= Solution
\b[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> is
$$
-\frac{\hbar^2}{2m}\psi''(x)-V_0\delta(x)\psi(x)=E\psi(x).
$$
The <wavefunction> is continuous at the origin. A finite jump in $\psi$ would produce a <derivative> of a <Dirac delta function> in $\psi''$, with no corresponding term in this potential. A jump in $\psi'$ does produce a delta contribution and is allowed. Integrate through a symmetric interval:
$$
-\frac{\hbar^2}{2m}[\psi'(\epsilon)-\psi'(-\epsilon)]
-V_0\psi(0)=E\int_{-\epsilon}^{\epsilon}\psi(x)\,dx.
$$
For continuous finite $\psi$, the right-hand side vanishes as $\epsilon\to0$. Consequently
$$
\boxed{\psi'(0^+)-\psi'(0^-)=-\frac{2mV_0}{\hbar^2}\psi(0).}
$$
This <derivative> jump is the boundary condition associated with the attractive <delta potential>.
\b[Unique <bound state>.] For $E<0$, write $\kappa=\sqrt{-2mE}/\hbar>0$. Away from the origin, $\psi''=\kappa^2\psi$. Normalizability excludes exponentials growing at either infinity, leaving
$$
\psi(x)=\begin{cases}A e^{\kappa x},&x<0,\\B e^{-\kappa x},&x>0.\end{cases}
$$
<Continuity> gives $A=B=C$. The <derivative> jump then becomes $-2\kappa C=-2mV_0C/\hbar^2$. For a nonzero state,
$$
\kappa=\frac{mV_0}{\hbar^2},\qquad
\boxed{E=-\frac{\hbar^2\kappa^2}{2m}=-\frac{mV_0^2}{2\hbar^2}.}
$$
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 using
$$
1=2|C|^2\int_0^\infty e^{-2\kappa x}\,dx=\frac{|C|^2}{\kappa}.
$$
Choosing the overall phase real and positive gives the <bound state in a delta potential>:
$$
\boxed{\psi_0(x)=\sqrt\kappa\,e^{-\kappa|x|},\qquad\kappa=\frac{mV_0}{\hbar^2}.}
$$
It peaks at the origin, has a downward cusp there, and decays exponentially on both sides. Its value, but not its slope, is continuous.
\b[Scattering amplitudes and fluxes.] For $E>0$, put $k=\sqrt{2mE}/\hbar$ and retain $\kappa=mV_0/\hbar^2$ as a strength parameter. A unit-amplitude wave incident from the left has
$$
\psi_L=e^{ikx}+r e^{-ikx},\qquad\psi_R=t e^{ikx}.
$$
<Continuity> gives $t=1+r$. The jump condition gives
$$
ik t-ik(1-r)=-2\kappa t,\qquad ikr=-\kappa(1+r).
$$
Solving,
$$
r=-\frac{\kappa}{\kappa+ik},\qquad t=\frac{ik}{\kappa+ik}.
$$
The <probability current> of a <plane wave> of amplitude $A$ is $\hbar k|A|^2/m$ in its propagation direction. Since the asymptotic <wavenumber> is the same on both sides, <reflection probability> and <transmission probability> are just $|r|^2$ and $|t|^2$:
$$
\boxed{R=\frac{\kappa^2}{k^2+\kappa^2}
=\frac{mV_0^2/\hbar^2}{2E+mV_0^2/\hbar^2},\qquad
T=\frac{k^2}{k^2+\kappa^2}
=\frac{2E}{2E+mV_0^2/\hbar^2}.}
$$
They satisfy $R+T=1$, expressing conserved <probability current> in this real potential.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-1-b6-delta-well.png]
{title=The normalized delta-well bound state and reflection/transmission probabilities as attraction increases}
The right-hand plot uses $\eta=\kappa/k=V_0\sqrt{m/(2E)}/\hbar$, 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.
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