Perturbation theory for a particle in a weak potential

Question

I have a basic question about quantum mechanics, maybe it has a basic answer.

Take a free particle in a quartic potential,

$L=\frac{1}{2}\dot{x}^2-\lambda x^4$

This is massless $\phi^4$ theory in 0+1 dimensions. Suppose that $\lambda$ is small. Then I want to know the spectrum of the theory. We expect a discrete spectrum with a small ground state energy. Is there a simple way to do perturbation theory here?

The reason I'm confused is because the free particle does not have a normalizable ground state. We could try to use ordinary first order perturbation theory, and the ground state energy is $\langle0|\lambda x^4|0\rangle$, but this is infinite. One other thing we could try is to regulate the free particle by adding a small mass $m$ or putting the theory in finite volume $L$. But then the answers only hold for $\lambda L^3\ll 1$ or $\lambda/m^3\ll 1$, so we can't remove the regulator completely.

For high energies we can use the WKB approximation, giving $E_n\sim \lambda^{1/3}n^{4/3}$. This doesn't work for the ground state though. In fact we could have guessed $E_n\sim \lambda^{1/3}$ by dimensional analysis. This alone suggests that perturbation theory is bad because the energies are nonanalytic in $\lambda$. (One can make the usual argument that the energies have to be nonanalytic in $\lambda$ by analytically continuing to negative $\lambda$ while analytically continuing the contour simultaneously.)

Are there any known approximation methods for this situation?

Edit: I figured out why there are no approximation methods for this theory. There is no small parameter! All the energies are proportional to $\lambda^{1/3}$, and the coefficients are pure numbers with no perturbation series. Maybe large N could help here.

Answer 1

Let me first clear up some misstatements in your question. If your $\lambda$ is positive then your Lagrangian describes a bound particle, not a free particle. Furthermore, it is a particle with mass=1. If it is instead a Lagrangian density, then it is massless $\phi^4$ theory in 0+1_D as you suggest. When you say you expect a discrete spectrum with a small ground state energy, I'm convinced that you are interested in the solution for a particle with mass=1 rather than the massless $\phi^4$ theory. Later, however, you speak of adding a small mass ($m$) which sounds like you are confusing the two possibilities. Let me eliminate this confusion by adding a general mass term to your Lagrangian: $$L=\frac{1}{2}m\dot x^2-\lambda x^4$$ The Hamiltonian for this system becomes: $$H=\frac{p^2}{2m}+\lambda x^4$$ Now let's consider a first order perturbation theory treatment of the ground state energy of this Hamiltonian. Let's choose the harmonic oscillator Hamiltonian for the unperturbed case: $$H_0=\frac{p^2}{2m}+\frac{k}{2}x^2$$ Since we know that the ground state energy of the harmonic oscillator is $\frac{\hbar}{2}\omega$ where $\omega=\sqrt{\frac{k}{m}}$, we can write the first order perturbation theory approximation to the ground state of H as: $$< u_0|H|u_0 >=\frac{\hbar}{2}\omega+\lambda<u_0|x^4|u_0>-\frac{m}{2}\omega^2<u_0|x^2|u_0>$$ where $u_o$ is the ground state wave function of the harmonic oscillator. It can be shown that for the harmonic oscillator ground state $$<u_0|x^4|u_0>=3<u_0|x^2|u_0>^2$$ This enables us to write: $$<u_0|H|u_0>=\frac{\hbar}{2}\omega+<u_0|x^2|u_0>(3\lambda<u_0|x^2|u_0>-\frac{m}{2}\omega^2)$$ Now by use of operator techniques or by direct integration it can be shown that $$<u_0|x^2|u_0>=\frac{\hbar}{2m\omega}$$ so that $$<u_0|H|u_0>=\frac{\hbar}{4}\omega+\frac{3\hbar^2\lambda}{4m^2\omega^2}$$

This approximation to the ground state energy of the quartic Hamiltonian is appropriate for any harmonic oscillator frequency $\omega$. As noted in the answer by @Urgie, however, this may not be a good approximation because the perturbation grows quite large for large $x$ whatever value we choose for $\omega$. In my comment on the question several days ago I alluded to coupling the harmonic oscillator approach to a variational principle. This would yield the best possible first order approximation based upon a harmonic oscillator wave function. This is achieved by equating the derivative of $<u_0|H|u_0>$ with respect to $\omega$ to 0 and then solving the resulting equation for $\omega$ as a function of $\lambda$. Following this procedure we get: $$\frac{d<u_0|H|u_0>}{d\omega}=\frac{\hbar}{4}-\frac{3}{2} \frac{\lambda \hbar^2}{m^2\omega^3}=0$$ Solving this we find the best value for $\omega$ $$\omega=\sqrt[3]{\frac{6\lambda\hbar}{m^2}}$$ This leads to a rather messy formula for the ground state energy which I will not bother to write down, but note that it satisfies your intuition that it should be proportional to $\lambda^{\frac{1}{3}}$.

Now I think it would be quite instructive for someone to solve this problem numerically (for the specific parameters $m=\lambda=\hbar=1$, for example) and see just how good this approximation is. If anyone takes me up on this, I will definitely upvote their answer.

Edit: Matthew provided a link in a comment below that permits the comparison I suggested above. Following this link, the gs energy for an exact (numerical) computation is 0.667985 (for the parameters I suggested above). This link also performs a semi-classical (Bohr-Sommerfeld) calculation resulting in 0.546267 for the same state (and same parameters). The result for the harmonic oscillator perturbation calculation (with variational method and same parameters) outlined in my answer above is 0.68142. The semi-classical calculation has an error of 18% while the HO first order perturbation calculation has an error of 2%. One last point to consider is the rate of convergence of the perturbation series for this approach. The HO (unperturbed) energy for this situation is 0.90856 so the first order correction is 0.22714 which is 94.4% of the full discrepancy (a rapid rate of convergence). This is a testament to the strength of perturbation theory coupled with variational method estimates for gs energies, even when the perturbation is not small in regions of interest.

Answer 2

From $L$ one obtains the Hamiltonian

\begin{equation*} L=\frac{1}{2}\dot{x}^{2}-\lambda x^{4}\Rightarrow H=\frac{1}{2}p^{2}+\lambda x^{4} \end{equation*} For this model $H$ has a compact resolvent $[z-H]^{-1}$ for $z$ outside the spectrum of $H$. This implies that $H$ has discrete spectrum with eigenvalues with finie degeneracy (see the volumes of Reed and Simon "Methods of modern mathematical physics "). In particular the free Hamiltonian is not a convenient zero order. Instead one might think of a harmonic potential in the zero order as mentioned by Lewis Miller. But the actual potential is not a small perturbation of the harmonic one and different methods are required.