Non-uniqueness of the Lagrangian

Question

The Hamiltonian for a simple harmonic oscillator is

$$H(p,q)=\frac{p^2}{2m}+\frac{k}{2}q^2$$

Then a possible Lagrangian corresponding to $H$ can be

1. $L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}q^2$

2. $L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}(q^2+3q^2\dot{q})$

3. $L=\frac{1}{2}m\dot{q}^2+\frac{k}{2}q^2$

4. $L=\frac{1}{2}m\dot{q}^2+\frac{k}{2}(q^2+3q^2\dot{q})$

We know by Legendre transformation that

$$H = \frac{\partial L}{\partial \dot{q}}\dot{q}-L=p\dot{q}-L$$

hence

$$L=p\dot{q}-H$$

Now since $H$ is given, by Hamilton movement equation we get

$$\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}$$

Substituting $p = m\dot{q}$ in $L=p\dot{q}-H$ we get the expression as in $(1)$. But this only determines one Lagrangian uniquely whereas another possible Lagrangian is given to be $(2)$. How to prove $(2)$ is also another Lagrangian for the given $H$? Any help will be appreciated.

Answer 1

The second Lagrangian is not equivalent to the given Hamiltonian. Let $$L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}q^2-\frac{3}{2}kq^2\dot{q}$$ Then the momentum is $$p=\frac{\partial L}{\partial \dot{q}}=m\dot{q}-\frac{3}{2}kq^2$$ Which means that the Hamiltonian is \begin{align} H &=p \dot{q}-L\\ &=m\dot{q}^2-\frac{3}{2}kq^2\dot{q}-\frac{1}{2}m\dot{q}^2+\frac{1}{2}kq^2+\frac{3}{2}kq^2 \dot{q}\\ &=\frac{1}{2}m\dot{q}^2+\frac{1}{2}kq^2 \end{align} And clearly: $$\frac{p^2}{2m}\neq \frac{1}{2}m\dot{q}^2$$

Answer 2

Considering

$$ L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}(q^2+3q^2\dot{q}) $$

we have $H = p\dot q - L$ and here $p = \frac{\partial L}{\partial \dot q} = m\dot q-\frac 32 k q^2$ so

$$ H(q,\dot q) = \frac 12(m \dot q^2+k q^2) $$

now eliminating $\dot q$ into

$$ H = \frac 12(m \dot q^2+k q^2)\\ p = m\dot q-\frac 32 k q^2 $$

we get

$$ H(p,q) = \frac{1}{8m}\left(2p+3kq^2)\right)^2+\frac k2 q^2\ne \frac{p^2}{2m}+\frac{k}{2}q^2 $$