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

Consider a fixed proton and one electron, omitting spin, fine structure and the Lamb shift. Let $e>0$ be the elementary charge, $a_0$ the Bohr radius in this fixed-nucleus model, and $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$. Because the electron's electric dipole moment is $\mathbf{d}=-e\mathbf{r}$, its electric-dipole interaction isThe first excited hydrogen eigenspace has a common unperturbed energy eigenvalue $E_2$. Write $|s⟩=|2,0,0⟩$ and $|p_m⟩=|2,1,m⟩$ in the hydrogen energy eigenbasis. The electric-dipole selection rules for hydrogen leave only the $s,p_0$ coupling within this eigenspace. Each diagonal matrix element vanishes by parity, but that does not make the entire perturbation vanish. Degenerate perturbation theory requires its eigenvalues in the whole eigenspace. With the usual real wavefunction phases the off-diagonal position matrix element is $⟨s|z|p_0⟩=-3a_0$.^[ref]

The factor three can be checked directly. Set $\rho =r/a_0$ and use normalized radial wavefunctions and spherical harmonics:The angular integral is $\int Y_{00}^{\ast }\cos \theta Y_{10}d\Omega =1/\sqrt{3}$. Including $z=r\cos \theta$ and the radial volume factor $r^2$ then givesReversing the quantum phase of $|p_0⟩$ reverses this sign, with no change to the predicted energy eigenvalues.

Thus the nontrivial two-dimensional block and its determinant equation areConsequently, for the displayed sign convention,The two remaining energy eigenstates $|p_{+1}⟩,|p_{-1}⟩$ have zero first-order shifts. The slopes correspond to opposite electric dipole moments, $⟨d_z⟩=+3e{a}_0$ and $-3e{a}_0$. The electric field selects combinations of opposite parity that already have these dipole expectations at leading order. This is why applying first-order nondegenerate perturbation theory separately to $2s$ and $2p_0$, and obtaining two zero diagonal entries, misses the splitting.

The approximation requires $e{a}_0|\mathcal{E}|$ small compared with the separation to other principal quantum numbers. For real hydrogen atoms, it also requires the relevant electric-dipole interaction to dominate the small splittings omitted here, including the Lamb shift. At sufficiently small electric fields a pre-existing nonzero $2s$–$2p$ separation must be retained before taking the limit. A simple two-state illustration with unperturbed separation $\delta >0$ has shifts relative to its meanFor $3e{a}_0|\mathcal{E}|\ll \delta$, the lower branch moves by $-(3e{a}_0\mathcal{E}{)}^2/\delta +O({\mathcal{E}}^4)$ relative to its zero-field value. At $\delta =0$ the splitting is linear. This interpolation explains the importance of the hydrogen energy-level degeneracy; it is not a complete model of the fine structure multiplet.

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.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

Each one-particle energy eigenstate has definite parity about the centre of the well, so $⟨n|x|n⟩=0$. Consequently, a doubly occupied orbital has $⟨x_1{x}_2⟩=0$. For distinct orbitals, the two off-diagonal exchange terms instead survive:The position matrix elements in a symmetric infinite square well, with the present width $a$, giveThe sign of $x_{12}$ depends on the orbital phase convention, while $D$ does not. The perturbation acts only on the spatial part and cannot mix orthogonal singlet states and triplet states. Within a triplet it is proportional to the identity on the spin factor. Thus first-order nondegenerate perturbation theory, with this trivial diagonal treatment of spin degeneracy, gives the shifts corresponding to the three rows in part (b):These formulae also apply for $J<0$, with the reordered levels described above. At $J=0$, they resolve the fourfold $5\epsilon$ level into a shifted triplet and singlet. The next doubly occupied orbital ${\Phi }_{22}^{+}$ still has zero first-order shift. The perturbative requirement is $|\alpha |a^2\ll \epsilon$ for mixing with remote orbital states; the spin-independent perturbation does not require division by the small singlet–triplet gap.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

In the normalized plane wave basis, the only nonzero potential matrix elements areThus a useful small parameter is $|\lambda |/E_L\ll 1$, which suppresses admixture of well-separated energy levels. First-order nondegenerate perturbation theory further requires $|\lambda |/2\ll |E_n-E_{n\pm 1}|$ for the coupled states. This latter condition fails near a half-integer $\alpha$ even for very small $\lambda$; there one must use degenerate perturbation theory in the nearly degenerate pair. Smallness is measured against the gaps to the other states, not the vanishing gap inside this pair.

