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An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

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.

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

For inviscid plane Couette flow $U=z$ between impermeable walls at $z=\pm 1$, the nonzero real streamwise wavenumber $\alpha$ gives no nontrivial smooth discrete vertical-velocity modes. Instead, the vorticity variable $q=(D^2-{\alpha }^2)w$ obeys $({\partial }_t+i\alpha z)q=0$. Localized vorticity sheets give generalized eigenfunctions with real frequencies ${\omega }_{\xi }=\alpha \xi$. Their closure forms a neutral continuous spectrum.

In 1926, Erwin Schrödinger constructed arrays from differential operators and complete orthogonal function systems, showing how their products reproduce the array calculus. His paper connected wave mechanics with matrix mechanics; the translation's sections 2–3 develop the product and commutation rules.^[ref]

The modern proof is a change of orthonormal basis. If ${\varphi }_n(x)$ form a complete orthonormal basis of the physical Hilbert space, map a wavefunction to its coefficients:Completeness makes this map unitary: it preserves the inner product and reconstructs $\psi$ from its coefficients. Inserting the complete orthonormal basis proves the matrix multiplication rule. The Schrödinger equation becomesFor a time-independent Hamiltonian operator, the Heisenberg picture transfers the same evolution to the observables, preserving measurement predictions. Unitary equivalence requires matching operator domains as well as formal entries. A finite selection of eigenfunctions gives a compression, not an exact replacement of the full Hilbert space; a continuous spectrum needs its spectral integrals. This equivalence concerns representations of specified operators, and does not remove the ordering question when quantizing an arbitrary classical expression, as Weyl ordering illustrates.

Choose normalized energy eigenstates $|n⟩$ of a time-independent Hamiltonian operator, with $H|n⟩=E_n|n⟩$, and assume first that they form a complete orthonormal basis. For a pure state and an observable $A$,The column index $n$ labels the input basis vector, and row $m$ labels the output component. Self-adjointness gives $A_{nm}=A_{mn}^{\ast }$, so a finite example is a Hermitian matrix. A matrix element can be negative or complex; by itself it is not generally a probability. The Born rule assigns measurement probabilities using the eigenstates of $A$, which need not be energy eigenstates. For an unbounded observable, these expressions require a suitable operator domain; a system with continuous spectrum also needs the corresponding spectral integrals.

Use the one-sided thermal conduction model encoded by the requested dispersion relation: neglect solid-side perturbation heat flux, take equal phase densities, and let the normal point from solid into liquid. Write the isotropic linear kinetic undercooling law using the solidification attachment mobility $\mathcal{M}>0$:Here $\gamma$ is solid-liquid surface energy, $\mathcal{K}$ is the sum of principal curvatures, and $T_m$ is absolute temperature. Positive curvature convex into liquid lowers the equilibrium freezing temperature through the Gibbs--Thomson relation. The planar base state has $\mathcal{K}=0$.

Set $\Delta T=T_m-T_{\infty }>0$ and defineThe paper's $S$ is the latent-to-sensible heat ratio, the reciprocal of the common Stefan number convention. Below, velocities, lengths and growth rates are dimensionless in these scales. In the travelling coordinate $\xi =X-Vt$, the liquid heat equation gives ${\theta }_0^{\prime \prime }+V{\theta }_0^{\prime }=0$, while the bounded solid steady field is constant. Matching a decaying liquid profile to the Stefan condition $SV=-{\theta }_0^{\prime }(0)$ givesThe kinetic undercooling condition is ${\theta }_i=1-V$. ThusThe speed is constant for $S<1$, determined by $S$ and the attachment coefficient in the specified scales. Thermal diffusivity sets the exponential decay length $\kappa /U$, not the selected speed. Including solid-side perturbation conduction would change the stability equation below.

For linear stability analysis, corrugate the interface to $\xi =\zeta e^{\sigma t+i\alpha y}$ and write the liquid perturbation as $A_1{e}^{-p\xi }{e}^{\sigma t+i\alpha y}$. The liquid heat equation requiresFor this graph the linearized curvature is ${\alpha }^2\zeta e^{\sigma t+i\alpha y}$. Evaluate the temperature on the displaced interface: its perturbation is $A_1+{\theta }_0^{\prime }(0)\zeta =A_1-SV\zeta$. The kinetic undercooling law therefore givesLikewise the derivative evaluated at the displaced surface is ${\theta }_0^{\prime }(0)+{\theta }_0^{\prime \prime }(0)\zeta -p{A}_1$, with ${\theta }_0^{\prime \prime }(0)=S{V}^2$. The Stefan condition at perturbation order readsElimination proves the one-sided kinetic solidification dispersion:The destabilizing term comes from a protrusion encountering colder liquid; capillarity and kinetic relaxation oppose it.

