# Domain of definition of a hamiltonian with delta(contact) potential

I am having a hard time making sense of the so-called "delta function potential well" in quantum theory. The Hamiltonian operator is defined as (with $$\mathscr D_H\subset \mathscr H=L^2(\mathbb R)$$) $$H:\mathscr D_H\rightarrow \mathscr H$$ $$\psi\mapsto H\psi$$ And $$(H\psi)(x):=-\frac{d^2}{dx^2}\psi(x)-\lambda\delta(x)\psi(x).$$ My job is to find the spectrum of this operator given a $$\lambda>0$$.

My problems are:

1. How do I construct $$\mathscr D_H$$?
2. What definition of the "delta function" is suitable for this kind of job?
• The wiki constructs explicit solutions. The domain would be functions with a kink at the origin like the eigenfunction shown there. Are you interested in showing that $H$ is essentially self-adjoint (in the rigorous meaning for unbounded operators)? Aug 24, 2019 at 19:49
• Yes, that would be what I want to show. Also, is that domain dense in $L^2(\mathbb R)$? Aug 25, 2019 at 7:26
• Here p.150 is one approach (for a finite interval). Aug 28, 2019 at 16:46
• What is your background in functional analysis? Do you have theorems that smooth functions are dense in $L^2$ (for finite intervals, at least)? Aug 30, 2019 at 3:38

When mathematical physicists work with point interactions, the “delta” in the potential is only a formal shorthand. What they actually mean are the corresponding boundary conditions $$\psi'(0+) - \psi'(0-) = -\lambda \, \psi(0) \: .$$ The Hamiltonian in your question is then rigorously defined as: $$\begin{gather*} \mathscr D_H = \big\{ \; \psi \in W^{1,2}(\mathbb R) \cap W^{2,2}(\mathbb R \setminus \{0\}) \;\; \big| \;\; \psi'(0+) - \psi'(0-) = -\lambda \, \psi(0) \; \big\} \: , \\[10pt] \big( H \psi \big)(x) = -\psi''(x) \quad \text{ for a.e. } \quad x \in \mathbb R \setminus \{0\} \: , \end{gather*}$$ where $$W^{k,p}$$ are the Sobolev spaces of $$k$$-times weakly differentiable $$L^p$$ functions.
You can check that this Hamiltonian is indeed self-adjoint: \begin{align*} ( H\psi, \varphi) &= \int_{-\infty}^{+\infty} \!\!\! -\overline{\psi''} \, \varphi \;=\; \int_{-\infty}^{0} \!\!\! -\overline{\psi''} \, \varphi \;+\; \int_{0}^{+\infty} \!\!\! -\overline{\psi''} \, \varphi \\[5pt] &= \big[-\overline{\psi'}\varphi \big]_{-\infty}^{0} + \big[ -\overline{\psi'}\varphi\big]_{0}^{+\infty} \;+\; \int_{-\infty}^{0} \!\!\! \overline{\psi'} \, \varphi' \;+\; \int_{0}^{+\infty} \!\!\! \overline{\psi'} \, \varphi' \\[5pt] &= \big[-\overline{\psi'}\varphi \big]_{-\infty}^{0} + \big[ -\overline{\psi'}\varphi\big]_{0}^{+\infty} + \big[-\overline{\psi}\varphi' \big]_{-\infty}^{0} + \big[ -\overline{\psi}\varphi'\big]_{0}^{+\infty} + \int_{-\infty}^{+\infty} \!\!\! -\overline\psi \, \varphi'' \\[5pt] &= \big[-\overline{\psi'}\!\varphi -\overline{\psi}\varphi' \big]_{-\infty}^{+\infty} - \lambda \, \overline\psi(0) \, \varphi(0) + \overline\psi(0) \, \big( \varphi'(0+) - \varphi'(0-) \big) + \!\int_{-\infty}^{+\infty} \!\!\!\!\! -\overline\psi \, \varphi'' \end{align*} which is equal to $$(\psi, H\varphi)$$ exactly iff $$\varphi \in \mathscr D_H$$. Really, the terms at $$\pm\infty$$ are zero, because functions from $$W^{1,2}$$ vanish at infinity (proof) and the terms at $$0\pm$$ exactly cancel out if $$\varphi$$ obeys the boundary condition.
This might seem unsatisfactory, because the rigorous formulation sidesteps the “multiplication by delta” operator. Therefore I will hint at a construction that follows the formal notation more closely. Choose your favourite space of test functions $$\mathcal D$$ (eg. smooth functions with compact support) and the space of continuous linear functionals on it $$\mathcal D'$$. Since $$\mathcal D$$ is naturally embedded in $$L^2(\mathbb{R})$$ and $$L^2(\mathbb{R})$$ is naturally embedded in $$\mathcal D'$$, we have the famous Gelfand sandwich $$\mathcal D \subset \mathcal H \subset \mathcal D'$$. Now we can define two operators. The first one is the distributional second derivative: $$\hat \Delta: \mathcal H \to \mathcal D' \: , \qquad \langle \hat \Delta \psi, \varphi \rangle = \int \psi \, \varphi'' \quad \forall \psi \in \mathcal H \quad \forall \varphi \in \mathcal D \: .$$ This is exactly how derivatives work on distributions (source). The second operator we define is the operator of multiplication by delta: $$\hat \delta: \big\{ \; \psi \in \mathcal H \; \big| \; \psi \text{ is continuous at } 0 \; \big\} \to \mathcal D' \: , \;\; \langle \hat \delta \psi, \varphi \rangle = \psi(0) \, \varphi(0) \;\, \forall \psi \in \mathcal H \;\, \forall \varphi \in \mathcal D \: .$$ Now we can rigorously define the Hamiltonian as $$\hat H = -\hat \Delta - \lambda \hat \delta$$ which is an operator from a subset of $$L^2(\mathbb{R})$$ to the distributions $$\mathcal D'$$. We could ask if there is a restriction of $$\hat H$$ to a self-adjoint operator on $$L^2(\mathbb{R})$$. Yes, there is – it is $$H = \hat H|_{\mathscr D_H}$$, as defined above!