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 $E = (n_x + 1/2) \hbar \omega_1+(n_y + 1/2) \hbar \omega_2+(n_z + 1/2) \hbar \omega_3$?


1 Answer 1


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
    Commented 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
    Commented 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 $\ddot Q_i=-\omega_i^2 Q_i$. $\endgroup$ Commented Aug 1, 2019 at 3:10

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