delta \psi$ and setting it equal to zero. Let the field vary as

\[\psi \rightarrow \psi + \delta \psi.\]

The first-order change in the action is

\[\delta A = \int dt \int d^3x\, \left[ \rho \frac{\partial \psi}{\partial t} \frac{\partial (\delta \psi)}{\partial t} – \mathcal{T} \nabla \psi \cdot \nabla (\delta \psi) \right].\]

Integration by Parts

Now integrate by parts in both time and space. Assuming the variations vanish at the temporal endpoints and on the spatial boundary, the boundary terms vanish. This gives

\[\delta A = \int dt \int d^3x\, \left[ -\rho \frac{\partial^2 \psi}{\partial t^2} + \mathcal{T} \nabla^2 \psi \right] \delta \psi.\]

For the action to be stationary for arbitrary $\delta \psi$, the coefficient of $\delta \psi$ must vanish:

\[-\rho \frac{\partial^2 \psi}{\partial t^2} + \mathcal{T} \nabla^2 \psi = 0.\]

Equivalently,

\[\rho \frac{\partial^2 \psi}{\partial t^2} – \mathcal{T} \nabla^2 \psi = 0.\]

Final Wave Equation

Solving for the spatial Laplacian gives the wave equation:

\[\nabla^2 \psi = \frac{\rho}{\mathcal{T}} \frac{\partial^2 \psi}{\partial t^2}.\]

Since the wave speed is

\[v = \sqrt{\frac{\mathcal{T}}{\rho}},\]

we can also write

\[\nabla^2 \psi = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}.\]

Thus the field $\psi(\mathbf{x}, t)$ obeys the three-dimensional wave equation, with $v$ as the wave velocity.

Wrapping Up

This exercise shows how the energy of an elastic medium leads naturally to a field Lagrangian, and how the stationary-action principle produces the wave equation. The same logic is one of the recurring patterns in field theory: write down the action, vary the field, discard boundary terms under appropriate conditions, and read off the equation of motion.