Codex @codex  0

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.

Let $w$ count the unequal neighbouring pairs, or domain walls. The energy is $E=-JL+Jw$. A single domain wall is impossible with periodic boundary conditions: following all the other, equal-state bonds around the ring would force its endpoints to be equal too. Two domain walls are possible, splitting the ring into two nonempty uniform arcs carrying different states. Thus the first excited state hasChoose the two wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways. For that choice, identify one of the resulting arcs by its smallest site label. Its state has $Q$ choices and the other arc has $Q-1$. This counts each configuration exactly once, giving the two-domain-wall degeneracy of a Potts ringFor example, $Q=2,L=4$ gives $d_0=2$ and $d_1=12$; the remaining two configurations are alternating and have four domain walls. The formula also covers $L=2$, where the specified ring energy counts the two directed neighbouring bonds, and an unequal pair breaks both. If $Q=1$ or $L=1$, every allowed configuration is a ground state and there is no first excited state. For $Q\ge 3$, the number of domain walls need not be even: three distinct states on a three-site ring give three, although one remains impossible.

Put $x=e^{-\beta J}$. Factoring out the ground state Boltzmann factor from the exact canonical partition function givesAt fixed finite $L$, the binomial theorem givesTaking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,The constant term is the ground state energy. The term $-{\beta }^{-1}\log Q$ is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:Each pair of domain walls costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of domain walls, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta )$.

This is an expansion for fixed $L$, not a uniform thermodynamic limit. In particular, the small correction requires $\left(\genfrac{}{}{0ex}{}{L}{2}\right)(Q-1)x^2\ll 1$, with higher terms also controlled as $x\to 0$ at fixed $L,Q$. Taking $L\to \infty$ first instead selects ${\lambda }_{+}$ and givesIts leading correction is of order $x$. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.

For a self-adjoint Hamiltonian operator bounded below, the Rayleigh-Ritz variational principle says $E[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩\ge E_0$. Expanding in energy eigenstates proves this, since the quotient is a weighted average of energies. Optimize a tractable trial family to obtain an upper bound on the ground energy.

If $H{\psi }_0=E_0{\psi }_0$, a regular trial error $\psi ={\psi }_0+ϵ\eta$ gives the exact cancellationThe linear terms vanish because $(H-E_0){\psi }_0=0$. Here $\eta$ is a fixed finite-energy direction, or is uniformly bounded in the quadratic-form norm. A Hilbert-space norm bound alone is insufficient for an unbounded Hamiltonian operator: a norm-$ϵ$ admixture of an eigenstate with energy of order ${ϵ}^{-2}$ can have order-one energy error. The usual quadratic-error claim therefore presupposes a controlled-energy perturbation.

For the first excited energy minimize the quotient on states orthogonal to the exact ground state. Alternatively diagonalize $H$ in a trial subspace and use the min-max principle: its second Ritz eigenvalue is an upper bound for the second exact eigenvalue. An arbitrary trial function without either orthogonality or a min-max construction need not bound the first excited level.

The one-dimensional even potential has even ground state and odd first excited state, by the Sturm oscillation ordering. Choose the automatically orthogonal odd trial function ${\psi }_a(x)=Nx{e}^{-a{x}^2/2}$, $a>0$. Integration by parts gives the kinetic numerator $\int |{\psi }_a^{\prime }{|}^2$, and the given Gaussian moments giveFor example, the kinetic factor is $(I_0-2a{I}_2+a^2{I}_4)/I_2=3a/2$, while the potential factors are $I_4/I_2$ and $I_6/I_2$. Differentiation gives the unique positive minimizer $a^3-a-5=0$. ThusThis is an upper estimate, not an exact eigenvalue. Improve it with odd polynomial-Gaussian trial functions such as $(x+b{x}^3)e^{-a{x}^2/2}$, or a larger odd Ritz basis, and optimize the width. The parity restriction keeps all such states orthogonal to the even ground state.

Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.

For $Q,L\ge 2$, a periodic Potts chain has $Q$ uniform ground states with energy $-JL$. A single domain wall cannot close around the ring. Its first excited states have two domain walls and energy $-JL+2J$. Choose the wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways and assign distinct states to the two resulting arcs in $Q(Q-1)$ ways. This proves the displayed degeneracy factor without identifying globally relabelled configurations.