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Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

As an estimate, model the confinement by an infinite square well of width $L$ and use the stipulated electron mass. Its energy eigenvalues are $E_j=j^2{\pi }^2{\hslash }^2/(2m{L}^2)$, $j=1,2,\dots$, so the gap controlling thermal occupation of a quantum well isThus $T\ll 130\mathrm{K}$ suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_{B}T)}$. For example $T\lesssim 28\mathrm{K}$ makes this ratio below one percent in this model. There is no sharp cutoff. The ground state energy itself corresponds to about $44\mathrm{K}$, but it is the excitation gap that enters the Boltzmann factor. Finite barrier height and a different effective mass would change the numerical scale; neither is supplied.

The one-dimensional time-independent Schrödinger equation isHere $m$ is the particle mass, $V(x)$ its potential energy, $E$ the stationary energy eigenvalue, and $\hslash =h/(2\pi )$ the reduced Planck constant. The spatial wavefunction is $\Psi$; the full stationary state is $\Psi (x)e^{-iEt/\hslash }$. Physical bound states satisfy the boundary conditions and wavefunction normalization $\int |\Psi {|}^{2}dx=1$. The operator on the left is the Hamiltonian operator, the sum of kinetic and potential energies.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

The infinite walls impose $\psi (0)=\psi (L)=0$, and the interior time-independent Schrödinger equation is ${\psi }^{\prime \prime }+k^2\psi =0$ with $E={\hslash }^2{k}^2/(2m)$. Its general solution is $A\sin kx+B\cos kx$. The first boundary condition gives $B=0$, and the second gives $kL=n\pi$. A nonzero normalizable state needs $n=1,2,\dots$; $n=0$ would give the zero function.

Wavefunction normalization, ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$, gives the infinite square well states and their energy eigenvalues:The wavefunctions vanish outside the well, and their full time dependence is ${\psi }_n(x)e^{-i{E}_{n}t/\hslash }$.

For the harmonic-oscillator number-state second moments, the number operator gives $⟨a^{†}a{⟩}_n=n$, while the canonical commutation relation gives $⟨a{a}^{†}{⟩}_n=n+1$. Terms with $a^2$ or $(a^{†}{)}^2$ have zero diagonal matrix elements, since they change the occupation number by two. ThereforeConsequently the potential energy and kinetic energy means areTheir sum is the oscillator energy eigenvalue $E_n=\hslash \omega (n+1/2)$. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.

For either number operator $\hat{N}={\hat{a}}^{†}\hat{a}$, positivity follows fromHence an eigenvalue $\nu$ cannot be negative. The commutatorsshow that $\hat{a}|\nu ⟩$, if nonzero, is an eigenstate with eigenvalue $\nu -1$. For a normalized eigenstate, $\Vert \hat{a}|\nu ⟩{\Vert }^2=\nu$.

If $\nu$ were not an integer, successive nonzero lowerings would eventually reach an eigenstate with $0<{\nu }^{\prime }<1$. Its next lowering would have nonzero norm but a negative eigenvalue, contradicting positivity. The ladder must therefore terminate at $\nu =0$, with $\hat{a}|0⟩=0$. Conversely, successive applications of the creation operator never vanish, becauseThe normalizable bottom state exists explicitly: solving $\hat{a}{\psi }_0=0$ gives ${\psi }_0(x)\propto \exp [-m\omega x^2/(2\hslash )]$. Thus the number operator has exactly the non-negative integer eigenvalues.

Since the two coordinate number operators commute, use the product eigenstates $|n_x,n_y⟩$. Their energy eigenvalues areAt fixed $N=n_x+n_y$, the allowed pairs are $(0,N),(1,N-1),\dots ,(N,0)$. Thus the energy level $E_N=(N+1)\hslash \omega$ has quantum degeneracy $N+1$, and the ground-state energy is $\hslash \omega$.

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.

Here $V(x)$ is potential energy, $E$ is the stationary-state total energy eigenvalue, and $\psi (x)$ is its spatial wavefunction. Wavefunction normalization gives $\int |\psi {|}^{2}dx=1$, and $|\psi {|}^{2}dx$ is the probability of finding the particle in $dx$.

A bound state requires $-V_0<E<0$. Inside the well the equation is ${\psi }^{\prime \prime }=-k^2\psi$, whereas outside it is ${\psi }^{\prime \prime }=q^2\psi$, withThe ground state is even and has no node, selecting $\cos (kx)$ inside. Normalizability selects decaying, not growing, exterior exponentials. Continuity of $\psi$ and ${\psi }^{\prime }$ givesThe first excited state, if present, is odd: a sine inside and opposite-sign decaying tails outside. Matching gives $-k_1\cot (k_{1}a/2)=q_1>0$. On its first odd branch, $\pi /2<k_{1}a/2<\pi$, hence $k_{1}a>\pi$. Since $k_1^2<2m{V}_0/{\hslash }^2$, existence requiresThis is the first odd bound-state threshold of a square well. There is no excited bound state below that threshold; equality gives a non-normalizable zero-energy limiting state, not an extra bound state.

