Qubit Physics

Starting from \(L(\dot{Q}, Q)\), based on the antisymmetry between the kinetic and potential energy in the Lagrangian, we manually introduce a negative sign when defining the canonical momentum \(\Phi\) corresponding to \(\dot{Q}\) to absorb this antisymmetry (otherwise, the resulting Hamiltonian would be negative): \[ \Phi \equiv -\frac{\partial \mathcal{L}}{\partial \dot{Q}} \implies \dot{\Phi} = -\frac{\partial \mathcal{L}}{\partial Q}. \] Next, we switch the Lagrangian to the \((\dot{\Phi}, \Phi)\) space, i.e., the momentum-space Lagrangian \(\bar{\mathcal{L}}\): \[ \bar{\mathcal{L}} \equiv \mathcal{L} + \frac{d}{dt} (\Phi Q) = \mathcal{L} + \dot{\Phi} Q + \Phi \dot{Q}. \]

Using differential techniques, we can prove that \(\bar{\mathcal{L}} = \bar{\mathcal{L}}(\dot{\Phi}, \Phi)\) is indeed a function of \((\dot{\Phi}, \Phi)\) space: \[ d\bar{\mathcal{L}} = d\mathcal{L} + d(\dot{\Phi} Q) + d(\Phi \dot{Q}) = \frac{\partial \mathcal{L}}{\partial \dot{Q}} d\dot{Q} + \frac{\partial \mathcal{L}}{\partial Q} dQ + Qd\dot{\Phi} + \dot{\Phi}dQ + \dot{Q}d\Phi + \Phi d\dot{Q}. \] Expanding further, since \(-\Phi = \frac{\partial \mathcal{L}}{\partial \dot{Q}}\), we have: \[ d\bar{\mathcal{L}} = -\Phi d\dot{Q} - \dot{\Phi} dQ + Q d\dot{\Phi} + \dot{\Phi} dQ + \dot{Q} d\Phi + \Phi d\dot{Q} = Q d\dot{\Phi} + \dot{Q} d\Phi. \]

The corresponding canonical momentum \(\Pi\) for \(\dot{\Phi}\) is: \[ \Pi = \frac{\partial \bar{\mathcal{L}}}{\partial \dot{\Phi}} = Q, \] which is the same as the original \(Q\). The corresponding Hamiltonian \(\bar{\mathcal{H}} \) is defined as: \[ \bar{\mathcal{H}} = \Pi \dot{\Phi} - \bar{\mathcal{L}} = Q\dot{\Phi} - \mathcal{L} - \dot{\Phi} Q - \Phi \dot{Q} = -\mathcal{L} - \Phi \dot{Q}. \] The final form is: \[ \bar{\mathcal{H}} (Q, \Phi) = -\mathcal{L} - \Phi \dot{Q}. \]

This allows us to simplify subsequent derivations by following these steps:

1. Write down \(L(\dot{Q}, Q)\).
2. Define \(\Phi \equiv -\frac{\partial \mathcal{L}}{\partial \dot{Q}}\) and establish the relationship between \(\dot{Q}\) and \(\Phi\).
3. Directly calculate the corresponding Hamiltonian: \[ \bar{\mathcal{H}} (Q, \Phi) = -\mathcal{L} - \Phi \dot{Q}, \] substituting \(\dot{Q}\) from the previous relationship.
4. Quantization condition: \[ [Q, \Phi] = i\hbar. \]

This eliminates the need for the cumbersome transformation \(L(\dot{Q}, Q) \to \bar{\mathcal{L}}(\dot{\Phi}, \Phi) \to \bar{\mathcal{H}} (Q, \Phi)\). Especially when introducing external coupling terms (e.g., in deriving how to drive and read quantum bits, external signals interacting with quantum bits, or coupling between two quantum bits), following the traditional approach would lead to extremely complex intermediate calculations, similar to the momentum shift \(p \to p - eA\) in an electromagnetic field.

Note: \[ \bar{\mathcal{H}} (Q, \Phi) = -\mathcal{L} - \Phi \dot{Q} = -\mathcal{L} + \frac{\partial \mathcal{L}}{\partial \dot{Q}} \dot{Q} = H(\Phi, Q). \] \(H(\Phi, Q)\) is the familiar Hamiltonian in classical mechanics. Moreover, \(\bar{\mathcal{H}} = H\) reaffirms the symmetry between kinetic and potential energy in the Hamiltonian.

Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.