Anharmonic Oscillator by Perturbation Theory

This worked example computes the leading perturbative corrections for the quartic anharmonic oscillator

$$H = \frac{P^2}{2m} + \frac12m\omega^2X^2 + \lambda X^4, \qquad \lambda\gt0.$$

In this worked problem, $\Delta E_n^{(r)}$ denotes the physical order-$r$ contribution and therefore includes the corresponding power of $\lambda$.

The unperturbed problem is the harmonic oscillator, and the perturbation is

$$V=\lambda X^4.$$

Required background. Ladder operators supply the matrix elements; second-order perturbation theory supplies the energy denominators.

Quartic Coupling and the Harmonic Reference

Use nondegenerate time-independent perturbation theory because the one-dimensional harmonic oscillator has isolated energy levels

$$E_n^{(0)} = \hbar\omega \left(n+\frac12\right).$$

The small parameter is the size of the quartic energy in the oscillator ground-state width compared with $\hbar\omega$:

$$g = \frac{\lambda\hbar}{m^2\omega^3}.$$

Low-order energy estimates require roughly $g(n+1)\ll1$. Approximating the eigenvector in the original oscillator basis is more restrictive: its leading mixing amplitudes scale as $g(n+1)^2$. Small relative energy changes do not automatically mean a small change of state.

Ladder-Operator Setup

Write

$$X = x_{\mathrm{zpf}} \left(a+a^\dagger\right), \qquad x_{\mathrm{zpf}} = \sqrt{\frac{\hbar}{2m\omega}}.$$

Then

$$X^4 = x_{\mathrm{zpf}}^4 \left(a+a^\dagger\right)^4.$$

The diagonal matrix element is

$$\langle n|X^4|n\rangle = x_{\mathrm{zpf}}^4 \left( 6n^2+6n+3 \right).$$

A short route is to apply $Q=a+a^\dagger$ twice:

$$Q^2|n\rangle=\sqrt{n(n-1)}|n-2\rangle+(2n+1)|n\rangle +\sqrt{(n+1)(n+2)}|n+2\rangle.$$

Because $Q$ is Hermitian, $\langle n|Q^4|n\rangle=\|Q^2|n\rangle\|^2$. Squaring the three orthogonal coefficients gives $n(n-1)+(2n+1)^2+(n+1)(n+2)=6n^2+6n+3$. This avoids treating noncommuting $a$ and $a^\dagger$ as ordinary numbers.

First-Order Energy Shift

The first-order correction is

$$\Delta E_n^{(1)} = \lambda\langle n|X^4|n\rangle.$$

Therefore

$$\Delta E_n^{(1)} = \lambda \left( \frac{\hbar}{2m\omega} \right)^2 \left( 6n^2+6n+3 \right).$$

Equivalently,

$$\Delta E_n^{(1)} = \lambda \frac{3\hbar^2}{4m^2\omega^2} \left( 2n^2+2n+1 \right).$$

For the ground state,

$$\Delta E_0^{(1)} = \frac{3\lambda\hbar^2}{4m^2\omega^2}.$$

Second-Order Ground-State Shift

For the ground state,

$$\Delta E_0^{(2)} = \sum_{r\ne0} \frac{ |\langle r|\lambda X^4|0\rangle|^2 }{ E_0^{(0)}-E_r^{(0)} }.$$

Parity implies that only even $r$ can contribute. Repeated ladder-operator action gives

$$X^4\lvert0\rangle = x_{\mathrm{zpf}}^4 \left( 3\lvert0\rangle + 6\sqrt2\,\lvert2\rangle + 2\sqrt6\,\lvert4\rangle \right).$$

Thus only $\lvert2\rangle$ and $\lvert4\rangle$ contribute to the sum:

$$\begin{aligned} \Delta E_0^{(2)} =& \lambda^2x_{\mathrm{zpf}}^8 \left[ \frac{|6\sqrt2|^2}{E_0^{(0)}-E_2^{(0)}} + \frac{|2\sqrt6|^2}{E_0^{(0)}-E_4^{(0)}} \right] \\ =& \lambda^2x_{\mathrm{zpf}}^8 \left[ \frac{72}{-2\hbar\omega} + \frac{24}{-4\hbar\omega} \right]. \end{aligned}$$

Therefore

$$\Delta E_0^{(2)} = - \frac{42\lambda^2x_{\mathrm{zpf}}^8}{\hbar\omega} = - \frac{21\lambda^2\hbar^3}{8m^4\omega^5}.$$

