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In local thermodynamic equilibrium, the ratio of two internal-level populations includes their energy-level degeneracies $g_i,g_j$ and their excitation-energy difference. Temperature-sensitive spectral line ratios can constrain the excitation temperature when line formation and radiative transfer are modelled consistently. Ionization fractions additionally require the Saha equation.

An energy level is degenerate when its eigenspace has dimension greater than one: several linearly independent energy eigenstates share the same energy eigenvalue. Any quantum superposition within that eigenspace is also an energy eigenstate with the same energy.

For a particle in a ring without applied flux, clockwise and anticlockwise momentum eigenstates have equal kinetic energy, except for the single zero-momentum state. A field or other perturbation may split such a pair if it distinguishes the two states. Degenerate perturbation theory identifies the combinations selected by a weak perturbation. The energy-level degeneracy counts independent states, rather than the number of different basis descriptions one can write for them.

An energy level is an allowed energy eigenvalue of a Quantum Hamiltonian. Its corresponding energy eigenstates satisfy $H|n⟩=E_n|n⟩$. Several independent energy eigenstates may share one energy level; their number is the energy-level degeneracy. A quantum superposition of different energy eigenstates need not have a definite energy, even though an ideal energy measurement yields one of the allowed outcomes.

Quantum confinement often produces discrete energy levels: the quantum harmonic oscillator has $E_n=\hslash \omega (n+1/2)$. An unconfined free particle instead has a continuous spectrum. In spectroscopy, a transition between two energy levels exchanges approximately $|E_f-E_i|=h\nu$ with a photon; recoil and external fields can affect the measured frequency. Selection rules determine which energy differences produce observable spectral lines.

The interior time-independent Schrödinger equation for a free particle isConfinement by the rectangular infinite well imposes Dirichlet boundary conditions: $\psi =0$ at $x=0,b$ and $y=0,\sqrt{5}b$. The wave function is square-integrable and may be normalized to unit probability on the rectangle.

Separation of variables gives products of the one-dimensional well states,The Hamiltonian is the sum of the independent $x$ and $y$ kinetic-energy operators, so the one-dimensional energies add. Put $C={\pi }^2{\hslash }^2/(2m{b}^2)$. ThenFor $n=1$, increasing $\ell$ gives $6C/5,9C/5,14C/5,21C/5$. The smallest energy with $n\ge 2$ is also $21C/5$, attained at $(2,1)$; all $n\ge 3$ give larger energies. Thus the ground and first three distinct excited energy levels areThe largest listed energy therefore has energy-level degeneracy $\boxed{2}$, since the two product eigenfunctions are linearly independent and no other pair solves $5n^2+{\ell }^2=21$.

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

With the specified ladder-operator normalization, $A_{+}^{†}=A_{-}$ andTherefore $\Vert A_{-}\psi {\Vert }^2=2m(E-\hslash \omega /2)\Vert \psi {\Vert }^2\ge 0$. This proves $\boxed{E\ge \hslash \omega /2}$, with equality exactly when $A_{-}\psi =0$. In position representation this equation is $\hslash {\psi }^{\prime }+m\omega x\psi =0$, whose general solution isFor a unit vector $|C|=(m\omega /\pi \hslash {)}^{1/4}$; its phase is arbitrary. The commutators $[H,A_{\pm }]=\pm \hslash \omega A_{\pm }$ make $A_{+}$ raise and $A_{-}$ lower energies by $\hslash \omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hslash \omega$. Conversely lowering any nonzero energy eigenvector must eventually terminate, since otherwise it would produce an energy below the lower bound. A terminating vector is annihilated by $A_{-}$ and hence has energy $\hslash \omega /2$. Thus no other eigenvalues occur.

Bilinearity of the two-dimensional commutators giveswhile all momentum-momentum and position-position commutators remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_{+}=(X_1+X_2)/\sqrt{2}$, $X_{-}=(X_1-X_2)/\sqrt{2}$ preserve the kinetic sum, andHence $\boxed{{\omega }_{+}=\omega ,{\omega }_{-}=3\omega }$ and $H=H_{+}+H_{-}$. Every cross-mode position/momentum pair commutes, so $[H_{+},H_{-}]=0$. On the tensor-product oscillator basis,For each $N\ge 0$, allowed pairs have $n_{-}=0,1,\dots ,\lfloor N/3\rfloor$ and $n_{+}=N-3n_{-}$. ThusEach pair contributes one product eigenvector; counting them accounts for the energy-level degeneracy.

At $7000\mathrm{K}$, $k_{B}T=0.603\mathrm{eV}$ andThe $2p$ ground term has fine-structure levels $J=1/2$ and $J=3/2$, with energy-level degeneracy $g_1=2$ and $g_2=4$. Detailed balance gives $C_{21}=C_{12}(g_1/g_2)e^{\Delta E/(k_{B}T)}=7.17\times 10^{-14}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The critical density is thereforeWith collisions and spontaneous decay both retained, $N_2/N_1=n_e{C}_{12}/(A_{21}+n_e{C}_{21})$. A Boltzmann distribution, $N_2/N_1\to 2e^{-\Delta E/(k_{B}T)}\simeq 1.97$, requires $n_e\gg n_{\mathrm{crit}}$, not merely equality. For example $n_e\gtrsim 3\times 10^8{\mathrm{m}}^{-3}$ brings the ratio within about ten per cent of that limit.

The three-dimensional isotropic harmonic oscillator has the same oscillator angular frequency $\omega$ in all three directions:The three independent quantum harmonic oscillators give energy eigenvalues $E_N=\hslash \omega (N+3/2)$, where $N=n_x+n_y+n_z$ and the three occupation numbers are nonnegative integers. The zero-point energy is the sum of the three one-dimensional contributions.

For fixed $N$, choosing all triples gives energy-level degeneracy $(N+1)(N+2)/2$. For example, $N=1$ has the three excitations $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Anisotropic oscillator frequencies generally split this degeneracy. The model is useful near a rotationally symmetric minimum of a smooth three-dimensional potential energy.