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While studying the lecture notes of my quantum mechanics course I came across something that seemed a bit odd. There we want to solve the Schrödinger equation for the potential $V(x)=V_0 \delta(x)$, where $\delta(x)$ represents the "delta-function". We therefore get the equation $$-\frac{\hbar^2}{2m}\psi''(x)+V_0\delta(x)\psi(x)=E\psi(x),$$ where the solutions are of the form $\psi_{\rm left}(x)=A\exp(ikx/\hbar)+B\exp(-ikx/\hbar)$ with the energy $E= k^2/2m$ for $x<0$ and analogous for $x>0$ we find $\psi_{\rm right}$. We know that $$\lim\limits_{x\to 0^-} \psi_{\rm left} = \lim\limits_{x\to 0^+}\psi_{\rm right}$$ should hold but also need a second condition. The professor then proceeded to write the following.

We integrate eq. (1) [the first equation in this post] form $-\epsilon$ to $\epsilon$ and get $$-\frac{\hbar^2}{2m}(\psi'(\epsilon)-\psi'(-\epsilon))+V_0\psi(0)=E\int^\epsilon_{-\epsilon} \psi(x) dx.$$ For $\epsilon \to 0$ we get $$\psi ^ { \prime } \left( 0 ^ { + } \right) - \psi ^ { \prime } \left( 0 ^ { - } \right) = \frac { 2 m V } { \hbar ^ { 2 } } \psi ( 0 ),$$ where $$ f \left( 0 ^ { + } \right) = \lim _ { x \rightarrow 0 \atop x > 0 } f ( x ) \quad f \left( 0 ^ { - } \right) = \lim _ { x \rightarrow 0 \atop x < 0 } f ( x ).$$

My main confusion here comes from the fact that apparently $$\int_{-\epsilon}^{\epsilon} \delta(x)\psi(x)dx = \psi(0).$$ Could someone explain this?


Notes on my background in math: In a course on mathematical physics we talked briefly about (tempered) distributions and there we saw the definition $$\delta[\varphi] =\int_\mathbb{R^n}\varphi(x)\delta(x)dx = \varphi(0),$$ for $\varphi \in \mathcal{S}(\mathbb{R}^n)$ and $$\mathcal { S } \left( \mathbb { R } ^ { n } \right) : = \left\{ \phi \in C ^ { \infty } \left( \mathbb { R } ^ { n } \right) \Big| \forall \alpha , \beta \in \mathbb { N } _ { 0 } ^ { n } : \sup _ { x \in \mathbb { R } ^ { n } } | x ^ { \alpha } D ^ { \beta } \phi ( x ) | < \infty \right\}.$$ I'm aware of the fact that in physics we often use the $\delta$-"function" rather sloppy and tend to brush of problematic arguments away with some kind of intuition (at least that's how it was presented to me up until this point) but I fail to see how it can be considered the same to integrate over an interval $[-\epsilon, \epsilon]$ and over $\mathbb{R}$ in this case.

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    $\begingroup$ For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $\Psi$ is continuous at $0$ then $\int_{-\epsilon}^\epsilon \delta(x) \Psi(x) \, dx \approx \int_{-\epsilon}^\epsilon \delta(x) \Psi(0) \, dx =\Psi(0) \int_{-\epsilon}^\epsilon \delta(x) \, dx = \Psi(0) \cdot 1$. $\endgroup$
    – littleO
    Commented Jan 4, 2019 at 19:59
  • $\begingroup$ In addition to @littleO 's comment, note that the delta function is zero outside the $[-\epsilon,\epsilon]$ interval, so integrating over $\mathbb R$ you can split into three parts: one less than $-\epsilon$ (integral $0$), one between $-\epsilon$ and $\epsilon$, and one interval from$\epsilon$ to $\infty$. The last one is also zero. $\endgroup$
    – Andrei
    Commented Jan 4, 2019 at 20:52
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    $\begingroup$ @FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one! $\endgroup$
    – Sito
    Commented Jan 5, 2019 at 16:24
  • $\begingroup$ Alright, just shifted my comment to the answer section! $\endgroup$ Commented Jan 5, 2019 at 16:55

3 Answers 3

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One may define $\delta$ as a measure on $\mathbb R$ which on some $A\subseteq\mathbb R$ evaluates to $\delta(A)=1$ if $0\in A$ and $\delta(A)=0$ else. Thus assuming $0\in A$, one gets $$ \int_{\mathbb R}\psi(x)\delta(x)\,dx=\int_{\mathbb R}\psi(x)\,d\delta(x)=∫_A\psi(x)\,d\delta(x)+∫_{\mathbb R\setminus A}\psi(x)\,d\delta(x) $$ where the second integral vanishes as $\mathbb R\setminus A$ has $\delta$-measure zero. By choosing $A=(-\varepsilon,\varepsilon)$ we obtain $$\psi(0)=\int_{\mathbb R}\psi(x)\delta(x)\,dx=\int_{-\varepsilon}^\varepsilon \psi(x)\delta(x)\,dx$$ for all $\varepsilon>0$.

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Define $\phi(x):=\psi(x)[|x|\le\epsilon]$, where the square bracket is an Iverson bracket. Then $\int_{-\epsilon}^\epsilon\delta(x)\psi(x)dx=\int_{\Bbb R}\phi(x)\delta(x)dx=\phi(0)=\psi(0)$.

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For another, possibly useful, perspective, an equation such as $(-\Delta+\delta)u=f$ can be construed as altering the Laplacian $\Delta$ by a "singular potential", meant to be a very-short-influence potential... idealized/simplified to Dirac's $\delta$.

Perhaps the simplest way to rigorize this idea is to determine the domain of this operator by requiring that it not map out of $L^2$. Since the potential $\delta$ really is the mapping $f\to \delta(f)\cdot \delta$, this requirement is that $\delta(f)=0$. This is a boundary-value condition.

That is, what may look like a crazy manipulation does have sense. :)

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