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In the Cohen-Tannoudji QM book they say that the potential $$V(x)=\frac{1}{2}m\omega^2(x_1-a)^2 + \frac{1}{2}m\omega^2(x_2+a)^2 + \lambda m\omega^2(x_1-x_2)^2$$ describes two classical coupled harmonic oscillator ($a, \lambda$ are parameters). I tried to come up with a physical system with this potential and I found it worked if I took the usual horizontal coupled spring system (wall - spring - mass - spring - mass - spring - wall) with all the masses and stiffnesses and rest lengths equal, except the center spring has rest length set to $0$ and stiffness $k'$ (Then $\lambda^2=k'/k)$ . Then the origin of $x$ axis is the middle point between the walls and $a$ is the distance from the origin to the equilibrium point of the left or right spring.

This is the only reasonable system to me because the potential predicts that in the equilibrium position of the system, the two mass are distance of $\frac{a}{1+4\lambda}$ apart, which I wouldn't understand if the center spring had a positive rest length.

Is that correct? Is there a more physical system with such a potential? Thanks.

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  • $\begingroup$ are the 1st and 2nd term only dependent on $x_1$? If yes just expand them out to simplify... $\endgroup$ – ZeroTheHero Aug 29 '20 at 16:45
  • $\begingroup$ sorry i fixed the typo $\endgroup$ – student_du_05 Aug 29 '20 at 16:46
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Ignore the first two terms for a second. Then the energy is minimized when $x_1=x_2$, i.e. the two masses are connected by a spring with zero rest length. Of course, springs in real life have a positive rest length, but it is useful to think about imaginary ideal springs with zero rest length (one less parameter in the problem).

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  • $\begingroup$ Yes but the meaning of $x_1$, $x_2$ could depend on the system. They could describe deviations from an equilibrium for instance $\endgroup$ – student_du_05 Aug 29 '20 at 16:28
  • $\begingroup$ Regardless, if $H= \lambda m \omega^2(x_1-x_2)^2$ then the energy is minimized when $x_1=x_2$. Do you agree? $\endgroup$ – Dwagg Aug 29 '20 at 18:03
  • $\begingroup$ sure. 15 characters. $\endgroup$ – student_du_05 Aug 29 '20 at 18:28

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