Schrödinger equation involving the Dirac-Delta

Question

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).

I've asked a related question on physics stackexchange. You can find the link in the comments.

Answer 1

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.

Answer 2

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.