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For the finite square well $V=-V_0$ on $|x|<a$ and $V=0$ outside, with $a,V_0>0$, there is always an even bound state, however small $V_0$ is. Define $\rho =a\sqrt{2m{V}_0}/\hslash$, the dimensionless decay constant $\alpha =a\sqrt{-2mE}/\hslash$, and interior oscillation constant $\beta =a\sqrt{2m(E+V_0)}/\hslash$. Wavefunction matching at a finite potential step gives $\alpha =\beta \tan \beta$ and ${\alpha }^2+{\beta }^2={\rho }^2$.

On $0<\beta <\pi /2$, these equations reduce to $\beta /\cos \beta =\rho$. The left side is continuous, strictly increasing from zero to infinity, with derivative $\sec \beta (1+\beta \tan \beta )>0$. The intermediate value theorem gives exactly one root on this first branch for every $\rho >0$. It has $0<\beta <\rho$, hence $\alpha >0$ and $-V_0<E<0$. The matched cosine interior and exponential tails therefore give a nonzero even square-integrable function with continuous first derivative, which can be normalized. As the well becomes shallow the binding energy approaches zero but remains negative; see shallow finite-square-well energy.

For a symmetric one-dimensional finite square well, nondegenerate bound states alternate even and odd parity, starting with an even ground state. The $n$th state counted from zero has $n$ interior nodes. Outside the well, it decays with inverse length $\sqrt{2m(U-E_n)}/\hslash$. Matching both the wavefunction and its first derivative at a finite step distinguishes it from the infinite-well approximation.

Consider an original one-dimensional effective mass model with $V(x)=0$ for $|x|<a$ and $V(x)=V_0>0$ outside. Take the same positive effective mass $m^{\ast }$ everywhere, an abrupt interface, and no interactions. The wavefunction represents a transverse envelope; free motion parallel to the layer is outside this calculation. A double heterostructure laser need not be thin enough for this quantization to dominate its operation.

Set $s=x/a$, $ϵ={\hslash }^2/(2m^{\ast }{a}^2)$, $z_0^2=V_0/ϵ$, $z^2=E/ϵ$ and $w=\sqrt{z_0^2-z^2}$. The time-independent Schrödinger equation becomesHere $u(s)$ denotes the dimensionless potential; a bound state has $0<z<z_0$. Matching the wavefunction and its derivative at $s=1$ givesFor the even bound state, the interior solution is $A\cos (zs)$ and the exterior solution is $A\cos z{e}^{-w(|s|-1)}$. For the odd bound state, replace $\cos (zs)$ by $\sin (zs)$ and $\cos z$ by $\mathrm{sgn}(s)\sin z$. The physical wavefunction is $\psi (x)=\varphi (x/a)/\sqrt{a}$.

Wavefunction normalization includes both exterior tails, not just the active layer. Direct integration givesThe first two terms are the interior integral; the last is the total exterior integral. Thus the interior probability is $P_{\mathrm{in}}=A^2[1\pm \sin (2z)/(2z)]$, with the upper sign for even parity.

Choose $z_0=2$. There is one even root in $(0,\pi /2)$ and one odd root in $(\pi /2,2)$; higher branches start beyond $z_0$. Each left-hand side increases on its stated interval while $\sqrt{4-z^2}$ decreases, proving uniqueness there. Numerically,These are calculated values of the chosen finite square well. The excited bound state lies closer to the barrier and has a longer exterior tail: being bound does not mean having unit probability inside the well.

This model demonstrates electronic confinement, not optical refractive index confinement or a complete laser. Unequal effective masses, band bending, interface disorder and interactions require a more detailed model. In particular, continuity of the ordinary derivative above uses the chosen constant effective mass assumption.

The infinite square well is an ideal confinement model with potential energy $V(x)=0$ for $0<x<L$ and infinite walls outside. The walls impose Dirichlet boundary conditions $\varphi (0)=\varphi (L)=0$ and a vanishing exterior wavefunction. Inside, the time-independent Schrödinger equation isThe first wall removes the cosine term; the second requires $\sin (kL)=0$, hence $k=n\pi /L$. The $n=0$ solution is identically zero and cannot satisfy wavefunction normalization. Each energy eigenstate has $n-1$ interior nodes of a standing wave, at $x=jL/n$ for $j=1,\dots ,n-1$. More curvature costs more kinetic energy, explaining the $n^2$ growth. The positive ground-state energy is a confinement effect: a constant zero-energy wavefunction cannot satisfy both walls and remain normalized.

For an electron with mass $m_e=9.1093837139\times 10^{-31}\mathrm{kg}$ in a box of length $L=1\mathrm{nm}$, use $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. These constants are from the 2022 CODATA adjustment. ^[ref]. Substitution givesDoubling $L$ quarters every energy eigenvalue; doubling $m$ halves them. This scaling makes the particle in a box useful for estimating confinement energies before attempting a more realistic calculation. A finite square well permits exterior tails for bound states and has only finitely many bound energy eigenstates. For an electron, the scalar box calculation assumes fixed spin and no spin-dependent interactions.

