Finite basis projector energies and variational electronic structure — Examples

The eigenvalues of $H$ are $0$ and $2$, with corresponding unit eigenvectors $e_1,e_2$ by [F1]. For a unit vector $v=(a,b)$, $v^{\ast }Hv=2\mathrm{\mid }b{\mathrm{\mid }}^2\ge 0$, and equality holds at $e_1$, so the full-space ground energy is $0$.

Every unit vector in $V_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$ is $e^{i\theta }{e}_2$ for some real $\theta$, and its energy is $(e^{i\theta }{e}_2{)}^{\ast }H(e^{i\theta }{e}_2)=2$. Thus the restricted minimum is $2>0$, strictly above the full finite-dimensional ground energy.

Each unit sphere in $K_j$ is closed and bounded in finite-dimensional real coordinates, hence compact by [F3]. The energy $v\mapsto v^{\ast }Hv$ is continuous, so [F4] gives a minimum on each sphere; diagonalizing the restrictions by [F2] gives the values $1,1,-1$ on $K_1,K_2,K_3$, respectively, and $-1$ for every $j\ge 3$.

The spaces satisfy $K_1\subset K_2\subset K_3$ and their infinite union is ${\mathbb{C}}^3$. The computed values are nonincreasing, with equality at the first enrichment, and converge to $-1={\inf }_{v\in {\mathbb{C}}^3,\mathrm{\parallel }v\mathrm{\parallel }=1}{v}^{\ast }Hv$. Applying [F1] to the infinite nested sequence of their unit spheres gives this same limit.

By [F2], the eigenvalues of $A$ are $-1$ and $1$, so its ground eigenvalue is $-1$. The vector $u=e_2$ is a unit eigenvector for $1$, whence $\mu =u^{\ast }Au=1$ and $(A-\mu I)u=0$. Thus $\rho =0$.

The general residual theorem [F1] therefore returns exact proximity to the spectrum, but this example's nearby eigenvalue is the excited value $1$, not the ground value $-1$. A zero residual still cannot supply spectral ordering information.

The vectors $q_{+}=(1,1)\mathrm{/}\sqrt{2}$ and $q_{-}=(1,-1)\mathrm{/}\sqrt{2}$ are orthonormal and satisfy $S_{\epsilon }{q}_{+}=(2-\epsilon )q_{+}$ and $S_{\epsilon }{q}_{-}=\epsilon q_{-}$. Both eigenvalues are positive, and their ratio is $(2-\epsilon )\mathrm{/}\epsilon$.

Let $S_{\epsilon }=L{L}^{\ast }$ be its Cholesky factorization. By [F1], $y=L^{\ast }c$ preserves the exact generalized eigenvalues and Rayleigh quotient. For any matrix $A$, $\mathrm{\parallel }A{\mathrm{\parallel }}_2^2={\max }_{\mathrm{\parallel }x{\mathrm{\parallel }}_2=1}{x}^{\ast }{A}^{\ast }Ax$, which equals the largest eigenvalue of the positive Hermitian matrix $A^{\ast }A$ by [F2]. Since $(L^{-1}{)}^{\ast }{L}^{-1}=L^{-\ast }{L}^{-1}=S_{\epsilon }^{-1}$, its eigenvalues are $1\mathrm{/}\epsilon$ and $1\mathrm{/}(2-\epsilon )$, so $\mathrm{\parallel }{L}^{-1}{\mathrm{\parallel }}_2={\epsilon }^{-1\mathrm{/}2}$. For $A=L^{-\ast }$, $A^{\ast }A=L^{-1}{L}^{-\ast }=L^{-1}{S}_{\epsilon }^{-1}L$, which is similar to $S_{\epsilon }^{-1}$; hence $\mathrm{\parallel }{L}^{-\ast }{\mathrm{\parallel }}_2={\epsilon }^{-1\mathrm{/}2}$ as well. The operator norm is attained on a unit vector in this finite-dimensional space, so a unit perturbation in a maximizing direction is amplified by this factor when converted back to coefficient coordinates.

As $\epsilon \downarrow 0$, the condition number and coordinate amplification diverge, although the exact whitening identities in [F1] continue to hold for each $\epsilon >0$. This is a conditioning observation; it gives no finite-precision energy-error estimate without a specified perturbation model.

