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Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

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.

The original diagram places hard walls at $0$ and $3L$, and the central plateau on $L<x<2L$ is exactly at the indicated ground-state energy. This equality determines the middle of the requested sketch. In the time-independent Schrödinger equation,the central interval has $V=E$, hence ${\psi }^{\prime \prime }=0$. The ground state can be chosen real and positive inside the walls by the nodeless theorem for a one-dimensional ground state. Reflection symmetry makes it even about $x=3L/2$, so its linear central segment must be constant.

The hard-wall boundary conditions are $\psi (0)=\psi (3L)=0$. On both outer intervals $V=0<E$, so the positive wavefunction has negative curvature and is sinusoidal. Wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to remain continuous at $L,2L$; the derivative therefore vanishes where each sinusoidal flank meets the flat middle. A normalized representative isEach side contributes $L/2$ and the plateau contributes $L$ to the squared norm before multiplying by $1/(2L)$. The corresponding consistency condition is $E={\pi }^2{\hslash }^2/(8m{L}^2)$; a numerical energy is not required for the sketch.

Thus the ground state at the top of a central rectangular barrier has no internal nodes, continuous value and slope, and a flat middle, rather than an exponential dip through a classically forbidden region. Multiplication by an overall constant phase changes none of these physical features.

Equation and boundary conditions. The delta potential has strength $V_0$ with units of energy times length, because the Dirac delta function has units of inverse length. The time-independent Schrödinger equation isThe wavefunction is continuous at the origin. A finite jump in $\psi$ would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no corresponding term in this potential. A jump in ${\psi }^{\prime }$ does produce a delta contribution and is allowed. Integrate through a symmetric interval:For continuous finite $\psi$, the right-hand side vanishes as $ϵ\to 0$. ConsequentlyThis derivative jump is the boundary condition associated with the attractive delta potential.

Unique bound state. For $E<0$, write $\kappa =\sqrt{-2mE}/\hslash >0$. Away from the origin, ${\psi }^{\prime \prime }={\kappa }^2\psi$. Normalizability excludes exponentials growing at either infinity, leavingContinuity gives $A=B=C$. The derivative jump then becomes $-2\kappa C=-2m{V}_{0}C/{\hslash }^2$. For a nonzero state,There is exactly one positive value of $\kappa$, so exactly one negative-energy bound state. An odd solution would have $\psi (0)=0$; the allowed decaying exponentials would then both vanish identically. Normalize the even solution usingChoosing the overall phase real and positive gives the bound state in a delta potential:It peaks at the origin, has a downward cusp there, and decays exponentially on both sides. Its value, but not its slope, is continuous.

Scattering amplitudes and fluxes. For $E>0$, put $k=\sqrt{2mE}/\hslash$ and retain $\kappa =m{V}_0/{\hslash }^2$ as a strength parameter. A unit-amplitude wave incident from the left hasContinuity gives $t=1+r$. The jump condition givesSolving,The probability current of a plane wave of amplitude $A$ is $\hslash k|A{|}^2/m$ in its propagation direction. Since the asymptotic wavenumber is the same on both sides, reflection probability and transmission probability are just $|r{|}^2$ and $|t{|}^2$:They satisfy $R+T=1$, expressing conserved probability current in this real potential.

The right-hand plot uses $\eta =\kappa /k=V_0\sqrt{m/(2E)}/\hslash$, proportional to $V_0$ at fixed incident energy. Then $R={\eta }^2/(1+{\eta }^2)$ rises monotonically from zero to one, while $T=1/(1+{\eta }^2)$ falls from one to zero; they are equal at $\eta =1$. Despite the absence of a classical forbidden barrier, an abrupt attractive interaction reflects a quantum wave. Increasing attraction strengthens the derivative mismatch and can approach complete reflection at fixed positive energy. At fixed $V_0$, increasing incident energy instead makes the well increasingly transparent. The negative-energy bound state and the positive-energy scattering states are different stationary solutions, not an absorptive loss channel.

Reduced radial equation. A spherical harmonic satisfies ${\hat{L}}^2{Y}_{\ell m}=\ell (\ell +1){\hslash }^2{Y}_{\ell m}$. Substitute $\psi =R{Y}_{\ell m}$ into the time-independent Schrödinger equation, divide out the angular function, and put $R=U/r$. Direct differentiation givesMultiplication by $r$ then gives the Radial Schrodinger equationHere $\mu =m_1{m}_2/(m_1+m_2)$ is the reduced mass, and the extra positive term is the centrifugal potential.

Boundary conditions. At the origin this finite central potential has no point interaction. The physical radial solution is regular there, requiring $U(0)=0$; for $\ell =0$, a nonzero constant $U(0)$ would give $R\sim 1/r$ and a point-source singularity of the three-dimensional Laplacian. Mere square integrability alone would not remove that singular solution. For a bound state, the radial probability integral is ${\int }_0^{\infty }|U{|}^{2}dr$, so the exterior solution must decay and $U(r)\to 0$ as $r\to \infty$.

