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I have a single-rectangular potential barrier in graphene. The transmission $T(E,\theta)$ is given by $$T(E,\theta) = \left[1+\left(\frac{V}{\hbar v_Fk_x}\right)^2\tan^2\theta\sin^2(k_xD)\right]^{-1}$$ where $k_x=\sqrt{(E-V)^2-E^2\sin^2\theta}/\hbar v_F$. I would like to plot conductance $G(E)$ versus energy $E$, with $E_{min}=-1 \, \mathrm{eV}, E_{max}=1.5 \, \mathrm{eV}$, using the Landauer-Buttiker approach: $$G(E)=\int_{-\pi/2}^{\pi/2}T(E,\theta)\cos\theta d\theta.$$ The constants are $V=200\,\mathrm{meV}$, $D=30\,\mathrm{nm}$ and $v_F = 10^8\,\mathrm{cm/s}$. More details about the model can be found in Physica E, 61 (2014) 118.

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  • $\begingroup$ What are the units of E? The units of $k_x$ must also be meV, so how does one create a dimensionless quantity from 1 meV * 30 nm? $\endgroup$ – Carl Woll Sep 26 '18 at 20:24
  • $\begingroup$ In the original article, the formula contains the Planck constant and the characteristic velocity $v_F=10^8 cm/s$. $\endgroup$ – Alex Trounev Sep 27 '18 at 5:18
  • $\begingroup$ Exactly, I considered $\hbar=v_F=1$ for simplicity. $\endgroup$ – Alex Nascimento Sep 27 '18 at 15:47
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We reconstruct the original formula from the paper with dimensional units, as a result, an additional dimensionless coefficient $k$ appears in the formula, and the units of measurement of energy $\mathrm{eV}$ and the length of $\mathrm{nm}$.

ℏ = 1.054572*^-27; 
vf = 1*^8;
d0 = 1*^-7;
eV = 1.6021766208*^-12;
k = eV*d0/ℏ/vf;

T[x_, E0_, V_, d_] := ((E0 - V)^2 - E0^2 Sin[x]^2)/((E0 - V)^2 - 
  E0^2 Sin[x]^2 + V^2 Sin[k*d Sqrt[(E0 - V)^2 - E0^2 Sin[x]^2]]^2 Tan[x]^2)

DensityPlot[
 T[x, E0, .2, 30], {E0, -.1, .5}, {x, -π/2 + 10^-2, π/2 - 10^-2}, 
 PlotPoints -> 100, PerformanceGoal -> "Quality", PlotRange -> All, 
 FrameLabel -> {"E, eV", "θ"}, PlotLabel -> T]

G[E0_, V_, d_] := 
NIntegrate[Cos[x]*T[x, E0, V, d], {x, -Pi/2, Pi/2}, 
  Method -> "DoubleExponential", PrecisionGoal -> 8, 
  MinRecursion -> 4]   


lst = Table[{E0, G[E0, .2, 30]}, {E0, -1, 1.5, .001}];

ListLinePlot[lst, AxesLabel -> {"E, eV", "G"}]

fig1

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  • 2
    $\begingroup$ This is pretty quick: With[{V = 1/5, d = 30}, Plot[NIntegrate[Cos[x] T[x, E0, V, d], {x, -Pi/2, Pi/2}, AccuracyGoal -> 8, Method -> "Trapezoidal"], {E0, -1, 3/2}]]. $\endgroup$ – J. M. is computer-less Sep 27 '18 at 7:32
  • $\begingroup$ @J. M. is somewhat okay. Thank you, very good solution of the problem. $\endgroup$ – Alex Trounev Sep 27 '18 at 8:53
  • $\begingroup$ Thank you so much, @AlexTrounev, You helped me beyond what I had asked! $\endgroup$ – Alex Nascimento Sep 27 '18 at 15:53
  • $\begingroup$ @Alex Nascimento You're welcome. I'm glad that this code turned out to be useful for you. $\endgroup$ – Alex Trounev Sep 27 '18 at 16:05
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The following is basically @Alex's answer, but making use of the units framework, and using a style more similar to the question.

First, define:

V = .2;
d = 30;

Next, determine the conversion factor using Ctrl-= to input the units (I used an image to see how it looks in a notebook):

enter image description here

(Alternatively, type in the FullForm equivalent):

conv = (Quantity["Electronvolts"]Quantity["Nanometers"])/(Quantity["ReducedPlanckConstant"]Quantity[10^8,"Centimeters"/"Seconds"])

1.5192675

(another shorter version that relies on Quantity interpretations)

Quantity["eV"] Quantity["nm"]/(Quantity["ℏ"] Quantity[10^8,"cm/s"])

1.519267

Then, define k and T:

k[e_, θ_] := Sqrt[(e-V)^2 - e^2 Sin[θ]^2]
T[e_,θ_] := (1 + V^2/k[e,θ]^2 Tan[θ]^2 Sin[k[e,θ] d conv]^2)^-1

Next, define G but restrict to only numeric arguments:

G[e_?NumericQ] := NIntegrate[T[e,θ] Cos[θ], {θ, -Pi/2, Pi/2}]

Finally, plot:

Plot[G[e], {e, -1, 1.5}, AxesOrigin->{0,0}]

enter image description here

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  • $\begingroup$ thanks for the great help, but I'd like to know where the unit conversion factor comes in the code. $\endgroup$ – Alex Nascimento Sep 27 '18 at 16:24
  • $\begingroup$ @AlexNascimento See update. $\endgroup$ – Carl Woll Sep 27 '18 at 16:33

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