# The quartic anharmonic oscillator

+ 4 like - 0 dislike
4235 views

What is the optimal way to split the potential $V(x)=\frac12 kx^2+ \frac 14 gx^4$ of the quartic oscillator of fixed mass $m$ into an exactly solvable potential $V_0(x)$ and a perturbation $W(x)=V-V_0(x)$ such that low order perturbation theory produces the most accurate dynamics at sufficiently long times?

The question makes sense both in the classical and in the quantum case. In the classical case I am interested in predicting for large $T$ the value of $x(t)$ for $t\in [T-dT,t+dT]$ from initial conditions for $x$ and $p$ at $t=0$, for large $T$ and small $d=0.01$, say, measuring the quality in terms of the maximal error, say. In the quantum case, the same for the expectation, givien that the wave function at $t=0$ is a coherent state with given position and momentum mean.

The simplest choice $V_0=\frac12 kx^2$ is clearly suboptimal as it can be superseded by an appropriate choice $V_0=\frac12 k(g)x^2$ where $k(g)\ne k$ for $g\ne 0$, with resulting interaction $W(x)=\frac12 \epsilon(g)x^2+ \frac 14 gx^4$, where $\epsilon(g)=k(g)-k\ne 0$ for $g\ne 0$.

For the time-independent case, a similar question can be asked; for the quantum case, see, e.g., this paper by Cohen and Kais,  J. Phys. A 19 (1986), 683. I am interested in the time-dependent case at very long times.

edited Nov 28, 2016

The question about optimal splitting needs some clarification. For example, we are interested in the most precise ground energy estimation solely $\tilde{E}_0(g)$. Or the energy $\tilde{E}_n (g)$ estimations without looking at the wave functions. Or the energy and wave function estimations for dynamics at a given $x$, etc.

If we look for $\tilde{E}_0(g)$ solely, then we can always pick up such a $k(g)$ that gives the exact ground energy $\tilde{E}_0(g)=E_0(g)$, it is clear. But the corresponding wave function $\tilde{\psi}_0(x)$ will be different from the exact one $\psi_0(x)$. One can note this from the WKB approximation for a given $g$. Thus, fitting the oscillator mass or the unperturbed frequency does not resolve the problem of "dynamics". One cannot replace an interacting case with a free case by simply changing the energy of the free case. In this sense, I would like to understand better what is meant by "dynamics at long times". What does one want to calculate?

I am for keeping $m$, $k$, and $g$ at their places, as in the original post.

Even with $k(g)=k+gs(g)$ the counting of degrees of freedom is still confusing to me. If you want to expand $k$ into a series why can the potential not be written as $\frac{1}{2}k(s) x^2+ gx^4$ with $g$ not appearing in any quadratic term?

@Dilaton: Because this will be the same potential as the given one only when $s=0$. I want to solve the original problem with the original potential, except that I rewrite it as a sum of a more general exactly solvable potential $V_0(x)$ and a correction term, rather than allowing only the ''bare'' split where $V_0(x)=\frac12 k x^2$. The Hamiltonian with $V_0(x)$ is then treated as the unperturbed system; and the correction term as the perturbation.

@VK: Yoe refer to the time-independent case; I am interested in the time-dependent case. I'll make the question more precise in a moment.

@VladimirKalitvianski yes, this is why I first overlooked that g is the quartic coupling, as I thought it should be the scale of the running k if anything ...

At fixed $m$ and $k$, the constant $g$ is the only parameter; hence the exactly solvable system defining $V_0(x)$ can only depend on $g$. Essentially, any dependence on $k$ and $m$ is suppressed in my notation. Another way to write it would be to introduce a new parameter $s$ and put $V_0(x):=\frac12 (k+gs) x^2$, so that $W(x):=g(x^4-sx^2)$. I am then asking for the optimal choice of $s$ given $g$, which would define $s=s(g)$ and then $k(g)=k+gs(g)$. If this is more intuitive I'll adapt my question accordingly.

