The Dirac Delta Function Potential

A delta potential, eg. $V_{0}\delta (x-a)\!$, is a special case of the finite square well, where the width of the well goes to zero and the depth of the well goes to infinity, while the produce of the height and depth remains constant. Additional information on the dirac delta function can be found here: Dirac Delta Function. For a delta potential, the wavefunction is still continuous across the potential, ie. $x=a\!$. However, the first derivative of the wavefunction is discontinous across the potential.

For a particle subject to an attractive delta potential $V(x)=-V_{0}\delta (x)\!$ the Schrodinger equation is

${\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}-V_{0}\delta (x)\psi (x)=E\psi (x)$

For $x\neq \!0$ the potential term vanishes, and all that is left is

${\frac {d^{2}\psi (x)}{dx^{2}}}+{\frac {2mE}{\hbar ^{2}}}\psi (x)=0$

A bound state(s) may exist when $E<0\!$, and $\psi (x)\!$ vanishes at $x=\pm \infty \!$. The bound state solutions are therefore given by:

$\psi _{1}(x)=Ae^{kx}x<0\!$

$\psi _{2}(x)=Be^{-kx}x>0\!$

where

The first boundary condition, the continuity of $\psi (x)\!$ at $x=0\!$, yields $A=B\!$.

The second boundary condition, the discontinuity of ${\frac {d\psi (x)}{dx}}\!$ at $x=0\!$, can be obtained by integrating the Schrodinger equation from $-\epsilon \!$ to $\epsilon \!$ and then letting $\epsilon \rightarrow 0\!$

Integrating the whole equation across the potential gives

$\int _{-\epsilon }^{\epsilon }{\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}dx+\int _{-\epsilon }^{+\epsilon }(-V_{0}\delta (x))\psi (x)dx=\int _{-\epsilon }^{\epsilon }E\psi (x)dx$

In the limit $\epsilon \rightarrow 0\!$, we have

${\frac {-\hbar ^{2}}{2m}}\left[{\frac {d\psi _{2}(0)}{dx}}-{\frac {d\psi _{1}(0)}{dx}}\right]+V_{0}\psi (0)=0$

which yields the relation: $2kA={\frac {2mV_{0}}{\hbar ^{2}}}A\!$.

Since we defined $k={\sqrt {\frac {2m|E|}{\hbar ^{2}}}}\!$, we have ${\sqrt {\frac {2m|E|}{\hbar ^{2}}}}={\frac {mV_{0}}{\hbar ^{2}}}\!$. Then, the energy is $E=-{\frac {mV_{0}^{2}}{2\hbar ^{2}}}\!$

Finally, we normalize $\psi (x)\!$:

$\int _{-\infty }^{\infty }|\psi (x)|dx=2|B|^{2}\int _{0}^{\infty }e^{-2kx}dx={\frac {|B|^{2}}{k}}=1$

so,

$B={\sqrt {k}}={\frac {\sqrt {mV_{0}}}{\hbar }}$

Evidently, the delta function well, regardless of its "strength" $V_{0}\!$, has one bound state:

$\psi (x)={\frac {\sqrt {mV_{0}}}{\hbar }}e^{\frac {-mV_{0}|x|}{\hbar ^{2}}}$

Similarly, for a delta potential of the form $V_{0}\delta (x-a)\!$, the discontinuity of the first derivative can be shown as follows:

The Schrodinger equation is

${\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}+V_{0}\delta (x-a)\psi (x)=E\psi (x)$

Integrating the whole equation across the potential gives

$\int _{a-\epsilon }^{a+\epsilon }{\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}dx+\int _{a-\epsilon }^{a+\epsilon }V_{0}\delta (x-a)\psi (x)dx=\int _{a-\epsilon }^{a+\epsilon }E\psi (x)dx$

In the limit $\epsilon \rightarrow 0\!$, we have

${\frac {-\hbar ^{2}}{2m}}\left[{\frac {d\psi (a^{+})}{dx}}-{\frac {d\psi (a^{-})}{dx}}\right]+V_{0}\psi (a)=0$

Hence the first derivative of the wave function across a delta potential is discontinuous by an amount:

${\frac {d\psi (a^{+})}{dx}}-{\frac {d\psi (a^{-})}{dx}}={\frac {2mV_{0}}{\hbar ^{2}}}\psi (a)$