Why is $L^2$ norm of the gradient called kinetic energy?

Question

I'm reading Lieb-Loss's book 'Analysis', chapter 7. The authors refer to the following integral:

$$\tag{1} \lVert \nabla f\rVert_2^2=\int_{\Omega}\lvert \nabla f(x)\rvert^2\, d^nx $$

as the kinetic energy (see pag. 172), without explanation. To what physical system are they referring to? My intuitions says that (1) would be more appropriately called potential energy, as we have discussed here.

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(Note: In a later paragraph the authors introduce a magnetic potential $\mathbf{A}$ and the so-called covariant derivative $\nabla+i\mathbf{A}$, remarking that after this introduction the kinetic energy integral must be replaced with

$$\int_{\mathbb{R}^n}\lvert (\nabla + i \mathbf{A})f(x)\rvert^2\, d^n x.$$

This induces me to think that they take as a model some kind of electromagnetic system.)

Thank you for your attention.

Answer 1

As Greg P mentions in a comment, this terminology is inspired by quantum mechanics.

1. $f(x)$ plays the role of the wave function $\psi(x)=\langle x|\psi\rangle$ in the position space representation.

2. The conjugate/canonical momentum operator $\hat{\bf p}$ is replaced by $\dfrac{\hbar}{i}{\bf \nabla}$ in the Schrödinger representation.

3. The kinetic/mechanical momentum operator $m\hat{\bf v}$ is replaced by $\hat{\bf p}+{\bf A}$ (if we absorb certain constants, such as the charge of the particles, into the definition of the magnetic potential ${\bf A}$.)

4. The expectation value $\langle\psi| \hat{K}|\psi\rangle$ of the non-relativistic kinetic energy operator $\hat{K}=\dfrac{m}{2}\hat{\bf v}^2$ is $$\langle\psi| \hat{K}|\psi\rangle ~=~\frac{m}{2} \int \mathrm{d}^n x~ |\hat{\bf v}\psi(x)|^2 ~=~\frac{1}{2m} \int \mathrm{d}^n x~ |(\hbar{\bf \nabla}-i{\bf A})\psi(x)|^2. $$