Apparently no one has asked why the word "amplitude" is used to refer to the complex numbers that determine the probability of a quantum object to be found in a certain state. The word amplitude to me means the distance between the lowest point on a graph and the highest point. But I don't know how to apply this use of the word to the amplitudes of a quantum object. Can you explain? I've been looking at various books and I suspect everyone just assumes the meaning is clear somehow because nobody seems to explain. If you could cite a source, that would be even more helpful. Thank you.

  • 3
    $\begingroup$ Possible line of reasoning: 1) A wavefunction looks like a wave 2) waves have amplitudes 3) finite-dimensional quantum states are kind of like discrete wavefunctions therefore 4) numbers describing quantum states are called amplitudes $\endgroup$
    – forky40
    Nov 10, 2023 at 1:57

2 Answers 2


It's an interesting question as to who first coined the term "amplitude" in the context of quantum mechanics, for the coefficients in wavefunctions - this forum suggests that it was Schrödinger himself (although I haven't found the primary source and it's probably in German).

Sine waves, such as AC signals or water waves or electromagnetic waves, are fully characterized in terms of their phase relative to time $t_0$ as well as their amplitudes, which, as you describe, is the difference between the maximum and the minimum. But certainly electrical engineers are familiar with Euler's formula relating sines and cosines to the exponential function - and they call the amplitudes the coefficient in front of the exponent. Schrödinger wrote down his equation, and saw that it was a wave equation - as @forky40 says it's a natural consequence to call the coefficients amplitudes.

Of note is Born's rule, which relates these amplitudes to the probability of finding the state in the corresponding eigenbasis. Born initially guessed that the probabilities are given by the absolute value of the amplitudes and then, in what's been described as one of the most famous footnotes in the history of science, corrected this to indicate that the probabilities are proportional to the squared absolute value.


Quantum mechanics has two different formulations which were later proven to be equivalent. Wave mechanics (Schrödinger) and matrix mechanics (Heisenberg). What we normally use is the matrix mechanics formulation where we talk about state vectors and linear combinations. The coefficients in a linear combination are then coefficients. Looking from the wave picture the basis vectors from matrix mechanics are waves (continuous formulation) and the coefficients just correspond to the amplitude if we formulate a wave as overlay/linear combination/superposition of waves. As Schrödinger came from the wave formulation and draw inspiration from classical electromagnetical waves using the term "amplitude" just made sense to him. And as Schrödinger and Heisenberg coined a lot of terms in QM the word Amplitude just was kept by later generations of scientists. Similar for Ansatz, eigenvector/eigenvalue (Eigenvektor) etc. The correct translation would be self vector or self value. Apparently linear algebra was not widespreadly known to all scientists back in the 1920s/1930s when QM came up.

  • $\begingroup$ What do you mean by "Similar for Ansatz, eigenvector/eigenvalue (Eigenvektor) etc. The correct translation would be self vector or self value."? Are you just saying that the German founders of QM liked to use linear-algebraic terms? Also, what is meant by "The coefficients in a linear combination are then coefficients" - this seems tautological? $\endgroup$ Nov 12, 2023 at 23:13
  • $\begingroup$ The word Ansatz is a German word which is directly imported and used in English as such. This is also true for the syllable Eigen (which means "self") in German. The correct translation in English would be self value or self vector (like in French vecteur propre and valeur propre). Instead English uses the word eigenvector or eigenvalue with the German syllable. $\endgroup$
    – sycramore
    Nov 14, 2023 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.