Solutions for zero orbital angular momentum. For $\ell =0$ and $-V_0<E<0$, defineThe interior equation is $U^{\prime \prime }+k^{2}U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U^{\prime \prime }-q^{2}U=0$; normalizability excludes the growing exponential. ThusThe constants cannot remain independent once the interface is matched. Since the potential step is finite, both $U$ and $U^{\prime }$ are continuous at $a$:Dividing the equations yields

Existence and number of s-wave bound states. Put $z=ka$, $y=qa$, and ${\rho }_0=a\sqrt{2\mu V_0}/\hslash$. ThenThe lowest branch lies on $\pi /2<z<\pi$. Along it, $y=-z\cot z>0$ and ${\rho }_0=z/\sin z$ increases continuously from $\pi /2$ to infinity. Therefore a first negative-energy solution exists precisely when ${\rho }_0>\pi /2$, orFurther branches begin at $z=3\pi /2,5\pi /2,\dots$ as $q\to 0^{+}$. Hence the bound-state thresholds of a spherical square well give exactly one negative-energy $\ell =0$ state whenThe upper equality is included: the second branch is then only at $E=0$, with constant exterior $U$, and is not normalizable. If threshold cases are excluded by convention, the commonly quoted open interval has a strict upper inequality; the physical negative-energy state count at equality is still one.

Deuteron estimate. Using $m_p{c}^2\simeq 938.27\mathrm{MeV}$, $m_n{c}^2\simeq 939.57\mathrm{MeV}$ and $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$ gives $\mu c^2\simeq 469.46\mathrm{MeV}$. At $a=2.1\mathrm{fm}$,The threshold for a second $s$ state is $9V_{\ast }\simeq 209\mathrm{MeV}$, far above the stipulated depth $33.5\mathrm{MeV}$. Thus this model has no radially excited s-wave bound state.

Checking other orbital angular momentum sectors matters before ruling out every excitation. At the first $p$-wave threshold, the regular interior reduced radial function is $r{j}_1(kr)$ and the zero-energy exterior function is proportional to $r^{-1}$. Matching their logarithmic derivatives requires $j_0(ka)=0$, whose first root is $ka=\pi$. HenceLarger $\ell$ only increases the centrifugal potential. Therefore no bound orbital excited states are expected in the specified central square-well model. Its spin dependence and tensor nuclear forces have not been specified; the calculation does not resolve the real deuteron's internal spin structure.

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.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

Equation and interpretation. The supplied equation is the time-independent Schrödinger equation for a nonrelativistic particle with mass $m$ in a prescribed potential energy $V(x)$. Equivalently,It is an eigenvalue problem for the Hamiltonian operator. A stationary energy state has full wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. By the Born rule, $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a small interval $dx$, with wavefunction normalization ${\int }_{-\infty }^{\infty }|\psi {|}^{2}dx=1$.

Infinite barriers. Infinite exterior potential energy excludes any finite-energy probability outside the interval. The infinite square well consequently has $\psi =0$ outside and the hard-wall boundary conditionsThe interior wavefunction must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.

Ground state. Inside the well $V=0$, so ${\psi }^{\prime \prime }+k^2\psi =0$, with $k=\sqrt{2mE}/\hslash$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\dots$. Thus the discrete energy eigenvalues are $E_n=n^2{\pi }^2{\hslash }^2/(2m{a}^2)$. The smallest, $n=1$, is the ground state. Sinceits normalized wavefunction, choosing an irrelevant overall phase to make it positive inside, is

Finite barriers. For a finite step and constant particle mass, both $\psi$ and ${\psi }^{\prime }$ are continuous at $x=0,a$. Wavefunction matching at a finite potential step follows by integrating the Schrödinger equation across a vanishingly small neighborhood of the step: a bounded potential cannot generate a finite jump in ${\psi }^{\prime }$. The bound-state exterior solutions decay rather than vanish identically:Normalizability excludes the growing exterior exponentials. The equivalent interior boundary conditions are ${\psi }^{\prime }(0)=\kappa \psi (0)$ and ${\psi }^{\prime }(a)=-\kappa \psi (a)$; neither boundary value is zero for the nontrivial ground state.

The ground state of a finite square well remains symmetric about $a/2$ and has no nodes. It bends smoothly into exponential tails at both walls, extending over a larger region than the infinite-well state. Its maximum after normalization is lower for the deep well considered here. When $V_w\gg E$, the penetration length $1/\kappa$ is short, so the two shapes are close except near the walls and in the exterior tails.

The finite curve is an illustrative exact matched solution with $V_w=10E_{1,\infty }$; no numerical wall height is specified by the question.

Energy comparison. The finite barriers relax the confinement, allowing a less rapidly varying interior wavefunction and therefore smaller kinetic energy. More explicitly, the even interior solution is proportional to $\cos [k(x-a/2)]$, and matching to the tails gives $k\tan (ka/2)=\kappa$. The lowest root lies at $0<ka<\pi$, while the infinite well has $ka=\pi$. HenceAs $V_w\to \infty$, the tails disappear and the finite-well energy approaches the infinite-well value from below. This comparison uses the same zero of potential energy inside both wells.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

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.

