The word mode pops up in many fields of physics, yet I can't remember ever encountering a simple but precise definition.

After having searched fruitlessly on this site as well, an easy to find place with (one or more) good answers seems in order.


Ideally, answers should give an intuitive and easy-to-remember definition of what a mode is, preferably in a general context. If limitation is necessary for a detailed answer, assume a context of theoretical physics, e.g. mode expansions in quantum field theory.

  • $\begingroup$ I have always thought the key is the energy: a mode is a... mode of oscillation whose energy cannot be transferred to other modes. As far as oscillations go, in a mode all points oscillate with the same frequency - and this is what lead to eigenvalues problem (or, if you like to see it the other way around, you find modes by solving eigenvalue problems). $\endgroup$
    – Peltio
    Oct 13, 2015 at 3:04
  • $\begingroup$ What's going on with this question? Do you need a more intuitive and less mathematical explanation? $\endgroup$
    – DanielSank
    Oct 19, 2015 at 16:34
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    $\begingroup$ Cool, and no worries. I periodically look through questions I've answered which have no accepted answers and ask the OP if they need more info. I figure either I can improve my answer or they just forgot :) $\endgroup$
    – DanielSank
    Oct 19, 2015 at 16:53

3 Answers 3


In a very mathematical sense, more often than not a mode refers to an eigenvector of a linear equation. Consider the coupled springs problem $$\frac{d}{dt^2} \left[ \begin{array}{cc} x_1 \\ x_2 \end{array} \right] =\left[ \begin{array}{cc} - 2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & - \omega_0^2 \end{array} \right] \left[ \begin{array}{cc} x_1 \\ x_2 \end{array} \right]$$ or in basis independent form $$ \frac{d}{dt^2}\lvert x(t) \rangle = T \rvert x(t) \rangle \, .$$ This problem is hard because the equations of motion for $x_1$ and $x_2$ are coupled. The normal modes are (up to scale factor) $$\left[ \begin{array}{cc} 1 \\ 1 \end{array} \right] \quad \text{and} \quad \left[ \begin{array}{cc} 1 \\ -1 \end{array} \right] \, .$$ These vectors are eigenvectors of $T$. Being eigenvectors, if we expand $\lvert x(t) \rangle$ and $T$ in terms of these vectors, the equations of motion uncouple. In other words

The set of normal modes is the vector basis which diagonalizes the equations of motion (i.e. diagonalizes $T$).

That definition will get you pretty far.

The situation is the same in quantum mechanics. The normal modes of a system come from Schrodinger's equation $$i \hbar \frac{d}{dt}\lvert \Psi(t) \rangle = \hat{H} \lvert \Psi \rangle \, .$$ An eigenvector of $\hat{H}$ is a normal mode of the system, also called a stationary state or eigenstate. These normal modes have another important property: under time evolution they maintain their shape, picking up only complex prefactors $\exp[-i E t / \hbar]$ where $E$ is the mode's eigenvalue under the $\hat{H}$ operator (i.e. the mode's energy). This was actually also the case in the classical system too. If the coupled springs system is initiated in an eigenstate of $T$ (i.e. in normal mode), then it remains in a scaled version of that normal mode forever. In the springs case, the scale factor is $\cos(\sqrt{\lambda} t)$ where $\lambda$ is the eigenvalue of the mode under the $T$ operator.

From the above discussion we can form a very physical definition of "mode":

A mode is a trajectory of a physical system which does not change shape as the system evolves. In other words, when a system is moving in a single mode, the positions of its parts all move with same general time dependence (e.g. sinusoidal motion with a single frequency) but may have different relative amplitudes.

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    $\begingroup$ Good answer. I think it is worth mentioning that you sometimes still call it a mode if you do perturbation theory and it was an eigenmode of the original unperturbed system. $\endgroup$
    – Timaeus
    Oct 12, 2015 at 16:53
  • $\begingroup$ @Timaeus Indeed I thought about how best to explain that. I suppose it's best to say "mode of the unperturbed system". Would you agree? $\endgroup$
    – DanielSank
    Oct 12, 2015 at 17:08

The Free Dictionary definition of "mode" in the context of physics is "any of numerous patterns of wave motion or vibration". However, this definition seems to be over-broad and imprecise. Mode can be divided into normal mode and quasi-normal mode. A normal mode is a time-independent oscillation where the frequency and shape of the wave are invariant with time. A quasi-normal mode is a perturbation of a field where the frequency and shape change with time.

Chapter 49 of the Feynman Lectures on Physics discusses modes as different results obtained when confining waves in various ways within some finite region.

In general, propagating waves are classified according to modes of propagation. Sound waves, for example, may result in various types of cyclical movement of particles as a wave passes through a medium. The mode can be determined by properties of the medium as well as by frequency of the wave.

Whenever you are dealing with an oscillation or vibration or other regular repetition of movement, you can classify it as exhibiting one or another mode of motion. When you have "collective motion of many individual particles", exhibiting wave like movement, classification according to modes may be an appropriate way to investigate and classify such phenomena.

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    $\begingroup$ "Mode" should refer specifically to oscillations who's shapes are time independent. Saying "any of numerous patterns of wave motion or vibration" seems to encompass far too much. $\endgroup$
    – DanielSank
    Oct 12, 2015 at 17:41
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    $\begingroup$ @Ernie Thanks for this answer. Collecting definitions from different sources is a great way to form your own intuitive picture. I especially liked the link to the Feynman lectures. His idea of a mode sounds like a classic Feynman: very descriptive and applied way of defining the word. However, the first definition, i.e. the one found in the free dictionary, is a great example to support the claim I made in my question: many definitions for the word mode are simply imprecise or even borderline vacuous. $\endgroup$
    – Casimir
    Oct 13, 2015 at 6:24

I will try it more intuitively. One of the most fundamental aspects of (not only) physical picture of the world is decomposing complexity into simpler parts. And it is even better when the pieces of the puzzle makes the picture whole without gaps or overlaps.

Such a thing (not only) for oscillations is the orthogonal modes. What does the words mean? "Mode" is "a possible way to do things" and "orthogonal" actually means that the pieces of the puzzle are not overlaping. By "cooperation" (e.g. linear combination, fourier coefficients...) of these "independent ways to do things" the complex motion is described. The point: we decompose an ugly oscillatory movement into some "nicer and more comprehensible parts".

The most famous example is coupled oscillations. If you look at the system by measuring the positions of the masses measured from solid wall, you will see a mess. But! If you see it as a combination of center of gravity and masses relative motion, then the simplicity occurs.


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