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Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrodinger 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 Schrodinger 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 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

The time-independent Schrodinger 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 $\eta =k\tan (ka)$.