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I'm interested in turning a quadratic + quartic oscillator Hamiltonian,

$$H = \frac{p^{2}}{2m} +\frac{kx^{2}}{2} + \lambda x^{4},$$

into a purely quadratic problem. One of the simplest things I can think of is replace the quartic term by

$$x^{4} \rightarrow \langle x^{2}\rangle x^{2}.$$

Has anyone come across such an approximation? If yes could you give me some reference to the work where this approximation has been used. Is there any other simple approximation which can turn the above Hamiltonian into a quadratic problem?

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Classical case

The Hamiltonian you have introduced is the one of a nonlinear oscillator. While for a linear oscillator period does not depend on the amplitude for this one it does. This means that quadratic approximation will always fail for some amplitudes.

If you want to describe oscillations of small amplitude you can just neglect the last term.

If you want to describe oscillations of some substantial amplitude you can use the trick with $\left<x^2\right>$. This approximation will work well for some range of amplitudes but will fail for others.

You have a potential of the following form: $$ V(x) = m \cdot \Omega^2(x) \cdot x^2; $$ and want to replace it with this one: $$ \overline{V}(x) = m \cdot \overline{\Omega}^2 \cdot x^2. $$ The average frequency for symmetric potential can be found with the following formula: $$ \frac{2\pi}{\overline{\Omega}} = \sqrt{2m}\int_0^A \frac{dx}{\sqrt{V(A)-V(x)}} \qquad (1) $$ where $A$ is the amplitude of the oscillations. This is just the integral over half of the period of one of the Hamilton's equations: $$ \frac{dx}{dt} = \frac{p(x)}{m} $$ where $p(x)$ is determined from the energy conservation law: $H(p,x)=\text{const}$.

Formula (1) gives the exact value of the frequency for amplitude $A$. If you don't want to deal with integrals you can suppose $\left<x^2\right>=\frac{1}{2}A^2$ (this is true for harmonic $x$). This is a less exact approximation.

Quantum case

The method discussed above can be used here for the higher states of the oscillator where quasiclassical approximation works well even for a nonlinear oscillator.

As for the lower states, the energy of the ground state is finite and if $\lambda$ is large enough you can not neglect it. You can find Perturbation theory useful in this case.

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Thanks for your prompt reply. "If you want to describe oscillations of some substantial amplitude you can use the trick with $⟨x^2⟩$. This approximation will work well for some range of amplitudes but will fail for others." Can you elaborate more on this statement. Is this a common approximation made to the non-linear problem? If you have some specific references they would be quite helpful. – Juzar Jan 13 '12 at 21:08
@Juzar, I have updated the post. I wouldn't say it is a common approximation but it is close to what you want to do. Common approximation is Perturbation theory: in general and in quantum mechanics – Maksim Zholudev Jan 14 '12 at 16:04
@ Maksim, I agree with your justification and even I do believe this approximation is not very common. My goal is to compare perturbation theory with this rudimentary approximation and see whether the answers show the same qualitative behaviour. I had never seen anyone use this before so I thought I could get some hints from here. Thanks once again for your informative post. If you have any particular reference that you have come across who employ this approximation do let me know. – Juzar Jan 14 '12 at 16:37
@Juzar, sorry, I have no references. I have never used or even seen this approach before. Everyone uses Perturbation theory. I agree with Ron. This looks very much like some kind of Hartree approximation since the potential depends on the state of the system, i.e. contains the term $\left<\Psi\right|x^2\left|\Psi\right>$ – Maksim Zholudev Jan 15 '12 at 13:41

This is an extraordinarily famous approximation, perhaps the most famous one, called the "Hartree Fock" or "Self-consistent field" approximation. It is the earliest and most famous approach to solving complicated self-interacting systems.

But in the case you are using it, it is very poor. The problem you state can be solved much better by Bohr-Sommerfeld quantization, or BKS, to a far better accuracy than a self-consistent field. What you do with the self-consisten field is replace the quartic by a best-fit quadratic potential, which will give completely wrong wavefunctions, and not good energy levels (BKS will certainly be better for everything).

The point of this approximation is that it works for many-body systems, where you expect that many-body interactions do take the form of a self-consistent field to a good approximation. This idea is verified in both atomic and nuclear physics.

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@ Ron, Hartree Fock is a mean-field theory in order to have a mean field I require other particles who generate this field. In the Hamiltonian I have pointed stated there is only one particle so I think the above approximation is not Hartree Fock. If I had more than one particles then I could have performed a Hartree Fock approximation. – Juzar Jan 14 '12 at 17:34
This is true, but the approximation is classified as Hartree Fock nonetheless. It is silly for a single particle--- the particle's oscillation frequency is modified by the mean field value of its squared oscillation amplitude, which is determined in turn from the energy level of the particle in the modified potential. This is self-consistent field. The approximation in field theory replaces particle-particle nonlinear scattering by particle scattering using the mean density, but its the same thing, in a more useful context. – Ron Maimon Jan 14 '12 at 23:41

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