For a positive integer $k$, cancellation of $U^{†}U=I$ gives $A^k=U{D}^k{U}^{†}$. The trace is invariant under matrix similarity, since $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, and the determinant is multiplicative. ThereforeThis includes zero eigenvalues without requiring $A$ to be invertible.

For any nonzero constant $C$, differentiating ${\psi }_0=C{e}^{-x^2/2}$ gives ${\psi }_0^{\prime \prime }=(x^2-1){\psi }_0$. HenceIt is indeed the lowest eigenvalue, since the factorization $H=(-d/dx+x)(d/dx+x)+1$ makes $⟨\psi ,H\psi ⟩\ge \Vert \psi {\Vert }^2$, with equality for this Gaussian. The variational bound exceeds the exact value by $\sqrt{10/7}-1\simeq 0.195229$, or about $19.52\%$.

The PDF's prefactor $1/\sqrt{2\pi }$ does not normalize this wavefunction: its squared norm is $1/(2\sqrt{\pi })$. The normalized ground state has prefactor ${\pi }^{-1/4}$. This normalization error does not affect the eigenvalue, because the equation is homogeneous in the wavefunction.

Put $C_a=\sqrt{15/(16a^5)}$, with $a>0$. Direct integration confirms the normalization:The kinetic energy and potential energy terms in the dimensionless Rayleigh quotient areThus ${\lambda }_0\le 5/(2a^2)+a^2/7$. Differentiating with respect to $a$ gives $-5/a^3+2a/7=0$, so the unique minimizing width is $a=(35/2{)}^{1/4}$. The two energy contributions are equal there, and the compact parabolic trial bound for the quantum harmonic oscillator isThe trial is continuous and zero at $\pm a$, so its weak first derivative is $-2C_{a}x$ inside and zero outside. The jumps of that derivative do not add delta terms to the first derivative, and its squared norm is finite. The energy-form calculation is therefore valid even though the trial does not belong to the operator domain of the harmonic oscillator. Applying the second-derivative operator naively and omitting its endpoint distributions would be incorrect.

For a regular self-adjoint problem with a lowest eigenvalue and a complete orthonormal eigenbasis in the weighted inner product, expand an admissible trial function as $\psi ={\sum }_j{c}_j{\psi }_j$. ThenThe Rayleigh-Ritz variational principle therefore provides the upper boundChoose a family satisfying the essential boundary conditions and finite energy, evaluate the quotient, and minimize over its parameters. Enlarging the trial family can only improve or preserve the bound; equality is attained if it includes an exact ground eigenfunction. For a finite linear trial space, this becomes a generalized matrix eigenvalue problem.

One suitable choice is homogeneous Dirichlet boundary conditions, $\psi (\alpha )=\psi (\beta )=0$, with variations vanishing there. Assume real $q$ and the stated positive $p,w$. Introduce a Lagrange multiplier $\mu$ for $N[\psi ]=1$. For real variations, integration by parts givesThe boundary term vanishes, so the fundamental lemma of the calculus of variations yieldsMultiplying this equation by $\psi$, integrating, and using the boundary conditions gives $Q[\psi ]=\mu N[\psi ]=\mu$. Thus the multiplier, the eigenvalue, and the constrained stationary value coincide. Conversely every normalized eigenfunction makes the first variation vanish for all tangent variations satisfying $\delta N=0$.

Homogeneous Neumann boundary conditions, $p{\psi }^{\prime }=0$ at both endpoints, or mixed Dirichlet and Neumann conditions also work with this energy form. General Robin boundary conditions require the corresponding boundary terms in the form; they cannot simply be inserted into the displayed unmodified integral.

The exponential power series gives $e^{-1/z}={\sum }_{m=0}^{\infty }(-1{)}^m/(m!z^m)$ for $z\ne 0$. Hence all singular terms areInfinitely many negative powers occur, so zero is an essential singularity.

Use the local coordinate $w=1/z$ at infinity. For positive integer $N$,The negative powers of this local coordinate are precisely $m=0,\dots ,N-1$. The singular part at infinity is thereforeInfinity is a pole at infinity of order $N$. In the original $z$ variable these singular terms are positive powers; it is the negative powers of $w=1/z$ that define the local principal part of a Laurent series. For $N=0$ the point at infinity is regular, and for negative integer $N$ it is a zero rather than a pole.

Put $w=z-i\pi$. Since $\sinh (w+i\pi )=-\sinh w$ and $\sinh w=w(1+w^2/6+O(w^4))$, the local Laurent series isThus all singular terms areThere is a third-order pole, with no second-order term.

The even extension is the triangular function $f(x)=(1-|x|)$ for $|x|<1$ and zero outside. Its Fourier transform is real and even:At zero the removable limiting value is one, also obtained directly by integrating the triangle. This is the Fourier transform of a triangular function.

