3
$\begingroup$

In de Broglie hypothesis, particles have wave nature. The question is does this wave have inertia? If so is it represented in the corresponding wave equation?

$\endgroup$
1
$\begingroup$

Let's first address the general question--do "waves" of any kind have (or can they have) inertia? I suppose here by "inertia" you mean "resistance to changes in velocity." This is certainly true of waves in, say, water--you've certainly felt resistance to your hand if you sweep it through water to make a wave; the destructive force of a tsunami is a more extreme example of the force a wave can transmit.

Even in classical electromagnetism, where the waves (light) are massless, they still have inertia. Consider, for a very direct example, a solar sail which reflects light from a source and acquires momentum from the rebounding light--we wouldn't expect this to happen if light waves had no inertia, since in that case presumably we could easily change their direction with no resistance.

Now, do quantum "matter waves" have inertia? This a more subtle question, because the wave should not be viewed as a "spread out" particle, but rather a probability amplitude for the particle to be measured as present at a particular point in space.

However, I think it should be clear that these waves do have inertia, by considering collision problems, just as we did for the previous two cases. We know that when two particles collide, they must conserve momentum. For instance, it is not possible that two particles of equal mass collide head-on and then both move in the same direction afterwards, and furthermore we know that this collision is extremely well-modeled quantum mechanically by Schrodinger's wave equations.

To get a more direct and mathematical answer, let's consider another definition of inertia, which is that an object has inertia if its velocity (and thus its momentum) remains unchanged unless acted on by an external force. For a massive, non-relativistic, particle, the Hamiltonian is given by: \begin{equation} H = \frac{p^2}{2m} + V(x) \end{equation} Here V(x) is our external potential ("force" is a difficult quantity to work with in quantum mechanics), $p$ is the momentum operator, and $m$ is the mass of the particle. Suppose there is no external force, so $V(x) = 0$ (or any constant really). Then the rate of change of the momentum operator using the Heisenberg picture (equivalent to the more traditional Schrodinger wave picture) is: \begin{equation} \frac{ \mathrm{d}{p}}{\mathrm{d} t} = \frac{i}{\hbar} \left[ H , p \right] = 0 \end{equation} This is because the momentum operator is only commuting with $p^2$, and operators always commute with their own squares. Therefore, momentum is unchanging, and we can see that the "principle of inertia" is indeed respected by the equations of quantum mechanics. This occurs in the equations essentially by the fact that the free-particle Hamiltonian commutes with momentum, and therefore if the particle is subject to this Hamiltonian the momentum will not change.

$\endgroup$
-4
$\begingroup$

TDSE

TISE

Here you have the time dependent and independent Schrodinger wave equations, respectively. These relate to the energy of particles, but the trident symbol, Psi is representative of the actual wave equation I believe you are referring to.

While De Broglie and schrodinger and others like them do describe particles as behaving like waves, they are referring primarily to probability functions, or wave equations describing where a particle is likely to be found. Meaning, in a given space, the particles and the "paths" they move in can be represented by waves, where you have crests (areas of high probability) and troughs (areas of low probability).

In short, particles CAN have mass and thus COULD have inertia, but the wave equation themselves do not as they are merely a representation of probability.

Hope this helps!

$\endgroup$
  • $\begingroup$ I'm sorry, but I have to downvote; there are several basic misunderstandings of QM in this answer. It's not "crests" and "troughs" that are high/low probability, as probability is given by the square of the wavefunction. It is not the case that "most particles move at or near the speed of light," and relativistic considerations are not the source of the uncertainty principle. You cite Schrodinger equations and say the equation doesn't have mass, but the $\hbar^2 / 2m$ term is precisely the mass and reflects the massive dispersion relation. $\endgroup$ – zeldredge Jul 13 '16 at 14:24
  • $\begingroup$ @zeldredge Sorry for the misunderstanding in the last statement, I meant the wave equations do not have inertia, but of course they have mass as you pointed out. I cited the Schrodinger equations as they are the ones I am most familiar with, but in retrospect they may not have been the best to cite as they are used for finding particle energy. Nonetheless, the symbol Psi represents the probability wave function as seen: wikimedia.org/api/rest_v1/media/math/render/svg/… This is the square of Psi and PSi*, yes but still uses wave and troughs. $\endgroup$ – Garrettfromhp Jul 13 '16 at 14:35
  • $\begingroup$ Also, I believe relativistic considerations are EXACTLY the source of the uncertainty principal, because particles move near the speed of light their positions are uncertain. But please, if I am completely incorrect, cite your sources. I am not trying to speak absentmindedly. $\endgroup$ – Garrettfromhp Jul 13 '16 at 14:37
  • 2
    $\begingroup$ It is better to put the equations inline using mathjax. $\endgroup$ – RedGrittyBrick Jul 13 '16 at 14:43
  • $\begingroup$ @RedGrittyBrick Thanks! That's actually very helpful. $\endgroup$ – Garrettfromhp Jul 13 '16 at 14:47

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.