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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.

A polarizing beam splitter converts a photon polarization qubit to two paths. Rotating one arm's photon polarization to match the other and recombining at a balanced beam splitter gives $P_1(\theta )=1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$. Measurements at phases zero and $\pi /2$ recover the transverse Bloch vector components; direct counts in the two separated arms recover the population difference. All three components, rather than a fringe alone, determine an arbitrary photon polarization density matrix.

// Target: quantum-theory.bigb

Interferometric visibility measures the contrast of a scanned interference fringe. For the mutually coherent, equally polarized two-beam model in wave interference, with fixed individual wave intensities and $I_1+I_2>0$, the constructive interference intensity and destructive interference intensity giveEqual nonzero wave intensities give $\mathcal{V}=1$; $I_1=1,I_2=0.25$ give $\mathcal{V}=0.8$. Unequal wave intensities can therefore reduce interferometric visibility even with perfect mutual optical coherence. Random phase differences or distinguishable wave polarizations can reduce it further; zero contrast alone does not identify the cause. Both intensities zero leave the ratio undefined.

For normalized paths with complex amplitudes $\alpha ,\beta$ and identical internal states, the output probability is $1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$, so $\mathcal{V}=2|\alpha \beta |$. A photon polarization fringe alone determines transverse quantum coherence but not the sign of the population difference.

// Target: optics.bigb