For the numerical well,The exterior probability of a symmetric finite-well state includes both tails:The threshold is $9.40\mathrm{eV}$, exceeding the well depth, so there is no excited bound state. The rounded supplied coefficients integrate to $0.9973$ rather than exactly one; this rounding does not change the quoted probability precision materially.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.

Choose the zero of energy so that the translation-invariant vacuum state has $P^{\mu }|\Omega ⟩=0$. A one-particle state of rest mass $m_{\psi }$ obeys the relativistic energy-momentum relationThe translation covariance of a quantum field in c remains valid for the full interacting four-momentum operator. Acting its exponential functions on the two momentum eigenstates givesHere $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. A different additive vacuum energy must be subtracted consistently from both energy eigenvalues before identifying the physical $p^0$.

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.

The issue is the square-integrable condition on the whole real line, not just decay at one end. Because the quantum harmonic oscillator potential is even, the even and odd parts of a square-integrable wavefunction are themselves square integrable and satisfy the same equation. It is therefore enough to examine each parity sequence separately.

For a nonterminating sequence, put $b_j=a_{2j+p}$, where $p=0$ or $1$. Its recurrence relation givesSince $ϵ$ is real for an energy eigenvalue, the ratio is eventually positive. After multiplying by one constant phase, the tail coefficients are all positive. Choose $0<\delta <1/2$. For sufficiently large $j$, the ratio is at least $(1-\delta )/(j+1)$, so for some $K>0$ the tail coefficients obey $b_j\ge K(1-\delta {)}^j/j!$. Consequently, for $\xi >0$, the magnitude of $f$ eventually exceeds a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of a fixed polynomial. Multiplication by $e^{-{\xi }^2/2}$ leaves exponential growth like $e^{(1/2-\delta ){\xi }^2}$, incompatible with the square-integrable condition. Thus each nonzero parity series must terminate. The parity argument also excludes cancellation between the two nonterminating sequences at both infinities. This proves Normalization forces Hermite-series termination.

If the terminating polynomial has degree $n$, its top nonzero coefficient requires $2n+1-ϵ=0$. Conversely, when $ϵ=2n+1$, choosing the seed of parity $n$ nonzero makes the recurrence relation terminate exactly at degree $n$. The opposite seed must be zero, since that sequence does not terminate. ThereforeThe resulting polynomials are proportional to the physicists' Hermite polynomials, and ${\psi }_n\propto H_n(\xi )e^{-{\xi }^2/2}$ is normalizable because any polynomial times this Gaussian function is square integrable. The two parity seeds describe the full solution space of the second-order differential equation; the preceding argument therefore exhausts the normalizable stationary states. Each eigenspace has dimension one. General time-dependent quantum states can of course be superpositions of these energy eigenstates.

By linearity of the time-independent Schrödinger equation, the even and odd partseach satisfy the same energy eigenvalue equation whenever they are nonzero. They obey $R{\psi }_{+}={\psi }_{+}$ and $R{\psi }_{-}=-{\psi }_{-}$, so they have definite parity. At least one is nonzero, and $\psi ={\psi }_{+}+{\psi }_{-}$. Each nonzero part can be normalized separately.

Therefore a basis of stationary solutions can be chosen to have definite parity, and solving the even and odd sectors loses no energy eigenvalues. This does not assert that every possible quantum superposition has definite parity; a superposition of the two sectors need not. For one-dimensional bound states in the usual regular setting, nondegeneracy of one-dimensional bound states further implies that each individual energy eigenstate has definite parity.

Define the reflected function $\varphi (x)=\psi (-x)$. The chain rule gives ${\varphi }^{\prime }(x)=-{\psi }^{\prime }(-x)$ and ${\varphi }^{\prime \prime }(x)={\psi }^{\prime \prime }(-x)$. Evaluating the time-independent Schrödinger equation for $\psi$ at $-x$ and using the even function identity $V(-x)=V(x)$ givesThus reflection preserves the energy eigenvalue. The same substitution $y=-x$ in the norm givesThis is the action of the parity operator $R\psi (x)=\psi (-x)$, which is a unitary operator. For a reflection-symmetric physical problem it preserves the Hamiltonian operator's operator domain as well as its differential expression.

The time-independent Schrödinger equation is the energy eigenvalue equation for the Hamiltonian operator:Here $\hslash$ is the reduced Planck constant. A corresponding stationary wavefunction evolves as $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, which follows by substitution in the Time-dependent Schrödinger equation. Its position probability density is time independent because the time factor has unit modulus. Physical energy eigenstates must also satisfy the system's boundary conditions; for a bound state, wavefunction normalization requires ${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1$.