Quantise the coupled harmonic oscillator

by diagonalising and then quantising

 * 1) Find the Lagrangian for a system of two identical oscillators with mass \(m\) displaced by \(x\) and \(y\) respectively coupled by a spring with spring constant \(\lambda\): \(\mathcal{L} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\dot{y}^2 - \frac{1}{2}m\omega^2x^2 - \frac{1}{2}m\omega^2y^2-\frac{1}{2}\lambda\left(y-x\right)^2\)
 * 2) Make the change of variables \(x=X+Y\) and \(y=X-Y\): \(\mathcal{L} = \frac{1}{2}m\left(\dot{X}+\dot{Y}\right)^2 + \frac{1}{2}m\left(\dot{X}-\dot{Y}\right)^2 - \frac{1}{2}m\omega^2\left(X+Y\right)^2 - \frac{1}{2}m\omega^2\left(X-Y\right)^2-\frac{1}{2}\lambda\left(\left(X-Y\right)-\left(X+Y\right)\right)^2\)
 * 3) Simplify: \(\mathcal{L} = m\dot{X}^2 + m\dot{Y}^2 - m\omega^2X^2 - m\omega^2Y^2 + \lambda Y^2\)
 * 4) Write in terms of \(X\) and \(Y\) and their derivatives: \(\mathcal{L} = \left[m\dot{X}^2 - m\omega^2X^2\right] + \left[m\dot{Y}^2 - \left(m\omega^2 + \lambda\right)Y^2\right]\)
 * 5) Find the ladder operators of the oscillator.
 * 6) Find the momenta conjugate to \(X\) and \(Y\), \(P\) and \(Q\): \(P_X = \frac{\partial \mathcal{L}}{\partial \dot{X}} = 2m\dot{X}\) \(Q_Y = \frac{\partial \mathcal{L}}{\partial \dot{Y}} = 2m\dot{Y}\)
 * 7) Write the standard ladder operators in terms of \(X\), \(P\), \(Y\) and \(Q\): \(\hat{A} = \sqrt{\frac{m\omega}{\hbar}} \left ( \hat{X} + i\frac{\hat{P}}{2m\omega} \right ) \) \(\hat{B} = \sqrt{\frac{m\omega\prime}{\hbar}} \left ( \hat{Y} + i\frac{\hat{Q}}{2m\omega\prime} \right ) = \sqrt{\frac{m\sqrt{\omega^2-2\lambda/m}}{\hbar}} \left ( \hat{Y} + i\frac{\hat{Q}}{2m\sqrt{\omega^2-2\lambda/m}} \right ) \)
 * 8) Write the position and momentum operators in terms of the ladder operators: \(\hat{X}=\frac{1}{2}\sqrt{\frac{\hbar}{m\omega}}\left(\hat{A} + \hat{A}^{\dagger}\right)\) \(\hat{P}=\frac{m\omega}{i}\sqrt{\frac{\hbar}{m\omega}}\left(\hat{A} - \hat{A}^{\dagger}\right)\) \(\hat{Y}=\frac{1}{2}\sqrt{\frac{\hbar}{m\omega\prime}}\left(\hat{B} + \hat{B}^{\dagger}\right)\) \(\hat{Q}=\frac{m\omega\prime}{i}\sqrt{\frac{\hbar}{m\omega\prime}}\left(\hat{B} - \hat{B}^{\dagger}\right)\)
 * 9) Find the Hamiltonian density \(\hat{\mathcal{H}} = \dot{\hat{X}}\hat{P} + \dot{\hat{Y}}\hat{Q} - \mathcal{L}\): \(\hat{\mathcal{H}} = \frac{1}{4m}\hat{P}^2 + m\omega^2 \hat{X}^2 + \frac{1}{4m}\hat{Q}^2 + m\omega\prime^2 \hat{Y}^2\)
 * 10) Write this in terms of ladder operators: \(\hat{\mathcal{H}} = \frac{-\hbar\omega}{4}\left(\hat{A}-\hat{A}^{\dagger}\right)^2 + \frac{\hbar\omega}{4}\left(\hat{A}+\hat{A}^{\dagger}\right)^2 + \frac{-\hbar\omega\prime}{4}\left(\hat{B}-\hat{B}^{\dagger}\right)^2 + \frac{\hbar\omega\prime}{4} \left(\hat{B}+\hat{B}^{\dagger}\right)\hat{Y}^2\)
