Skip to main content

plotting - how to plot the dispersion curve in first brillouin zone?


Those who know some solid state physics should know what the first brillouin zone is. How do I plot the dispersion relation in the 1st brillouin zone so that the curves can 'fold back'?


For instance the code I have now is:


OmegaP=8; 
epM[Omega_]:=1-OmegaP^2/Omega^^2;
epD[Omega_]:= 1;

e2alpha[Omega_]:=(epM[Omega]-epD[Omega])/(epM[Omega]+epD[Omega])

ContourPlot[{Exp[0.5*ak]==e2alpha[Omega],Exp[0.5*ak]==-e2alpha[Omega]},{ak,0,0.5},{Omega,0.05,8}];

I have rescaled the axis so that 1st Brillouin zone is 0-0.5, 2nd Brillouin zone is 0.5-1.0, and so on. The above above only generates the lower part of the whole dispersion relation. If I extend the range of ak to, say {ak,0,2.5}, then I get a lot more missing curves, but how to I 'fold' the part from 0.5 to 2.5 into the 1st Brillouin zone?


PS: Please click here (last page, Fig.16) for an image showing what the '1st Brillouin zone' is.



Answer



The "folding in" of the reduced zone scheme is achieved by translating the center of the energy function, placing one center on each brillouin zone edge:


Plot[
{x^2, (x - Pi)^2, (x + Pi)^2, (x - 2 Pi)^2, (x + 2 Pi)^2},
{x, -Pi/2, Pi/2},
AspectRatio -> GoldenRatio,

PlotStyle -> Black
]

Mathematica graphics


I plotted here a function proportional to $x^2$ as would be the case for free electrons. I used AspectRatio only to compensate for not having the right constant of proportionality, $\frac{\hbar^2}{2m}$. The centers were taken to be $\pm \pi n$ but should be $\pm \frac{\pi n}{a}$ for the actual case, where $a$ is the lattice constant.


To plot an arbitrary number of energy bands without having to type in the functions manually you can generate them using Table:


Plot[
Flatten@Table[{(x + Pi n)^2, (x - Pi n)^2}, {n, 0, 10}],
{x, -Pi/2, Pi/2},
AspectRatio -> GoldenRatio,

PlotStyle -> Black
]

For your dispersion relation, maybe something like this could work:


eqns = Flatten@Table[{
Exp[0.5*(ak - n Pi)] == e2alpha[Omega],
Exp[0.5*(ak - n Pi)] == -e2alpha[Omega],
Exp[0.5*(ak + n Pi)] == e2alpha[Omega],
Exp[0.5*(ak + n Pi)] == -e2alpha[Omega]
}, {n, 0, 5}];


ContourPlot[Evaluate[eqns], {ak, 0, 0.5}, {Omega, 0.05, 8}]

Mathematica graphics


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...