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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 1 hour ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
$endgroup$
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 1 hour ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
$endgroup$
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 1 hour ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
$endgroup$
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 1 hour ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
edited 1 hour ago
Michael E2
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
edited 1 hour ago
Michael E2
151k12203483
edited 1 hour ago
Michael E2
151k12203483
edited 1 hour ago
Michael E2
151k12203483
151k12203483
asked 1 hour ago
Hazoor ImranHazoor Imran
263
asked 1 hour ago
Hazoor ImranHazoor Imran
263
asked 1 hour 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}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 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}]
answered 1 hour 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
27 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 mins 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 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 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}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 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}]
answered 1 hour 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 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
60.7k585171
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 mins ago
add a comment |
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 mins ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 mins ago
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
11 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}]
answered 1 hour 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
27 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 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}]
answered 1 hour 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
27 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 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}]
answered 1 hour 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 1 hour ago
Michael E2Michael E2
151k12203483
answered 1 hour ago
Michael E2Michael E2
151k12203483
answered 1 hour ago
Michael E2Michael E2
151k12203483
answered 1 hour 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
27 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 mins 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
27 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 mins 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
27 mins 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
27 mins 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
19 mins 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
19 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 mins ago
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
12 mins ago
add a comment |
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