# De Broglie waves and standing waves

I have a conceptual misunderstanding/confusion, for which I will give two examples to illustrate my problem.

Example 1: In Bohr's atom, de Broglie describes the atomic electrons as waves, where the length of the orbit is given by an integer number of wavelengths $$2\pi r = n\lambda$$ Here, the electron is described as a standing wave around the nucleus.

$$\lambda$$, being the de Broglie wavelength, is given by $$\lambda = \frac{h}{mv}$$ where $$v$$ is the particle's velocity.

Example 2: In solid state physics, the electron's matter wave undergoes a strong reflection (Bragg's reflection) on the boundaries of the Brillouin Zone. So the incoming and the reflected waves overlap and produce, again, a standing wave.

From the definition of the standing wave, it has group velocity $$v_g = 0$$. The velocity of a particle is given by its wave's group velocity.

My question is, how is de Broglie's matter-wave idea still valid in these situations of standing waves of zero group velocity.

• In quantum mechanics, the wave associated with a particle does not describe the trajectory of the particle, but rather, it describes the probability distribution of the particle’s position. Therefore, even though the group velocity of the standing wave is zero, this does not imply that the electron itself is stationary.
– user391340
Commented Jan 26 at 18:18

Finally, a point about terminology. In physics the term 'group velocity' is usually defined by the equation $$v_g = \frac{d \omega}{d k}$$ where $$\omega = 2\pi f$$ (where $$f$$ is the frequency) and $$k = 2 \pi/\lambda$$ (where $$\lambda$$ is the wavelength). The group velocity is the velocity that a wavepacket would travel at if there were a wavepacket. (Here a wavepacket is a type of wave motion involving a small range of frequencies and a finite length of the wave). You are correct to say that a standing wave just stays at one place but the term "group velocity" does not refer to the motion of a standing wave.
• You can always write $\sin(kx) = (\exp(ikx) - \exp(-ikx))/2i$, which means a standing wave can always be written as a combination of travelling waves. In the potential well if you look at any given energy eigenstate in terms of momentum, you find it does not have simply zero momentum, but rather a superposition of momenta in both directions. Commented Jan 27 at 12:12