Anharmonic Oscillator

A bent oscillator

The harmonic oscillator is exactly solvable, but no real potential is exactly parabolic. A more realistic model adds a small quartic term:

The unperturbed levels are the familiar $E_n^{(0)} = \hbar\omega(n + \tfrac12)$, and we compute how the quartic term shifts them. This is the canonical worked example of perturbation theory because the unperturbed states are known and the matrix elements are computed cleanly with ladder operators.

Ladder operators do the work

Write position in terms of the lowering and raising operators,

with $\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle$ and $\hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$. The first-order shift needs the diagonal element of $\hat{x}^4$:

Expanding $(\hat{a}+\hat{a}^\dagger)^4$ gives sixteen terms; only those with equal numbers of raising and lowering operators survive the diagonal element $\langle n|\cdots|n\rangle$. Collecting them yields the standard result

Working in units where $\hbar/(2m\omega) = 1$, the first-order energy shift is simply

Reading the correction

A few observations:

• The shift is positive (for $\lambda > 0$): a quartic wall is steeper than a parabola at large $|x|$, raising every level.
• It grows with $n$ like $n^2$: highly excited states sample the steep quartic region more, so they are pushed up faster. This is also a warning sign — for fixed $\lambda$ the correction eventually stops being small, and perturbation theory fails for high enough $n$.
• The ground state ($n = 0$) shifts by $E_0^{(1)} = 3\lambda$ in these units, the smallest shift of all the levels.

Try it

For the quartic anharmonic oscillator with $\lambda = 1$ (and units where $\hbar/(2m\omega) = 1$), compute the first-order energy shift of the ground state $E_0^{(1)} = \lambda\,\langle 0|\hat{x}^4|0\rangle$ using $\langle n|\hat{x}^4|n\rangle = 3(2n^2 + 2n + 1)$, and return it as a number.

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