How would quantum mechanics explain doppler effect?

And just for curiosity, is there any effect similar to doppler effect occuring at quantum level?

  • $\begingroup$ Perhaps related here. $\endgroup$
    – hhh
    Sep 5, 2012 at 13:55

2 Answers 2


The (non-relativistic) Doppler effect is the result of a Galilean transformation, and non-relativistic quantum mechanics is invarient under Galilean transformations so systems described with QM automatically show Doppler effects. There isn't any sense in which QM has to "explain" the Doppler effect.

The same applies to relativistic (Lorentz) transformations, though here you'd need to use quantum field theory rather than the Schrodinger equation. Because QFT is invarient under Lorentz transformations you automatically get relativistic Doppler effects.


This is something that originally motivated quantum mechanics, and gave Planck's quantization law as natural. If you have a moving light source, and the stationary light source emits a radio-pulse of frequency $\omega$, if you boost the thing so that it is moving in the direction of the outgoing pulse, the frequency and energy have the same transformation law.

The reason is that both the energy and the frequency make four-vectors (or four-covectors if you prefer), since the wave-number $k_\mu$ is dotted with $x^\mu$ to make the phase of the wave, and the energy and momentum transform together as a four-vector. This means that the quantum condition

$$ E = \hbar \omega$$

is consistent with relativity (and therefore with the Doppler shift) only if

$$ p =\hbar k$$

and this is how Einstein deduced that photons have momentum that obeys deBroglie's relation. DeBroglie deduced that this is true of matter particles, and that the old-quantum condition

$$\int p dx = n h$$

is the condition that there are an integer number of wavelengths that fit along the classical orbit.

In quantum mechanics, the Doppler shifts changes the energy and momentum of the quanta and it changes the wavenumber and frequency, but it doesn't change the ratio of the energy to the frequency, or the momentum to the wavenumber, so the photon picture makes sense.

  • $\begingroup$ hi @Ron in your answer physics.stackexchange.com/a/29361/1814 . "Galilean invariance of the Schrodinger equation: this part is not done in any book, I think only because Dirac omitted it from his. It is essential to know how to boost wavefunctions. Since Feynman derives the Szchrodinger equation from a tight-binding model (a lattice approximation), the Galilean invariance is not obvious at all." -> I think you need to add the fact that specifically, we are doing a lorentz boost to the canonical momentum in the wavefubction? $\endgroup$
    – pcr
    Jun 6, 2012 at 6:40
  • 1
    $\begingroup$ @pcr: There is no wavefunction in my answer here--- this is about the consistency of quanta. When you do Lorentz boosts of wavefunctions, you just do a naive Lorentz boost inserting an additional quadratic phase factor. This can be understood from the path integral, or by an infinitesimal boost generator. Regarding relativistic wavefunctions for particles, the concept is barely sensible, since the particles go back in time. You can define it on asymptotic particles, or perhaps using a proper-time formalism, but the results are not physically obvious because of the back and forth in time. $\endgroup$
    – Ron Maimon
    Jun 6, 2012 at 8:17

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