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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.

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    $\begingroup$ There is a near infinite literature on Schrodinger operators with highly singular potentials, you might want to do a literature search. $\endgroup$ Jul 16, 2015 at 15:43
  • $\begingroup$ There is no problem multiplying delta with a function continuous in 0. Delta doesn't need infinitely differentiable test functions (unless you want to define derivatives of all orders). $\endgroup$
    – md2perpe
    Jul 23, 2015 at 19:22

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(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.

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    $\begingroup$ Thank you. On the first line, you mean $-1/2-\epsilon$ in the exponent, right? A reference discussing these specific types of issues would be great. $\endgroup$ Jul 16, 2015 at 17:42
  • $\begingroup$ Ooops, yes, I'll change that sign on the epsilon. The general, basic things about Sobolev spaces are treated in most or many books on PDEs, especially linear ones. Folland's book (or Tata lectures), Brezis' book, are two that do this sort of thing. The question of multiplication is an immediate corollary, then, since $H^{+s}$ and $H^{-s}$ (in various contexts) are in duality, so "pair" to $L^1$, at least. The Sobolev imbedding business is treated in those sources, certainly. Googling "Sobolev space" should give lots of useful results, too. $\endgroup$ Jul 16, 2015 at 18:26
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    $\begingroup$ The product of a continuous function $y$ and Dirac $\delta$ "function" is $y(0)\delta$, a measure, not a locally integrable function. It's Carlo Beenakker's answer that is right, isn't it? $\endgroup$ Jul 21, 2015 at 8:34
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    $\begingroup$ Perhaps it's useful to point out that there has been an ongoing controversy (and a lot of ink has been wasted) on what $-d^2/dx^2 +\delta'$ "should" mean. I guess what I'm trying to say in this context is that this kind of thing isn't rocket science; you make some rigorous definition and try to convince others that this is a good interpretation of what you're trying to make sense of (that said, there is agreement on how to make sense of $\delta$ here). $\endgroup$ Jul 21, 2015 at 21:29
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    $\begingroup$ @ChristianRemling, yes, indeed, thanks... and to give an example-objection: one way to interpret the (mathematical) difficulty in interpreting $\delta'$ as a "potential" in analogous fashion is that it is not inside $H^{-1}$, and it itself cannot be "legally" applied to solutions to $(\Delta-\lambda)u=\delta'$, since these solutions will only be in $H^{1/2-\epsilon}$, not $H^{3/2+\epsilon}$, etc. $\endgroup$ Jul 21, 2015 at 21:56
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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 $\lambda=\pm 1/2$.

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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

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  • $\begingroup$ Exactly. Dirac's marvelous intuition in the late 1920s was nicely rigorized in one fashion by (B.Levi-) Sobolev's spaces by the 1930s, as promoted and amplified by Gelfand-et-al in the "Generalized Functions" 6 volumes. $\endgroup$ Jul 20, 2015 at 22:07

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