Result

The ground-state energy through second order is

$$E_0 = \frac12\hbar\omega + \frac{3\lambda\hbar^2}{4m^2\omega^2} - \frac{21\lambda^2\hbar^3}{8m^4\omega^5} + O(\lambda^3).$$

In terms of

$$g=\frac{\lambda\hbar}{m^2\omega^3},$$

this is

$$\frac{E_0}{\hbar\omega} = \frac12 + \frac34g - \frac{21}{8}g^2 + O(g^3).$$

Validity Discussion

The first correction is small compared with the unperturbed ground energy when

$$g\ll1.$$

The second-order term is smaller than the first when, roughly,

$$\frac{21g^2/8}{3g/4} = \frac72g \ll1.$$

This numerical estimate is not a rigorous bound, but it is a useful warning: even for the ground state, the second-order coefficient is sizable.

For high $n$, the first-order shift grows like $gn^2$ while the unperturbed energy grows like $n$. A low-state approximation is therefore not uniform in excitation number. At fixed $n$, the quartic series is asymptotic rather than convergent; the large-order analysis is due to Bender and Wu. The Gaussian calculation offers an upper bound at any positive coupling, and the numerical benchmark compares both methods.

Cross-Checks

The correction has the right dimensions: $\lambda X^4$ is an energy.

The first-order shift is positive for $\lambda\gt0$, as expected for a positive added potential. The second-order shift is negative because excited-state denominators satisfy

$$E_0^{(0)}-E_r^{(0)}\lt0.$$

Parity is respected: $X^4$ connects the ground state only to even oscillator states.

Exercises

1. Extend the second-order calculation to an arbitrary level $n$. Use the action of $Q^2$ above to find the four off-diagonal matrix elements of $Q^4$, and show that

$$\frac{\Delta E_n^{(2)}}{\hbar\omega} =-\frac{g^2}{8}(34n^3+51n^2+59n+21).$$

Why does the answer grow as $n^3$, although individual terms in the second-order sum grow as $n^4$? Compare the resulting energy criterion with the criterion for small eigenvector mixing.

Solution

Applying $Q^2$ a second time gives

$$\begin{aligned} \langle n-4|Q^4|n\rangle&=\sqrt{n(n-1)(n-2)(n-3)},\\ \langle n-2|Q^4|n\rangle&=(4n-2)\sqrt{n(n-1)},\\ \langle n+2|Q^4|n\rangle&=(4n+6)\sqrt{(n+1)(n+2)},\\ \langle n+4|Q^4|n\rangle&=\sqrt{(n+1)(n+2)(n+3)(n+4)}. \end{aligned}$$

Omit any channel with a negative state label. Since $\lambda x_{\mathrm{zpf}}^4/(\hbar\omega)=g/4$, the energy denominators give

$$\begin{aligned} \frac{\Delta E_n^{(2)}}{\hbar\omega} =\frac{g^2}{16}\Bigg[& \frac{n(n-1)(n-2)(n-3)}{4} +\frac{(4n-2)^2n(n-1)}{2}\\ &-\frac{(4n+6)^2(n+1)(n+2)}{2} -\frac{(n+1)(n+2)(n+3)(n+4)}{4}\Bigg]. \end{aligned}$$

Expansion gives the stated cubic polynomial and recovers $-21g^2/8$ at $n=0$. Lower intermediate levels contribute with positive denominators and upper levels with negative denominators; their leading $n^4$ terms cancel. The surviving $g^2n^3$ contribution is smaller than the first-order $gn^2$ shift when $gn\ll1$ at large $n$.

There is no analogous cancellation in the squared norm of the first-order state correction: it sums positive squared mixing amplitudes. Each allowed amplitude is of order $gn^2$, so closeness to the original number state needs the stronger condition $gn^2\ll1$. These are low-order diagnostics for different observables, not convergence guarantees for the asymptotic series.

2. Why does a cubic perturbation $\lambda X^3$ give no first-order energy shift for oscillator eigenstates?

Solution

Harmonic oscillator eigenstates have definite parity. The operator $X^3$ is odd under parity, so

$$\langle n|X^3|n\rangle=0.$$

The first formal nonzero shift is second order. A purely cubic addition makes the potential unbounded below in one direction, however, so it has no true ground-state bound spectrum on the whole line. The coefficients then describe a local metastable expansion or a model with additional stabilizing terms.

References

• C. M. Bender and T. T. Wu, “Anharmonic oscillator,” Physical Review 184, 1231-1260, 1969.
• C. M. Bender and T. T. Wu, “Anharmonic oscillator. II. A study of perturbation theory in large order,” Physical Review D 7, 1620-1636, 1973.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.