1
$\begingroup$

Quantum mechanically, a quartic anharmonic oscillator with potential $$V(x)=\frac{1}{2}m\omega^2x^2+\lambda x^4$$ is dealt with perturbation theory- the approximate energies $E_n$ and energy eigenstates $|\phi_n\rangle$ are obtained using the time-independent perturbation theory. Classically, the problem amounts to solving the trajectory $x(t)$. At this point, one is stuck with a nonlinear differential equation which cannot be solved in closed analytical form. How do we go about solving such problems, classically? Do we use similar perturbation technique to obtain corrections to the trajectory order by order? Any suggestions? In short. I am curious about the classical behaviour of this system.

$\endgroup$
  • 2
    $\begingroup$ physics.stackexchange.com/q/78415 $\endgroup$ – AHusain Jul 12 '19 at 13:46
  • $\begingroup$ @AHusain Classically the problem is exactly solvable? $\endgroup$ – mithusengupta123 Jul 12 '19 at 14:28
  • 1
    $\begingroup$ There is a 2d phase space and one conserved quantity. That's why. It gives t as a elliptic function of x and turning that the other way around may trouble you. $\endgroup$ – AHusain Jul 12 '19 at 14:59
  • $\begingroup$ You can integrate the equations numerically. $\endgroup$ – nicoguaro Jul 12 '19 at 18:17
1
$\begingroup$

This problem is discussed as an example of secular perturbation theory in

José, J. and Saletan, E., 2000. Classical dynamics: a contemporary approach.

For first order in $\epsilon$, the (rationalized) equation of motion $$ \ddot{x}+\omega_0^2 x+\epsilon x^3=0 $$ has the approximate solution $$ x(t)=a \cos(\omega_0 t)-\epsilon \frac{a^3}{8\omega_0^2} \left(3\omega_0t \sin\omega_0 t + \frac{1}{4}(\cos\omega_0 t- -\cos(3\omega_0t))\right) $$ with the secular term (linear in $t$) appearing explicitly. One can get rid of the secular term using Poincaré-Lindstedt theory, i.e. by introducing corrections to the unperturbed frequency $\omega_0$.

The same problem is used in José and Saletan as an example of canonical perturbation theory (or Hamilton-Jacobi perturbation theory), where the solution is to find successive canonical transformations. using the unperturbed canonical variables $$ \phi_0=\arctan(m\omega_0 q/p)\, ,\qquad J_0=\frac{1}{2}\left(\frac{p^2}{2\omega_0}+m\omega_0 q^2\right)\, . $$ The first order correction to the frequency is $\omega\approx\omega_0+\epsilon \frac{3a^2}{8\omega_0}$ when the initial conditions are $p(0)=0, q(0)=a$.

This problem, also as an example of the canonical perturbation approach, is similarly discussed in Example 8.3 of

Percival, I.C. and Richards, D., 1982. Introduction to dynamics. Cambridge University Press.

$\endgroup$
  • 1
    $\begingroup$ Yes, but note that these textbook exercises are in some sense artificial (and only presented as pedagogical examples), because the problem is completely integrable (in terms of elliptic functions). This also means that the easiest way to get the perturbative expansion is to take the exact answer and expand it out in powers of $\epsilon$. $\endgroup$ – Thomas Jul 12 '19 at 18:09
  • $\begingroup$ @Thomas isn't this what you would do anyways to get something other than a purely numerical value for the elliptic integral? $\endgroup$ – ZeroTheHero Jul 12 '19 at 18:12
  • $\begingroup$ Not really, I think. Canonical perturbation theory beyond leading order is very tedious, whereas there are fast and efficient ways of computing elliptic functions (based on the fact that they are double-periodic in the complex plane). $\endgroup$ – Thomas Jul 12 '19 at 18:19
  • $\begingroup$ @ZeroTheHero Can you tell whether there is an exact solution? The pendulum problem, I think, has an exact solution even when the angular amplitude is large. $\endgroup$ – mithusengupta123 Jul 12 '19 at 18:48
  • $\begingroup$ @mithusengupta123 there is an exact solution in terms of elliptic integrals, and this is useful for numerical work, but when it comes to analytical work I do not know of any solution expressible in terms of "simple" functions. $\endgroup$ – ZeroTheHero Jul 12 '19 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.