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 \begin{equation} T=\frac{p_{i}^{2}+p_{j}^{2}}{2m} \,. \label{eq:CS_kinE} \end{equation} and potential energy, \begin{equation} 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} \end{equation}

The Hamiltonian of coupled harmonic oscillator thus reads as

\begin{equation} \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} \end{equation} 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 \begin{equation} \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} \,. \end{equation} 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 $$ \dot{\mathbf{x}}=C\mathbf{x} $$ \begin{equation} \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} \end{equation}

We can factorize matrix C as \begin{equation} R^{-1}\dot{\mathbf{x}}=DR^{-1}\mathbf{x} \,, \label{eq:HE_diagMot} \end{equation} where \begin{equation} \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} \,, \end{equation} is the matrix whose columns are the eigenvector of C, and \begin{equation} \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} \,, \end{equation} is the diagonal matrix whose diagonal elements are the corresponding eigenvalues. \begin{equation} \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} \,. \end{equation}

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.

  • $\begingroup$ Way you don’t use the eigen values and eigen vectors to diagonalized your matrix? $\endgroup$
    – Eli
    Commented Aug 3, 2019 at 13:54
  • $\begingroup$ 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. $\endgroup$ Commented Aug 3, 2019 at 14:09
  • $\begingroup$ Please define all quantities. Do $i$ and $j$ refer to the two oscillators? What is $\tilde q$ (why is there a ~)? $\endgroup$
    – DanielSank
    Commented Aug 3, 2019 at 16:45
  • $\begingroup$ Unfortunately yes. But even if not, I would be just curious why it's different :/ $\endgroup$ Commented Aug 4, 2019 at 7:02
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Aug 12, 2019 at 18:22

1 Answer 1


with: $$A=\left[ \begin {array}{cccc} 0&\vartheta &0&0\\ - \vartheta &0&\kappa&0\\ 0&0&0&\vartheta \\\kappa&0&-\vartheta &0\end {array} \right] $$

I get:

$$D= \left[ \begin {array}{c} i\sqrt {\vartheta }\sqrt {\vartheta +\kappa} \\ -i\sqrt {\vartheta }\sqrt {\vartheta +\kappa} \\ \sqrt {\vartheta }\sqrt {-\vartheta +\kappa} \\ -\sqrt {\vartheta }\sqrt {-\vartheta +\kappa} \end {array} \right] $$ 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 $$R^{-1}\,A\,R=D$$

  • $\begingroup$ @Frobenius I use my own program (Maple) to do the calculations, so you did a benchmark for my program! $\endgroup$
    – Eli
    Commented Aug 12, 2019 at 18:34

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