Noise in Eigenvalues plot
5
$begingroup$
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 Apr 21 at 21:11
Michael E2
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
$endgroup$
add a comment |
5
$begingroup$
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 Apr 21 at 21:11
Michael E2
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
$endgroup$
add a comment |
5
5
5
$begingroup$
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 Apr 21 at 21:11
Michael E2
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
$endgroup$
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 Apr 21 at 21:11
Michael E2
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
edited Apr 21 at 21:11
Michael E2
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
edited Apr 21 at 21:11
Michael E2
151k12203483
edited Apr 21 at 21:11
Michael E2
151k12203483
edited Apr 21 at 21:11
Michael E2
151k12203483
151k12203483
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
asked Apr 21 at 20:58
Hazoor ImranHazoor Imran
363
363
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
7
$begingroup$
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 Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
add a comment |
4
$begingroup$
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 Apr 21 at 21:17
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
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
7
$begingroup$
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 Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
add a comment |
7
$begingroup$
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 Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
add a comment |
7
7
7
$begingroup$
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 Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
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 Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
answered Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
answered Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
answered Apr 21 at 21:16
Henrik SchumacherHenrik Schumacher
61.2k585171
61.2k585171
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
add a comment |
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
Thanks @ Henrik Schumacher
$endgroup$
– Hazoor Imran
Apr 21 at 22:17
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Apr 21 at 22:40
add a comment |
4
$begingroup$
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 Apr 21 at 21:17
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
add a comment |
4
$begingroup$
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 Apr 21 at 21:17
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
add a comment |
4
4
4
$begingroup$
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 Apr 21 at 21:17
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 Apr 21 at 21:17
Michael E2Michael E2
151k12203483
answered Apr 21 at 21:17
Michael E2Michael E2
151k12203483
answered Apr 21 at 21:17
Michael E2Michael E2
151k12203483
answered Apr 21 at 21:17
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
$begingroup$
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
Apr 21 at 22:01
$begingroup$
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
Apr 21 at 22:01
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
@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
Apr 21 at 22:09
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
$begingroup$
Thanks @ Michael E2, Yes this work.
$endgroup$
– Hazoor Imran
Apr 21 at 22:16
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Post as a guest
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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