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.

(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.

  • 4
    $\begingroup$ In quantum mechanics, the gradient operator represents momentum (to within a constant factor). That is why they would call the square of the gradient the kinetic energy (momentum squared, to within a constant factor). That is quite general and not confined to any particular system. But yes, the vector potential is added to deal with the electromagnetic field. $\endgroup$
    – Greg P
    Commented Apr 10, 2012 at 2:13
  • $\begingroup$ @Greg P, Wouldn't the kinetic energy be proportional to the expectation value of the Laplacian operator? I was thinking of exactly what you have said, but was unsure since the OP has a different expression. $\endgroup$ Commented Apr 10, 2012 at 8:17
  • $\begingroup$ @Vijay: integrate by parts, then you see that the expectation value of the (negative) Laplacian operator is precisely the $L^2$ norm of the gradient. $\endgroup$ Commented Apr 10, 2012 at 9:30
  • $\begingroup$ @Willie Wong, Yes I can see that. But that sort of assumes that $f(x)$ or $\nabla f(x)$ vanishes on the boundary. If this is indeed what the OP had in mind, then fine. $\endgroup$ Commented Apr 10, 2012 at 10:16
  • $\begingroup$ I don't think this is necessarily quantum mechanical. This term is called kinetic energy in every fluid dynamics book as well. $\endgroup$
    – MBN
    Commented Apr 10, 2012 at 10:53

1 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 $\frac{\hbar}{i}{\bf \nabla}$ 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 $\hat{K}=\frac{m}{2}\hat{\bf v}^2$ is $$\langle\psi| \hat{K}|\psi\rangle ~=~\frac{m}{2} \int d^n x~ |\hat{\bf v}\psi(x)|^2 ~=~\frac{1}{2m} \int d^n x~ |(\hbar{\bf \nabla}-i{\bf A})\psi(x)|^2. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.