0
$\begingroup$

Consider the Hamiltonian $H = -\Delta + V$ where $V$ is the potential conrresponding to an infinite square well:

$$V(x) = \begin{cases}0,&\text{if } 0, \leq x \leq L;\\\infty,&\text{otherwise}.\end{cases}$$

We take

$$\mathcal{D}(H) = \{f\in H^2[0,L] : f(0) = f(L) = 0\}$$

so that $H$ is self-adjoint.

The corresponding eigenvalue problem is $H\psi = \lambda\psi$ for functions $\psi\in L^2[0,L]$, i.e.

$$-\frac{\operatorname d^2}{\operatorname dx^2}\psi = \lambda\psi.$$

The solutions are

$$\psi_n(x) = \sin\Big(\frac{\pi n}{L}x\Big),\quad\lambda_n = \frac{\pi^2n^2}{L^2}$$

so the point spectrum of $H$ is

$$\sigma_p(H) = \{\pi^2n^2/L^2 : n\in\Bbb N\}$$

Question Why is the point spectrum the whole spectrum in this case?

$\endgroup$
0
$\begingroup$

Notice that the eigenvectors $\{\psi_n\}_{n\in\Bbb N}$ form an orthogonal basis of $L^2[0,L]$. Hence, we can write any $f\in L^2[0,L]$ as a series

$$f = \sum_{n=1}^\infty \alpha_n\psi_n$$.

For any $\lambda \notin \sigma_p(H)$ we have

$$(H-\lambda\Bbb 1)f = \sum_{n=1}^\infty \alpha_n(\lambda_n-\lambda)\psi_n$$

So, as $\lambda \not=\lambda_n$,

$$(H-\lambda\Bbb 1)^{-1}f = \sum_{n=1}^\infty \alpha_n\frac{1}{(\lambda_n-\lambda)}\psi_n$$

As $1/(\lambda_n-\lambda) = L^2/(\pi^2n^2-L^2\lambda) \to 0$, the inverse operator is bounded. Hence, $\lambda$ cannot be in the continuous spectrum.

Since $H$ is self-adjoint, the residual spectrum is empty, so the point spectrum is the whole spectrum.

$\endgroup$

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.