$V(x)$ is the particle's potential energy as a function of position, and $E$ is its total energy eigenvalue in this time-independent Schrödinger equation. Solving it with the boundary conditions determines the allowed stationary states and energy levels.

The wavefunction is a probability amplitude: the Born rule gives position probability density $|\psi (x){|}^2$, normalized to unit integral for one particle. $\psi$ itself need not be real or positive and is not a classical matter-density profile. The full stationary-state time dependence is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, leaving $|\Psi {|}^2$ time independent.

The one-dimensional time-independent Schrödinger equation isFor the infinite square well on $[0,a]$, $V=0$ inside and the wave function is zero outside. The boundary conditions are $\psi (0)=\psi (a)=0$. For positive integer $n$, ${\psi }_n=A\sin (n\pi x/a)$ satisfies them, and ${\psi }_n^{\prime \prime }=-(n\pi /a{)}^2{\psi }_n$. Substitution gives the energy levelsNormalization requires $1={\int }_0^a|{\psi }_n{|}^{2}dx=A^{2}a/2$. The requested real positive constant is therefore

The general time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$. For a spherically symmetric wave function, ${\nabla }^2\psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$; the Coulomb attraction between the proton and electron is $V=-q^2/(4\pi {ϵ}_{0}r)$. These supply respectively the kinetic and potential terms in the displayed radial equation. Rearranging identifiesas the Bohr radius.

Substitute $\psi =(A+Br)e^{-\lambda r}$ into the rearranged equation and put $ϵ=2mE/{\hslash }^2$. Equating the coefficients of $r,1,r^{-1}$ after cancelling the exponential givesA nonzero state has $A\ne 0$. If $B=0$, these identities give $\lambda =1/a$ and $ϵ=-1/a^2$. If $B\ne 0$, they give $\lambda =1/(2a)$, $B=-A/(2a)$, and $ϵ=-1/(4a^2)$. Thus the two independent eigenfunctions and their corresponding energies areThey are eigenfunctions at different energies, rather than two independent solutions at one fixed energy.

For the higher-energy state, radial probabilities include the spherical volume factor $4\pi r^2$. The normalization constant and $4\pi$ cancel in the ratio. Using the supplied integral at $b=a$ and at infinity,HenceThis calculation fixes the denominator as $8e$ in the source's unparenthesized expression; interpreting it as a product with $e$ would give a negative probability.

Inside the box, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$. The zero endpoint values eliminate the cosine term and require $ka=n\pi$. There is no nonzero zero-energy solution with both endpoints zero. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, soInsert the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take its inner product with ${\Psi }_m$. Orthonormality gives $i\hslash {\dot{c}}_m=E_m{c}_m$ for each $m$, and conversely these equations make every expansion coefficient of the residual zero. Thus $c_n(t)=c_n(0)e^{-i{E}_{n}t/\hslash }$ in the usual Hilbert-space or strong-solution class appropriate to the expansion.

For the specified state,The modes are real, so the interference term is a cosine of the energy difference. Writing $s_n=\sin (n\pi x/a)$,Its integral remains one because the cross term integrates to zero. The probabilities of the two energy outcomes are fixed at $9/25$ and $16/25$, even though the position probability density oscillates.

Inside the infinite square well, the time-independent Schrödinger equation and Dirichlet boundary conditions areand the wavefunction vanishes outside the interval. For a nonzero state, integration by parts givesThe strict inequality follows because a function with zero derivative and zero endpoint values is zero. Put $k=\sqrt{2mE}/\hslash$. The general solution is $A\sin kx+B\cos kx$; the first boundary condition gives $B=0$, and the second gives $ka=n\pi$, $n=1,2,\dots$. Since ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, the normalized energy eigenstates and eigenvalues areThe normalization fixes the magnitude of the coefficient; its arbitrary global phase has been chosen real and positive.

The dimensionless coordinate for the quantum harmonic oscillator gives $d/dx=\sqrt{m\omega /\hslash }d/d\xi$. The time-independent Schrödinger equation therefore becomesFor $\psi =e^{-{\xi }^2/2}f$, direct differentiation yieldsCancelling the nonzero exponential reduces the equation to the Hermite differential equation

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 ${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1$.

In either exterior region of the finite square well, the time-independent Schrödinger equation reduces to ${\psi }^{\prime \prime }=k^2\psi$, since $E<0$. Its general solution is a linear combination of $e^{kx}$ and $e^{-kx}$. On the left, $e^{-kx}$ grows as $x\to -\infty$ and cannot be a square-integrable function, so its coefficient must be zero. On the right, the excluded growing term is $e^{kx}$. Hence the surviving tails are $A{e}^{kx}$ for $x\le -a$ and $D{e}^{-kx}$ for $x\ge a$.

Inside the well, $E+V_0>0$ and the equation becomes ${\psi }^{\prime \prime }=-{\kappa }^2\psi$, whose general solution is $B\cos \kappa x+C\sin \kappa x$. ThusThe coefficients are initially independent; wavefunction matching at a finite potential step imposes the remaining relations. Here the paper's $k$ is the exterior decay constant and $\kappa$ is the interior oscillation constant.