What Are These Energy Band Graphs Actually Showing?

AI Thread Summary
The discussion centers on understanding energy band graphs, specifically their axes and the meaning of negative curves. The graphs represent energy as a function of the electron's wave vector, with the upper and lower graphs indicating conduction and valence bands, respectively. The choice of zero energy is arbitrary, which explains the presence of negative values in the curves. Additionally, the relationship between electron momentum and wave vector is clarified, emphasizing that momentum could also be represented on the x-axis. The conversation highlights the complexities of interpreting energy dispersion in band theory.
WolfOfTheSteps
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Homework Statement



Here is the problem:
http://img413.imageshack.us/img413/748/64my8.th.jpg

The Attempt at a Solution



I got the effective mass. That's trivial. My question is, what are these graphs actually of? The horizontal and vertical axes are not labeled.

Is it an energy dispersion? If so, how come some of the curves are negative? All throughout the book, there is not a single energy dispersion curve with negative branches!

It can't be the effective mass vs K, because the second derivative of E with respect to k is not a parabola. And besides, the question "which band has the higher effective mass" would not make much sense. I know that on the dispersion curve, the narrower bands have lower effective mass. But the negative bands are just throwing me off.

I'm really frustrated. Please help if you can!

Much thanks!
 
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It represents energy as a function of wave vector of the electron. The upper and lower graphs are the conduction and valence bands of the material, respectively. The choice of where zero energy is chosen is arbitrary, really.

also, note that the electron momentum is proportional to the wave vector. this is simply because k = 2pi/lambda, but lambda = h/p, where p is momentum. therefore, k = 2pi/(h/p) = 2(pi)p/h. Or, p = (h-bar)k. So, you could have momentum on the x-axis instead of wave vector, if you like.
 
Thanks, leright!
 
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