Mixing
What is the difference between the three types of bosonic reservoirs : sub-ohmic, ohmic and super-ohmic?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I want to ask what is the difference between the three types of bosonic reservoirs that we use in the theory of quantum decoherence: sub-ohmic, ohmic and super-ohmic. I know that there is a parameter "s" that changes in the mathematical relation of the spectral density; for $s = 1$, the reservoir is of ohmic type, for $0 < s < 1$, the reservoir is sub-ohmic and for $s > 1$, the reservoir is called super-ohmic.
What is the physical meaning of each type?
quantum-mechanics quantum-information open-quantum-systems
edited Jul 15 at 21:55
glS
5,70031652
asked Jul 14 at 14:19
Abdellah Salaoui
192
migrated from math.stackexchange.com Jul 14 at 20:00
This question came from our site for people studying math at any level and professionals in related fields.
add a comment |Â
up vote
3
down vote
favorite
I want to ask what is the difference between the three types of bosonic reservoirs that we use in the theory of quantum decoherence: sub-ohmic, ohmic and super-ohmic. I know that there is a parameter "s" that changes in the mathematical relation of the spectral density; for $s = 1$, the reservoir is of ohmic type, for $0 < s < 1$, the reservoir is sub-ohmic and for $s > 1$, the reservoir is called super-ohmic.
What is the physical meaning of each type?
quantum-mechanics quantum-information open-quantum-systems
edited Jul 15 at 21:55
glS
5,70031652
asked Jul 14 at 14:19
Abdellah Salaoui
192
migrated from math.stackexchange.com Jul 14 at 20:00
This question came from our site for people studying math at any level and professionals in related fields.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I want to ask what is the difference between the three types of bosonic reservoirs that we use in the theory of quantum decoherence: sub-ohmic, ohmic and super-ohmic. I know that there is a parameter "s" that changes in the mathematical relation of the spectral density; for $s = 1$, the reservoir is of ohmic type, for $0 < s < 1$, the reservoir is sub-ohmic and for $s > 1$, the reservoir is called super-ohmic.
What is the physical meaning of each type?
quantum-mechanics quantum-information open-quantum-systems
edited Jul 15 at 21:55
glS
5,70031652
asked Jul 14 at 14:19
Abdellah Salaoui
192
I want to ask what is the difference between the three types of bosonic reservoirs that we use in the theory of quantum decoherence: sub-ohmic, ohmic and super-ohmic. I know that there is a parameter "s" that changes in the mathematical relation of the spectral density; for $s = 1$, the reservoir is of ohmic type, for $0 < s < 1$, the reservoir is sub-ohmic and for $s > 1$, the reservoir is called super-ohmic.
What is the physical meaning of each type?
quantum-mechanics quantum-information open-quantum-systems
edited Jul 15 at 21:55
glS
5,70031652
asked Jul 14 at 14:19
Abdellah Salaoui
192
edited Jul 15 at 21:55
glS
5,70031652
edited Jul 15 at 21:55
glS
5,70031652
edited Jul 15 at 21:55
glS
5,70031652
5,70031652
asked Jul 14 at 14:19
Abdellah Salaoui
192
asked Jul 14 at 14:19
Abdellah Salaoui
192
asked Jul 14 at 14:19
Abdellah Salaoui
192
192
migrated from math.stackexchange.com Jul 14 at 20:00
This question came from our site for people studying math at any level and professionals in related fields.
migrated from math.stackexchange.com Jul 14 at 20:00
This question came from our site for people studying math at any level and professionals in related fields.
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
This is the quantum dissipation.
Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level.
Its main purpose is to derive the laws of classical dissipation from QM view.
The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved.
The solution is to split the system into two:
the QM, where the dissipation works
and the environment or bath, where the energy will flow.
The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles.
In the harmonic bath model, the good description of the dissipation is the bath spectral function.
