Somebody came to me and asked the following:

Why is the multiset of eigenvalues called spectrum?

I cannot find the reason anywhere.

  • $\begingroup$ I usually compare the spectrum of an operator with the spectrum of light, these are the building blocks allowing to fully understand their effects. $\endgroup$
    – Surb
    Jul 24, 2017 at 6:17

3 Answers 3


Not too surprisingly, it seems to come from physics. I'm sure others here can provide much better detail, but here is a quote from a small article called "Favorite Eigenvalue Problems" in SIAM News, Volume 44, Number 10, December 2011 by Nick Trefethen:

Eigenvalues played a role in a great coincidence of scientific history. Physicists saw spectral lines in the light from stars; indeed, an unfamiliar line led to the discovery of helium. Later, Hilbert defined spectra of operators. Not until Schrödinger, decades on, was it understood that physicists’ spectra were exactly a case of mathematicians’ spectra, with each line corresponding to the difference in energy of two eigenstates. And it was these same spectral lines that led to the discovery of the red shift and the expanding universe.

  • 2
    $\begingroup$ This answer is the basic idea. The Schrödinger equation in Quantum Mechanics is the key. The mathematical spectrum of the Schrödinger operator for the hydrogen atom matched the physical emission spectrum of hydrogen. $\endgroup$ Feb 8, 2018 at 21:52

Newton introduced the word "spectrum" in sciences from 1666. The terminology of the eigen-elements of a matrix is fixed by Jordan in 1870 (in fact, for a while, mathematicians use "proper value" (Jordan) or "eigenvalue" (Hilbert)).

In 1907, functional analysis appears; in 1910, Weyl gives his definition of what would later be called a variant of the "essential spectrum".

Then comes the equivalence between the Heisenberg's matricial mechanics and the Schrodinger's equation (1926). Yet, the word spectrum still does not exist in the sense that interests us.

Note that a spectrum (for its future standard definition) of a Banach operator may be discrete or continuous (as $\phi:f(x)\in C^0([0,1])\rightarrow xf(x)\in C^0([0,1]))$. Therefore, the so called spectrum has some points in common with the spectrum of physicists. Moreover these two notions give each one a lot of information about the studied object.

Anyway, in english, "spectrum" is used -for operators- from 1948. Since in finite dimension, the spectrum reduces to the set of eigenvalues, the word "spectre" is used in France -for the matrices- from 1964; on the other hand, "spectrum" is pronounced faster than "the set of eigenvalues"!!


Eigenstate of a bounded operator $\hat{\cal L}$ on a complex Banach space is a vector $v_i$ such that: $$ \hat{\cal L}v_i=\lambda_i v_i, $$ where the complex number $\lambda_i$ is the eigenvalue associated with the eigenvector. Mathematical spectrum is the multiset of the operator eigenvalues.

The physical spectrum is somewhat different. To start with, the physical operators are self adjoint, so that all eigenvalues are real. The eigenvalues of Hamiltonian build together the energy spectrum, which is however not the observable physical spectrum.

The latter is not the energy eigenvalues themselves but differences between them: $\hbar\omega_{ij}=|\lambda_i-\lambda_j|$. Assume that a physical system has $n$ low energy eigenvalues the rest being separated by so large gap that the transitions to these higher energy states cannot be excited. What one would observe are however not the $n$ energy eigenvalues, but $\frac{n(n-1)}{2}$ spectral lines corresponding to transitions between the states, provided that all differences between eigenvalues are distinct and all transitions are allowed. The latter is however rarely the case. For linear harmonic oscillator for example the differences between the closest energy eigenvalues are the same and only transitions between these states are allowed. Thus the observable spectrum would consist of a single line $\hbar\omega$, whereas the energy spectrum is $\lambda_n=\left(n+\frac{1}{2}\right)\hbar\omega$.


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