6
$\begingroup$

Is there anyone who studied on the book "Solvable Models In Quantum Mechanics" by Albeverio? I don't succed in understanding the proof of page 116 about the eigenvalues of the Hamiltonian with point interactions. In particular I don't understand the proof of the fact that eigenvalues conrespond to the zero of the determinant of the matrix of the Hamiltonian with N point interxations: $$A(k)=\bigg[\bigg(\alpha_j-\frac{ik}{4\pi}\bigg)\delta_{jj'}-\tilde{f}(y_j-y_{j'})\bigg]_{jj'}$$ where $\tilde{f}(x)$ is $\tilde{f}(x)=\frac{e^{ik|x|}}{4\pi|x|}$ se $x\neq 0$ and $0$

$\endgroup$
1
  • 7
    $\begingroup$ Dear Fawkes: first of all, welcome to MO. Most probably, not very many MO users have read page 116 of Albeverio's book. Therefore, gien the way you formulated your question, not very many people will be able to help you. However, if you phrased your question a little bit differently, you might have much more chances of getting some help. First of all: could you maybe state the result whose proof you do not understand? Second: could you try to sketch the proof, and indicate the precise place where you're no longer able to follow the argument? $\endgroup$ Commented Mar 5, 2013 at 22:10

1 Answer 1

3
$\begingroup$

I don't know if this will satisfy, but here is a `physics proof' of the above formula. The wavefunction satisfies the Lippmann--Schwinger equation (with units mass $m=1/2$ and $\hbar=1$)

$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} - \int d^3\mathbf{r}'\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|} V(\mathbf{r}')\psi_{\mathbf{k}}(\mathbf{r}'),$$

where $V(\mathbf{r})=\sum_i V_i(\mathbf{r})$ is the potential due to the scatterers. Now a single point scatterer has the (angle independent) scattering amplitude

$$f_i^{(1)}=-\frac{a_i}{1+ia_ik},$$

where $a_i$ is the scattering length (i.e. the `effective range' of the potential is zero). Recall that the scattering amplitude is defined by the asymptotic behavior as $r\to\infty$

$$\psi_{\mathbf{k}}(\mathbf{r}) \to e^{i\mathbf{k}\cdot\mathbf{r}} + f(\theta,\phi)\frac{e^{ikr}}{r}$$

Imagine iterating the Lippmann--Schwinger equation. Since we know the answer for a single scatterer, we can sum up repeated scattering off the same scatterer in the resulting (Born) series to give the scattering amplitude above, only keeping track of when we switch to a different scatterer

$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}+4\pi\sum_i \tilde f(\mathbf{r}-\mathbf{r}_i) f_i^{(1)}\psi_{\mathbf{k}}(\mathbf{r}_i),$$

where $\tilde f$ is the function you defined above. $\psi_{\mathbf{k}}(\mathbf{r}_i)$ on the right hand side should be understood in the asymptotic sense i.e. as one approaches the $i^{\text{th}}$ scatterer. This is for the scattered wave. If you want a bound state, it should be a solution of the homogenous equation for the wavefunction at each of the $N$ scatterers i.e. without any incoming plane wave. Then the appropriate determinant must vanish, which is just the condition stated in your question (with the identification $\alpha_i=-(4\pi a_i)^{-1}$).

I believe this is called the Foldy--Lax method in scattering theory.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .