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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 4 hours ago
Michael E2
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
$endgroup$
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 4 hours ago
Michael E2
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
$endgroup$
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 4 hours ago
Michael E2
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
$endgroup$
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 4 hours ago
Michael E2
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
edited 4 hours ago
Michael E2
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
edited 4 hours ago
Michael E2
151k12203483
edited 4 hours ago
Michael E2
151k12203483
edited 4 hours ago
Michael E2
151k12203483
151k12203483
asked 4 hours ago
Hazoor ImranHazoor Imran
313
asked 4 hours ago
Hazoor ImranHazoor Imran
313
asked 4 hours ago
Hazoor ImranHazoor Imran
313
313
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 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours 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 4 hours 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours 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 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours 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}]
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours 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}]
answered 4 hours 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 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
60.7k585171
60.7k585171
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours ago
add a comment |
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 hours 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 4 hours 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours 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 4 hours 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours 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 4 hours 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 4 hours ago
Michael E2Michael E2
151k12203483
answered 4 hours ago
Michael E2Michael E2
151k12203483
answered 4 hours ago
Michael E2Michael E2
151k12203483
answered 4 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours 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
3 hours 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
3 hours 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
3 hours 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
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
3 hours ago
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