Degenerate perturbation theory

First-order degenerate perturbation theory diagonalizes the perturbing operator within each degenerate eigenspace of the unperturbed Hamiltonian.

Let $X$ be a unitary operator with $X^2=I$, and let $P$ be the orthogonal projection operator onto an unperturbed energy eigenspace. Suppose $X$ acts as the same sign $\epsilon =\pm 1$ throughout that eigenspace and $XVX=-V$. For any $u,v$ in the eigenspace, preservation of the inner product gives $⟨u,Vv⟩=⟨Xu,XVv⟩=-⟨\epsilon u,V\epsilon v⟩=-⟨u,Vv⟩$. Hence every projected matrix element vanishes, so $PVP=0$ and degenerate perturbation theory gives zero first-order energy corrections. A zero diagonal alone would be insufficient. For equal spins, the exchange parity of two coupled identical spins is constant on each total-spin sector, while their spin-projection difference changes sign under exchange.

A perturbative unitary conjugation eliminating couplings between chosen separated energy sectors. For $H=H_0+V$ and an anti-Hermitian $S$ satisfying $[S,H_0]=-V$ at leading order, the transformed Hamiltonian contains $\frac{1}{2}[S,V]$ at second order. Exactly resonant coupled states must be retained in a common block. The method was used for the Anderson-to-Kondo relation by John Robert Schrieffer and P. A. Wolff.

A perturbative generator dividing by an unperturbed energy difference cannot eliminate a nonzero coupling between exactly degenerate states. Those states must remain in the retained degenerate block. Small denominators also invalidate an expansion when the actual hopping matrix element, including occupation enhancement, is not small in comparison.

If site-transfer energies are $\mathcal{E}(\Delta )$, the left-ordered raising-hop generator has coefficient $f(\Delta )=t/[\mathcal{E}(\Delta )-\mathcal{E}(\Delta -2)]$. Nonlinear occupation energies generally make the effective second-order shifts and spin-exchange amplitudes depend on the occupation sector.

With $X={\sum }_s{a}_s^{†}{b}_s$, $Y=X^{†}$ and a real coefficient $f$, the fixed-occupancy second-order operator is $t[f(\Delta )XY-f(\Delta +2)YX]$. Normal ordering gives $XY=N_a+P$ and $YX=N_b+P$, where $P={\sum }_{s{s}^{\prime }}{a}_s^{†}{a}_{s^{\prime }}{b}_{s^{\prime }}^{†}{b}_s$ exchanges internal states without changing site totals.

A real perturbation with zero diagonal and every off-diagonal element $t$ has one eigenvalue $2t$ on the symmetric vector and a twofold eigenvalue $-t$ on its orthogonal complement. It is $t(J_3-I)$, where $J_3$ is the all-ones matrix. Diagonal entries alone do not give first-order energies in a degenerate subspace.

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

If the perturbation is already diagonal in a basis of a degenerate eigenspace, those basis vectors receive its diagonal entries as first-order shifts.

When a perturbation preserves each unperturbed degenerate block and has no inter-block coupling, diagonalizing each block can give the exact spectrum.

For oscillators of frequencies $1$ and $2$, the perturbationpreserves $N_A+2N_B$. On the energy-$9/2$ basis $|3,0⟩,|1,1⟩$, it is the matrixso the symmetric and antisymmetric combinations are exact eigenstates with shifts $\pm \sqrt{6}\lambda$.