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Noise in Eigenvalues plot
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Noise in Eigenvalues plot
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$begingroup$
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
edited 50 mins ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
$endgroup$
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
edited 50 mins ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
$endgroup$
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
edited 50 mins ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
$endgroup$
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 50 mins ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
edited 50 mins ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
edited 50 mins ago
Michael E2
151k12203483
edited 50 mins ago
Michael E2
151k12203483
edited 50 mins ago
Michael E2
151k12203483
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
213
asked 1 hour ago
Hazoor ImranHazoor Imran
213
asked 1 hour ago
Hazoor ImranHazoor Imran
213
213
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}]
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
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}]
answered 44 mins ago
Michael E2Michael E2
151k12203483
$endgroup$
$begingroup$
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 min ago
add a comment |
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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}]
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
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}]
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
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}]
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
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 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 45 mins ago
Henrik SchumacherHenrik Schumacher
60.7k585171
60.7k585171
add a comment |
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}]
answered 44 mins ago
Michael E2Michael E2
151k12203483
$endgroup$
$begingroup$
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 min 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}]
answered 44 mins ago
Michael E2Michael E2
151k12203483
$endgroup$
$begingroup$
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 min 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}]
answered 44 mins ago
Michael E2Michael E2
151k12203483
$endgroup$
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 44 mins ago
Michael E2Michael E2
151k12203483
answered 44 mins ago
Michael E2Michael E2
151k12203483
answered 44 mins ago
Michael E2Michael E2
151k12203483
answered 44 mins 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 min 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 min 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 min 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 min ago
add a comment |
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