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For a spherical square well of radius $a$, depth $V_0$ and reduced mass $\mu$, the regular zero-orbital angular momentum solution has reduced radial wavefunction $U=A\sin kr$ inside and $U=B{e}^{-qr}$ outside. Wavefunction matching at a finite potential step gives $k\cot (ka)=-q$, where $k^2=2\mu (V_0+E)/{\hslash }^2$ and $q^2=-2\mu E/{\hslash }^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\dots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.

For the first $\ell =1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto r{j}_1(kr)$, where $j_1$ is a Spherical Bessel function. Matching logarithmic derivatives gives $1+z{j}_1^{\prime }(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hslash$. Its first threshold is $z=\pi$, so $V_{p,0}={\pi }^2{\hslash }^2/(2\mu a^2)$. Higher values of orbital angular momentum increase the positive centrifugal potential, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.

Consider hard walls at $x=0,3L$, zero potential energy on $0<x<L$ and $2L<x<3L$, and a central rectangular barrier whose height equals the ground state energy $E$. The Time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=0$ on the central interval. Reflection symmetry and the nodeless theorem for a one-dimensional ground state require an even, positive state about $3L/2$, hence a constant central segment. Wavefunction matching at a finite potential step makes the derivatives vanish at $L,2L$. The side segments are proportional to $\sin (\pi x/(2L))$ and $\sin (\pi (3L-x)/(2L))$. Their common plateau value is $1/\sqrt{2L}$ after wavefunction normalization, and $E={\pi }^2{\hslash }^2/(8m{L}^2)$. Equality of barrier height and energy is essential: a higher barrier gives positive curvature of a positive wavefunction in the middle, while a lower barrier gives negative curvature.

For $V(x)=0$ on $0<x<a$ and $V(x)=V_w>0$ outside, the lowest bound state is even about $a/2$, has no nodes, and has $0<E<V_w$. Write $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_w-E)}/\hslash$. The interior wavefunction is proportional to $\cos [k(x-a/2)]$; the exterior tails decay exponentially. Wavefunction matching at a finite potential step gives the displayed condition. Its lowest root has $0<ka<\pi$, so $E<{\pi }^2{\hslash }^2/(2m{a}^2)$, the infinite square well ground state energy. As $V_w$ increases, the penetration length $1/\kappa$ tends to zero and that limiting energy is approached from below.

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.

SetAll are real and $k_0>k_1>k_2>0$. Suppressing a common time quantum phase $e^{-iEt/\hslash }$, chooseThe rightmost region has no left-moving term because no wave is incident from the right. The shifted quantum phase convention makes $F$ the transmitted amplitude at the second interface.

For finite potential steps and equal mass, wavefunction matching at a finite potential step requires continuity of both $\psi$ and ${\partial }_x\psi$. Explicitly,A discontinuous wavefunction would introduce an inadmissible singular kinetic term, while integrating the stationary Schrödinger equation across a finite step shows that the derivative cannot jump. These conditions also keep the probability current continuous, so a finite interface neither creates nor destroys probability. The derivative rule would need modification for a delta-function potential or a different effective mass; neither occurs here.

Inside the barrier, the stationary Schrödinger equation gives an exponentially decaying wavefunction with inverse penetration lengthThe amplitude falls approximately as $e^{-\kappa a}$, so the leading exponential estimate for the transmission probability isFor comparison, wavefunction matching at a finite potential step at both edges gives the rectangular-barrier tunnelling probabilityThe difference is the interface prefactor: for an opaque barrier it is $16E(V_0-E)/V_0^2=2.56$. Both give a transmission of order $10^{-6}$; the first is the requested exponential estimate.

The first two probability densities areBoth vanish at the walls. The ground state has one maximum at $L/2$ and no interior node. The first excited state has a node at $L/2$ and equal maxima at $L/4$ and $3L/4$.

For a finite square well whose exterior potential is $V_0>E_n$, the exterior wavefunction decays as $e^{-\kappa d}$, where $\kappa =\sqrt{2m(V_0-E_n)}/\hslash$ and $d$ is distance from the relevant edge. Wavefunction matching at a finite potential step makes the wavefunction and, for constant mass, its first derivative continuous. Thus the probability density is nonzero at the edges and has evanescent tails outside, the interior wavelength increases, and the energy levels are lower than for infinite walls at the same positions. Reflection symmetry still gives an even ground state and odd first excited state about the center, when both are bound. A shallow well may not support that excited bound state at all.

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.

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.

For a positive step potential and left incidence, take $E\to V_0^{+}$ at fixed incident probability amplitude $A$. The transmitted wavenumber tends to zero while the incident wavenumber tends to $k=\sqrt{2m{V}_0}/\hslash$. Wavefunction matching at a finite potential step gives reflected amplitude $A$ and transmitted amplitude $2A$, producing the spatial wavefunction in the title. The left probability density is $4|A{|}^2{\cos }^{2}kx$ and the right density $4|A{|}^2$. The transmitted probability current still tends to zero because its velocity tends to zero. Over adjacent intervals of length $2\pi /k$, the right integrated density is twice the left. These ratios use finite local integrals of a non-normalizable stationary scattering state.