# Coupled oscillators in Hamiltonian formalism - problem with diagonalization

I have a problem with simple coupled oscillator system. I tried to solve single oscillator with Hamiltonian, and then coupled system of two, but when I try to put coupling constant $$k^\prime=0$$ in my second solution, I don't get same result. I tried to diagonalize my problem and find solution in new coordinates.

Coupled harmonic oscillator can be described by kinetic energy $$$$T=\frac{p_{i}^{2}+p_{j}^{2}}{2m} \,. \label{eq:CS_kinE}$$$$ and potential energy, $$$$V=\frac{k_{i}+k^{\prime}}{2}\tilde{q}_{i}^{2}+\frac{k_{j}+k^{\prime}}{2}\tilde{q}_{j}^{2}-k^{\prime}\tilde{q}_{i}\tilde{q}_{j} \,, \label{eq:CS_potE}$$$$

The Hamiltonian of coupled harmonic oscillator thus reads as

$$$$\mathcal{H}(\tilde{q},\tilde{p})=\frac{\tilde{p}_{i}^{2}+\tilde{p}_{j}^{2}}{2m}+\frac{k_{i}+k^{\prime}}{2}\tilde{q}_{i}^{2}+\frac{k_{j}+k^{\prime}}{2}\tilde{q}_{j}^{2}-k^{\prime}\tilde{q}_{i}\tilde{q}_{j} \,. \label{eq:CS_hamiltonianTilde}$$$$ To make this calculations easier to read, lets asume $$k_i=k_j$$. Then, with an introduction of new coordinate and momentum \begin{alignat}{1} q_{k}=\sqrt{m\nu_{k}}\tilde{q}_{k} \,, \label{eq:CS_tildeQ} \\ p_{k}=\frac{\tilde{p}_{k}}{\sqrt{m\nu_{k}}} \,, \label{eq:CS_tildeP} \end{alignat} where $$\nu$$ is frequency of an oscillator spring constant $$k+k^\prime$$ and mass $$m$$, so that harmonic oscillator Hamiltonian becomes $$$$\label{eq:CS_hamiltonian} \mathcal{H}(q,p)=\frac{\nu_{i}}{2}\left[q_{i}^{2}+p_{i}^{2}\right]+\frac{\nu_{j}}{2}\left[q_{j}^{2}+p_{j}^{2}\right]-\kappa q_{i}q_{j} \,.$$$$ where $$\kappa=\frac{k^\prime}{m\sqrt{\nu_{i}\nu_{j}}}$$.

From the canonical relations we find that \begin{alignat}{1} \dot{q_{i}}&=\nu_{i} p_{i} \,, \label{eq:HE_motionCoord1} \\ \dot{p_{i}}&=-\nu_{i} q_{i}+\kappa q_{j} \,, \label{eq:HE_motionMom1} \\ \dot{q_{j}}&=\nu_{j} p_{j} \,, \label{eq:HE_motionCoord2} \\ \dot{p_{j}}&=-\nu_{j} q_{j}+\kappa q_{i} \,. \label{eq:HE_motionMom2} \end{alignat} We can represent this set of equations in matrix form as $$\dot{\mathbf{x}}=C\mathbf{x}$$ $$$$\frac{d}{dt} \begin{pmatrix} q_{i}\\ p_{i}\\ q_{j}\\ p_{j} \end{pmatrix} = \begin{pmatrix} 0 & \nu_{i} & 0 & 0\\ -\nu_{i} & 0 & \kappa & 0\\ 0 & 0 & 0 & \nu_{j}\\ \kappa & 0 & -\nu_{j} & 0 \end{pmatrix} \begin{pmatrix} q_{i}\\ p_{i}\\ q_{j}\\ p_{j} \end{pmatrix} \label{eq:HE_motionMatrix}$$$$

