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Noise in Eigenvalues plot
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Problem with plotting eigenvaluesHow to overlay ListPlot on a ContourPlot with correct range?Trying to find intersection of 3 functions graphicallySome glitch in the Plot: Two approaches for plotting give different resultsDEigenvalues with Robin B.C. sign problemHow can I add a custom color function and a custom mesh to a 3D parametric plot?How do I plot $y=8 sin(2 pi / 3)$?Plotting eigenvalues in one plot for three different parametersEigenvalues of a non-Hermitian complex periodic potentialHow to compute eigenvalues of a large symbolic matrix?
$begingroup$
I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.
A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
H[d_, λ_, β_, m_] :=
a (Sin[x] A1 + Sin[ky] A2) + A3 β +
d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
Sin[ky] A6 + λ Sin[z] A7+m*A8;
ky = 0;
a = 1;
b = 1;
t = 1.5;
α = 0.3;
Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]

Any help will be highly appreciated.
plotting eigenvalues
$endgroup$
add a comment |
$begingroup$
I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.
A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
H[d_, λ_, β_, m_] :=
a (Sin[x] A1 + Sin[ky] A2) + A3 β +
d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
Sin[ky] A6 + λ Sin[z] A7+m*A8;
ky = 0;
a = 1;
b = 1;
t = 1.5;
α = 0.3;
Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]

Any help will be highly appreciated.
plotting eigenvalues
$endgroup$
add a comment |
$begingroup$
I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.
A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
H[d_, λ_, β_, m_] :=
a (Sin[x] A1 + Sin[ky] A2) + A3 β +
d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
Sin[ky] A6 + λ Sin[z] A7+m*A8;
ky = 0;
a = 1;
b = 1;
t = 1.5;
α = 0.3;
Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]

Any help will be highly appreciated.
plotting eigenvalues
$endgroup$
I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.
A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
H[d_, λ_, β_, m_] :=
a (Sin[x] A1 + Sin[ky] A2) + A3 β +
d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
Sin[ky] A6 + λ Sin[z] A7+m*A8;
ky = 0;
a = 1;
b = 1;
t = 1.5;
α = 0.3;
Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]

Any help will be highly appreciated.
plotting eigenvalues
plotting eigenvalues
edited 2 hours ago
Michael E2
151k12203483
151k12203483
asked 2 hours ago
Hazoor ImranHazoor Imran
263
263
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.
You can use Max to plot the largest eigenvalue:
Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]

Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":
Plot3D[
Eigenvalues[
H[0.1, 0.5, 0.7, 0], -1,
Method -> "Arnoldi", "Criteria" -> "RealPart"
],
x, - Pi, Pi, z, 0, 2 Pi]
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
add a comment |
$begingroup$
Not sure why you pick the 4th element, but maybe this will help:
ev4 = Eigenvalues[H[p, q, r, s]][[4]] /.
Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
Plot3D[ev4, x, -π, π, z, 0, 2 π]

$endgroup$
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value-0.5on the right hand side to something bigger. For exampleContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.
You can use Max to plot the largest eigenvalue:
Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]

Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":
Plot3D[
Eigenvalues[
H[0.1, 0.5, 0.7, 0], -1,
Method -> "Arnoldi", "Criteria" -> "RealPart"
],
x, - Pi, Pi, z, 0, 2 Pi]
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
add a comment |
$begingroup$
By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.
You can use Max to plot the largest eigenvalue:
Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]

Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":
Plot3D[
Eigenvalues[
H[0.1, 0.5, 0.7, 0], -1,
Method -> "Arnoldi", "Criteria" -> "RealPart"
],
x, - Pi, Pi, z, 0, 2 Pi]
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
add a comment |
$begingroup$
By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.
You can use Max to plot the largest eigenvalue:
Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]

Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":
Plot3D[
Eigenvalues[
H[0.1, 0.5, 0.7, 0], -1,
Method -> "Arnoldi", "Criteria" -> "RealPart"
],
x, - Pi, Pi, z, 0, 2 Pi]
$endgroup$
By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.
You can use Max to plot the largest eigenvalue:
Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]

Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":
Plot3D[
Eigenvalues[
H[0.1, 0.5, 0.7, 0], -1,
Method -> "Arnoldi", "Criteria" -> "RealPart"
],
x, - Pi, Pi, z, 0, 2 Pi]
answered 2 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
60.7k585171
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
add a comment |
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
40 mins ago
add a comment |
$begingroup$
Not sure why you pick the 4th element, but maybe this will help:
ev4 = Eigenvalues[H[p, q, r, s]][[4]] /.
Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
Plot3D[ev4, x, -π, π, z, 0, 2 π]

$endgroup$
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value-0.5on the right hand side to something bigger. For exampleContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
add a comment |
$begingroup$
Not sure why you pick the 4th element, but maybe this will help:
ev4 = Eigenvalues[H[p, q, r, s]][[4]] /.
Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
Plot3D[ev4, x, -π, π, z, 0, 2 π]

$endgroup$
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value-0.5on the right hand side to something bigger. For exampleContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
add a comment |
$begingroup$
Not sure why you pick the 4th element, but maybe this will help:
ev4 = Eigenvalues[H[p, q, r, s]][[4]] /.
Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
Plot3D[ev4, x, -π, π, z, 0, 2 π]

$endgroup$
Not sure why you pick the 4th element, but maybe this will help:
ev4 = Eigenvalues[H[p, q, r, s]][[4]] /.
Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
Plot3D[ev4, x, -π, π, z, 0, 2 π]

answered 2 hours ago
Michael E2Michael E2
151k12203483
151k12203483
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value-0.5on the right hand side to something bigger. For exampleContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
add a comment |
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value-0.5on the right hand side to something bigger. For exampleContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value
-0.5 on the right hand side to something bigger. For example ContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].$endgroup$
– Michael E2
1 hour ago
$begingroup$
@HazoorImran Yes, but set the value
-0.5 on the right hand side to something bigger. For example ContourPlot[ev4 == 2, x, -[Pi], [Pi], z, 0, 2 [Pi]].$endgroup$
– Michael E2
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
1 hour ago
add a comment |
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