Time-independent perturbation theory

Time-independent perturbation theory expands the eigenvalues and eigenstates of $H_0+V$ in powers of a small perturbation $V$.

This is the stationary branch of perturbation theory.

For reduced mass $\mu$ and Coulomb strength $a$, put $a_q={\hslash }^2/(\mu a)$. The perturbation $br$ shifts the $1s,2s,2p$ energies by $3b{a}_q/2,6b{a}_q,5b{a}_q$, respectively. These values follow by integrating $r^3|R_{nl}{|}^2$. Because the perturbation is central, angular orthogonality makes its matrix diagonal inside each Coulomb energy eigenspace. It splits the accidental degeneracy between different orbital angular momenta while preserving magnetic degeneracy.

For perturbation $V_0{e}^{-{\beta }^2{x}^2}$ and oscillator inverse length $\alpha =\sqrt{m\omega /\hslash }$, direct Gaussian expectation values give $\Delta E_0=V_0\alpha /({\alpha }^2+{\beta }^2{)}^{1/2}$ and $\Delta E_1=V_0{\alpha }^3/({\alpha }^2+{\beta }^2{)}^{3/2}$. A narrow perturbation samples the odd state's node and suppresses its shift by two extra powers of $\alpha /\beta$. A broad perturbation approaches a constant energy offset for both states.

Perturbation theory for an isolated eigenvalue with a one-dimensional eigenspace. First-order nondegenerate perturbation theory gives the diagonal perturbation expectation, while second-order nondegenerate perturbation theory accounts for coupling to other unperturbed eigenstates through energy denominators. Symmetry can permit its use within one block even if another, uncoupled symmetry block has the same unperturbed eigenvalue.

For the normalized sine eigenfunctions of a particle in a box on $(0,a)$, multiplying the lowest one by $ϵ\cos (2\pi x/a)$ gives $ϵ({\varphi }_3-{\varphi }_1)/2$. Thus the first-order diagonal shift is $-ϵ/2$, and the only second-order coupling is to level three. With $E_3=9E_1$, its contribution is $({ϵ}^2/4)/(E_1-E_3)=-{ϵ}^2/(32E_1)$, giving the displayed expansion. This simple selection rule makes the full second-order perturbation theory sum a single term.

The ratio $|⟨k|V|i⟩/(E_i-E_k)|$ measures a leading admixture between unperturbed eigenstates. Small matrix elements compared with relevant gaps justify a nondegenerate expansion locally. A small overall perturbation does not suffice near a coupled degeneracy, where the nearly degenerate block needs diagonalization.

For a differentiable normalized eigenstate of a differentiable self-adjoint Hamiltonian, differentiate $H\psi =E\psi$ and pair with $\psi$. The terms involving ${\psi }^{\prime }$ cancel because $H$ is self-adjoint, leaving the stated formula. It turns energy derivatives into observable expectations. A smoothly chosen eigenbranch and suitable operator-domain regularity are required, particularly at degeneracies or crossings.

The first-order energy correction is the coefficient of the perturbation parameter in the expansion of an energy eigenvalue.

Let ${\sigma }^2=\hslash /(2m\omega )$ and $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ be the quantum harmonic oscillator ground-state probability density. Put $p(a)={\int }_{-a}^{a}g(x)dx$. The first excited-state density is $x^{2}g(x)/{\sigma }^2$. Integration by parts, using $g^{\prime }=-xg/{\sigma }^2$, gives the first-order differenceThe bracket grows from zero to one, since its derivative is $2a^{2}g(a)/{\sigma }^2$. Thus the first-order predicted level gap for $H_0+\lambda H^{\prime }$ is $\hslash \omega [1-\lambda \{p(a)-2ag(a)\}/2]$. For every $\lambda >2$ it becomes negative at sufficiently large $a$.

For $\lambda >1$ the full potential has a central barrier and symmetric minima at $\pm a$. The unperturbed even ground state samples the central barrier more strongly than the odd excited state, whose node reduces its shift. This explains the perturbative tendency, but the negative predicted gap is a failure diagnostic of the first-order truncation. The exact one-dimensional ground state remains nodeless and below the first excited state; in a high symmetric barrier the actual pair has a small positive double-well tunneling splitting.

An energy eigenvalue is nondegenerate when its eigenspace is one-dimensional. Its normalized eigenstate is then unique up to a complex phase.

For a normalized eigenstate $|n⟩$ belonging to a nondegenerate energy eigenvalue,

For a quantum harmonic oscillator perturbed by $\lambda X^4$, write $X=\sqrt{\hslash /(2m\omega )}(A+A^{†})$. Sincefirst-order nondegenerate perturbation theory gives

For a normalized nondegenerate eigenstate $|n⟩$,For a ground state, every denominator is negative, so the correction is nonpositive.

For a nondegenerate lowest state, every excited-state denominator $E_0-E_k$ is negative. Since each numerator $|V_{k0}{|}^2$ is nonnegative, their sum is nonpositive, and strictly negative if a coupled higher state exists. The statement assumes the perturbative expansion and appropriate sum or spectral integral are defined.

Forthe exact energies areThe perturbation series around $\lambda =0$ has radius $\Delta /2$, set by the branch points $\lambda =\pm i\Delta /2$.

Suppose one state $|0⟩$ couples to degenerate states $|1⟩,\dots ,|m⟩$ with amplitudes $v_1,\dots ,v_m$, while the degenerate states do not couple to each other. The normalized bright state proportional to ${\sum }_j{v}_j|j⟩$ is the only combination coupled to $|0⟩$. Every orthogonal combination is a dark state and remains an exact eigenstate at the degenerate energy. The remaining spectrum comes from a two-dimensional Hamiltonian whose off-diagonal entry is $({\sum }_j|v_j{|}^2{)}^{1/2}$.