We can factorize matrix C as $$$$R^{-1}\dot{\mathbf{x}}=DR^{-1}\mathbf{x} \,, \label{eq:HE_diagMot}$$$$ where $$$$\label{eq:HE_eigenvectors} R=\frac{1}{2} \begin{pmatrix} i\sqrt{\frac{\nu}{\nu-\kappa}}&-i\sqrt{\frac{\nu}{\nu-\kappa}}&-i\sqrt{\frac{\nu}{\nu+\kappa}}&i\sqrt{\frac{\nu}{\nu+\kappa}}\\ 1&1&-1&-1\\ i\sqrt{\frac{\nu}{\nu-\kappa}}&-i\sqrt{\frac{\nu}{\nu-\kappa}}&i\sqrt{\frac{\nu}{\nu+\kappa}}&-i\sqrt{\frac{\nu}{\nu+\kappa}}\\ 1&1&1&1 \end{pmatrix} \,,$$$$ is the matrix whose columns are the eigenvector of C, and $$$$\label{eq:HE_eigenvalues} D= \begin{pmatrix} -i\sqrt{\nu^2-\kappa\nu}&0&0&0\\ 0&i\sqrt{\nu^2-\kappa\nu}&0&0\\ 0&0&-i\sqrt{\nu^2+\kappa\nu}&0\\ 0&0&0&i\sqrt{\nu^2+\kappa\nu} \end{pmatrix} %= %\begin{pmatrix} %-i\nu\sqrt{1-\frac{\kappa}{\nu}}&0&0&0\\ %0&i\nu\sqrt{1-\frac{\kappa}{\nu}}&0&0\\ %0&0&-i\nu\sqrt{1+\frac{\kappa}{\nu}}&0\\ %0&0&0&i\nu\sqrt{1+\frac{\kappa}{\nu}} %\end{pmatrix} \,,$$$$ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues. $$$$\label{eq:HE_eigenvectorsT} R^{-1}= \frac{1}{2} \begin{pmatrix} -i\sqrt{\frac{\nu-\kappa}{\nu}}&1&-i\sqrt{\frac{\nu-\kappa}{\nu}}&1\\ i\sqrt{\frac{\nu-\kappa}{\nu}}&1&i\sqrt{\frac{\nu-\kappa}{\nu}}&1\\ i\sqrt{\frac{\nu+\kappa}{\nu}}&-1&-i\sqrt{\frac{\nu+\kappa}{\nu}}&1\\ -i\sqrt{\frac{\nu+\kappa}{\nu}}&-1&i\sqrt{\frac{\nu+\kappa}{\nu}}&1 \end{pmatrix} \,.$$$$

With an introduction of new quantities $$\xi$$ defined as $$R^{-1}\mathbf{x}=\Xi$$ $$\begin{pmatrix} \xi_{I}\\ \xi_{II}\\ \xi_{III}\\ \xi_{IV} \end{pmatrix} =\frac{1}{2} \begin{pmatrix} -i\sqrt{\frac{\nu-\kappa}{\nu}}q_{i}-i\sqrt{\frac{\nu-\kappa}{\nu}}q_{j}+p_{i}+p_{j}\\ i\sqrt{\frac{\nu-\kappa}{\nu}}q_{i}+i\sqrt{\frac{\nu-\kappa}{\nu}}q_{j}+p_{i}+p_{j}\\ i\sqrt{\frac{\nu+\kappa}{\nu}}q_{i}-i\sqrt{\frac{\nu+\kappa}{\nu}}q_{j}-p_{i}+p_{j}\\ -i\sqrt{\frac{\nu+\kappa}{\nu}}q_{i}+i\sqrt{\frac{\nu+\kappa}{\nu}}q_{j}-p_{i}+p_{j} \end{pmatrix}$$ which are complexly conjugated, so that \begin{alignat*}{4} \xi_{i}&=\xi_{I} \hspace{0.25cm}&\text{and,}\hspace{0.25cm} &\xi_{i}^{\ast}&=\xi_{II}\,,\\ \xi_{j}&=\xi_{III} \hspace{0.25cm}&\text{and,}\hspace{0.25cm} &\xi_{j}^{\ast}&=\xi_{IV}\,, \end{alignat*} our equations of motions become \begin{alignat}{1} \label{eq:HE_motXi1} \dot{\xi}_{i}&=-i\sqrt{\nu^2-\kappa\nu}\xi_{i} \,, \\ \label{eq:HE_motXi1Ast} \dot{\xi}_{i}^{\ast}&=i\sqrt{\nu^2-\kappa\nu}\xi_{i}^{\ast} \,, \\ \label{eq:HE_motXi2} \dot{\xi}_{j}&=-i\sqrt{\nu^2+\kappa\nu}\xi_{j} \,, \\ \label{eq:HE_motXi2Ast} \dot{\xi}_{j}^{\ast}&=i\sqrt{\nu^2+\kappa\nu}\xi_{j}^{\ast} \,. \end{alignat} which can be easily solved. We can represent our coordinates and momentum with newly established quantities as $$\mathbf{x}=R\Xi \,,$$ which now reads as \begin{alignat}{1} q_{i}&=\frac{i}{2}\left[\sqrt{\frac{\nu}{\nu-\kappa}}(\xi_{i}-\xi_{i}^\ast)-\sqrt{\frac{\nu}{\nu+\kappa}}(\xi_{j}-\xi_{j}^\ast)\right] \,, \label{eq:HE_MotCoord1nsSol}\\ q_{j}&=\frac{i}{2}\left[\sqrt{\frac{\nu}{\nu-\kappa}}(\xi_{i}-\xi_{i}^\ast)+\sqrt{\frac{\nu}{\nu+\kappa}}(\xi_{j}-\xi_{j}^\ast)\right] \,, \label{eq:HE_MotCoord2nsSol}\\ p_i&=\xi_i+\xi_{i}^\ast-(\xi_j+\xi_{j}^\ast) \,, \label{eq:HE_MotMom1nsSol}\\ p_j&=\xi_i+\xi_{i}^\ast+\xi_j+\xi_{j}^\ast \,. \label{eq:HE_MotMom2nsSol} \end{alignat}

