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In QM, it is said that perturbation theory can be used in the case in which the total Hamiltonian is a sum of two parts, one whose exact solution is known and an extra term that contains a small parameter, $\lambda$ say. We can obtain the solution of the full Hamiltonian as a systematic expansion in terms of that small parameter. Now take as a specific example

$$ H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}kx^2+\lambda x^3\equiv H^0+\lambda x^3$$

Is there an intuitive way to see in what sense the $\lambda$ term can be smaller or bigger than the unperturbed Hamiltonian?

Or is there someway to estimate the size of every term in the Hamiltonian, based on which we can decide the parameter space for $\lambda$ (which will consequently make our perturbation method valid)? how one can even estimate the size of a derivative!?

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It may be interesting for you to note that the splitting is not unique. In fact, given the quartic oscillator, it is well-known that splitting it into the quadratic part plus the quartic part is not the optimal way to split. Much better results are obtainable by variational perturbation theory, which chooses the interaction in such a way as to minimize sensitivity of the results. This gives a different split (and even splits depending on what precisely you want to determine).

Stevenson, P.M., Optimized perturbation theory, Physical Review D 23 (1981), 2916.

With variational perturbation theory, one can get convergent expansions for the quartic oscillator rather than asymptotic expansions only. See, e.g., Phys. Lett. A 173, 332 (1993); Physical Review A 59 (1999), 102.

Your problem, the cubic oscillator, is treated with variational perturbation theory in quant-ph/9502027.

See also

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In general you cannot decide whether perturbation theory gives a good approximation (say to the energy levels) just by looking at the Hamiltonian; this can also depend on the state whose energy you want to calculate. Take for example the anharmonic oscillator, $$ H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac12m\omega^2x^2+\lambda x^4. $$ (I do not use your example with a cubic term since such Hamiltonian is not bounded from below and does not have bound states. Here, perturbation theory gives you metastable states with a long lifetime: particle in such a state will be localized in the potential well for a long time but eventually it will tunnel through the potential barrier and escape.) The higher excited state you look at, the larger values of $x$ can the particle reach, and thus the worse the perturbative approximation to the energy levels will be. In fact, for highly excited states the term proportional to $x^2$ can be neglected and one can then show that the energy of the $n$-th excited state scales asymptotically as $\lambda^{1/3}n^{4/3}$. Such dependence on $\lambda$ obviously cannot be obtained at any finite order of perturbation theory.

This example of the anharmonic oscillator has been studied a lot in literature. (Among others, the perturbative corrections to the energy levels have been calculated to all orders in perturbation theory.) It is well known that the perturbation series does not converge no matter how small $\lambda$ is. This is easy to understand intuitively. A power series always converges on a circle in the complex plane, in particular if it converges for some $\lambda$, it must converge for $-\lambda$ as well. However, for negative $\lambda$ this Hamiltonian does not even have a ground state. Therefore, the perturbation series has a zero radius of convergence.

Fortunately, all is not lost. The perturbation series in this example (and many others, especially in quantum field theory) is asymptotic. This roughly speaking means that if you start calculating perturbative corrections to the energy levels, the successive terms initially seem to converge - and adding more terms makes sense. However, from certain order on the corrections start to grow. What is the value of the "critical" order then depends on the size of the coupling $\lambda$.

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What you can generally do is first do some variable substitutions so that all your constants are dimensionless.

Because, at the moment, you'd have $\lambda$ with units of Energy per Length cubed, but $k$ would have units Energy per Length squared and that makes it difficult to compare.

Exactly how you do that doesn't really matter, so one way would be to introduce a variable $\xi = x / \sqrt{k}$. Then you have

$$H = -\frac{\hbar^2}{2mk} \frac{d^2}{d\xi^2} + \frac{1}{2} \xi^2 + \frac{\lambda}{k^{2/3}} \xi^3$$

Now if $\lambda / k^{2/3} \ll 1/2$ you know that $\lambda$ is a small parameter and perturbation theory should work well. But if $\lambda / k^{2/3} \sim 1/2$, you would expect perturbation theory to fail.

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After changing variables, $\lambda/k^{2/3}$ is not dimensionless! It has dimension $1/\zeta$ – Revo Feb 13 '12 at 19:36
Also how about the size of the term that contains a derivative? – Revo Feb 13 '12 at 19:40
Oops, you are right. I'll go over it again tonight or tomorrow – Lagerbaer Feb 14 '12 at 2:30

Strictly speaking, there is an intuitive way to see in what sense the λ term is always bigger than the unperturbed Hamiltonian :-) The spectrum of the Hamiltonian with the λ term ($H_\lambda$) is not bound from below (you may just draw the relevant potential to understand that), so it has no discrete spectrum, unlike the Hamiltonian without the λ term ($H_0$), only continuous spectrum. If the initial state is similar to the ground state of $H_0$, it will eventually tunnel in the direction $x\rightarrow -\infty$ (assuming $\lambda>0$). So in some sense the perturbation theory is always unsatisfactory, and the relevant series always diverges. However, if $\lambda$ is small enough, the tunneling time is extremely large, so a truncated asymptotic series of perturbation theory gives a very good approximation. So I would suggest the following, rather arbitrary, criterium: if for a given $\lambda$ the second correction of perturbation theory is much smaller than the first correction for the state you are interested in, then application of perturbation theory is justified.

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