Choose the argument interval $-3\pi /2<\theta <\pi /2$, which makes the logarithm real on the positive real axis and puts its cut on the positive imaginary axis. Parametrize $z=R{e}^{i\theta }$. The constant term $\ln R$ contributes $\ln R\oint dz=0$, leavingEquivalently the cut-crossing point is $iR$, so the circle integral of a logarithm with a radial branch cut is $2\pi i(iR)=-2\pi R$. The different cut changes where the discontinuity lies on the circle; it is not merely an additive constant change of branch along the entire contour.

On the unit circle use $z=e^{i\theta }$ with $0<\theta <2\pi$. The specified branch has $\mathrm{Log}z=i\theta$ there and $dz=i{e}^{i\theta }d\theta$. ThereforeThis is the circle integral of a logarithm with a radial branch cut, using the one-sided values at the cut crossing. The logarithm is discontinuous at that crossing, so the Cauchy integral theorem for a function holomorphic around a closed contour does not apply.

The difference quotient at zero isFor $a>1$, its absolute value is at most $|h{|}^{a-1}$. Given $\epsilon >0$, set $\delta ={\epsilon }^{1/(a-1)}$. Then $0<|h|<\delta$ makes the difference quotient have absolute value below $\epsilon$. Thus $f_a^{\prime }(0)=0$.

Conversely, along the second sequence the difference quotient is exactly zero. Any finite derivative would consequently have to be zero. Along the first positive sequence, however, the difference quotient equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\epsilon =1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite derivative exists. We obtainIn particular, $0<a\le 1$ gives continuity without differentiability.

For $a>0$, the fact that sine is a bounded function gives $|f_a(x)-f_a(0)|\le |x{|}^a$. Given $\epsilon >0$, choose $\delta ={\epsilon }^{1/a}$. If $|x|<\delta$, then $|f_a(x)-f_a(0)|<\epsilon$. This proves continuity directly from its definition.

For $a\le 0$, use the first sequence from part (b), for which $f_a(x_n)=x_n^a$. When $a=0$ these values are one; when $a<0$ they tend to $+\infty$. In either case, for $\epsilon =1/2$ every neighbourhood of zero contains a sufficiently late $x_n$ with $|f_a(x_n)-f_a(0)|\ge 1/2$. This contradicts the definition of continuity. Hence

The function $f$ is differentiable at $x_0$ if there exists a finite real number $L$ such thatThen $f^{\prime }(x_0)=L$. The puncture $0<|x-x_0|$ makes the difference quotient defined; requiring a finite real $L$ distinguishes an ordinary derivative from an infinite slope.

The function $f$ is continuous at $x_0$ ifThe choice of $\delta$ may depend on $\epsilon$ and $x_0$, but must work for every $x$ in that neighbourhood. This is the epsilon-delta definition of continuity.

Let $J=A/a$, so $J^2=-I$. The even and odd powers in the matrix exponential giveHere the two scalar sums are the Taylor series for cosine and sine. ConsequentlyWith the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by $-a$; the parameter in the form printed in the paper is $\theta =a$.

The defining condition for an antisymmetric matrix forcesIts dot product with $(x,y{)}^T$ is zero, and its Euclidean norm is $|a|\sqrt{x^2+y^2}$. Thus it combines a right-angle rotation with a uniform dilation by $|a|$. More precisely, the rotation is clockwise for $a>0$ and anticlockwise for $a<0$. A direction-independent description is a clockwise right-angle rotation followed by multiplication by the signed scalar $a$.

A second application givesGeometrically this is a half-turn combined with a dilation by $a^2$. Because $A$ is nonzero, $a\ne 0$.

Choose the unnormalized eigenvectorsTheir linear combination giving the specified vector isApplying the reciprocal eigenvalues of the inverse matrix therefore yieldsIn Cartesian coordinates, this is $(-1/3,2/3{)}^T$, and direct matrix multiplication gives $S\mathbf{y}=(1,0{)}^T$.

Since every eigenvalue is nonzero, $S$ is an invertible matrix. Multiplying $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ by the inverse matrix givesHence the eigenvectors are unchanged and the corresponding eigenvalues are $1/{\lambda }_i$.

Induction on $p$ gives $S^p{\mathbf{u}}_i={\lambda }_i^p{\mathbf{u}}_i$, including $p=0$. The absolutely convergent matrix exponential series may therefore be applied term by term to each eigenvector:ThusFor a repeated eigenvalue, every nonzero vector in its eigenspace remains an eigenvector. Because the real exponential function is injective, distinct real eigenvalues of $S$ do not merge under this transformation.