Derive the quantised Hamiltonian corresponding to the charged Klein-Gordon equation


 * 1) Set the Klein-Gordon equation, \(m^2c^4\phi = \hbar^2c^2\nabla^2\phi - \hbar^2\partial_t^2\phi\), in natural units: \(m^2\phi = \nabla^2\phi - \partial_t^2\phi\)
 * 2) Show that \(\mathcal{L}=\frac{1}{2}\left(\partial^{\mu}\phi\right)^{\dagger}\left(\partial_{\mu}\phi\right)-\frac{1}{2}m^2\phi^{\dagger}\phi\) is the Lagrangian representation of the Klein-Gordon equation.
 * 3) Separate the time and space derivatives: \(\mathcal{L}=\frac{1}{2}\left(\left(\partial_t\phi\right)^{\dagger}\left(\partial_t\phi\right) - \left(\nabla\phi\right)^{\dagger}\left(\nabla\phi\right)\right)-\frac{1}{2}m^2\phi^{\dagger}\phi\)
 * 4) Treating \(\phi\) and \(\phi^{\dagger}\) as independent fields, input these Lagrangians into the Euler-Lagrange equation, \(\frac{\partial \mathcal{L}}{\partial\phi} - d_t\frac{\partial \mathcal{L}}{\partial\dot{\phi}} - \nabla\frac{\partial \mathcal{L}}{\partial\nabla\phi} = 0\) (and its conjugate): \(\frac{\partial}{\partial\phi}\frac{-1}{2}m^2\phi^{\dagger}\phi\ - d_t\frac{\partial}{\partial\dot{\phi}}\frac{1}{2}\dot{\phi}^{\dagger}\dot{\phi}\ - \nabla\frac{\partial}{\partial\nabla\phi}\frac{-1}{2}\left(\nabla\phi^{\dagger}\phi\right) = 0\) \(\frac{\partial}{\partial\phi^{\dagger}}\frac{-1}{2}m^2\phi^{\dagger}\phi\ - d_t\frac{\partial}{\partial\dot{\phi}^{\dagger}}\frac{1}{2}\dot{\phi}^{\dagger}\dot{\phi}\ - \nabla\frac{\partial}{\partial\nabla\phi^{\dagger}}\frac{-1}{2}\left(\nabla\phi^{\dagger}\phi\right) = 0\)
 * 5) Solve and sum together the two equations: \(- m^2\phi - \ddot{\phi} + \nabla^2\phi=0\)
 * 6) Find the equivalent Hamiltonian.
 * 7) Find the canonical momentum: \(\Pi = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = i\hbar\phi^{\dagger}\) \(\Pi^{\dagger} = \frac{\partial \mathcal{L}}{\partial \dot{\phi}^{\dagger}} = \dot{\phi}\)
 * 8) Solve for the Hamiltonian density, \(\mathcal{H}=\Pi\dot{\phi} + \Pi^{\dagger}\dot{\phi}^{\dagger} - \mathcal{L}\): \(\mathcal{H}=2\Pi^{\dagger}\Pi + \nabla\phi^{\dagger}\nabla\phi + m^2\phi^{\dagger}\phi\)
 * 9) Integrate over \(x\) to find the Hamiltonian: \(\hat{\mathbf{H}}=\int dx\left(2\Pi^{\dagger}\Pi + \nabla\phi^{\dagger}\nabla\phi + m^2\phi^{\dagger}\phi\right)\)
 * 10) Input the general solutions, \(\phi(x,t)=\int \frac{dk}{(2\pi)^3 2\omega_k}\left[\hat{a}_k(t)e^{-ik\cdot x} + \hat{b}_k^{\dagger}(t)e^{ik\cdot x}\right]\) \(\Pi(x,t)=\int \frac{dk}{(2\pi)^3 2\omega_k}i\omega_k\left[\hat{a}_k(t)e^{-ik\cdot x} + \hat{b}_k^{\dagger}(t)e^{ik\cdot x}\right]\) where \(\omega_k = k^2 + m^2\) into the Hamiltonian: \(\hat{\mathbf{H}}=\frac{1}{2}\,\int dx\int \int dk dq \frac{1}{(2\pi)^6 4\omega_k\omega_q}\left(\omega_k\omega_q \left(\hat{a}_k(t)e^{-ik\cdot x} - \hat{b}_k^{\dagger}(t)e^{ik\cdot x}\right)\left(\hat{a}_q(t)e^{-iq\cdot x} - \hat{b}_q^{\dagger}(t)e^{iq\cdot x}\right)\right.\) \( + \left(-ik\hat{a}_k(t)e^{-ik\cdot x} + \hat{b}_k^{\dagger}(t)e^{ik\cdot x}\right)\left(iq\hat{a}_q(t)e^{-iq\cdot x} - iq\hat{b}_q^{\dagger}(t)e^{iq\cdot x}\right)\) \(\left. m^2\left(\hat{a}_k(t)e^{-ik\cdot x} + \hat{b}_k^{\dagger}(t)e^{ik\cdot x}\right)\left(\hat{a}_q(t)e^{-iq\cdot x} + \hat{b}_q^{\dagger}(t)e^{iq\cdot x}\right) \right)\)
 * 11) Simplify: \(\hat{\mathbf{H}}=\int\frac{dk}{2(2\pi)^3}\left[\hat{a}_k\hat{a}_k^{\dagger} + \hat{b}^{\dagger}\hat{b}\right]\)