# A Deriving the Lagrangian from the Hamiltonian operator

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1. Mar 7, 2017

### redtree

In classical mechanics, the Hamiltonian and the Lagrangian are Legendre transforms of each other. By analogy, in quantum mechanics and quantum field theory, the relationship between the Hamiltonian and the Lagrangian seems to be preserved. Where can I find a derivation of the Lagrangian operator as the Legendre transform of the Hamiltonian operator in quantum mechanics? Or a similar derivation for the Lagrangian density and Hamiltonian density in quantum field theory via the Legendre transform?

2. Mar 8, 2017

### redtree

Figured it out. No need to respond.

3. Mar 8, 2017

### Useful nucleus

https://arxiv.org/abs/0806.1147

Did you find some other interesting resources ? Please share.

4. Mar 8, 2017

### redtree

Thanks for the reply. I was already familiar with that article. Good resource, but incomplete; fails to mention the supernum/infinium aspect of the Legendre transform. Better: https://www.andrew.cmu.edu/course/33-765/pdf/Legendre.pdf

I figured out the question on my own.

One interesting note:

Given:

$\mathcal{L}(\vec{v},\vec{q},t)=\sup \left[ \langle \vec{p},\vec{v} \rangle - \textbf{H}(\vec{p},\vec{q},t)\right]$
$=\sup \left[ \vec{p}\cdot \vec{v} -\left( \frac{\vec{p}^2}{2 m}+ V\right)\right]$
$=\sup \left[ m\vec{v}\cdot\vec{v} -\left( \frac{m^2\vec{v}^2}{2 m}+ V\right)\right]$
$=\sup \left[ m\vec{v}^2 -\left( \frac{m\vec{v}^2}{2}+ V\right)\right]$
$=\sup \left[ \frac{m\vec{v}^2}{2}- V\right]$
And:
$\mathcal{S}=\int_{t_1}^{t_n} \mathcal{L}(\vec{v},\vec{q},t)dt$

Question: Why is the action $\mathcal{S}$ minimized (and not maximized), given that $\textbf{H}(\vec{p},\vec{q},t)$ is a convex function and therefore $\mathcal{L}(\vec{v},\vec{q},t)$ is a supernum (not an infimum)?