At $S=0$, $V=1$ and the base liquid has no thermal contrast generated by latent heat. The interface obeys $\boxed{\sigma =-\Gamma {\alpha }^2}$: positive surface energy damps all nonzero modes, while $\Gamma =0$ is neutral at this order. Here the thermal perturbation amplitude vanishes, so the auxiliary decay exponent is immaterial to that pure kinetic relaxation. At $S=1$, $V=0$ and the formal limit givesFor $\Gamma >0$ it has a negative real branch $-{\alpha }^2<\sigma <0$ and no growing branch; at $\Gamma =0$ the interface branch is neutral. Strictly, $V=0$ is a singular endpoint of the travelling profile: no finite-width stationary exponential connects ${\theta }_i=1$ to the far-field value zero. These stability statements describe the limit of the travelling-front model.

For $0<S<1$, assume $\sigma =O({\alpha }^2)$ at small wavenumber. ExpandSubstituting into the dispersion relation cancels $S\sigma /V$ on both sides and givesAt the capillary stability threshold for a kinetic freezing front $\Gamma =S/V$, continuing the same expansion gives $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$: the nonzero long waves are damped even at threshold.

For large wavenumber, multiply the dispersion relation by $p$ and eliminate $\sigma$ using the thermal equation. The exact algebraic form isIt prevents the uncontrolled replacement $p\simeq \alpha$ when $\sigma$ itself is of order ${\alpha }^2$. In the usual small-capillarity range the leading results areFor the second result, set $p=\sqrt{1-\Gamma }\alpha +d+O({\alpha }^{-1})$ in the cubic; its quadratic-order balance gives $2d-V=S\Gamma /(1-\Gamma )$. Thus positive capillarity stabilizes sufficiently short waves. If also $\Gamma \ll 1$ and $|\sigma |\ll {\alpha }^2$, the useful reduced approximation isits finite terms should not be treated as a uniform expansion for arbitrary fixed $\Gamma$.

For completeness, large dimensionless capillarity has different balances. At $\Gamma =1$, $p\sim S^{1/3}{\alpha }^{2/3}$ and $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to p_{\ast }\equiv S/(\Gamma -1)>V/2$, so $\sigma =-{\alpha }^2+p_{\ast }^2-V{p}_{\ast }+O({\alpha }^{-2})$. At $\Gamma =1+2S/V$, the real branch reaches $p=V/2$: for $S<1/3$ its growth rate is $-{\alpha }^2-V^2/4+O({\alpha }^{-4})$, and for $S=1/3$ that endpoint is exact. For $S>1/3$ at this boundary, or for $\Gamma >1+2S/V$, the selected real-$p$ normal mode ceases to exist at sufficiently large wavenumber; the full diffusion problem includes a thermal continuous spectrum and cannot be assigned a real isolated interface growth rate by this ansatz there. Indeed, writing the bulk thermal perturbation as ${\theta }_1=e^{-V\xi /2}u$ reduces its heat equation to $u_t=u_{\xi \xi }-({\alpha }^2+V^2/4)u$, whose half-line diffusion modes have decay rates $-{\alpha }^2-V^2/4-k^2$ for $k\ge 0$. The thermal branch endpoint is $p=V/2$. Loss of a damped isolated interface mode is not a morphological instability.

The approximate instability condition is already clear from the long-wave coefficient: $\Gamma <S/V$, equivalently $S>\Gamma /(1+\Gamma )$. In fact it is exact for growing modes of the present one-sided model. At $\sigma =0$, let $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, so ${\alpha }^2=p_0(p_0-V)$. The right side of the dispersion relation becomesFor real $\sigma \ge 0$, the residual $F=\sigma (1+S/p)-SV+\Gamma {\alpha }^2+S{V}^2/p$ is strictly increasing, sinceTherefore a unique positive root exists precisely when $\Gamma <SV/p_0^2$. For $0<\Gamma <S/V$, the unstable band isThere are no unstable oscillatory roots hidden by this real-root test. Indeed, if $p=a+ib$ and $\mathrm{Re}\sigma \ge 0$, the thermal equation implies $a\ge V$. For $b\ne 0$, the imaginary parts of the two equations would require $2a-V=S(V-\Gamma {\alpha }^2)/|p+S{|}^2\le SV<V$, contradicting $2a-V\ge V$. Thus growing modes have real $p$ and real $\sigma$.

The sketch shows zero capillarity approaching the positive plateau $SV$, weak capillarity giving a positive band followed by short-wave damping, and threshold/stronger capillarity giving damping at every nonzero wavenumber. The zero-wavenumber mode is a translation. As $S\to 1$ with fixed positive $\Gamma$, both the unstable bandwidth and the positive growth rates shrink to zero; the small-wavenumber expansion is not uniform in that limit.

Impermeability requires $\hat{w}(-1)=\hat{w}(1)=0$. Let $c=\omega /\alpha$ with real $\alpha \ne 0$, and set $a=|\alpha |$. If $c$ is nonreal, or real outside the channel, the Rayleigh equation for inviscid shear flow gives ${\hat{w}}^{\prime \prime }-a^2\hat{w}=0$ everywhere. The two wall conditions then force $\hat{w}=0$. In fact there is no nonzero globally smooth eigenfunction for any $c$: the complete family is the inviscid Couette continuous spectrum.

