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?
  • 1
    $\begingroup$ 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)? $\endgroup$ Aug 24, 2019 at 19:49
  • $\begingroup$ Yes, that would be what I want to show. Also, is that domain dense in $L^2(\mathbb R)$? $\endgroup$ Aug 25, 2019 at 7:26
  • $\begingroup$ Here p.150 is one approach (for a finite interval). $\endgroup$ Aug 28, 2019 at 16:46
  • $\begingroup$ What is your background in functional analysis? Do you have theorems that smooth functions are dense in $L^2$ (for finite intervals, at least)? $\endgroup$ Aug 30, 2019 at 3:38

1 Answer 1


Rigorous formulation

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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.