# Question 3.5: (a) Consider a one-dimensional potential well approximated b......

(a) Consider a one-dimensional potential well approximated by a delta function in space so that $V(x)=−bδ(x = 0)$. Show that there is one bound state for a particle of mass $m$, and find its energy and eigenstate.

(b) Show that any one-dimensional delta function potential with $V(x)=±bδ(x = 0)$ always introduces a kink in the wave function describing a particle of mass $m$.

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(a) Here we seek the eigenfunction and eigenvalue of a particle in a delta-function potential well, $V(x)=±bδ(x = 0)$, where $b$ is a measure of the strength of the potential. Due to the symmetry of the potential we expect the wave function to be an even function. The delta function will introduce a kink in the wave function. Schrödinger’s equation for a particle subject to a delta-function potential is

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

Integrating between $x = 0−$  and $x = 0+$

$\int\limits_{x=0-}^{x=0+}{\left(\frac{d^{2} }{dx^{2} } \psi (x)\right) dx} =\int\limits_{x=0-}^{x=0+}{\left(\frac{-2m}{\hbar ^{2} }(E+b\delta (x=0))\psi (x) \right) dx}$

results in

$\psi ^{\prime } (0+)-\psi ^{\prime } (0-)=\frac{-2mb}{\hbar ^{2} } \psi (0)$

where we have defined $d\psi /dx=\psi ^{\prime }$. One now notes that for an even function the spatial derivative is odd, so that $\psi ^{\prime } (0+)=-\psi ^{\prime } (0-)$, and one may write

$\psi ^{\prime } (0+)-\psi ^{\prime } (0-)=2\psi ^{\prime } (0+)=\frac{-2mb}{\hbar ^{2} } \psi (0)$

This is our boundary condition.

The Schrödinger equation is $(-\hbar ^{2}/2m )d^{2} \psi (x)/dx^{2} =E\psi (x)$ for $x > 0$ and $x < 0$. For a bound state we require $\psi (x)→0$ as $x→\infty$ and $\psi (x)→0$, as $x→-\infty$ so the only possibility is

$\psi (x)=Ae^{-k\left|x\right| }$

where $k=\sqrt{2m\left|E\right|/\hbar ^{2} }$. Substituting this wave function into Schrödinger’s equation gives $-\hbar ^{2} k^{2} /2m=E$, but from our previous work

$2\psi ^{\prime } (0+)=-2Ake^{-kx} \mid _{x=0} = \frac{-2mb}{\hbar ^{2} } \psi (0) = \frac{-2mb}{\hbar ^{2} }Ae^{-kx} \mid _{x=0}$

so that $k=mb/\hbar ^{2}$. Hence, we may conclude that

$E=\frac{-mb^{2} }{2\hbar^{2} }$

is the energy of the bound state.

To find the complete expression for the wave function we need to evaluate $A$. Normalization of the wave function requires

$\int\limits_{x=-\infty }^{x=\infty }{\left|\psi (x)\right|^{2}dx } =1$

and this gives $A=\sqrt{k}$ so that $\psi (x)=\sqrt{mb/\hbar ^{2} } e^{-\frac{mb}{\hbar ^{2} } \left|x\right| }$.

The spatial decay (the $e^{-1}$ distance) for this wave function is just $\Delta x=1/k=\hbar ^{2} /mb$. Physically, it is reasonable to use the delta-function potential approximation when the spatial extent of the potential is much smaller than the spatial decay of the wave function. This might occur, for example, in the description of certain single-atom defects in a crystal.

(b) It follows from (a) that any delta function in the potential of the form $V(x)=±bδ(x = 0)$ introduces a kink in the wave function. To show this one integrates Schrödinger’s equation for a particle subject to a delta-function potential

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

between $x = 0−$ and $x = 0+$ in such a way that

$\int\limits_{x=0-}^{x=0+}{\left(\frac{d^{2} }{dx^{2} } \psi (x)\right) dx}=\int\limits_{x=0-}^{x=0+}{\left(\frac{-2m}{\hbar ^{2} }(E\mp b\delta (x))\psi (x) \right)dx }$

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

So, for finite $\psi (0)$ and $b$ there is always a difference in the slope of the wave functions at $\psi (0+)$ and $\psi (0-)$, and hence a kink, in the wave function about $x = 0$. We may conclude that a delta function potential introduces a kink in the wave function.

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