# Delta distribution and the Schrödinger equation

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.

• 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 $\int_{-\epsilon}^\epsilon \delta(x) \Psi(x) \, dx \approx \int_{-\epsilon}^\epsilon \delta(x) \Psi(0) \, dx =\Psi(0) \int_{-\epsilon}^\epsilon \delta(x) \, dx = \Psi(0) \cdot 1$. – littleO Jan 4 '19 at 19:59
• 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. – Andrei Jan 4 '19 at 20:52
• @FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one! – Sito Jan 5 '19 at 16:24
• Alright, just shifted my comment to the answer section! – Frederik vom Ende Jan 5 '19 at 16:55

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