Make mathematical sense of the Dirac well Potential Equation

Question

A classical problem in quantum mechanics involving the Dirac Delta function is given by $$ y''+(\delta(x)-\lambda^2)y=0. $$ Then, to find ''bound states'', you solve on the right and find the converging solution as $x\rightarrow \infty$, then solve on the left similarly. Assume continuity of the solution $y$. The jump condition on $y'$ is found by integrating from $-\epsilon$ to $\epsilon$ and taking $\epsilon\rightarrow 0$. The only value of $\lambda$ giving a bound state is then found to be $1/2$. The corresponding state is continuous everywhere and differentiable everywhere except at zero.

My issue with this is that to make this solution mathematically correct, one has to make sense of the multiplication of a continuous non-differentiable function by the delta Dirac function. In a previous question of mine, Alan U. Kennington suggested to use Radon measures. However, I believe there must be something simpler that makes sense of the problem above by mathematically making sense of multiplying a continuous function by a Dirac delta function. I am not necessarily looking for a complete answer but rather a reference that discusses the problem.

The reason I am asking this question is because I am facing a third order equation with coefficients involving the Dirac delta function. I am able to find a solution but I would like to make my computations more mathematically sound. I already asked this question on Math Stack Exchange but did not get an answer.

Answer 1

(The previous version of this attempted answer was too hasty, resulting in some silly nonsense in part...)

The rigorous interpretations of differential equations or other operator equations with unbounded operators, and/or with rough coefficients, involves issues of self-adjoint extensions of restrictions of symmetric operators, so, in the end, the apparent question of "multiplying" things that aren't meant to be multiplied is illusory.

The case of "singular potential" (an example of "exactly solvable model") such as $H=-\Delta+\delta$ on $\mathbb R$ does appear to be asking to multiply possibly-not-differentiable functions by $\delta$, which is problemmatical. But that "problem" is completely parallel to the "problem" of applying $\Delta$ to functions insufficiently differentiable but that their images under $\Delta$ are outside $L^2(\mathbb R)$, although possibly in some Sobolev space. Thus, the $\delta$ can/should be interpreted as the operator $\delta\otimes\delta$ on test functions given by $(\delta\otimes\delta)u=\delta(u)\cdot \delta$. (So, yes, for continuous $u$ this produces a multiple of $\delta$ by $u(0)$). But since this maps outside $L^2$, it is not a legitimate unbounded operator unless $\delta(u)=0$. So the proper domain of the operator $\delta\otimes\delta$ is test functions vanishing at $0$. This also works for $-\Delta+\delta$ (or, more properly, $-\Delta+\delta\otimes\delta$).

That is, the effect of the "operator" $\delta$ is to impose a boundary condition. At the same time, the Friedrichs extension $T$ of that restriction of $-\Delta+\delta$ does have some "exotic" features: $Tu=f$ if and only if $-\Delta u=f+c\cdot \delta$ for some constant $c$, and the domain of $T$ is $H^2+\mathbb C\cdot u_o$ where $u_o\in H^1$ is a solution of $(-\Delta+\lambda_o) u_o=\delta$ for some $\lambda_o$ off the real line. But this formulation skirts the issue of literal pointwise multiplication.

Yes, this seems to define away the issue, but it's not really so. For two point-charges on $\mathbb R$, there are "exotic" eigenfunctions: for example, ($-\Delta+\lambda)u=\delta_a+\delta_b$ has some $L^2$ solutions that meet the implied boundary conditions $u(a)=0=u(b)$, namely, all the fragments of sines and cosines vanishing at the endpoints. So, although $-\Delta$ has no $L^2$ eigenfunctions, the two-point singular potential variant does have, but/and the "multiplication" occurring is not literal pointwise at all.

Yes, solutions are in $H^1$ and the singular potentials are in $H^{-1}$ so application of the latter to the former is legitimate. (My earlier remark about this product as somehow pointwise were both silly and irrelevant, in fact.)

A genuine issue arises in higher dimensions, because Dirac deltas are not in $H^{-1}$ any more. In two dimensions, there are other (less canonical, perhaps) self-adjoint extensions of the restriction of $-\Delta$ to $u$ such that $\delta(u)=0$, and the generic ones of these allow singular potentials in somewhat worse Sobolev spaces. But, still, as the dimension goes up, the problem of maintaining the above interpretation seems to require re-interpreting point-charges as something else, otherwise one does not really quite have a self-adjoint operator on a Hilbert space.

Answer 2

If you wish to avoid delta functions, you could just Fourier transform, $f(k)=\int_{-\infty}^\infty e^{ikx}y(x)dx$. The differential equation then transforms into

$$(k^2+\lambda^2) f(k)=y(0)$$

Divide both sides of the equation by $k^2+\lambda^2$, integrate over $k$ and you arrive at

$$\int_{-\infty}^\infty f(k)dk = y(0)\int_{-\infty}^\infty \frac{1}{k^2+\lambda^2}dk$$ $$\Rightarrow 2\pi y(0)=\frac{\pi}{|\lambda|}y(0)$$

Since $y(0)\neq 0$ (otherwise $y(x)=0$ for all $x$), you can conclude that $\lambda=\pm 1/2$.

Answer 3

Probably the correct type of space for y is actually not a Hilbert space but a Rigged Hilbert space. Here's a paper about Rigged Hilbert spaces in quantum physics: http://arxiv.org/pdf/quant-ph/0502053v1.pdf