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I believe this is a mathematical question, but everytime I expose my question to a mathematician the answer is 'because they do this in Quantum Mechanics'. So here I am: a very ignorant person in Physics.

For a given real-valued and bounded potential $V(x)$, consider the Schrödinger operator $$L=-\partial_x^2 + V(x).$$ I have this definition for spectrum:

  • The discrete spectrum is the set of isolated eigenvalues of $L$ with finite algebraic multiplicity.
  • The continuous spectrum is the set of $\lambda\in\mathbb{C}$ such that the solutions of $Lu=\lambda u$ are bounded.

I completely understand the discrete spectrum as it is a subset of the point spectrum of functional analysis, but I fail to understand the continuous.

I know that if we define $L$ from $H^2(\mathbb{R})$ into $L^2(\mathbb{R})$ this is a self-adjoint operator, so residual spectrum is empty and the spectrum is in the real line. I know as well that from self-adjointness we can completely characterize the spectrum in terms of Weyl functions and then define the spectrum in terms of discrete + continuous in the sense of Weyl.

But why bounded here? Where is this coming from? For example, there is this classical paper by Barry Simon that defines the spectrum in terms of bounded solutions. Or a more recent paper by Dmitry Pelinovsky in which he also considers bounded for a different self-adjoint operator.

Why is it enough to require boundedness of the eigenfunctions for the continuous spectrum? Is this equivalent to Weyl's criterium?

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    $\begingroup$ Because the solutions are interpreted in terms of probability, which is bounded? $\endgroup$
    – Roger V.
    Jan 27, 2023 at 14:47
  • $\begingroup$ Where is your definition from? $\endgroup$
    – J. Murray
    Jan 27, 2023 at 15:39
  • $\begingroup$ @J.Murray the paper I mentioned by Barry Simon (S in page 2), but more generally it is the definition of spectrum for integrable equations that every (!!!) paper uses. The paper by Pelinovsky (page 4, right after eq 3.3) has the definition of Lax spectrum, which is the same as the continuous spectrum. $\endgroup$
    – MsWynfled
    Jan 27, 2023 at 15:43

1 Answer 1

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The interpretation of the wavefunction1 $\psi(x)$ (you called it $v$) is that its absolute square is the probability density of the particle to be at position $x$. The particle must be somewhere, so $$ \int_{-\infty}^{\infty} |\psi(x)|^2 dx \overset{!}{=} 1$$ i.e. the wave function must be normalised. This implies a bounded $\psi(x)$.

1 I think limiting ourselves to the 1D position-dependent case is sufficient here.

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  • $\begingroup$ But this means that the wavefunction (my eigenvector) is already square integrable, which does not make much sense to me. For example, if $L=i\partial_x$ is the momentum operator, the eigenvectors are of the form $\Psi(x) = K\exp(-i\lambda x)$, that is, they are not square integrable if $K\neq 0$. If we are requiring square integrability the spectrum is empty, but we know this is not the case. In fact, if we assume that $\Psi$ is only bounded, then the (continuous) spectrum is the entire real line (as it should be). $\endgroup$
    – MsWynfled
    Jan 27, 2023 at 15:55
  • $\begingroup$ The wavefunction of a single particle must be square integrable. Plane waves are not proper (physical) wavefunctions when describing individual particles. Linear combinations thereof can be. The uncertainty principle gives the same answer. A wavefunction $\exp(-i\lambda x)$ would have zero momentum uncertainty and is thus unphysical. $\endgroup$
    – noah
    Jan 27, 2023 at 16:05
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    $\begingroup$ Oh, I see. I think I understand where it comes from, at least from a very naïve physical point of view. Thanks a lot. $\endgroup$
    – MsWynfled
    Jan 27, 2023 at 16:07

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