For every $\theta$, $u_{\theta }^{\ast }{u}_{\theta }=1$, so [F2] gives $P_{\theta }\in {\mathcal{P}}_1$. Multiplication yields $\mathrm{tr}(H{P}_{\theta })={\cos }^2\theta -{\sin }^2\theta =\cos (2\theta )$.

Near $\theta =\pi \mathrm{/}4$, the left branch of $\mathrm{\mid }\cos (2\theta )\mathrm{\mid }$ has derivative $-2$ at $\pi \mathrm{/}4$, while the right branch has derivative $2$. The restriction is not differentiable there; by [F3], a $C^2$ ambient functional composed with the smooth curve $P_{\theta }$ would be differentiable. Hence the candidate cannot satisfy the $C^2$ requirement in [F1].

For Hermitian $A,B$, $\mathrm{tr}(HA)$ and $\mathrm{tr}(HB)$ are real, and $⟨G(A),B{⟩}_F=2\mathrm{tr}(HA)\mathrm{tr}(HB)=⟨A,G(B){⟩}_F$. Thus $G$ is self-adjoint and real-linear; the model's quadratic term is $\frac{1}{2}\mathrm{tr}(PG(P))=\mathrm{tr}(HP{)}^2$.

For an arbitrary unit vector $u=(u_1,u_2)\in {\mathbb{C}}^2$, put $s=\mathrm{\mid }{u}_2{\mathrm{\mid }}^2$. Then $P=u{u}^{\ast }\in {\mathcal{P}}_1$ by [F1], $\mathrm{tr}(HP)=\mathrm{\mid }{u}_1{\mathrm{\mid }}^2-\mathrm{\mid }{u}_2{\mathrm{\mid }}^2=1-2s$, and $E(P)=(1-2s{)}^2$. A neighborhood of $P_0=e_1{e}_1^{\ast }$ is given by $0\le s<1\mathrm{/}2$; every distinct projector there has $s>0$ and hence $(1-2s{)}^2<1=E(P_0)$. Thus $P_0$ is a strict local maximum in the full complex projector domain.

At $P_0$, $F(P_0)=G(P_0)=2H$, which is diagonal and therefore commutes with $P_0$. By [F3], $P_0$ is stationary, while step 2.1 shows it is not a local minimum.

If $X^{\ast }X=I_r$, then $P=X{X}^{\ast }$ is Hermitian and $P^2=X(X^{\ast }X)X^{\ast }=P$, with range equal to the column space of $X$ and therefore rank $r$. For unitary $Q$, $(XQ)(XQ{)}^{\ast }=XQ{Q}^{\ast }{X}^{\ast }=P$. Thus $E(X{X}^{\ast })$ is unitary-invariant and descends to complex Grassmann by [F1].

For every horizontal $Z$, [F2] and [F4] yield the complex Grassmann gradient $2(I-P)F(P)X$. This formula uses the real derivative of the Hermitian-matrix energy, as required for a real-valued objective on complex coordinates.

Conversely, let $P$ be any rank-$r$ orthogonal projector. Choose an orthonormal basis of its $r$-dimensional range by [F7] and collect its vectors as the columns of $X$. Then $X^{\ast }X=I_r$ and $X{X}^{\ast }=P$, so the parameterization covers exactly the model's projector domain. By [F5], the polar-retracted trial remains a complex orthonormal frame, so its projector remains Hermitian, idempotent, and rank $r$; rejection leaves the old projector unchanged by [F6].

Fix any finite ${\mathrm{\Delta }}_{\max }>0$. The set of pairs $(X,\eta )$ with $X^{\ast }X=I_r$, $X^{\ast }\eta =0$, and $\mathrm{\parallel }\eta {\mathrm{\parallel }}_F\le {\mathrm{\Delta }}_{\max }$ is closed and bounded in a finite-dimensional real coordinate space, hence compact by [F8]. The polar retraction is smooth on it, and [F1]–[F2] make the pullback energy $C^2$ along the horizontal tangent bundle. Its fiber Hessian is continuous on this compact set, so its operator norm has a finite maximum by [F9]. Integrating this Hessian along each fiber segment gives a uniform Lipschitz bound for the pullback gradient. Choose the model operator $B_x=0$, which is self-adjoint and uniformly bounded by $M_B=0$. Hence all regularity and model hypotheses in [F6] hold, and a finite cap at least the theorem's trial bound suffices.