The spectral function shows constraints in the Ci, and when it is in the form of
J ( É ) = η É
then the classical view of dissipation can be shown to be Ohmic.
A more generic form is J ( É ) ∠É^s
In this case, when s>1 the dissipation is super-Ohmic.
If s<1, it is called sub-Ohmic.
The EM field is under some circumstances super-Ohmic.
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is the quantum dissipation.
Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level.
Its main purpose is to derive the laws of classical dissipation from QM view.
The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved.
The solution is to split the system into two:
the QM, where the dissipation works
and the environment or bath, where the energy will flow.
The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles.
In the harmonic bath model, the good description of the dissipation is the bath spectral function.
The spectral function shows constraints in the Ci, and when it is in the form of
J ( É ) = η É
then the classical view of dissipation can be shown to be Ohmic.
A more generic form is J ( É ) ∠É^s
In this case, when s>1 the dissipation is super-Ohmic.
If s<1, it is called sub-Ohmic.
The EM field is under some circumstances super-Ohmic.
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
add a comment |Â
up vote
0
down vote
This is the quantum dissipation.
Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level.
Its main purpose is to derive the laws of classical dissipation from QM view.
The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved.
The solution is to split the system into two:
the QM, where the dissipation works
and the environment or bath, where the energy will flow.
The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles.
In the harmonic bath model, the good description of the dissipation is the bath spectral function.
The spectral function shows constraints in the Ci, and when it is in the form of
J ( É ) = η É
then the classical view of dissipation can be shown to be Ohmic.
A more generic form is J ( É ) ∠É^s
In this case, when s>1 the dissipation is super-Ohmic.
If s<1, it is called sub-Ohmic.
The EM field is under some circumstances super-Ohmic.
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is the quantum dissipation.
Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level.
Its main purpose is to derive the laws of classical dissipation from QM view.
The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved.
The solution is to split the system into two:
the QM, where the dissipation works
and the environment or bath, where the energy will flow.
The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles.
In the harmonic bath model, the good description of the dissipation is the bath spectral function.
The spectral function shows constraints in the Ci, and when it is in the form of
J ( É ) = η É
then the classical view of dissipation can be shown to be Ohmic.
A more generic form is J ( É ) ∠É^s
In this case, when s>1 the dissipation is super-Ohmic.
If s<1, it is called sub-Ohmic.
The EM field is under some circumstances super-Ohmic.
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
This is the quantum dissipation.
Quantum dissipation is the branch of physics, that studies the QM processes of irreversible loss of energy viewed at the classical level.
Its main purpose is to derive the laws of classical dissipation from QM view.
The main problem is to show the irreversible loss of energy, but QM usually uses Hamiltonian, where the total energy is conserved.
The solution is to split the system into two:
the QM, where the dissipation works
and the environment or bath, where the energy will flow.
The simplest way to model the bath is with infinite number of harmonic oscillators, in QM a set of bosonic particles.
In the harmonic bath model, the good description of the dissipation is the bath spectral function.
The spectral function shows constraints in the Ci, and when it is in the form of
J ( É ) = η É
then the classical view of dissipation can be shown to be Ohmic.
A more generic form is J ( É ) ∠É^s
In this case, when s>1 the dissipation is super-Ohmic.
If s<1, it is called sub-Ohmic.
The EM field is under some circumstances super-Ohmic.
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
answered Jul 16 at 18:08
Ãrpád Szendrei
2,427419
2,427419
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
add a comment |Â
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Please use mathjax instead of pasting images of equations.
– DanielSank
Jul 16 at 22:04
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
Also, please do not use a new line for every sentence. English writing is organized into paragraphs. Keep sentences that are related to the same idea in one single paragraph.
– DanielSank
Jul 16 at 22:05
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f417668%2fwhat-is-the-difference-between-the-three-types-of-bosonic-reservoirs-sub-ohmic%23new-answer', 'question_page');
);
Post as a guest
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
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
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