# Interpretation of velocity in the de Broglie wavelength of an electron in a crystal

The de Broglie wavelength of a free electron is $$\lambda = h/mv$$ whre m is the free electron mass, and v is the velocity. Often in introductory solid state physics literature (review articles, lower-level textbooks etc.) I've seen an analogous quantity $$\lambda = h/m^*v$$, where $$m^*$$ is the effective mass, used to try and articulate the "free" electron nature of electrons in bands. Often it is accompanied with a cartoon equating this crystalline de Broglie wavelength with the wavelength of the complex exponential envelope function in Bloch's theorem. This de Broglie wavelength is not a quantity I work with regularly and I'm unsure how much it is useful, versus a pedagogical device.

A few specific questions:

• How does one correctly interpret the velocity in the solid state case? One may simply obtain velocity from the standard definition of kinetic energy and the dispersion relation i.e. $$\frac{1}{2}m^*v^2 = \hbar^2k^2/2m^* \rightarrow v = \hbar k/m$$, but is this the right velocity?
• The above implies that the wavelength is then $$2\pi m^*/k$$. However, taken at face value, Bloch's theorem gives the envelope as simply $$\exp(ikx)$$ which has $$\lambda = 2\pi/k$$, differing by a factor of $$m^*$$. Given that $$m^*$$ is a crude way to incorporate the complicated periodic potential of the crystal into a simple quantity, it makes sense that it enters into the wavelength in an analogous fashion to the refractive index for light waves where $$\lambda = 2\pi n/k$$. But is that interpretation sound?
• Finally, can we extend these interpretations to nonparabolic bands e.g. Kane models, where the notion of effective mass becomes more nuanced (drift-diffusion interpretation, band curvature interpretation, etc.)?

Thanks.

• The textbooks should cover this. The appropriate quantity is $\nabla_kE$ Commented Feb 1 at 0:00
• This should give you some intuition about the de Broglie wavelength - The more general uncertainty principle, regarding Fourier transforms Commented Feb 1 at 0:09
• I think you answered your own question with the last point. In some cases it is a useful approximation and in many others it's not. I don't think it works well in case of scattering dominated phenomena, like conduction, does it? My last solid state physics class was many decades ago. Commented Feb 1 at 2:58
• Thank you @naturallyInconsistent, you are correct (but missing a factor of $1/\hbar$), and applying this to a parabolic band gives the same answer I have in bullet one, $v = \hbar k/m$. I am curious more about the physical interpretation of this velocity, and what analogy (if any) can be drawn with the velocity of a free electron. Commented Feb 1 at 3:25