The main problem is that solution of single oscilator gives me

\begin{alignat}{1} q_{j}&=\frac{1}{\sqrt{2}}\left[\xi_{j}+\xi_{j}^{\ast}\right] \,, \label{eq:S_coordXi}\\ p_j&=\frac{i}{\sqrt{2}}\left[\xi_{j}-\xi_{j}^{\ast}\right] \,, \label{eq:S_momXi} \end{alignat}

which are with $$k^\prime=0$$ diferent solutions in signs and complexity. And I am desperately trying to find a mistake. Thanks in advance to everyone.

• Way you don’t use the eigen values and eigen vectors to diagonalized your matrix? – Eli Aug 3 '19 at 13:54
• I factorized matrix with eigen vectors. Matrix R is the matrix whose columns are the eigenvector of C, and D is the diagonal matrix whose diagonal elements are the corresponding eigenvalues. – Tomáš Červeň Aug 3 '19 at 14:09
• Please define all quantities. Do $i$ and $j$ refer to the two oscillators? What is $\tilde q$ (why is there a ~)? – DanielSank Aug 3 '19 at 16:45
• Unfortunately yes. But even if not, I would be just curious why it's different :/ – Tomáš Červeň Aug 4 '19 at 7:02
• Hi Frobenius. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Aug 12 '19 at 18:22

with: $$A=\left[ \begin {array}{cccc} 0&\vartheta &0&0\\ - \vartheta &0&\kappa&0\\ 0&0&0&\vartheta \\\kappa&0&-\vartheta &0\end {array} \right]$$
$$D= \left[ \begin {array}{c} i\sqrt {\vartheta }\sqrt {\vartheta +\kappa} \\ -i\sqrt {\vartheta }\sqrt {\vartheta +\kappa} \\ \sqrt {\vartheta }\sqrt {-\vartheta +\kappa} \\ -\sqrt {\vartheta }\sqrt {-\vartheta +\kappa} \end {array} \right]$$ and
$$R= \left[ \begin {array}{cccc} -1/2\,{\frac {\sqrt {2}\sqrt {\vartheta } }{\sqrt {2\,\vartheta +\kappa}}}&-1/2\,{\frac {\sqrt {2}\sqrt { \vartheta }}{\sqrt {2\,\vartheta +\kappa}}}&1/2\,{\frac {\sqrt {2} \sqrt {\vartheta }}{\sqrt {\kappa}}}&1/2\,{\frac {\sqrt {2}\sqrt { \vartheta }}{\sqrt {\kappa}}}\\ {\frac {-1/2\,i \sqrt {2}\sqrt {\vartheta +\kappa}}{\sqrt {2\,\vartheta +\kappa}}}&{ \frac {1/2\,i\sqrt {2}\sqrt {\vartheta +\kappa}}{\sqrt {2\,\vartheta + \kappa}}}&1/2\,{\frac {\sqrt {2}\sqrt {-\vartheta +\kappa}}{\sqrt { \kappa}}}&-1/2\,{\frac {\sqrt {2}\sqrt {-\vartheta +\kappa}}{\sqrt { \kappa}}}\\ 1/2\,{\frac {\sqrt {2}\sqrt {\vartheta } }{\sqrt {2\,\vartheta +\kappa}}}&1/2\,{\frac {\sqrt {2}\sqrt { \vartheta }}{\sqrt {2\,\vartheta +\kappa}}}&1/2\,{\frac {\sqrt {2} \sqrt {\vartheta }}{\sqrt {\kappa}}}&1/2\,{\frac {\sqrt {2}\sqrt { \vartheta }}{\sqrt {\kappa}}}\\ {\frac {1/2\,i\sqrt {2}\sqrt {\vartheta +\kappa}}{\sqrt {2\,\vartheta +\kappa}}}&{\frac {- 1/2\,i\sqrt {2}\sqrt {\vartheta +\kappa}}{\sqrt {2\,\vartheta +\kappa} }}&1/2\,{\frac {\sqrt {2}\sqrt {-\vartheta +\kappa}}{\sqrt {\kappa}}}& -1/2\,{\frac {\sqrt {2}\sqrt {-\vartheta +\kappa}}{\sqrt {\kappa}}} \end {array} \right]$$
so $$R^{-1}\,A\,R=D$$