Excited-State Targeting

Excited-state targeting must prevent a solver or optimizer from collapsing into states already found. It must also distinguish the physical state being followed from its position in an energy-sorted list. These requirements become central near degeneracies and avoided crossings, where a small energy error can coexist with a large error in state identity.

Helpful background. Sparse eigensolvers covers spectral targeting and Rayleigh–Ritz supplies subspace bounds.

Orthogonality constraints and block methods

For a Hermitian matrix with known orthonormal lower eigenvectors $u_0,\ldots,u_{k-1}$, minimize the Rayleigh quotient subject to

$$x^\dagger x=1,\qquad u_j^\dagger x=0.$$

Equivalently use the projector $P=I-\sum_{j<k}u_ju_j^\dagger$ and restrict the operator to its range. Do not diagonalize $PHP$ on the whole original space without accounting for the zero eigenvalues introduced on the removed subspace.

Approximate lower states make the projector imperfect relative to the exact Hamiltonian. Their errors can contaminate higher states; refine them and monitor every mutual overlap. A block Rayleigh–Ritz calculation often handles a cluster more naturally, solving for the entire subspace at once. For a nonorthogonal representation, the constraints are $X^\dagger SX=I$.

Symmetry decomposition can remove unwanted states exactly. The lowest state in an odd-parity sector is an excited state of the full problem, but need not be the globally first excited state. State the sector whenever reporting an energy.

Penalties and interior targets

A penalty operator

$$H_\lambda=H+\lambda\sum_{j<k}u_ju_j^\dagger$$

raises the already known levels by $\lambda$ when those states are exact. Choose $\lambda$ large enough to move them above the desired state. Excessively large penalties worsen spectral scaling and can make optimization stiff; they do not repair inaccurate projectors.

Shift-invert targets eigenvalues near a specified energy, independent of how many lower levels exist. The folded-spectrum objective,

$$\langle x|(H-\sigma I)^2|x\rangle,$$

also favors levels nearest $\sigma$, but squares the spectral width and can worsen conditioning. Equidistant levels become degenerate in the folded operator. Always recover energies and residuals using $H$, not only the transformed objective.

Tracking states through a crossing

For consecutive parameter values $\lambda$ and $\lambda+\delta\lambda$, compare the overlap matrix

$$M_{ij}=\langle u_i(\lambda)|u_j(\lambda+\delta\lambda)\rangle.$$

Large overlaps guide assignments, but individual-vector matching becomes ambiguous inside a nearly degenerate cluster. Track its projector or align the two bases by an SVD of $M$. A smooth phase convention can set neighboring overlaps real and positive when they are nonzero. It cannot turn a genuine crossing into a unique physical labeling.

For

$$H(\lambda)=\begin{pmatrix}\lambda&g\\g&-\lambda\end{pmatrix},$$

with real $\lambda$ and $g$, the energies are $\pm\sqrt{\lambda^2+g^2}$. At $g=0$, diabatic basis states cross at $\lambda=0$, and the energy-sorted ground vector changes basis label abruptly. At $g\neq0$, the eigenvectors rotate over a scale $|\lambda|\sim|g|$. A parameter grid much coarser than that scale can misidentify the branch even with accurate diagonalization.

What constitutes a verified excitation

Report energy gaps together with absolute energies, state or subspace residuals, orthogonality, and symmetry labels. Converge the basis for both states: cancellation of their errors may make one gap look unusually stable. Transition matrix elements require consistently normalized states and appropriate phase or gauge handling. A dark transition can be a selection rule or a state-labeling error; inspect the operator and sectors before interpreting it.

Exercises

For exact energies $E_0=0$, $E_1=1$, $E_2=3$, what penalty makes the first excited state the unique ground state of $H_\lambda=H+\lambda|0\rangle\langle0|$?

Solution

The modified energies are $\lambda,1,3$. The desired state is uniquely lowest for $\lambda>1$. At $\lambda=1$ it is degenerate with the penalized original ground state. There is no advantage in choosing an arbitrarily enormous penalty.

References

• Bai, Z., J. Demmel, J. Dongarra, A. Ruhe, and H. van der Vorst, eds., Templates for the Solution of Algebraic Eigenvalue Problems: A Practical Guide, SIAM, 2000, online text — invariant subspaces and interior eigenvalues.
• Saad, Y., Numerical Methods for Large Eigenvalue Problems, revised ed., SIAM, 2011 — block iteration, deflation, and spectral transformations.