 * 11) Simplify: \(\hat{\mathcal{H}} = \hbar\omega\left(\hat{A}^{\dagger}\hat{A} + \frac{1}{2}\right) + \hbar\omega\prime\left(\hat{B}^{\dagger}\hat{B} + \frac{1}{2}\right)\)

by quantising and then diagonalising

 * 1) Find the Lagrangian for a system of two identical oscillators with mass \(m\) displaced by \(x\) and \(y\) respectively coupled by a spring with spring constant \(\lambda\): \(\mathcal{L} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\dot{y}^2 - \frac{1}{2}m\omega^2x^2 - \frac{1}{2}m\omega^2y^2-\frac{1}{2}\lambda\left(y-x\right)^2\)
 * 2) Find the momenta conjugate to \(x\) and \(y\): \(p_x = \frac{\partial \mathcal{L}}{\partial \dot{x}}=m\dot{x}\) \(q_y = \frac{\partial \mathcal{L}}{\partial \dot{y}}=m\dot{y}\)
 * 3) Identify the ladder operators for this Hamiltonian: \(\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} + i\frac{\hat{p}}{m\omega} \right ) \) \(\hat{a}^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{x} - i\frac{\hat{p}}{m\omega} \right ) \) \(\hat{b} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{y} + i\frac{\hat{q}}{m\omega} \right ) \) \(\hat{b}^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} \left ( \hat{y} - i\frac{\hat{q}}{m\omega} \right ) \)
 * 4) Write the position and momentum operators in terms of the ladder operators: \(\hat{x}=\frac{1}{2}\sqrt{\frac{2\hbar}{m\omega}} \left ( \hat{a} + \hat{a}^{\dagger} \right ) \) \(\hat{p}=\frac{m\omega}{2i}\sqrt{\frac{2\hbar}{m\omega}} \left ( \hat{a} - \hat{a}^{\dagger} \right ) \) \(\hat{y}=\frac{1}{2}\sqrt{\frac{2\hbar}{m\omega}} \left ( \hat{b} + \hat{b}^{\dagger} \right ) \) \(\hat{q}=\frac{m\omega}{2i}\sqrt{\frac{2\hbar}{m\omega}} \left ( \hat{b} - \hat{b}^{\dagger} \right ) \)
 * 5) Write the Hamiltonian in terms of the ladder operators.
 * 6) Find the Hamiltonian density \(\hat{\mathcal{H}} = \dot{\hat{X}}\hat{P} + \dot{\hat{Y}}\hat{Q} - \mathcal{L}\): \(\hat{\mathcal{H}}=\frac{\hat{p}_x^2}{2m} + \frac{\hat{q}_x^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 + \frac{1}{2}m\omega^2\hat{y}^2 + \frac{1}{2}\lambda\left(\hat{y}-\hat{x}\right)\)
 * 7) Substitute in the expressions for the position and momentum operators in terms of the ladder operators: \(\hat{\mathcal{H}}=\frac{\hbar\omega}{2}\left(\hat{a}^{\dagger}\hat{a} + \hat{a}\hat{a}^{\dagger}\right) + \frac{\hbar\omega}{2}\left(\hat{b}^{\dagger}\hat{b} + \hat{b}\hat{b}^{\dagger}\right) + \frac{1}{4}\frac{\lambda\hbar}{m\omega}\left(\hat{a} + \hat{a}^{\dagger} -\hat{b} - \hat{b}^{\dagger} \right)\)
 * 8) Use matrices to simplify: \(\hat{\mathcal{H}}=\begin{pmatrix}\hat{a}^{\dagger} & \hat{a} & \hat{b}^{\dagger} & \hat{b}\end{pmatrix} \left[\frac{\hbar\omega}{2}\mathbb{I} + \frac{1}{4}\frac{\lambda\hbar}{m\omega}\begin{pmatrix}1&1&-1&-1 \\ 1&1&-1&-1 \\ -1&-1&1&1 \\ -1&-1&1&1\end{pmatrix}\right]\begin{pmatrix}\hat{a} \\ \hat{a}^{\dagger} \\ \hat{b} \\ \hat{b}^{\dagger}\end{pmatrix}\)
 * 9) Diagonalise.
