Wavelength of matter waves at speed equal to zero.

aleemudasir

According to de-broglie's equation λ=h/p, so a/c to this equation what would be wavelength of particle at zero speed?

Bill_K

aleemudasir, I think you know the answer, but I'll say it anyway. Infinity. As p gets smaller and smaller, the deBroglie wavelength gets longer and longer.

Since an actual object will be in practice confined to some finite volume, this implies that p can never go all the way to zero. And so a real object can never be completely at rest.

Drakkith

Isn't this ignoring the relation f=E/h, which includes the energy content of the particle? (Of which the energy content includes rest mass)

aleemudasir

Yeah I knew that it would be infinity, but what I wanted to know was physical meaning of this, lets assume we have a ball of mass 1 kg with v=0, so how would would define the de-broglie's wavelength physically to a stationary observer looking at the ball.

Bill_K

You cannot have a ball of mass 1 kg with v = 0. This would imply that you have absolutely no idea where it is, it's just as likely to be somewhere on alpha centauri as it is on the table. If you do observe the ball sitting on the table then its deBroglie wavelength λ must be at least as small as the table, implying that the object has a small nonzero velocity ~ h/mλ.

aleemudasir

This is the point which I don't understand, lets assume I am an observer and I am at origin and ball(mass=1kg) isn't changing its co-ordinates with respect to origin, what does that mean, doesn't it mean, the body is at rest w.r.t origin?

If it means yes, then its dx/dt=0, which means v=0 and by equation λ=h/mv, λ should be infinity, which means I shouldn't be able to see the ball as I see it on the floor and it should be spread over an infinite area but that doesn't happen instead it is spread over a finite and small area!

If it means no, then how is the velocity ≠0?

chill_factor

a ball is not a quantum mechanical object. instead we can think about it as a collection of polymer molecules which are quantum mechanical and use statistical physics.

there are N (where N is a gigantic number like 10^23) polymer molecules bound to their equilibrium positions within the ball by electrostatic van der Waals forces. due to this potential, their wavefunctions are confined to finite volumes with a finite momentum. they oscillate about those equilibrium positions due to thermal energy (but also due to the zero point energy). because this motion is thermal, it is approximately uniform in magnitude (follows the Boltzmann distribution in this case) and random in direction, and thus the net displacement is zero for a very large collection of molecules.

The net force is what we perceive to be a motionless ball.

Let me just say it again: a ball is made up of polymers, a ball is not quantum mechanica, the polymer molecules that make up the ball are quantum mechanical.

aleemudasir

Thanks!
I got that.

aleemudasir

But what if instead of a ball having mass m I have an electron, and the rest of the problem is same as described earlier!

hefty

Hi Chill Factor,
Technically speaking, It's also true that a small ball has a chance to behave as a quantum object, right?
Meaning that it may totally disappear and appear somewhere else for example. (Probably We need to wait longer than the 1 zillion times the life of our universe, but it may happen, right?

Regards
Hefty

Whovian

Electrons don't have temperature. Temperature is the measure of the average kinetic energy of particles in an object that doesn't contribute to its velocity. In the case of an electron, the kinetic energy (in an arbitrary reference frame) all contributes to the electron's total velocity, therefore it's 0. (Or at least, I think. This is a bit less rigorous than I would like, but it works.)

aleemudasir

Does that mean electrons will never have zero velocity, as temperature can't be 0K(because at 0K volume becomes zero)?
What if I use another force to keep the electron stationary? Can I do that?
If yes, then what will happen according to λ=h/mv as said in the actual problem?
If no, then why?

scijeebus

There's all these hints in nature and physics that x/0 = infinity, why doesn't math just consider that and say it does equal infinity unless there is evidence to provide otherwise?
Individual particles can have temperature, the average would just be x/1. Kinetic energy is equivalent to thermal energy, so...

Not only that, but can't something have a velocity of 0 to a specific frame of reference? Look around you: can you see everything rotating on the surface of the Earth? You shouldn't be able to because of your frame of reference.

That "can" happen, but the reason it's improbable is because the ball isn't a quantum object itself, it is comprised of many quantum mechanical object's who's probability densities practically go to 0 at macroscopic distances.
Not only that, but treating a quantum particle merely as a wave itself is sort of out-dated. You might want to look into quantum field theory where particles can be described using the harmonic oscillations of fields.

Halcyon-on

poor de Broglie!

The solution is very simple, of course, treating quantum particles as waves. In the semiclassical limit m >> p the spatial wavelength tends to infinity lambda = h / p, the time periodicity tends to zero T = h /E. If you plot the modulo square of a matterwave in the semiclassical limit exp[-i (p^2 t/ 2 M - p x) / hbar] you will see that the wave is spatially localized inside the compton wavelength lambda-c = h /M c which tands to zero in this limit. In this way you obtain the classical pointlike particle and many other intersting things...

jtbell

A particle represented by a "pure" sinusoidal wave with an exactly precise wavelength (momentum), whether zero or any other value, is not physically possible. A real electron or other particle has to be represented as a wave packet: a superposition of waves with a range of momenta.

The closest thing you can get to a "stationary" electron in QM is a wave packet whose component waves span a range of momentum centered on p = 0. Half of the component waves travel in one direction, half travel in the opposite direction. (For simplicity I'm considering only one-dimensional motion here.) If the distribution of momentum is symmetric around p = 0, each wave with a particular +p is matched by a wave with a corresponding -p. In that case I'd expect the wave function of the packet to be basically a standing wave like you get with a vibrating string, with an overall amplitude that has a maximum at some location and decreases towards zero as you go away from that location in either direction. The spatial width is inversely proportional to the "momentum width" via the Heisenberg uncertainty principle.