# Normalizability of continuous and discrete spectrum

• B
• Sunny Singh
In summary: But the theorem is certainly true, and it's a necessary condition for the existence of observables. In summary, this is not an obvious fact.
Sunny Singh
TL;DR Summary
I need a proof which says that continuous eigenvalue spectrum of hermitian operators are non-normalizable
I was reading introduction to quantum mechanics by DJ Griffiths and while discussing the formalism of quantum mechanics, he says that if for a hermitian operator, the eigenvalues are continuous, the eigenfunctions are non-normalizable whereas if the eigenvalues are discrete, then they can be properly normalized but still both span the hilbert space. He doesn't really explain why this is so or maybe i missed something. Can anyone please help me see why this is so obvious?

Sunny Singh said:
Summary: I need a proof which says that continuous eigenvalue spectrum of hermitian operators are non-normalizable

I was reading introduction to quantum mechanics by DJ Griffiths and while discussing the formalism of quantum mechanics, he says that if for a hermitian operator, the eigenvalues are continuous, the eigenfunctions are non-normalizable whereas if the eigenvalues are discrete, then they can be properly normalized but still both span the hilbert space. He doesn't really explain why this is so or maybe i missed something. Can anyone please help me see why this is so obvious?

It's not obvious. In linear algebra you have the finite dimensional spectral theorem, which guarantees an orthonormal basis of eigenvectors for any normal matrix (which includes Hermitian ones).

For infnite dimensional spaces, you have the same property for compact, self-adjoint operators.

In QM it's assumed that the Hermitian operators associated with observables have this property.

A continuous spectrum of eigenvalues takes you out of the realm of traditional linear algebra and into the shadowy realm of "rigged Hilbert spaces". In which case the eigenfunctions won't be the usual normalisable functions, but something more exotic.

Again, I would simply take it as a postulate of QM that the eigenfunctions of an operator representing a continuous observable have this property.

Sunny Singh
Sunny Singh said:
I was reading introduction to quantum mechanics by DJ Griffiths and while discussing the formalism of quantum mechanics, he says that if for a hermitian operator, the eigenvalues are continuous, the eigenfunctions are non-normalizable whereas if the eigenvalues are discrete, then they can be properly normalized
By definition, a number ##E## belongs to the discrete spectrum of an operator ##H## if there is a normalizable eigenvector ##\psi## with ##H\psi=E\psi##. The full spectrum is the set of ##E## such that ##E-H## is not invertible. Except in the accidental case where a discrete eigenvalue is embedded in the continuum (which is unstable under perturbations and leads to resonances, in physics responsible for particle decay) the continuous spectrum is disjoint from the discrete spectrum, hence there are no associated normalizable eigenstates.
Sunny Singh said:
but still both span the hilbert space
This is misleading. Given a fixed Hermitian operator, the discrete eigenstates span the full Hilbert space only if the spectrum is discrete only, and the continuum eigenstates span the full Hilbert space only if the spectrum is continuous only. Both these cases occur in the same Hilbert space but for different operators.

Sunny Singh, vanhees71 and WWGD
Sunny Singh said:
Summary: I need a proof which says that continuous eigenvalue spectrum of hermitian operators are non-normalizable

I was reading introduction to quantum mechanics by DJ Griffiths and while discussing the formalism of quantum mechanics, he says that if for a hermitian operator, the eigenvalues are continuous, the eigenfunctions are non-normalizable whereas if the eigenvalues are discrete, then they can be properly normalized but still both span the hilbert space. He doesn't really explain why this is so or maybe i missed something. Can anyone please help me see why this is so obvious?

There is no "B" level proof that the eigenvectors of an operator with continuous spectrum are not in the Hilbert space. One needs (big chunks of) the full machinery of functional analysis to write a comprehensive proof.

Sunny Singh, vanhees71 and Keith_McClary

## What is the difference between continuous and discrete spectrum?

The continuous spectrum refers to a set of values that are uncountable, while the discrete spectrum refers to a set of values that are countable. In the context of normalizability, this means that the continuous spectrum contains an infinite number of possible values, while the discrete spectrum contains a finite number of possible values.

## Why is it important to consider the normalizability of both continuous and discrete spectrum?

In quantum mechanics, the wave function must be normalized in order to accurately describe the behavior of a physical system. This means that the integral of the wave function squared must equal 1. If the spectrum is not normalized, the wave function will not accurately represent the system and predictions about its behavior will be incorrect.

## How is the normalizability of continuous and discrete spectrum determined?

The normalizability of a spectrum is determined by calculating the integral of the wave function squared over the entire spectrum. If the integral equals 1, then the spectrum is considered normalizable. If the integral is not equal to 1, then the spectrum is not normalizable.

## Can a spectrum be both continuous and discrete?

Yes, a spectrum can be both continuous and discrete. This is known as a mixed spectrum. In this case, there are both uncountable and countable values within the spectrum. The normalizability of a mixed spectrum must be considered for both the continuous and discrete parts separately.

## What are some examples of systems with normalizable continuous and discrete spectrum?

One example is the particle in a box system, where the particle is confined to a finite region. The continuous spectrum represents the energy levels of the particle, while the discrete spectrum represents the allowed values for the position of the particle within the box. Another example is the hydrogen atom, where the continuous spectrum represents the energy levels of the electron, while the discrete spectrum represents the allowed values for the angular momentum of the electron.

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