Delta well: bound state and scattering

A bound state has energy $E<0$. Let $\kappa=\sqrt{-2mE}/\hbar>0$. Away from the origin, the stationary equation reads $\phi"=\kappa^2\phi$. Normalisability requires retaining the decaying exponential on each side:

$$\phi(x)=Ae^{-\kappa x}\ (x>0),\qquad\phi(x)=Be^{\kappa x}\ (x<0).$$

Continuity at $0$ requires $A=B$, and hence $\phi(x)=Ae^{-\kappa|x|}$. The derivatives are $\phi'(0^+)=-\kappa A$ and $\phi'(0^-)=\kappa A$. The discontinuity condition gives

$$-2\kappa A=-\frac{2mg}{\hbar^2}A,\qquad\text{that is}\qquad\kappa=\frac{mg}{\hbar^2}=\kappa_0 .$$

There is only one possible value of $\kappa$, and therefore only one bound state, with energy

$$E_0=-\frac{\hbar^2\kappa_0^2}{2m}=-\frac{mg^2}{2\hbar^2}.$$

The normalisation $1=|A|^2\int_{\R}e^{-2\kappa_0|x|}dx=|A|^2/\kappa_0$ gives $A=\sqrt{\kappa_0}$, and $\phi_0(x)=\sqrt{\kappa_0}\,e^{-\kappa_0|x|}$. The bound state is even. An odd state would have to vanish at $0$ by continuity; however, it would also have to be of the form $Ae^{-\kappa x}$ for $x>0$, which does not vanish at $0$ unless $A=0$. There is therefore no odd bound state: unlike the finite square well, which may possess several, the delta well has only one bound state, whatever its depth $g$.

We have

$$P\Bigl(|x|>\frac{1}{\kappa_0}\Bigr)=2\kappa_0\int_{1/\kappa_0}^\infty e^{-2\kappa_0x}dx=e^{-2}\simeq0{,}14 .$$

The length $1/\kappa_0=\hbar^2/(mg)$ is the characteristic size of the bound state: the deeper the well, the more localised the particle. Note that, for this negative energy, the entire line apart from the origin is classically forbidden: the particle lies entirely in a region where a classical particle could not be found.

On the left, there is the incident wave $e^{ikx}$, whose amplitude is set to one, and a reflected wave $e^{-ikx}$. On the right, there is only a transmitted wave travelling towards $+\infty$; a term in $e^{-ikx}$ would describe a wave arriving from the right, which is not the situation under consideration. Continuity at $0$ gives $1+r=t$. The derivatives are $\phi'(0^-)=ik(1-r)$ and $\phi'(0^+)=ikt$, and the discontinuity condition reads

$$ikt-ik(1-r)=-2\kappa_0t .$$

Replacing $t$ by $1+r$, we obtain $ik(1+r)-ik(1-r)=2ikr=-2\kappa_0(1+r)$, hence $r(ik+\kappa_0)=-\kappa_0$ and

$$r=-\frac{\kappa_0}{\kappa_0+ik},\qquad t=1+r=\frac{ik}{\kappa_0+ik}.$$

Since the potential vanishes on both sides, the incident, reflected and transmitted waves have the same wavenumber, and hence the same speed, so the current ratios reduce to the squared moduli of the amplitudes:

$$R=|r|^2=\frac{\kappa_0^2}{\kappa_0^2+k^2},\qquad T=|t|^2=\frac{k^2}{\kappa_0^2+k^2},\qquad R+T=1 .$$

The equality $R+T=1$ expresses the conservation of probability current. In terms of the energy, with $k^2=2mE/\hbar^2$ and $\kappa_0^2=m^2g^2/\hbar^4$,

$$T=\frac{1}{1+\kappa_0^2/k^2}=\frac{1}{1+\dfrac{mg^2}{2\hbar^2E}}=\frac{1}{1+|E_0|/E}.$$

At low energy, $E\ll|E_0|$, the transmission tends to zero: a slow particle is almost completely reflected, even by an attractive well, which is a purely wave-mechanical effect with no classical counterpart. At high energy, $E\gg|E_0|$, the transmission tends to one, and the well becomes transparent. Note that the bound-state energy provides the energy scale for scattering. Finally, $R$ and $T$ depend only on $\kappa_0^2$, and hence on $g^2$: the result is the same for a repulsive barrier $+g\delta(x)$. Only the phases of $r$ and $t$ change.

The denominator $\kappa_0+ik$ vanishes for $k=i\kappa_0$. For this value, the wave $e^{ikx}=e^{-\kappa_0x}$ decays to the right, and $e^{-ikx}=e^{\kappa_0x}$ decays to the left: if the solution is divided by $t$ and $t$ is allowed to tend to infinity, the incident wave disappears and $e^{-\kappa_0|x|}$ remains, precisely the bound state. The corresponding energy, $E=\hbar^2k^2/(2m)=-\hbar^2\kappa_0^2/(2m)$, is the bound-state energy. This is a general result in scattering theory: bound states correspond to the poles of the transmission amplitude located on the positive imaginary axis of the complex $k$ plane.