Discussion
If the Hamiltonian of the dynamical system is
\[ \begin{aligned} &H(q_{1}, q_{2}, p_{1}, p_{2}) \\ &= p_{1}^{2} + p_{2}^{2} - k\, q_{1}^{2}, \end{aligned} \]
then the Lagrangian is given by :
Answer: C
The Lagrangian is \(L = \dfrac{\dot{q}_{1}^{2}}{4} + \dfrac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2}\).
Hamilton's equation \(\dot{q}_{i} = \partial H / \partial p_{i} = 2p_{i}\) gives \(p_{i} = \dot{q}_{i}/2\). Then the Legendre transform \(L = \sum p_{i}\dot{q}_{i} - H\) gives
\[ \begin{aligned} L &= \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{2} - \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{4} + kq_{1}^{2} \\ &= \frac{\dot{q}_{1}^{2}}{4} + \frac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2} \end{aligned} \]
The sign of \(kq_{1}^{2}\) flips because it enters H as part of the potential (\(H = T + V\), \(L = T - V\) with \(V = -kq_{1}^{2}\)). Options with \(-kq_{1}^{2}\) miss that, and those with denominators 2 or 1 use the wrong relation between \(p\) and \(\dot{q}\).
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