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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 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?

my2cts
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Guerlando OCs
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    Ordinary mechanical waves carry momentum, too. See "Water waves" in https://www.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. – probably_someone Jun 06 '18 at 21:59
  • 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 $p=\gamma mv$ – PM 2Ring Jun 06 '18 at 22:11
  • @PM2Ring Given that this is a quantum mechanics course (i.e. non-relativistic) that probably won't be important for a while. – probably_someone Jun 06 '18 at 22:20
  • 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? – Alfred Centauri Jun 06 '18 at 22:37
  • 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: https://en.wikipedia.org/wiki/Momentum_operator – Steven Sagona Jun 06 '18 at 23:17
  • @StevenSagona isn't the wave function about finding the probability of the particle being at $x$ in time $t$? Why did you say veloctity? – Guerlando OCs Jun 07 '18 at 00:21
  • @AlfredCentauri that was the only thing I could think of, but it's wrong. I don't know how to think of an wavefunction for an electron. I know what a probability distribution is, but I don't know: why a wavefunction is complex and what it represents, what really is momentum in a wave. – Guerlando OCs Jun 07 '18 at 00:29
  • @stevensagona I completely understood what you said about the particle having velocity. It really has. And it also has mass, so given a wave I can easily calculate velocity and then multiply by $m$. However, there is the concept of the momentum of a wave. What is it? I cannot understand – Guerlando OCs Jun 07 '18 at 05:40
  • You could try to derive what this property is (the "momentum of the wave" in the wavefunction) similar to how momentum is derived for the wave of an E-field (https://physics.stackexchange.com/questions/114908/how-do-the-electric-or-magnetic-fields-contain-momentum), but this is not what is conventionally meant by the momentum wavefunction in quantum mechanics – Steven Sagona Jun 07 '18 at 17:13
  • @probably_someone in that PDF, it still didn't explain what exactly is the momentum of a wave, neither where did the momentum $P=hk$ came from. In fact it just assumes that momentum = energy/phase and applies to several waves (classical and quantum) – Guerlando OCs Jun 08 '18 at 01:08
  • Which MIT lecture? What does it say precisely? – my2cts May 24 '21 at 12:15

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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.

Árpád Szendrei
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  • What's free? I agree that this is the momentum of a particle, but what about the momentum of a wave? Why it is $P = hk$? – Guerlando OCs Jun 08 '18 at 01:09
  • 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. – Árpád Szendrei Jun 08 '18 at 01:13
  • 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. – Árpád Szendrei Jun 08 '18 at 01:14
  • $p=hk$ is not for classical waves. $k$ is planck's constant – Guerlando OCs Jun 08 '18 at 01:14
  • you are talking about the wave function. – Árpád Szendrei Jun 08 '18 at 01:14
  • and the wave packet. that is what I wrote my equation for. If you would like the integral type equations, please read here: http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/WaveFunction.pdf – Árpád Szendrei Jun 08 '18 at 01:15
  • The whole point of this thing is that "we have constucted a wave function in the form of a wave packet that is particle-like in nature" – Árpád Szendrei Jun 08 '18 at 01:18
  • "we are in the position to make a reasonable guess at a mathematical expression for the wave associated with a particle." – Árpád Szendrei Jun 08 '18 at 01:20
  • And so they are doing a wavefunction for a particle-like wavepacket and using De Broglie for Energy E and momentum p and wave of frequency f and wavelength λ. This is where k=p/h comes from. – Árpád Szendrei Jun 08 '18 at 01:23
  • Where you are confused is that you think it is just for a wave. They are instead using a particle-like wavepacket and from that they are trying to use the particle-like momentum on the entire wave. – Árpád Szendrei Jun 08 '18 at 01:32