How to handle purely imaginary Hamiltonians
Suppose I have a system of complex ODE's of the form
$$ idot{mathbf{c}}(t)=mathbf{f}(mathbf{c}(t))$$
and I can write down a Hamiltonian such that each ODE can be written as
$$dot{c}_j=frac{partialmathcal{H}}{partial c_j^*}$$ for each $j$ where * denotes complex conjugate. As a very simple example, the Hamiltonian
$$ mathcal{H}=-ileft(|c_0|^2+|c_1|^2right)
$$
leads to the equations
$$idot{c}_0=c_0, qquad idot{c}_1=c_1. $$
Questions:
What are the conjugate momenta for this system (Are they just the complex conjugates)? Also, is there any way to transform this problem (via action-angle coordinates/madelung transform) to one where the Hamiltonian is purely real or where the system evolves under real dynamics? Is an imaginary Hamiltonian even an issue if I want to analyze a much more complicated non-linear system of this type using canonical perturbation or bifurcation theory?
differential-equations dynamical-systems classical-mechanics hamilton-equations
edited 2 days ago
Cosmas Zachos
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
migrated from physics.stackexchange.com Jul 19 '17 at 19:30
This question came from our site for active researchers, academics and students of physics.
add a comment |
Suppose I have a system of complex ODE's of the form
$$ idot{mathbf{c}}(t)=mathbf{f}(mathbf{c}(t))$$
and I can write down a Hamiltonian such that each ODE can be written as
$$dot{c}_j=frac{partialmathcal{H}}{partial c_j^*}$$ for each $j$ where * denotes complex conjugate. As a very simple example, the Hamiltonian
$$ mathcal{H}=-ileft(|c_0|^2+|c_1|^2right)
$$
leads to the equations
$$idot{c}_0=c_0, qquad idot{c}_1=c_1. $$
Questions:
What are the conjugate momenta for this system (Are they just the complex conjugates)? Also, is there any way to transform this problem (via action-angle coordinates/madelung transform) to one where the Hamiltonian is purely real or where the system evolves under real dynamics? Is an imaginary Hamiltonian even an issue if I want to analyze a much more complicated non-linear system of this type using canonical perturbation or bifurcation theory?
differential-equations dynamical-systems classical-mechanics hamilton-equations
edited 2 days ago
Cosmas Zachos
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
migrated from physics.stackexchange.com Jul 19 '17 at 19:30
This question came from our site for active researchers, academics and students of physics.
2
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
2
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
1
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39
add a comment |
Suppose I have a system of complex ODE's of the form
$$ idot{mathbf{c}}(t)=mathbf{f}(mathbf{c}(t))$$
and I can write down a Hamiltonian such that each ODE can be written as
$$dot{c}_j=frac{partialmathcal{H}}{partial c_j^*}$$ for each $j$ where * denotes complex conjugate. As a very simple example, the Hamiltonian
$$ mathcal{H}=-ileft(|c_0|^2+|c_1|^2right)
$$
leads to the equations
$$idot{c}_0=c_0, qquad idot{c}_1=c_1. $$
Questions:
What are the conjugate momenta for this system (Are they just the complex conjugates)? Also, is there any way to transform this problem (via action-angle coordinates/madelung transform) to one where the Hamiltonian is purely real or where the system evolves under real dynamics? Is an imaginary Hamiltonian even an issue if I want to analyze a much more complicated non-linear system of this type using canonical perturbation or bifurcation theory?
differential-equations dynamical-systems classical-mechanics hamilton-equations
edited 2 days ago
Cosmas Zachos
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
Suppose I have a system of complex ODE's of the form
$$ idot{mathbf{c}}(t)=mathbf{f}(mathbf{c}(t))$$
and I can write down a Hamiltonian such that each ODE can be written as
$$dot{c}_j=frac{partialmathcal{H}}{partial c_j^*}$$ for each $j$ where * denotes complex conjugate. As a very simple example, the Hamiltonian
$$ mathcal{H}=-ileft(|c_0|^2+|c_1|^2right)
$$
leads to the equations
$$idot{c}_0=c_0, qquad idot{c}_1=c_1. $$
Questions:
What are the conjugate momenta for this system (Are they just the complex conjugates)? Also, is there any way to transform this problem (via action-angle coordinates/madelung transform) to one where the Hamiltonian is purely real or where the system evolves under real dynamics? Is an imaginary Hamiltonian even an issue if I want to analyze a much more complicated non-linear system of this type using canonical perturbation or bifurcation theory?
differential-equations dynamical-systems classical-mechanics hamilton-equations
differential-equations dynamical-systems classical-mechanics hamilton-equations
edited 2 days ago
Cosmas Zachos
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
edited 2 days ago
Cosmas Zachos
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
edited 2 days ago
Cosmas Zachos
1,560520
edited 2 days ago
Cosmas Zachos
1,560520
edited 2 days ago
Cosmas Zachos
1,560520
1,560520
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
asked Jul 18 '17 at 21:59
garserdt216garserdt216
346114
346114
migrated from physics.stackexchange.com Jul 19 '17 at 19:30
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jul 19 '17 at 19:30
This question came from our site for active researchers, academics and students of physics.