Near $\alpha =1/2$, keep $n=0,1$. The resulting Hamiltonian operator matrix and its eigenvalues areIn particular,For positive $\lambda$, the lower quantum state at the crossing is the symmetric combination of $n=0,1$, and the upper state is the antisymmetric combination. A negative $\lambda$ exchanges these combinations but leaves the ordered energy eigenvalues unchanged.

This cosine-potential gap in a quantum ring appears in the middle panel above: each lowest-branch crossing opens into an avoided crossing of gap $|\lambda |$ to first order. Away from these crossings, the diagonal matrix elements vanish, so the leading shifts are second order. At integer $\alpha$, the two first-excited plane waves differ by two units of $n$ and therefore have no first-order coupling; their smaller splitting begins at second order. The numerical plot includes that splitting.

The unperturbed infinite square well is invariant under the parity operator. Its one-dimensional energy eigenvalues are nondegenerate, so each energy eigenstate has definite parity. Its position expectation value therefore vanishes: $|{\varphi }_n(x){|}^2$ is even, while $x$ is odd. The first-order nondegenerate perturbation theory correction is consequentlyThe electric field does mix opposite-parity quantum states, and second-order nondegenerate perturbation theory contains two factors of $eEx$. Its leading contribution is thus proportional to $E^2$: this is a quadratic Stark effect. More generally, applying the parity operator changes $H(E)$ into $H(-E)$, so their corresponding energy eigenvalues agree. An analytic expansion about zero electric field contains only even powers of $E$. We assume $e|E|a$ is small compared with the unperturbed excitation scale.

Insert the expansions into the energy eigenvalue equation $\hat{H}{\Psi }_i=E_i{\Psi }_i$. Through first order,The zeroth-order coefficients reproduce part (i). Equating the coefficients of $\lambda$ and moving them to one side givesTerms involving ${\hat{H}}^{(2)}$ first enter at order ${\lambda }^2$ and do not affect this equation of first-order nondegenerate perturbation theory.

Write $|\uparrow ⟩,|\downarrow ⟩$ for the two $S_3$ eigenstates, of eigenvalues $\pm \hslash /2$. The spin-spin interaction can be evaluated usingFor $|\beta |\gg |\alpha |$, take $H_0=\beta (S_3^{(1)}-S_3^{(2)})/\hslash$ and $V=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$. The unperturbed energy eigenvalues are $\beta ,-\beta ,0,0$ on $|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$, respectively. For an isolated normalized energy eigenstate $|n⟩$, first-order nondegenerate perturbation theory gives $\Delta E_n^{(1)}=⟨n|V|n⟩$. The two antiparallel spins have expectation $-\alpha /4$, soFor the zero-energy degenerate eigenspace, degenerate perturbation theory requires diagonalizing the restriction of $V$. On the ordered basis $(|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩)$ it is $(\alpha /4)I$: the ladder operators cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so the remaining two levels are exactly $\alpha /4$, retaining their twofold quantum degeneracy.

For $|\alpha |\gg |\beta |$, take $H_0=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hslash$. The spin-one-half triplet state basis and spin-one-half singlet state areThe identity ${\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}=({\mathbf{J}}^2-3{\hslash }^2/2)/2$ gives unperturbed energy eigenvalues $E_t^{(0)}=\alpha /4$ and $E_s^{(0)}=-3\alpha /4$. Direct action givesConsequently the restriction of $V$ to the entire triplet degenerate eigenspace is zero, so no first-order splitting is missed. The singlet expectation is also zero. In second-order nondegenerate perturbation theory, $\Delta E_n^{(2)}={\sum }_{r\ne n}|⟨r|V|n⟩{|}^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective matrix is ${\beta }^2|t_0⟩⟨t_0|/\alpha$, already diagonal in this basis. Thus, through second order,The last two expressions have higher corrections beginning at $O({\beta }^4/|\alpha {|}^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the energy denominators above.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

For a normalized eigenstate $|n⟩$ with a nondegenerate energy eigenvalue, first-order nondegenerate perturbation theory and second-order nondegenerate perturbation theory give

Let $\Delta =E_{+}-E_{-}>0$. Hermiticity givesso in the ordered basis $(|-⟩,|+⟩)$,For a nondegenerate unperturbed level, the first-order nondegenerate perturbation theory formulas areThe diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections areThus, through linear order,Normalization changes only at quadratic order.

The unperturbed ground state is $|0,0⟩$ with energy $\hslash \omega$. Sincethe first-order energy correction vanishes and first-order nondegenerate perturbation theory gives