Let's assume we are given a potential for coupled harmonic oscillator:

$$U = \frac{k_1(x_1^2 +x_3^2)+k_2 x^2+k_3 (x_1x_2 + x_2x_3)}{2}$$

If I solve the normal modes of the oscillator I get the frequency

$$\omega_1 = \sqrt{\frac{k_1}{m}}$$ $$\omega_2 = \sqrt{\frac{k_1+k_2}{m}}$$ $$\omega_3 = 0$$

After i have done studying this problem, I wanted to study this in terms of Quantum mechanics. where we can write the full Hamiltonian:

$$H = \frac{1}{2} m\dot x_1^2 + \frac{1}{2} m\dot x_2^2 + \frac{1}{2} m\dot x_3^2 +\frac{k_1(x_1^2 +x_3^2)+k_2 x^2+k_3 (x_1x_2 + x_2x_3)}{2}$$

What would be the energy Eigenvalue of the system? Can I follow the study (i mean the frequencies that we have found already) and write down the energy equation?

Will the energy Eigen value be $E = (n_x + 1/2) \hbar \omega_1+(n_y + 1/2) \hbar \omega_2+(n_z + 1/2) \hbar \omega_3$?


2 Answers 2


Presumably you will have a kinetic part in $\dot{x}_3^2\sim p_3^2$. Assuming that much the transformation from the $x_i$'s to the generalized coordinates $Q_i$'s that decouple the equations of motion should also bring your Hamiltonian to the form of a simple sum $$ H=\sum_i \frac{P_i^2}{2m}+\frac{1}{2}m\omega_i^2 Q_i^2 $$ from which you can find (easily) the eigenvalues.

  • $\begingroup$ Thanks, I have updated that accordingly. $\endgroup$
    – user193422
    Aug 1, 2019 at 1:25
  • $\begingroup$ One question, Do you mean $\dot Q_i= x_1, x_2, x_3$? and also did you mean that the energy eigen value would be: $E = (n_x + 1/2) \hbar \omega_1+(n_y + 1/2) \hbar \omega_2+(n_z + 1/2) \hbar \omega_3 $ $\endgroup$
    – user193422
    Aug 1, 2019 at 1:41
  • 1
    $\begingroup$ no I mean that the normal coordinates $Q_i$ will be linear combinations of the original coordinates $x_j$ so that the new coordinates satisfy $\ddot Q_i=-\omega_i^2 Q_i$. $\endgroup$ Aug 1, 2019 at 3:10

more details :

The equations of motion in your case are:

$$\ddot{\vec{q}}+\frac{\partial U}{\partial \vec{q}}=0\tag 1$$

where $\vec{q}$ is the vector of generalized coordinates and $U$ is the potential energy

if $\frac{\partial U}{\partial \vec{q}}$ is a linear function of $\vec{q}$ you can write equation (1) as:

$$\ddot{\vec{q}}+C\,\vec{q}=0\tag 2$$

where $C$ is a quadratic constant matrix :

$C=\frac{\partial}{\partial \vec{q}}\left(\frac{\partial U}{\partial \vec{q}}\right)$

to transformed equation (2) to diagonal shape with the transformation matrix $T$, we calculate the eigenvalues and the eigenvectors of the matrix $C$.

with $\vec{q}=T\,\vec{Q}$ and $T=\left[\vec{E}_{v1}(\lambda_1)\,,\vec{E}_{v2}(\lambda_2)\,,\ldots\right]$ the transformation matrix. ,where $\lambda_i$ are the eigenvalues of the matrix $C$ and $\vec{E}_{vi}$ are the eigenvectors. we obtain for equation (2)

$$T\ddot{\vec{Q}}+C\,T\,\vec{Q}=0\tag 3$$

multiply equation (3) with $T^T$ we get:

$$T^T\,T\ddot{\vec{Q}}+T^T\,C\,T\,\vec{Q}=0\tag 4$$

because $T^T\,T=I$ unity matrix and $T^T\,C\,T=\text{Diag}\left[\lambda_1\,,\lambda_2\,,\ldots\,,\lambda_n\right]$ diagonal shape , we get



$U=1/2\,{\it k1}\, \left( {x_{{1}}}^{2}+{x_{{3}}}^{2} \right) +1/2\,{\it k2}\,{x_{{2}}}^{2}+1/2\,{\it k3}\, \left( x_{{1}}x_{{2}}+x_{{2}}x_{{3} } \right) $


$C=\left[ \begin {array}{ccc} {\it k1}&1/2\,{\it k3}&0 \\ 1/2\,{\it k3}&{\it k2}&1/2\,{\it k3} \\ 0&1/2\,{\it k3}&{\it k1}\end {array} \right] $


$\vec{\lambda}=\left[ \begin {array}{c} {\it k1}\\ 1/2\,{\it k1}+1 /2\,{\it k2}+1/2\,\sqrt {{{\it k1}}^{2}-2\,{\it k2}\,{\it k1}+{{\it k2 }}^{2}+2\,{{\it k3}}^{2}}\\ 1/2\,{\it k1}+1/2\,{\it k2}-1/2\,\sqrt {{{\it k1}}^{2}-2\,{\it k2}\,{\it k1}+{{\it k2}}^{2}+2 \,{{\it k3}}^{2}}\end {array} \right] $


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