# Schrödinger equation involving the Dirac-Delta

I am taking a course on quantum mechanics and I try to understand the time-independent Schrödinger-equation with the Delta-potential: $$\frac{-\hslash^2}{2m}\psi''(x)-V_0\delta(x)\psi(x)=E\psi(x)$$ where $$V_0,m,\hslash>0$$ and $$\psi$$ and $$E$$ are unknown. I can somehow mimic the procedure for solving the equation but I do not understand what I am doing there. The main problem is that I do not understand what $$\delta(x)$$ exactly means. I know that $$\delta$$ is a distribution which can for example be defined as being a functional or as being a measure. However, I don't understand why $$\delta$$ can take $$x$$ as an argument. I guess it is some abuse of notation, right? So, here are my questions:

What does $$\delta(x)$$ in the above equation mean? Or more general: What does $$\delta(x)$$ mean if it occurs in an ODE? How is a solution for such an ODE defined? If $$\delta(x)$$ is abuse of notation, what would be the correct notation for such an equation?

Of course, I know the "physicist's explanation" of $$\delta(x)$$ (i.e.: it can be used to describe a potential $$V(x)=0$$ for $$x \neq 0$$ and $$V(0)=-\infty$$). Unfortunately, this does not help me. I am looking for a mathematical precise explanation. A reference to a textbook would also be great (a mathematics book - not a physics book).

• – Thomas Fritsch Jun 17 '19 at 17:59
• You can always use Fourier transform (to momentum space) to get rid of delta function. You will get an integral equation perfectly defined everywhere – Yuriy S Jun 17 '19 at 18:19
• See this answer for the solution using Fourier transform – Yuriy S Jun 17 '19 at 18:27

1. The symbol $$\delta$$ is actually defined by the formal property that for any test function $$\phi$$, i.e., a smooth function $$\phi : \mathbb{R} \to \mathbb{R}$$ that vanishes outside some finite interval $$[a,b]$$, $$\int_{-\infty}^\infty\delta(x)\,\phi(x)\,\mathrm{d}x = \phi(0).$$ Thus, from a rigorous mathematical standpoint, the Dirac delta is actually the linear transformation $$\{\text{test functions on \mathbb{R}}\} \to \mathbb{R}, \quad \phi \mapsto \phi(0);$$ more generally, any such linear transformation $$T$$, which we call a distribution, can be symbolically viewed as defining a “generalised function” $$\tau$$ by setting $$\int_{-\infty}^\infty \tau(x) \, \phi(x) \, \mathrm{d}x := T(\phi)$$ for any test function $$\phi$$. Note that any ordinary (integrable) function $$\tau$$ defines a distribution $$T$$ by reading this same equation right to left. This allows you, for instance, to differentiate distributions to your heart’s content via integration by parts, in a way that perfectly generalises ordinary differentiation: formally, for any test function $$\phi$$ $$\int_{-\infty}^\infty \tau^\prime(x)\,\phi(x)\,\mathrm{d}x = -\int_{-\infty}^\infty \tau(x)\, \phi^\prime(x)\,\mathrm{d}x = -T(\phi^\prime),$$ so we can rigorously define the derivative of the distribution $$T$$ to be the distribution $$\mathrm{D}T$$ given by $$\mathrm{D}T(\phi) := -T(\phi^\prime)$$ for any test function $$\phi$$. Note that all these considerations were first developed by Laurent Schwartz in the 1940s precisely to make rigorous sense of Dirac’s formal intuition.
2. Let $$\phi$$ be any test function. If we assume that $$\psi$$ is a smooth solution of your time-independent Schrödinger equation, then by the above formal integration by parts \begin{align} \int_{-\infty}^\infty E\,\psi(x) \,\phi(x) \, \mathrm{d}x &= \int_{-\infty}^\infty \left(-\frac{\hbar^2}{2m}\psi^{\prime\prime}(x)-V_0\delta(x)\,\psi(x) \right)\phi(x)\,\mathrm{d}x\\ &= -\frac{\hbar^2}{2m}\int_{-\infty}^\infty\psi^{\prime\prime}(x)\,\phi(x)\,\mathrm{d}x - V_0 \int_{-\infty}^\infty \delta(x) \, \psi(x) \, \phi(x)\,\mathrm{d}x\\ &= -\frac{\hbar^2}{2m}\int_{-\infty}^\infty \psi(x) \, \phi^{\prime\prime}(x)\,\mathrm{d}x - V_0 \,\psi(0)\, \phi(0). \end{align} Thus, following the above disucssion, we can rigorously define a function $$\psi$$ to be a distributional or weak solution of your time-independent Schrödinger equation if there exists $$E \in \mathbb{R}$$, such that for every test function $$\phi$$, $$-\frac{\hbar^2}{2m}\int_{-\infty}^\infty\psi(x)\,\phi^{\prime\prime}(x)\,\mathrm{d}x - V_0\,\psi(0)\,\phi(0) = E \int_{-\infty}^\infty \psi(x) \, \phi(x)\,\mathrm{d}x.$$ This notion of distributional solution, which is completely consistent with the differential calculus of distributions, even turns out to be useful when considering perfectly ordinary looking differential equations.

Essentially, the Delta distribution makes you evaluate the wave function $$\psi (x)$$ and the potential at $$x=0$$, in this case.

The Delta distribution is indeed somewhat of an abuse of notation, but it is usually interpreted as meaning the evaluation of a function at the point the makes the argument of the Delta function equal 0. For instance, $$\delta (x-a)\psi (x)$$ would mean you evaluate the function $$\psi (x)$$ at $$x=a$$.

In your case, the potential is a scalar, I assume, and it is multiplied by the value of the wave function at $$x=0$$.

Of course, this explanation is not so mathematically precise, but it is the way I learned it in a physics context. This might not please more mathematical-rigour-driven folks, but the question was asked with respect to a physical context, so that's my two cents on it.

Also, see this question for solving the SE with a Dirac distribution.

Edit: I've found the book by Hoskins, Delta Functions: Introduction to Generalised Functions to be a good supplement to the physics resources I had on the Dirac Delta distribution.

• Thanks for the reference. I'll give it a try. – russoo Jun 17 '19 at 18:15