For each $\xi \in (-1,1)$, permit a localized vorticity sheet. Let $G_a$ be the Dirichlet Green function satisfying $(D^2-a^2)G_a(z,\xi )=\delta (z-\xi )$. An explicit generalized eigenfunction iswhere $z_{<}=\min (z,\xi )$ and $z_{>}=\max (z,\xi )$. It vanishes at both walls, is continuous at $z=\xi$, and has derivative jump $[{\partial }_z{G}_a{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$. Consequentlyusing the Dirac delta multiplication identity. These are vorticity-sheet eigenfunctions of inviscid Couette flow, understood as generalized eigenfunctions, not as a discrete smooth Sturm-Liouville eigenfunction expansion.

To see completeness, define the vorticity variable $q=(D^2-a^2)w$. Its evolution is $({\partial }_t+i\alpha z)q=0$, so for any admissible initial vorticity $q_0$,The homogeneous Dirichlet problem for $D^2-a^2$ has only the zero solution, so this inversion reconstructs every initial vertical-velocity field in its usual function space. The generalized frequencies fill the interval with endpoints $\pm \alpha$; the endpoint values are understood as the closure of the continuous spectrum.

For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hslash {\partial }_i{A}_j$. Therefore the mechanical momentum satisfiesThe displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^{†}]=1$ andThe free longitudinal motion gives a continuous spectrum in $k_z$, and the transverse Landau levels have guiding-centre degeneracy of a Landau level. For arbitrary sign of $qB$, use $|qB|$ in the energy and $a=({\pi }_x+i\mathrm{sgn}(qB){\pi }_y)/\sqrt{2\hslash |qB|}$. The question's normalizable wavefunctions likewise assume $qB>0$.

For the stated symmetric-gauge wavefunction, $|\psi {|}^2\propto r^{2M}{e}^{-qB{r}^2/(2\hslash )}$, whose peak for $M>0$ is at $r_M^2=2\hslash M/(qB)$. Thus the largest useful angular index is about $qB{R}^2/(2\hslash )$, givingThe $M=0$ state should also be included in the usual lowest Landau level; its omission by “positive integer” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hslash /(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the wavefunctions make this a large-flux estimate, not an exact finite-disc count.

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

For a nonrelativistic spinless particle in a scalar potential, the Time-dependent Schrödinger equation isAn initial wavefunction together with boundary conditions specifies the evolution. The Hamiltonian operator must be self-adjoint on the appropriate domain for closed-system evolution to be unitary. The equation is linear, so a superposition of solutions is another solution. Its complex wavefunction predicts probability amplitudes; the Born rule separately supplies the connection to measurement probabilities.

For a time-independent potential, separation $\psi (\mathbf{r},t)=\varphi (\mathbf{r})e^{-iEt/\hslash }$ gives the time-independent Schrödinger equation $H\varphi =E\varphi$. Normalizability and boundary conditions select the allowed energies of bound states; scattering states instead use incoming and outgoing boundary conditions and generalized normalization. A general state is a superposition of energy eigenstates, with a spectral integral as well if there is a continuous spectrum. Different energies accumulate different phases, allowing time-dependent position densities and oscillatory expectation values even though a single energy eigenstate has a stationary density.

For a real scalar potential, combining the equation with its complex conjugate yields probability conservation:With no current escaping the boundary, $\int |\psi {|}^2{d}^{3}r$ is constant. Matching conditions at a finite potential step follow from the differential equation and enforce the flux balance used in B6.

The Schrödinger equation is a dynamical postulate of nonrelativistic quantum mechanics, not a consequence of classical mechanics alone. The substitutions $E\mapsto i\hslash {\partial }_t$ and $\mathbf{p}\mapsto -i\hslash \nabla$ motivate its free-particle form from the dispersion relation, but do not derive the measurement postulates. The scalar equation omits intrinsic spin and relativistic particle creation. Spinful or relativistic problems require the corresponding extended theory. Nor does its deterministic unitary evolution, by itself, prescribe conditioning on a particular observed measurement outcome.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.

For a time-independent Hamiltonian operator, the time-evolution operator is $U(T)=e^{-iHT/\hslash }$, with $T=t-t^{\prime }$. Insert its resolution of identity in normalized energy eigenstates:Since $⟨x|\alpha ⟩={\phi }_{\alpha }(x)$, the retarded quantum propagator in the question isThe Heaviside step function imposes the chosen retarded convention. If the Hamiltonian operator has a continuous spectrum, the spectral sum includes its continuum integral; it must not be restricted to bound states. The spectral resolution of a quantum propagator also shows that $K$ tends to $\delta (x-x^{\prime })$ as $T\downarrow 0$. With the step factor it obeys $(i\hslash {\partial }_t-H)K=i\hslash \delta (T)\delta (x-x^{\prime })$; the corresponding Green function with a unit delta source is $K/(i\hslash )$.

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.