For the ground state, the probability of finding the particle in the middle half of the box is fully computable:Thus half the available length contains about $81.8\%$ of the probability. The figure contrasts the signed wavefunctions with their nonnegative probability densities: a negative lobe is an amplitude with opposite quantum phase, not negative probability.

An equal quantum superposition of the first two energy eigenstates gives a simple example of moving density:Its density period is $T=h/(E_2-E_1)\approx 3.66608\mathrm{fs}$ for the electron box above. At $t=0$ the cross term enhances the left side; at $T/4$ it vanishes; at $T/2$ it enhances the right side. More quantitatively,which takes the values $0.924413$, $0.5$, and $0.075587$ at those times. Orthogonality keeps the total probability one throughout. An energy measurement still returns $E_1$ or $E_2$ with equal probabilities: changing position density arises from relative quantum phase, not changing populations. A mixed state describing a statistical mixture with those populations has the constant probability density $({\varphi }_1^2+{\varphi }_2^2)/2$; it lacks this matter-wave interference term.

Assume type-I confinement of both carriers, and first use an infinite square well as an explicit estimate. For thickness $L=5$ nm and each carrier mass $0.2m_e$, confinement measured from the corresponding well band edge isThus the first three infinite-well levels are approximately $75$, $301$ and $677$ meV for each carrier. Interband envelope overlap favors matching quantum numbers when the electron and hole well profiles are the same. Ignoring exciton binding energy initially, their absorption thresholds areThe first two lie in the requested optical window. Two-dimensional subband joint densities of states produce step-like onsets, with excitonic peaks just below them, rather than only a bulk three-dimensional onset. Confinement strengthens electron–hole binding: the ideal strict-two-dimensional hydrogen ground binding is $4R^{\ast }\simeq 40$ meV, while a finite-width well interpolates toward the bulk value near $10$ meV. The $5$ nm width is comparable to the bulk exciton radius, so neither limiting binding value is exact. A schematic based on the infinite-well estimate therefore has first exciton peaks around $1.73$–$1.76$ eV and $2.18$–$2.21$ eV, followed by their subband continua; the $2.3$ eV barrier absorption is another feature near the top of the requested window.

The barrier gap alone does not specify the finite confinement energies. Under type-I alignment it supplies only $\Delta E_c+\Delta E_v\simeq 2.3-1.62=0.68$ eV. It does not give their division or the barrier effective masses, and a larger gap by itself does not prove type-I alignment. Infinite-well levels are consequently estimates, not uniquely determined measured levels.

For a concrete conditional comparison, assigning equal $0.34$ eV electron and hole barriers and taking the same mass on both sides gives finite square well levels $44.1$, $169.4$ and $331.7$ meV per carrier. These follow from $z\tan z=\sqrt{z_0^2-z^2}$ for even levels and $-z\cot z=\sqrt{z_0^2-z^2}$ for odd levels, with $z=kL/2$, $z_0=(L/2)\sqrt{2m^{\ast }{V}_0}/\hslash$. The corresponding unbound-pair optical thresholds would be $1.708$, $1.959$ and $2.283$ eV before exciton shifts. This example is deliberately conditional and demonstrates why band offsets must not be invented to make a unique finite square well spectrum. The requested schematic below states which approximation it uses.

No: ordering the attractive potentials does not impose the proposed ordering of reflection probabilities. Take $k=\pi /(4a)$ and let $M$ denote the particle mass. Choose $V_2=0$ everywhere andThese satisfy every stated ordering and support condition. The incoming energy is ${\hslash }^2{k}^2/(2M)$; the interior wavenumber in the finite square well is $q=\sqrt{k^2-2M{V}_1/{\hslash }^2}=3k$.

Here is an explicit matching calculation. For a general well of width $L=2a$, take exterior waves $e^{ik(x+a)}+r{e}^{-ik(x+a)}$ on the left and $t{e}^{ik(x-a)}$ on the right. Shifting the incident phase in this way does not change the reflection probability. Continuity of the wavefunction and its first derivative gives the propagation relationInverting it and adding the two equations after dividing the derivative equation by $ik$ yieldsThe incident and reflected exterior waves have the same speed, so their flux ratio isFor the chosen parameters $qL=3\pi /2$ and $(q^2-k^2{)}^2/(4k^2{q}^2)=16/9$. ThusThe zero potential has no reflected component. This proves the failure with explicit probabilities. More generally the same expression vanishes at $qL\in \pi \mathbb{Z}$, explaining resonant transmission through a square well and why increasing attractive well depth does not monotonically control reflection.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

The time-independent Schrödinger equation gives the even interior wavefunction $\psi =A\cos (\alpha x)$, with ${\alpha }^2=2m(U+E)/{\hslash }^2$. Outside, normalizability of a bound state requires $\psi =C{e}^{-\beta (|x|-l)}$, with ${\beta }^2=-2mE/{\hslash }^2$.

