Worked Examples

Consider a quantum harmonic oscillator perturbed by a quartic anharmonic term. The full Hamiltonian is:

The unperturbed system is the harmonic oscillator with well-known energy levels $E_n^{(0)} = \hbar\omega(n + \tfrac{1}{2})$ and ladder operator algebra.

The position operator in terms of ladder operators is:

Therefore the perturbation becomes:

We need $E_0^{(1)} = \langle 0|\hat{H}'|0\rangle = \lambda(\hbar/2m\omega)^2\langle 0|(\hat{a}+\hat{a}^\dagger)^4|0\rangle$.

Expanding $(\hat{a}+\hat{a}^\dagger)^4$, only terms with equal numbers of $\hat{a}$ and $\hat{a}^\dagger$ survive (to return to $|0\rangle$). Using normal ordering and the commutation relation $[\hat{a},\hat{a}^\dagger]=1$:

Evaluating each term carefully using $\hat{a}|0\rangle = 0$ and $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$:

Therefore the first-order energy correction is:

The anharmonic term raises the ground state energy, consistent with the positive quartic potential adding confinement.

For the second-order correction, we need matrix elements $\langle m|\hat{x}^4|0\rangle$ for $m \neq 0$. The operator $\hat{x}^4$ can connect $|0\rangle$ to $|2\rangle$ and $|4\rangle$ only (by selection rules from ladder operators):

Applying the second-order formula:

As expected, the second-order correction is negative (lowering the energy) and proportional to $\lambda^2$.

The Stark effect describes the splitting of atomic energy levels in an external uniform electric field $\vec{\mathcal{E}} = \mathcal{E}_0\hat{z}$. The perturbation Hamiltonian is:

For hydrogen, the first-order correction vanishes for all states with definite parity (since $z$ is odd under parity, $\langle nlm|z|nlm\rangle = 0$ for all states with well-defined $l$). This is the quadratic Stark effect:

For the hydrogen ground state (1s), the exact second-order result is:

where $\alpha_d = \frac{9}{2}a_0^3$ is the static electric polarizability of hydrogen. This energy shift is always negative and quadratic in the field strength.

An atom placed in a weak magnetic field $\vec{B} = B\hat{z}$ experiences the perturbation:

In the coupled basis $|j, m_j\rangle$, the first-order energy correction involves the Lande g-factor:

where the Lande g-factor is:

and $\mu_B = e\hbar/(2m_e) \approx 5.79 \times 10^{-5}$ eV/T is the Bohr magneton. Each $j$-level splits into $2j+1$ equally spaced sublevels.

The convergence of the perturbation series depends critically on the ratio of perturbation matrix elements to energy level spacings. Several important scenarios arise:

• Good convergence: When $|\langle m|\hat{H}'|n\rangle| \ll |E_n^{(0)} - E_m^{(0)}|$ for all states, low-order perturbation theory gives excellent results.
• Asymptotic series: The perturbation series for the anharmonic oscillator ($\lambda x^4$) actually diverges for any $\lambda > 0$! The series is asymptotic: the first few terms approximate well, but adding more terms eventually makes the result worse. The reason is that for $\lambda < 0$, the potential is unbounded below, so the series has zero radius of convergence.
• Near-degeneracy: When two levels are close in energy, the corresponding terms in the sum become very large. The cure is to treat the near-degenerate subspace exactly (quasi-degenerate perturbation theory).

Practical Rule of Thumb

A useful criterion for when to trust perturbation theory to order $k$: the $(k+1)$-th order correction should be significantly smaller than the $k$-th order correction. If you compute:

then the perturbation series is converging well at order $k$, and the truncation error is roughly the size of the last term kept.