# What does it mean for an 'electron wave' to have momentum?

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?

• 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. – probably_someone Jun 6 '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 6 '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 6 '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 6 '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: en.wikipedia.org/wiki/Momentum_operator – 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.

• 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 8 '18 at 1:09
• 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 8 '18 at 1:13
• 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 8 '18 at 1:14
• $p=hk$ is not for classical waves. $k$ is planck's constant – Guerlando OCs Jun 8 '18 at 1:14
• you are talking about the wave function. – Árpád Szendrei Jun 8 '18 at 1:14