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Suppose there is a Lagrangean $L = \frac{1}{2} m c \dot x - \frac{1}{2}kx^2$ where $c$ is a constant to keep the dimensions right.

The conjugate momentum is then $ p_x = \frac{\partial L} {\partial \dot x} = \frac{1}{2}mc$ which is independent of $\dot x$. So, apparently there is a difficulty of writing $\dot x = \dot x (p_x, x)$. Then how do I use the Legendre transformation $ H = p_x \dot x - L$ to get the Hamiltonian?

This might seem like a unphysical situation. But on chapter 13,vproblem 4 on Classical Mechanics, 3rd edition, Goldstein, Poole, Safko there is a similar example. Given the Lagrangean density $$\mathcal{L} = \frac{h^2}{8 \pi^2 m} \nabla\psi \cdot \nabla\psi^* + V \psi^* \psi + \frac{h}{4 \pi i} (\psi^* \dot \psi - \psi \dot \psi ^*)$$

When you try to calculate the conjugate momenta for $\psi$, you get $\pi_{\psi} = \frac{\partial \mathcal{L}} {\partial \dot \psi} = \frac{h}{4 \pi i} \psi^*$ which is independent of $\dot \psi $ or $\dot \psi^*$. So, apparently there is a difficulty of writing $\dot \psi = \dot \psi(\pi_\psi, \pi_{\psi^*}, \psi, \psi^*)$.

Then how do I perform the Legendre transformation from Lagrangean density $\mathcal{L}$ to Hamiltonian density $\mathcal{H}$?

Qmechanic
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Mockingbird
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  • Possible duplicates: https://physics.stackexchange.com/q/15242/2451 and links therein. – Qmechanic Jan 25 '20 at 03:26
  • @Qmechanic So you mean just identity the term $ p_{\psi} \dot \psi $ term in $\mathcal{L}$ and write the rest of the $\mathcal{L}$ as $-\mathcal{H}$ ? – Mockingbird Jan 26 '20 at 17:21
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    That's the simplest method. (Note that Goldstein's Lagrangian has an unconventional overall minus sign -- the corresponding Hamiltonian is negative definite.) – Qmechanic Jan 26 '20 at 18:41

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