The Dirac Delta Function Potential

Quantum Mechanics A

Schrödinger Equation The most fundamental equation of quantum mechanics; given a Hamiltonian ${\mathcal {H}}$, it describes how a state $|\Psi \rangle$ evolves in time.

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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. For a delta potential, the wavefunction is still continuous across the potential, ie. $x=a\!$. However, the first derivative of the wavefunction is discontinuous across the potential.

For a particle subject to an attractive delta potential $V(x)=-V_{0}\delta (x)\!$ the Schrödinger 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 Schrödinger 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0\delta(x-a)\!} , the discontinuity of the first derivative can be shown as follows:

The Schrödinger equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon \rightarrow 0 \!} , we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}=\frac{2m V_0}{\hbar^2}\psi(a)}

Problem

Problem 1

External Link

Additional information on the dirac delta function