Time-independent Schrödinger equation ($H\psi =E\psi$)

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

For a real, sufficiently smooth, nonvanishing wavefunction of a particle with constant mass, the time-independent Schrödinger equation determines the potential energy through the displayed logarithmic-curvature formula. An additive choice of energy origin fixes the remaining constant. For $\psi =e^{-a{x}^4}$ with $a>0$ and $V(0)=0$, it gives $E=0$ and $V=({\hslash }^2/2m)(16a^2{x}^6-12a{x}^2)$. At a node the quotient must instead be interpreted by a limit; an arbitrary trial wavefunction can produce singularities rather than a regular potential.

A classically forbidden region would require negative classical kinetic energy, but a quantum wavefunction need not vanish there. For constant potential energy, it is $A{e}^{qx}+B{e}^{-qx}$ with $q=\sqrt{2m(V-E)}/\hslash$. Boundary matching determines both coefficients. Beside a hard wall at $D$, a nonzero forbidden-region solution is proportional to $\sinh [q(D-x)]$, and therefore has no interior nodes. Across a finite barrier, both exponentials are generally needed; a pure decaying exponential is appropriate only when the other boundary condition permits it.

For a particle of fixed energy $E$, a classically allowed region has positive classical kinetic energy $E-V(x)$. For a constant potential energy, the stationary wavefunction is a combination of $e^{ikx}$ and $e^{-ikx}$, where $k=\sqrt{2m(E-V)}/\hslash$ is the wavenumber. In a varying potential energy this local value underlies the WKB approximation away from turning points.

For a finite jump in a one-dimensional potential energy, with constant mass and no delta potential, both the wavefunction and its first spatial derivative are continuous. Integrating the time-independent Schrödinger equation across a vanishing interval makes the jump in ${\psi }^{\prime }$ vanish, since the integral of $(V-E)\psi$ vanishes. A jump in $\psi$ itself would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no matching term in the equation. The second derivative can jump. This matching allows transmission and quantum tunnelling; a finite step is not a hard wall.

In an even, piecewise constant potential, integrate outward from the centre with either even boundary conditions ${\psi }^{\prime }(0)=0$ or odd boundary conditions $\psi (0)=0$. Match the wavefunction and its derivative at every finite step and impose exponential decay outside. The resulting discrete energies alternate in parity, with the fourth bound state odd and containing three nodes of a wavefunction. There is no hard-wall zero at a finite step; the second derivative changes according to ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$.

For equal-mass one-dimensional scattering between regions with real wavenumbers $k_0>k_2>0$, insert a layer of width $a$ and wavenumber $k_1$. For a purely right-moving incident wave of amplitude $A$ and no reflection, continuity at the first edge gives interior amplitudes $C=A(1+k_0/k_1)/2$ and $D=A(1-k_0/k_1)/2$. If $k_{1}a=\pi /2$, continuity at the second edge gives a transmitted edge amplitude $F=i(k_0/k_1)A$ and the matching condition $k_1^2=k_0{k}_2$. Thus $F/A=i\sqrt{k_0/k_2}$ and the transmitted probability current equals the incident current, despite their unequal amplitudes.

For potentials $0,V_1,V_2$ with $E>V_2>V_1>0$, the condition is $(E-V_1{)}^2=E(E-V_2)$, or $E=V_1^2/(2V_1-V_2)$, requiring $V_2/2<V_1<V_2$. It is the quantum-wave counterpart of a quarter-wave impedance transformer: the intermediate wavenumber is the geometric mean and the two reflected amplitudes cancel.

A Schrödinger operator is a differential operator of the formwhere $\Delta$ is the Laplace operator and $V$ is a potential function. Its eigenvalue problem is the time-independent Schrödinger equation.

A self-adjoint operator of this form with $a={\hslash }^2/(2m)$ supplies the Quantum Hamiltonian of a particle in the potential $V$.

Matching a wavefunction and its derivative across finite potential discontinuities, together with decay or wall conditions, permits only discrete bound-state energies.