 * 10) Make the change of variable \(\hat{C}=\frac{\left(\hat{a}+\hat{b}\right)}{\sqrt{2}}\) and \(\hat{D}=\frac{\left(\hat{a}-\hat{b}\right)}{\sqrt{2}}\): \(\begin{pmatrix}\hat{a}^{\dagger} & \hat{a} & \hat{b}^{\dagger} & \hat{b}\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}\hat{C}^{\dagger} & \hat{C} & \hat{D}^{\dagger} & \hat{D}\end{pmatrix}\begin{pmatrix}1&0&1&0 \\ 0&1&0&1 \\ 1&0&-1&0 \\ 0&1&0&-1\end{pmatrix}\) \(\begin{pmatrix}\hat{a}^{\dagger} \\ \hat{a} \\ \hat{b}^{\dagger} \\ \hat{b}\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&0&1&0 \\ 0&1&0&1 \\ 1&0&-1&0 \\ 0&1&0&-1\end{pmatrix}\begin{pmatrix}\hat{C} \\ \hat{C}^{\dagger} \\ \hat{D} \\ \hat{D}^{\dagger}\end{pmatrix}\)
 * 11) Substitute this into the Hamiltonian: \(\hat{\mathcal{H}}=\begin{pmatrix}\hat{C}^{\dagger} & \hat{C} & \hat{D}^{\dagger} & \hat{D}\end{pmatrix} \left[\frac{\hbar\omega}{2}\mathbb{I} + \frac{1}{2}\frac{\lambda\hbar}{m\omega}\begin{pmatrix}0&0&0&0 \\ 0&0&0&0 \\ 0&0&1&1 \\ 0&0&1&1\end{pmatrix}\right]\begin{pmatrix}\hat{C} \\ \hat{C}^{\dagger} \\ \hat{D} \\ \hat{D}^{\dagger}\end{pmatrix}\)
 * 12) Simplify: \(\hat{\mathcal{H}}=\begin{pmatrix}\hat{C}^{\dagger} & \hat{C}\end{pmatrix}\frac{\hbar\omega}{2}\begin{pmatrix}\hat{C} \\ \hat{C}^{\dagger}\end{pmatrix} + \begin{pmatrix}\hat{D}^{\dagger} & \hat{D}\end{pmatrix}\left(\frac{\hbar\omega}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} + \frac{\lambda\hbar}{2m\omega} \begin{pmatrix}1&1 \\ 1&1\end{pmatrix}\right)\begin{pmatrix}\hat{D} \\ \hat{D}^{\dagger}\end{pmatrix}\)
 * 13) Make a second change of variable, \(\begin{pmatrix}\hat{B} \\ \hat{B}^{\dagger}\end{pmatrix} = \begin{pmatrix}\alpha & \beta \\ \beta & \alpha \end{pmatrix}\begin{pmatrix}\hat{D} \\ \hat{D}^{\dagger}\end{pmatrix}\), where \(\alpha + \beta = \sqrt{\frac{\sqrt{\omega^2+2\lambda/m}}{\omega}}\) and \(\alpha^2 - \beta^2 = 1\): \(\begin{pmatrix}\hat{D} \\ \hat{D}^{\dagger}\end{pmatrix} = \begin{pmatrix}\alpha & -\beta \\ -\beta & \alpha \end{pmatrix}\begin{pmatrix}\hat{B} \\ \hat{B}^{\dagger}\end{pmatrix}\)
 * 14) Substitute this expression into the Hamiltonian: \(\hat{\mathcal{H}}=\begin{pmatrix}\hat{C}^{\dagger} & \hat{C}\end{pmatrix}\frac{\hbar\omega}{2}\begin{pmatrix}\hat{C} \\ \hat{C}^{\dagger}\end{pmatrix} + \begin{pmatrix}\hat{B}^{\dagger} & \hat{B}\end{pmatrix}\frac{\hbar \sqrt{\omega^2+2\lambda/m}}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}\hat{D} \\ \hat{D}^{\dagger}\end{pmatrix}\)
 * 15) Simplify, setting \(\omega\prime = \sqrt{\omega^2+2\lambda/m}\): \(\hat{\mathcal{H}} = \hbar\omega\left(\hat{C}^{\dagger}\hat{C} + \frac{1}{2}\right) + \hbar\omega\prime\left(\hat{B}^{\dagger}\hat{B} + \frac{1}{2}\right)\)