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An excited state is an energy eigenstate whose energy exceeds the ground state energy.

The ground-state energy is the smallest energy eigenvalue of a Hamiltonian operator.

An energy eigenvalue is nondegenerate when its eigenspace is one-dimensional. Its normalized eigenstate is then unique up to a complex phase.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:

For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.

For $0<E<U_0$, the Time-independent Schrodinger equation and the wall conditions givewhere $k=\sqrt{2mE}/\hslash$ and $l=\sqrt{2m(U_0-E)}/\hslash$. Continuity of $\psi$ and ${\psi }^{\prime }$ at the finite potential step givesDividing and rearranging yields the bound-state quantization condition

The unitary time evolution isFirst suppose that time reversal $T$ is linear and unitary. Differentiatingat $t=0$ givesand linearity permits the scalar $i$ to pass through $T$. HenceIf $H|E⟩=E|E⟩$, thenThus every energy has its negative as an energy. The Coulomb Hamiltonian has arbitrarily large positive energies in its continuum, so this symmetry would force energies arbitrarily far below zero. It would have no lowest energy eigenvalue and therefore no stable ground state. This is the unitary time reversal reverses the energy spectrum obstruction.

Now let $T$ be antiunitary. It is conjugate linear, soDifferentiating the same intertwining equation givesThereforeTime reversal now maps an energy eigenstate to another state with the same energy:The spectrum is preserved rather than reflected through zero, so the lower-bounded Coulomb spectrum and its stable ground state are compatible with time-reversal symmetry. This is why antiunitary time reversal preserves the energy spectrum.

Let $H$ be the one-dimensional Hamiltonian operator and letbe its discrete energy eigenvalues. Expanding a normalized trial wavefunction in an orthonormal energy basis,givesThus the Rayleigh-Ritz variational principle givesfor every nonzero admissible trial wavefunction. One chooses a parameterized family and minimizes the quotient.

When $V(x)=V(-x)$, parity commutes with $H$. In one dimension the nondegenerate ground state is even and the first excited state is odd. Restricting the trial family to odd wavefunctions makes every trial state orthogonal to the ground state, soThis is the odd-state variational principle for an even potential.

Inside the infinite square well, the time-independent Schrodinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are