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.)?


  • $\begingroup$ The textbooks should cover this. The appropriate quantity is $\nabla_kE$ $\endgroup$ Commented Feb 1 at 0:00
  • $\begingroup$ This should give you some intuition about the de Broglie wavelength - The more general uncertainty principle, regarding Fourier transforms $\endgroup$
    – mmesser314
    Commented Feb 1 at 0:09
  • $\begingroup$ 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. $\endgroup$ Commented Feb 1 at 2:58
  • $\begingroup$ 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. $\endgroup$
    – intraband
    Commented Feb 1 at 3:25

1 Answer 1


The relation between (de Broglie) wavelength and wave vector of a sinusoidal wave is always πœ† =2πœ‹/π‘˜. Thus your formula πœ† = 2πœ‹π‘šβˆ—/π‘˜ cannot be correct. The effective mass in it makes it dimensionally incompatible. In general, at any k, in a crystal band structure with Bloch waves and a given dispersion relation E(π‘˜)/ℏ = πœ”(k) the effective quantum mechanical electron velocity is given by the group velocity dπœ”(k)/dk. The phase velocity of the wave is πœ”(k)/k.The phase velocity is the wave velocity relevant for the wavelength. Your quasi-classical formula (effective mass approximation) for the electron velocity v is correct, it gives the group velocity of the electron near the conduction band minimum.

The effective mass model approximates the Bloch wave dispersion relation at extrema, in particular at the minimum of the conduction band and the maximum of the valence band where it is mostly used. In the Kane model approximation, the effective mass ist not constant, you have to use the group velocity in this formula for the electron velocity.


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