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In the stationary-state terminology used here, an excited state is a normalized energy eigenstate with energy eigenvalue $E>E_0$, where $E_0$ is an attained ground-state energy. Its excitation cost above the ground state is $\Delta E=E-E_0>0$. Several energy eigenstates can share one excited energy eigenvalue through degenerate energy levels. An arbitrary quantum superposition with energy expectation value above $E_0$ need not be a single stationary excited state; its energy content is described by its Born rule probabilities or, for a continuum, its scalar spectral measure.

An excited state can be a stationary state: the second infinite square well mode is an immediate example. Spontaneous photon emission is a transition caused by coupling matter to the electromagnetic field; it is absent from an isolated matter-only Hamiltonian operator. The relevant energy eigenstate of that isolated model is therefore not automatically an energy eigenstate of the full interacting system. This distinction explains how an ideal stationary excited state can approximate a state with a finite experimental lifetime. See electric-dipole interaction for the coupling used in the simplest optical model.

For a one-photon transition between stationary internal states, neglecting recoil, conservation of energy requiresHere $h$ is the Planck constant and $\nu$ is the photon frequency. Energy matching alone does not guarantee a transition: in the electric-dipole interaction approximation its strength also involves the matrix element $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$, where $\mathbf{d}$ is the electric dipole moment operator and $\mathbf{ϵ}$ specifies photon polarization. A zero matrix element gives an electric-dipole selection rule. Thus an absent optical line does not prove the corresponding energy eigenvalue is absent; a different interaction can couple different states. ^[ref]. The Franck-Hertz experiment gives a particularly instructive comparison between collision excitation and photon emission.

The first excited state terminology assumes a least energy eigenvalue above $E_0$. Such a next level is not guaranteed by quantum mechanics: a continuous spectrum can start immediately above the spectral bottom, or energy eigenvalues can accumulate there. For example, a Hamiltonian operator with energy eigenvalues $0,\epsilon ,\epsilon /2,\epsilon /3,\dots$, where $\epsilon >0$, has a ground-state energy zero but no least positive energy eigenvalue: every candidate $\epsilon /n$ is exceeded downward by $\epsilon /(n+1)$. A free particle on the whole line has neither a ground state nor a first stationary excited state. Each use of a discrete excitation ladder therefore needs a specified Hamiltonian operator and its boundary conditions.

For a faithful representation of $C(K)$ by integration against a regular projection-valued measure $P$, every nonempty open subset $U$ satisfies $P(U)\ne 0$. Choose a nonzero continuous function supported inside $U$, using compact Hausdorff normality. If $P(U)=0$, its spectral integral vanishes, contradicting faithfulness. Regularity is understood through the scalar spectral measures.

Use the usual unital-subalgebra convention $I_H\in A$. The spectral theorem for a commutative operator algebra states that the character space ${\Phi }_A$, equipped with the Gelfand topology, is a compact Hausdorff space, and there is a unique regular projection-valued measure $E$ on its Borel sigma-algebra satisfyingwhere $\hat{a}(\chi )=\chi (a)$ is the Gelfand transform. The map $a\mapsto \hat{a}$ is an isometric unital star-isomorphism from $A$ onto $C({\Phi }_A)$ by the Commutative Gelfand--Naimark theorem.

More explicitly, every $E(B)$ is an orthogonal projection, $E(\varnothing )=0$, $E(B\cap C)=E(B)E(C)$, and for disjoint Borel sets $B_n$,with convergence in the norm of the Hilbert space. Regularity means that each scalar spectral measure $B\mapsto ⟨E(B)v,w⟩$ is a regular Borel measure. The integral identity then implies $\Vert a\Vert ={\sup }_{\chi \in {\Phi }_A}|\hat{a}(\chi )|$. No separability assumption on $H$ is needed.

First construct the spectral measure of a normal operator and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^{\ast }(I,T)$ generated by $T$ and $T^{\ast }$ is commutative because $T$ is a normal operator. The Gelfand transform identifies $A$ with $C({\Phi }_A)$. By spectral permanence for C-star algebras,is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal{B}(H)$. It is injective because a character is determined by its values on $T,T^{\ast }$ and $I$, and $\chi (T^{\ast })=\overline{\chi (T)}$. Thus $h$ is a homeomorphism. Apply part (a) and setThis is a projection-valued measure with $P(K)=I$ andThe associated continuous functional calculus is isometric: $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }$ for $u\in C(K)$.

If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq \mathbb{C}$. If $P(U)=0$, then $u(T)={\int }_{K}udP=0$, contradicting $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }>0$. ThereforeThis proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.

Now take an isolated point $\lambda \in K$. Its singleton is relatively open, so $P(\{\lambda \})\ne 0$. Multiplication of spectral integrals givesThus its range is contained in the eigenspace $\ker (T-\lambda I)$. Conversely, for $v$ in that kernel, the positive scalar spectral measure ${\mu }_v(B)=⟨P(B)v,v⟩$ satisfiesIt follows that ${\mu }_v(K\setminus \{\lambda \})=0$, and hence $\Vert (I-P(\{\lambda \}))v{\Vert }^2=0$. ConsequentlyThe identification of this projection with the eigenspace holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].