4
$\begingroup$

In Landau, Lifshitz, "Quantum Mechanics, non-relativistic theory" in $\S18$ "The fundamental properties of Schrödinger's equation" the following is said about potential $U(x,y,z)$ in a footnote:

it must be mentioned that, for some particular mathematical forms of the function $U(x,y,z)$ (which have no physical significance), a discrete set of values may be absent from the otherwise continuous spectrum.

For reference, in Russian version the wording is

надо, однако, оговориться, что при некоторых определенных видах функции $U(x,y,z)$ (не имеющих физического значения) из непрерывного спектра может выпадать дискретный набор значений.

I wonder, what are the examples of such mathematical forms of potential?

I've previously posted this question on Math.SE, even tried offering a bounty, but apparently no one knows the answer there.

$\endgroup$
3
  • 2
    $\begingroup$ in a tight-binding model on a bipartite lattice, there is a chiral symmetry that prevents $E=0$ from being an eigenvalue (the density of states of the continuous spectrum vanishes in the limit $E\rightarrow 0$); but I would guess that the sentence in L&L is a "typo" in выпадать: "absent from" --> "added to", as it has been changed in some translations. $\endgroup$ Commented Jan 6, 2015 at 16:34
  • $\begingroup$ The highlighted statement (in English, I don't read Russian) doesn't make mathematical sense if taken at face value. Attempts at interpretation are provided in the answers below. In my experience, when physicists say "continuous spectrum", more often than not, they are referring to what mathematicians would call the essential spectrum. $\endgroup$ Commented Jan 6, 2015 at 18:53
  • $\begingroup$ Maybe this makes sense in the rigged Hilbert space approach ? $\endgroup$
    – jjcale
    Commented Jan 6, 2015 at 19:45

2 Answers 2

4
$\begingroup$

The spectrum of an operator is always a closed set. But perhaps they are defining the "continuous spectrum" to be all points of the spectrum that are not in the point spectrum (i.e. not eigenvalues). Then there can be eigenvalues surrounded by continuous spectrum. The classic example of this in a Schrödinger operator is due to Wigner and von Neumann. See e.g. this recent paper of Milivoje Lukic.

$\endgroup$
1
  • $\begingroup$ Or the embedded eigenvalues could be dense, even for a potential that almost (but not quite) decays as $O(1/x)$ (first example is due to Naboko 1986). Or there could be dense pure point spectrum, with decay $O(x^{-1/2+\epsilon})$ of the potential. $\endgroup$ Commented Jan 6, 2015 at 18:58
4
$\begingroup$

Consider the (stationary) Schrödinger equation $$ -\Delta u+ Vu=0, $$ or the differential inequality $\vert\Delta u \vert\le \vert V u\vert$, where $V$ is some "potential" function. The following unique continuation property holds and is optimal with respect to the exponents involved. Take $$\vert V(x)\vert \le \frac{C}{\vert x\vert^2}. $$ Then if $\vert\Delta u \vert\le \vert V u\vert$ and $u$ is flat at $0$, i.e. is such that $$ \int_{\vert x\vert\le R}\vert u(x)\vert^{2} dx= O(R^N),\quad\text{for all $N\in \mathbb N$}, $$ then $u$ vanishes identically. You can prove using that unique continuation property that no eigenvalue is embedded in the continuous spectrum for the operator $-\Delta + V$. Various other conditions could replace our example above such that $L^{dimension/2}_{loc}$.

$\endgroup$
0

You must log in to answer this question.

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