2
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
2
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
1
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39
add a comment |
2
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
2
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
1
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39
2
2
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
2
2
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
1
1
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39
add a comment |
1 Answer
1
active
oldest
votes
Well, yes, this is imaginary time dynamics, that is, you absorb -i into t,
for a new variable $beta=t/i$ and keep the Hamiltonian real/Hermitean; you run your Poisson brackets, QM, or whatever as usual.
That is to say, a purely imaginary Hamiltonian is essentially a pure real one, once you have analytically continued your time to a pure imaginary parameter, and, as and when opportunities present themselves, all your conventional answers from real time, with care.
Very often, in statistical mechanics, thermodynamics, QFT, β is essentially the inverse temperature and the exponentials of βH are real functions of partition functions, etc... instead of unitary evolution operators, exponentials of -itH.
Cf. Garaschuk 2010.
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2364253%2fhow-to-handle-purely-imaginary-hamiltonians%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Well, yes, this is imaginary time dynamics, that is, you absorb -i into t,
for a new variable $beta=t/i$ and keep the Hamiltonian real/Hermitean; you run your Poisson brackets, QM, or whatever as usual.
That is to say, a purely imaginary Hamiltonian is essentially a pure real one, once you have analytically continued your time to a pure imaginary parameter, and, as and when opportunities present themselves, all your conventional answers from real time, with care.
Very often, in statistical mechanics, thermodynamics, QFT, β is essentially the inverse temperature and the exponentials of βH are real functions of partition functions, etc... instead of unitary evolution operators, exponentials of -itH.
Cf. Garaschuk 2010.
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
add a comment |
Well, yes, this is imaginary time dynamics, that is, you absorb -i into t,
for a new variable $beta=t/i$ and keep the Hamiltonian real/Hermitean; you run your Poisson brackets, QM, or whatever as usual.
That is to say, a purely imaginary Hamiltonian is essentially a pure real one, once you have analytically continued your time to a pure imaginary parameter, and, as and when opportunities present themselves, all your conventional answers from real time, with care.
Very often, in statistical mechanics, thermodynamics, QFT, β is essentially the inverse temperature and the exponentials of βH are real functions of partition functions, etc... instead of unitary evolution operators, exponentials of -itH.
Cf. Garaschuk 2010.
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
add a comment |
Well, yes, this is imaginary time dynamics, that is, you absorb -i into t,
for a new variable $beta=t/i$ and keep the Hamiltonian real/Hermitean; you run your Poisson brackets, QM, or whatever as usual.
That is to say, a purely imaginary Hamiltonian is essentially a pure real one, once you have analytically continued your time to a pure imaginary parameter, and, as and when opportunities present themselves, all your conventional answers from real time, with care.
Very often, in statistical mechanics, thermodynamics, QFT, β is essentially the inverse temperature and the exponentials of βH are real functions of partition functions, etc... instead of unitary evolution operators, exponentials of -itH.
Cf. Garaschuk 2010.
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
Well, yes, this is imaginary time dynamics, that is, you absorb -i into t,
for a new variable $beta=t/i$ and keep the Hamiltonian real/Hermitean; you run your Poisson brackets, QM, or whatever as usual.
That is to say, a purely imaginary Hamiltonian is essentially a pure real one, once you have analytically continued your time to a pure imaginary parameter, and, as and when opportunities present themselves, all your conventional answers from real time, with care.
Very often, in statistical mechanics, thermodynamics, QFT, β is essentially the inverse temperature and the exponentials of βH are real functions of partition functions, etc... instead of unitary evolution operators, exponentials of -itH.
Cf. Garaschuk 2010.
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
edited 18 hours ago
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
answered 2 days ago
Cosmas ZachosCosmas Zachos
1,560520
1,560520
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2364253%2fhow-to-handle-purely-imaginary-hamiltonians%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
(VERY VAGUE) I think that purely imaginary Hamiltonians correspond to "gradient flows" (by which I mean the dynamical systems having the form $dot{x}=-nabla V(x)$). My (rough) intuition is that Hamiltonian systems are "conservative" while gradient systems are "dissipative". Putting an imaginary unit in a Hamiltonian should amount to transforming it into a dissipation function. I hope that, searching for the keywords given in this post, you can find something useful.
– Giuseppe Negro
Jul 27 '17 at 15:10
2
This is not really a "Hamiltonian" in a usual physics sense, as "Hamiltonian" is usually Hermitian (and thus purely real) in the quantum sense, and producing a conserved flow in classical settings. With the imaginary factors you no longer have the same kind of conserved quantity. If you want to study the dynamics, one option is to treat $c$ and $c^*$ as separate real variables. (Of course you could also do Real(c) and Imag(c), but that usually works out less cleanly.) Then you have 4 real variables, and c_0 / c_0* are conjugate; and c_1 / c_1* are conjugate. But your dimension is doubled.
– Alex Meiburg
Jul 27 '17 at 20:21
1
Something related that you might be interested in is PT-symmetric quantum mechanics. There, one considers non-Hermitian Hamiltonians that nonetheless have an entirely real spectrum. In the latter aspect they are of course different from what you propose, but they might be interesting.
– Cyclone
Sep 11 '17 at 21:39