Continuity of the wavefunction and its derivative at $x=l$ gives $C=A\cos (\alpha l)$ and $-\beta C=-\alpha A\sin (\alpha l)$. A nonzero solution cannot have $\cos (\alpha l)=0$, so division givesAt the formally allowed endpoint $E=-U$, the even interior solution is constant; derivative matching would force the exterior amplitude to vanish, leaving the zero solution. Nontrivial bound states therefore have $-U<E<0$. The matching formula is the even-state condition for a finite square well.

PutThe time-independent Schrödinger equation has the formsThe first region contains the incident and reflected waves, while the final region contains only the transmitted wave.

Continuity of $\psi$ and ${\psi }^{\prime }$ at both edges of the finite square well gives the standard transmission coefficientWhen $V_0=3E$, one has $K=2k$, and thereforeHence the transmission probability is

The time-independent Schrödinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.

The second infinite-well state hasIts maxima are at $L/4$ and $3L/4$, with nodes at the walls and center. The density is zero outside.

If the second bound state exists in a symmetric finite square well, odd parity preserves its central node and two equal lobes. However, the density is nonzero at the walls and has exterior tails proportional to $e^{-2\kappa |x-x_{\mathrm{wall}}|}$. Normalization includes those tails, and internal peak positions and heights change. Finite barriers change penetration and energy; symmetry preserves the central node. The original wells figure compares the two normalized second-state densities.

Choose zero potential for $0<x<L$. In an infinite square well,The first three levels have ratios $1:4:9$, with zero, one and two interior nodes; the wavefunctions vanish outside.

For a finite square well with exterior potential $V_0$, bound states have $0<E<V_0$, sinusoidal interiors and tails with $\kappa =\sqrt{2m(V_0-E)}/\hslash$. Matching wavefunction and derivative for equal masses giveswhen parity is taken about the well center. Corresponding energies fall below the infinite-barrier values, and the wavefunctions spread beyond the walls. There are finitely many bound states; a shallow well need not possess three. States above $V_0$ form a scattering continuum.

The parity operator acts as $P\psi (x)=\psi (-x)$, with $P^2=I$. For an even potential energy $[H,P]=0$, so energy eigenstates can be chosen even or odd, with parity eigenvalues $+1$ and $-1$. One-dimensional bound states are nondegenerate; the ground state is nodeless and even.

For the finite square well, square integrability excludes the growing exterior exponential, while even parity excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails andContinuity at $a$ gives $B\cos ka=A{e}^{-qa}$ and $-Bk\sin ka=-qA{e}^{-qa}$. Their ratio yieldsThe conditions at $-a$ follow from the same parity.

For a narrow weak well, $ka\ll 1$, hence $q\simeq k^{2}a\simeq 2m{V}_{0}a/{\hslash }^2$ andThis is self-consistent because $|E_0|/V_0\simeq 2m{V}_0{a}^2/{\hslash }^2\ll 1$. It is the shallow finite-square-well energy, equivalently the attractive delta potential limit with integrated strength $2a{V}_0$. An arbitrarily weak attractive well of this form still has its even bound ground state.

The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi /2<ka<\pi$. At the binding threshold $q\to 0$, sois necessary and sufficient for the first odd bound state. Equality gives a zero-energy nonnormalizable threshold solution, not an additional bound state.

Choose the centre of the finite square well as the origin and its edges as $\pm a$. The first three bound wavefunctions have even, odd, even parity, with respectively zero, one and two interior nodes. They oscillate inside and decay exponentially outside, with ${\kappa }_n=\sqrt{2m(U-E_n)}/\hslash$. Both the wavefunction and its first derivative are continuous at each finite potential step. The higher bound states have longer external decay lengths because they lie closer to the barrier height. Overall sign is arbitrary.

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.

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.

Write $\rho =a\sqrt{2m{V}_0}/\hslash >0$. On the first tangent branch, $0<\beta <\pi /2$, the matching relations are equivalent toThe function $g(\beta )=\beta /\cos \beta$ is continuous and strictly increasing there:The intermediate value theorem therefore gives a unique first-branch root for every $\rho >0$. At this root $\beta <\rho$ and $\alpha =\beta \tan \beta >0$, so the energy eigenvalueobeys $-V_0<E<0$. This proves the even bound state in every attractive square well result.

To verify all the requested properties, take the resulting even wavefunction with exterior tails written in terms of their boundary values:The matching equation makes both the values and first derivatives agree at the endpoints. The exterior tails are square-integrable because $k>0$, and the interior interval is finite. More explicitly,so choosing $B$ as the positive inverse square root of the bracket gives wavefunction normalization. Thus an even, continuously differentiable, normalizable stationary solution always exists, however small the positive depth.

A quantum well confines a particle in one spatial direction while allowing motion along the layer. In a semiconductor heterostructure, band offsets can create that confinement. The following finite square well model makes the resulting discrete transverse energy levels explicit.

For an attractive finite square well of width $a$, outside wave number $k$ and inside wave number $q$, matching the wavefunction and its first derivative at both endpoints gives the displayed transmission probability. It equals one when $qa$ is an integer multiple of $\pi$: the internal reflections cancel. For $q=2k$ the coefficient of ${\sin }^2(qa)$ is $9/16$.