I'm studying quantum physics from MIT lectures and there's a concept that they alredy start with: momentum of a wave.

Given the wave-particle duality, I can kinda imagine that momentum is possible to define, since the electron has mass and it's travelling somehow as a wave. So the only possible interpretation for momentum of a wave that I can think of is:

By saing that a wave has momentum $p$ we're actually saying that an electron with mass $m$ will have velociy $v = m/p$ in that wave (since $p=mv$).

Is my definition at least near of what it's supposed to mean?

  • 1
    $\begingroup$ Ordinary mechanical waves carry momentum, too. See "Water waves" in math.nyu.edu/faculty/peskin/papers/wave_momentum.pdf. No quantum physics required for this particular subject - rather, it's entirely a classical (though somewhat nontrivial) concept. $\endgroup$ – probably_someone Jun 6 '18 at 21:59
  • $\begingroup$ BTW, $p=mv$ is just a low speed approximation (unless you're using relativistic mass, which you shouldn't do). The full relativistic version is $p=\gamma mv$ $\endgroup$ – PM 2Ring Jun 6 '18 at 22:11
  • $\begingroup$ @PM2Ring Given that this is a quantum mechanics course (i.e. non-relativistic) that probably won't be important for a while. $\endgroup$ – probably_someone Jun 6 '18 at 22:20
  • $\begingroup$ An 'electron wave' in this context probably refers to the wavefunction for an electron which is not a material wave in space and time but, rather, a complex valued 'probability amplitude wave' in configuration space. Are you picturing the electron as a point particle with definite velocity embedded in some kind of wave in ordinary space? $\endgroup$ – Alfred Centauri Jun 6 '18 at 22:37
  • $\begingroup$ if you're willing to accept that a particle can have a velocity, then momentum is just m times that velocity. Since in quantum mechanics there's a distribution (wavefunction) associated with finding different velocities, to find the momentum distribution then you just multiply that velocity distribution at m. If you want a more precise definition of momentum (in terms of position), it's defined here: en.wikipedia.org/wiki/Momentum_operator $\endgroup$ – Steven Sagona Jun 6 '18 at 23:17

Here you mean kinetic energy. Since it is a free particle.

E^2 = (pc)^2 + (mc^2)^2

\begin{align} E^2 &= \left(\frac{hfc}{v}\right)^2 + (mc^2)^2 \\ K &= -mc^2 + \sqrt{\left({hfc}/{v}\right)^2 + (mc^2)^2} \end{align}

It is the momentum of the free electron wave.

  • $\begingroup$ What's free? I agree that this is the momentum of a particle, but what about the momentum of a wave? Why it is $P = hk$? $\endgroup$ – Guerlando OCs Jun 8 '18 at 1:09
  • $\begingroup$ This means that you cannot use this equation for a bound electron around a nucleus in a certain energy level as per QM. This equation is only for a free electron wave packet. That is for the equation I wrote for a free electron wave packet. What you wrote, P=hk is for a classical wave, like sound or water waves. $\endgroup$ – Árpád Szendrei Jun 8 '18 at 1:13
  • $\begingroup$ Substituting p = hk, this becomes vg = p/m. i.e. the packet is indeed moving with the velocity of a particle of momentum p, as suspected. This is a result of some significance, i.e. we have constructed a wave function of the form of a wave packet which is particle-like in nature. $\endgroup$ – Árpád Szendrei Jun 8 '18 at 1:14
  • $\begingroup$ $p=hk$ is not for classical waves. $k$ is planck's constant $\endgroup$ – Guerlando OCs Jun 8 '18 at 1:14
  • $\begingroup$ you are talking about the wave function. $\endgroup$ – Árpád Szendrei Jun 8 '18 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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