+ 1 like - 0 dislike

I. Classical case. Let us have a particle with the initial position $x(0)$ and the initial velocity $v(0)$ with the corresponding uncertainties denoted with corresponding $\delta_x$ and $\delta_v$. For simplicity, let us consider a free motion. Then the solution is known: $$x(t) = x(0)\pm \delta_x +v(0)t\pm \delta_v t\qquad(1).$$ Thus, the position absolute uncertainty $\delta x(t)$ grows with time as $\pm\delta_v t$. Now, if we put this particle in a box $x\in [0,L]$, then starting from a certain time the particle position uncertainty will become larger than the box size and the particle position will become completely uncertain within the box. All we may say is that $x\in[0,L]$ with a constant probability distribution $f(X)=1/L$.

Something similar I expect from a classical linear oscillator "dynamics": its position uncertainty becomes limited with the turning points with some oscillator-like probability density between them.

II. QM. So I jump to the quantum mechanical case where you propose to play with the linear oscillator frequency (energy) via "better" choice of $k(g)$. Let us admit that we managed to guess the function $k(g)$ giving the exact eigenvalue $E_0(g)$ with help of a “free oscillator”. Using $\tilde{E}^{(0)}_0(g) = E_0(g)$ does not prevent us from unnecessary perturbative eigenvalue corrections (I called such series “blank”): $W_{00}= \langle [k(g)-k]x^2/2+g x^4/4\rangle_{00}\ne 0$, $W_{nm}\ne 0$. The whole series starting from the “first order” $W_{00}$ results in principle (i.e., when correctly summed up) into zero since no corrections to $\tilde{E}^{(0)}_{00}(g)= E_0(g)$) should exist, but each term in each order is not zero since $k(g)$ is a non trivial function of $g$ and one does not expands it in powers of $g$. In other words, a truncated perturbative series is a non-zero-valued polynomial. I encountered such a case too, many years ago. Despite such a series is unnecessary for the eigenvalue (it spoils the good initial approximation), the corresponding series for the eigenfunction is useful since it modifies a wrong linear oscillator wave function into a anharmonic oscillator wave function. I cannot say what a long-time dynamics can be predicted for a superposition of such states.

By the way, in my practice I encountered only one one case when knowing the exact eigenvalue determined the exact eigenfunction: it was the case of piece-wise behaviour of the Sturm-Liouville problem coefficients (a multi-layer system). Then one can write exactly (analytically) the solution within each layer, something like this: $\psi_E(x)=A_i \sin[\sqrt{E} (x-x_i)]+B_i \cos[\sqrt{E}(x-x_i)], \; i=1,2,3,…$, where $i$ is the layer number. Injecting the exact eigenvalue in this analytical formula gives the exact eigenfunction, but for that to be possible, one has to be able to express the eigenfunction as an analytical function of $E$. In particular, I used it for describing a long-time dynamics of heat conduction of a multi-layer body.

If one uses the regular perturbation theory, one obtains here a series with the zero-radius of convergency. For summing (partially) its terms, many approaches have been proposed (including mine ;-), see also here (Отступление.)). I am not an expert in this field, but my practice shows that any available information about the exact solution may help construct reasonable approximations.

answered Nov 29, 2016 by (102 points)
edited Nov 29, 2016

Both in the classical case and in the quantum case one may assume that the initial conditions are exact. The question is not about uncertainty propagation but about which choice of the splitting for second-order perturbation theory, say, will introduce the least approximation error at long times. It is clear that the errors will increase with time and are at short times negligible, but the maximal error at a given large time depends depends on $k(g)$ and it should be possible to work out the best choice. The divergence of the asymptotic does not matter; all approximations are calculated at fixed order.

Do you have the answer to your question or just an idea?

I believe the answer for the classical case is somewhere in the literature, but I don't have a reference and didn't do the analysis myself. I don't know the answer in the quantum case. But I'd be very surprised if $k(g)=k$ independent of $g$ would be optimal.

In the classical case the amplitude may be very small and then only the linear oscillator potential matters (the quartic term is negligible), so the $g$-independent $k$ is the best choice for small amplitude oscillations.