[Physics] About de Broglie relations, what exactly is $E$? Its energy of what

quantum mechanics

Well, you may know de Broglie relations, here they are:
$$
E = h\nu, \quad\quad p = \frac{h}{\lambda}
$$

My question is simple:
What exactly is $E$? Is it the total energy? Maybe only kinetic energy? Other thing?


Testing consistence.
$$
p = \frac{h}{\lambda} = \frac{h\nu}{c} = \frac{E}{c}\quad\implies\quad
E = pc
$$

This used only the de Broglie relations. The equation $E = pc$ holds for a photon. But de Broglie relations is not about photons, is it? Its about anything. Matter and waves. Its about matter waves! Lets then start generalizing it to matter.


Hypothesis 01: $E$ is the relativistic total energy.

A reasonable hypothesis. Wikipedia will also agree. This leads to a matter wave with phase speed $c/\beta$, and with group velocity of $v$. This explicitly means, I cannot go non-relativistic, stating that $E = 1/2 m v^2$. The least I can do to approximate things up is having:
$$
E \approx m_0c² + \frac{1}{2}m_0v^2
$$

i.e., the energy is the kinetic energy (which I approximated), plus the rest energy. However, if I am too slow, then $E\approx m_0 c^2$. I then have:
$$
m_0 c^2 = pc \quad\implies\quad
m_0 c = p
$$

Odd momentum, isn't it? If we go fully relativistic:
$$
E = m_0 c^2 = pc = \gamma m_0 vc \quad\implies\quad
c = \gamma v
$$

Just odd. Just odd. And of course, wrong. Hypothesis disproven. Evidently thus, wikipedia is wrong. A different way to clearly see this, its $E = \sqrt{(m_0 c^2)^2 + (pc)^2}$ from relativity is evidently different from $E = pc$ from de Broglie relations (derived in the beginning). Therefore, $E$ cannot be the total energy of the particle, and hypothesis is proven wrong.


Hypothesis 02: $E$ is the kinetic energy.

Some books lies in here (chapter 3 if you most know). We now have $E = \Delta m c^2$, which for low speeds resumes to $E = 1/2 mv^2$. Now, phase velocity is $v/2$ and group velocity is $v$. In addition:
$$
E = \frac{1}{2}mv^2 = pc = mvc \quad\implies\quad v = 2c
$$

And in general, i.e., in the relativistic case:
$$
E = \Delta m c^2 = pc = \gamma m_0vc \quad\implies\quad
\Delta m c = (m – m_0)c = p = mv
$$

Wrong! Hypothesis disproven.


Hypothesis 03: $E$ has nothing to do with energy of the matter

A reasonable assumption given hypothesis 01 and 02 disproven. Its natural then to assume $E$ is the energy of the matter wave (associated with the wave only), which is identical to the quantity $h\nu$.

Now.. what is phase velocity? Should be $E/p$. We use $E = pc$ from beginning to finally show the phase velocity: $v_p = c$. Now, the group velocity:
$$
v_g = \frac{\partial\omega}{\partial k}
= \frac{\partial E}{\partial p}
= \frac{\partial}{\partial p}\left(pc\right) = c
$$

So.. $v_p = v_g = c$. How unusual… all matter has speed $c$. Wrong!


Well.. well… what in the world is $E$ after all?

Best Answer

  • For your initial check, you used the formula $c = \nu \lambda$, giving the incorrect formula $E = pc$. This is wrong, matter waves don't travel at the speed of light; instead, $\nu \lambda$ is equal to the phase velocity $v_p$. You then use $E = pc$ in all the other derivations, making them all wrong.
  • Hypothesis #1 is correct for relativistic quantum mechanics.
  • Hypothesis #2 is correct for nonrelativistic quantum mechanics.
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