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An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

Here $\alpha =e^{2\pi i/3}$. The missing-trace matrix has determinant $-\sqrt{3}k\Delta (k)$. Its nonzero zeros are the eigenvalues in the parametrization $Av=i{k}^{3}v$ of $A=-d^3/d{x}^3$ with $v(0)=v(L)=v^{\prime }(L)=0$. Integration by parts gives $\mathrm{Re}⟨v,Av⟩=-|v^{\prime }(0){|}^2/2$. No nonzero zero lies in $\mathrm{Re}(-i{k}^3)\le 0$: on its boundary the determinant equation reduces, after rotation, to an equality between a unit-modulus exponential and $\cosh u-i\sqrt{3}\sinh u$, whose modulus exceeds one unless $u=0$. The zero at the origin is removable in the original boundary problem. There are nevertheless infinitely many genuine negative eigenvalues of $A$, obtained from $e^{-3s/2}=2\cos (\sqrt{3}s/2+\pi /3)$, $s>0$, with $k=-is/L$. Absence of contour poles in the Fokas method does not imply absence of discrete spectrum of the operator.

Expanding the matrix determinant from part (b) givesThe finite-interval Airy spectral determinant obeys $\Delta (\alpha k)={\alpha }^2\Delta (k)$. At the origin,This zero comes from coalescence of the three rotated equations; it is removable in the underlying boundary system. Indeed, $v^{\prime \prime \prime }=0$ with $v(0)=v(L)=v^{\prime }(L)=0$ has only the zero quadratic solution.

For $k\ne 0$, a zero of $\Delta$ is equivalent to the existence of a nonzero combination $v(x)={\sum }_{j=0}^2{c}_j{e}^{i{\alpha }^{j}kx}$ satisfying the homogeneous endpoint conditions. This can be checked by expanding the determinant of those three endpoint equations. Such a function satisfies $v^{\prime \prime \prime }=\omega (k)v$, so it is an eigenfunction of $A$ with eigenvalue $-\omega (k)$. Integration by parts yieldsso every nonzero zero has $\mathrm{Re}\omega (k)\ge 0$. None lies on the boundary $\mathrm{Re}\omega =0$: by a cube-root rotation it suffices to take $K=kL$ real. With $u=\sqrt{3}K/2$, the determinant equation becomesThe right-hand side has squared modulus $1+4{\sinh }^{2}u>1$ for $K\ne 0$, whereas the left-hand side has modulus one. Therefore there are no nonremovable spectral poles in $\overline{D}$, and the contour representation has no discrete residue contribution from that domain.

Taken literally as a claim about the operator's discrete spectrum, however, the printed assertion is false. To see this constructively, put $k=-is/L$, $s>0$. ThenThus zeros occur whenBetween successive positive cosine maxima and adjacent negative minima the difference changes sign. There are consequently infinitely many positive roots, producing genuine negative eigenvalues $-s^3/L^3$ of $A$. The third-order resolvent maps into $H^3(0,L)$, whose inclusion into $L^2(0,L)$ is compact, so these are indeed discrete spectrum of an operator with compact resolvent. The valid conclusion is absence of poles in the reconstruction sectors, not absence of operator eigenvalues.

Put $T=A-zI$. We use the following closed-range lemma:Indeed,Away from this kernel, zero is separated from the spectrum of the positive self-adjoint operator $T^{\ast }T$ exactly whenfor some $c>0$. This is equivalent to closed range. Zero is then absent from the essential spectrum exactly when $\dim \ker T<\infty$. Similarly,because $\ker T^{\ast }=(\mathrm{ran}T{)}^{\perp }$ measures the cokernel.

Applying the lecture definitions of the three essential spectra of a closed operator now givesand

If $A$ is normal, so is $T$. The spectral theorem givesand maps the spectral mass of $T$ at $0$ exactly to the spectral mass of $T^{\ast }T$ at $0$. Thus zero is isolated with finite multiplicity for $T^{\ast }T$ exactly when $z$ is an isolated eigenvalue of finite multiplicity for $A$. Consequently

Normality is essential. Let $U{e}_j=e_{j+1}$ be the unilateral shift and take $A=U^{\ast }$. Its spectrum is the closed unit disk, so $0$ is not in the discrete spectrum. Butwhose zero eigenvalue is isolated and simple. Hence $0\in {\mathrm{Sp}}_{\mathrm{d}}(A^{\ast }A)$ while $0\notin {\mathrm{Sp}}_{\mathrm{d}}(A)$.