0
$\begingroup$

Is it valid to treat a matter wave as a light wave with wavelength equal to the de Broglie wavelength of the matter wave? Either way please can you explain why?

$\endgroup$
  • 4
    $\begingroup$ No, because the group velocity of the matter wave will always be less than $c$, while the group velocity of light is always equal to $c$. $\endgroup$ – John Rennie May 6 '15 at 10:05
2
$\begingroup$

It's not valid to treat light and matter waves alike. Why? Apart from the obvious reason that they are not the same (we can after all see light distinctly from matter), the two have different equations of motion - the (non-relativistic) "matter wave" obeys the famous Schrodinger equation, while classical light waves go around according to Maxwell's equations (or the standard wave equation).

There are many differences because of this:

  1. A 'matter wave' is (unavoidably) complex valued, unlike a light wave, where the electric and magnetic fields are always real valued.
  2. The frequency of a matter wave goes as the inverse square of its wavelength, and that of a light wave goes as the inverse of its wavelength. This means that while light of all wavelengths has the same speed $c = \nu\lambda$, the speed of a matter wave depends essentially on its wavelength/frequency. (In addition to John Rennie's point - the phase velocity of a matter wave is half its group velocity, while they are the same for light).
  3. There can always be a time when there is no light in the universe, as light waves can be absorbed and emitted; but the total "intensity" of a matter wave must always be a constant for all time (this follows from the probability interpretation).
  4. A rather famous example is the case of refraction: when light goes into a region where it slows down, it bends towards the normal; when matter goes into a region of higher potential (where it slows down), it bends away from the normal to the surface of the region: this is because in the latter case, only the component of velocity along the normal reduces. This is once again because the wavelength increases with increasing speed for a light wave, but decreases with increasing speed for a matter wave. In fact, as the dependence of wavelength on speed is reciprocal between the two cases, matter waves obey the "inverse" of Snell's law.

Finally, there is only ever one electromagnetic field in space at any point of time, and all these waves are disturbances on that field. These disturbances all add up. When there are many ("entangled") particles, however, the wave equation itself is different (a many-particle Schrodinger equation) and it is incredibly hard to express the wave function as a simple "sum" of many matter waves, one for each particle. Instead, matter waves are not waves in real space anymore, but simply a function of various properties of a material system.

$\endgroup$
  • $\begingroup$ "The frequency of a matter wave goes as the inverse square of its wavelength, and that of a light wave goes as the inverse of its wavelength. " how do you conclude this? see physics.stackexchange.com/q/62522 $\endgroup$ – anna v May 6 '15 at 13:14
  • $\begingroup$ @annav For a light wave, $\lambda\nu = c$, a universal constant, right from the wave equation. Similarly, for a matter wave, $\lambda^2\nu = \frac{h}{2m}$. I'm not talking about a fixed group velocity here. Am I missing something else? $\endgroup$ – AV23 May 6 '15 at 15:49
  • $\begingroup$ from the link in my comment it should be the velocity on the right or lamdanu. the photon and the particle are consistent when mass=0 . I do not see where you get the lamda squared, and this at the limit m=0 does not give 1 (where c=1). I get lnu=sqrt(m^2+p^2)/p $\endgroup$ – anna v May 6 '15 at 16:02
  • $\begingroup$ it may be correct, if one substitutes but it is not the analogue of the photon relation. $\endgroup$ – anna v May 6 '15 at 16:07
  • $\begingroup$ @annav I guess I should have mentioned in the answer that I am considering the non-relativistic case, with "matter wave" referring to the single-particle wavefunction in the position representation. I will edit to include this. $\endgroup$ – AV23 May 6 '15 at 16:20
0
$\begingroup$

I think this is very commonly confused topic about this wave-particle thing. Let me clarify. Like the post above indicates, from Maxwell's Equations, we find evidence, in a vacuum, for the electric and magnetic fields to travel together at the speed of light, perpendicular to each other. This is simply the solution of the well known equations which govern the electric and magnetic fields. We call this an electromagnetic "light" wave, referring to light. Now, at the same time, we find evidence we don't really know where particles are, such as electrons. They actually have a probability of being in space. The evolution of this probability is governed by Schrödinger's equation. I refrain from using "matter wave" and instead use probability. Thus the evolution of the probability of a particle and a light wave are totally different things, talking about different objects, governed by different equations and unnecessarily confused in our terminology.

P.S. When we speak of wave-particle duality, be careful to use in the right context. When we speak of this term, we mean the nature of light to behave as a particle and an electromagnetic wave, nothing more: 1. light -> evidence suggests light exists as both a particle (photon) and an electromagnetic wave. We are not talking about a probability here. 2. particles -> particles have a probability of being in space. Forget using this term "wave-particle duality" when speaking about particles. They fundamentally have a probability of existing in space and you can never know where both their position and velocity are at any moment.

Hope you have a good one!

$\endgroup$
0
$\begingroup$

Actually both are analogous. When you quantize a single mode which is simply an oscillation associated with a single photon, you adopt the Lorentz model where you imagine your single photon as a mass on a spring, then you translate your quantum mechanical Hamiltonian that includes both position and momentum to phase space where the electric and magnetic fields will be your new canonical variables equipped with the same uncertainty relation that governs the relation between position and momentum. You can also interpret the canonical variables associated with your single photon as an AM and FM signals with the same uncertainty relation described before, since you can always switch back and forth between them using fourier transform.

$\endgroup$

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