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The squared sine has integral $a$ over the interval of length $2a$, since it contains an integer number of half-wavelengths. A real positive phase convention therefore givesInside the infinite square well, the Time-independent Schrodinger equation is $-{\hslash }^2{\psi }_n^{\prime \prime }/(2m)=E_n{\psi }_n$. The wavenumber is $\pi n/(2a)$, soThe stationary states vanish at both walls and form an orthonormal basis of the square-integrable state space. A common arbitrary phase on each basis function does not alter any energy or measurement probability.

For a bound state, $0<E<U_0$: the Hamiltonian operator is nonnegative, and an energy at or above the exterior potential does not give a nonzero normalizable wavefunction. The Time-independent Schrodinger equation and the hard wall at zero therefore giveContinuity of the wavefunction and its derivative at the finite step gives $k\cot (ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2m{U}_0}/\hslash$. The matching equation becomesIt requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi ,(j+1)\pi )$ for $j=0,1,\dots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\mathrm{sgn}(\sin z)\cos z)/{\sin }^{2}z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one bound state if and only if $z_0>(j+1/2)\pi$. These are the bound-state thresholds for a square well with one hard wall.

Exactly one normalizable bound state exists if and only ifThe printed strict upper inequality misses the threshold case. At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa =0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one bound state at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.

A bound state has $E<0$ and $\kappa =\sqrt{-2mE}/\hslash >0$, since outside the delta potentials a square-integrable solution must decay exponentially. The wavefunction is continuous at each delta: a jump would produce a delta derivative in ${\psi }^{\prime \prime }$ with no matching term in the equation. Integrating the Time-independent Schrodinger equation across either $x_0=\pm a$ givesThe symmetry of the symmetric double-delta potential permits even and odd solutions. Up to a nonzero normalization factor, the even candidate is $\cosh (\kappa x)$ for $|x|\le a$, and $\cosh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its derivative jump at $a$ givesThe odd candidate is $\sinh (\kappa x)$ inside and $\mathrm{sgn}(x)\sinh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its jump givesParity ensures the jump at $-a$ as well. Put $t=\kappa a$ and $L=2m\lambda a/{\hslash }^2$. The continuous strictly increasing function $t(1+\tanh t)$ ranges from $0$ to infinity, so there is exactly one even bound state for every $L>0$. The continuous strictly increasing function $t(1+\coth t)$ ranges from $1$ to infinity, so the odd bound state of a symmetric double-delta potential exists if and only if $L>1$. Strict increase follows from the allowed monotonicity facts plus the strictly increasing term $t$. ThereforeAt equality the odd limiting solution has $\kappa =0$ and is not square integrable, so it is not a bound state. No nonnegative energy can supply another full-line square-integrable state, since its exterior solutions are oscillatory or affine.

For the quantum harmonic oscillator, the Time-independent Schrodinger equation iswith $m,\omega ,\hslash >0$. The given dimensionless coordinate yields ${\psi }_{\xi \xi }+(2E/(\hslash \omega )-{\xi }^2)\psi =0$. Substituting the Gaussian factor givesBy the supplied normalizability fact, $f$ is a nonzero polynomial. If its degree is $n$, comparison of the leading coefficient forces $2E/(\hslash \omega )-1=2n$. Conversely, at this value the coefficient recurrenceproduces a degree-$n$ polynomial with parity $n$: start with a nonzero constant for even $n$ or a nonzero linear coefficient for odd $n$, and the recurrence terminates at degree $n$. This is a multiple of the Hermite polynomial. Its product with the Gaussian is square integrable and can be normalized. Therefore all and only the allowed energies are

For a finite potential energy discontinuity, the stationary state $\psi$ and its first derivative are continuous at $a$. Integrating the Time-independent Schrodinger equation across a shrinking interval about $a$ gives continuity of ${\psi }^{\prime }$; a jump in $\psi$ would produce a distributional derivative incompatible with a finite potential energy.

For $E>V_0$, put $k=\sqrt{2mE}/\hslash$, $q=\sqrt{2m(E-V_0)}/\hslash$ and use unit incident amplitude:There is no wave incident from the right. The matching equations $1+r=t$, $k(1-r)=qt$ give $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. Comparing reflected and incident probability currents gives the reflection coefficientThe transmitted fraction is $T=(q/k)|t{|}^2=4kq/(k+q{)}^2$, so $R+T=1$.

For $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and retain only the decaying right-hand solution $t{e}^{-\kappa (x-a)}$. Matching gives $r=(k-i\kappa )/(k+i\kappa )$, which has modulus one. The evanescent wave carries no transmitted probability current. At $E=V_0$, the bounded right-hand solution is constant and matching again gives $r=1$. ThusClassically, a particle is wholly transmitted for $E>V_0$ and wholly reflected for $E<V_0$; there is no classical partial reflection above the step. At the threshold it has zero right-hand speed and no transmitted flux, a marginal case requiring a convention about motion exactly on the discontinuity. The quantum result approaches $R=0$ as $E/V_0\to \infty$.

In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the Gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.

For $0<E<U_0$, the Time-independent Schrodinger equation and the wall conditions givewhere $k=\sqrt{2mE}/\hslash$ and $l=\sqrt{2m(U_0-E)}/\hslash$. Continuity of $\psi$ and ${\psi }^{\prime }$ at the finite potential step givesDividing and rearranging yields the bound-state quantization condition

PutThe Time-independent Schrodinger 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

A Schrodinger 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 Schrodinger equation.