How to decrease number of noises in stochastic differential equation












0












$begingroup$


I have a Quantum SDE containing both white and color noises (open quantum system).
$$ dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{X}^+_1 hat{c})rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$
Here A is a constant, $hat{c}^dagger$ and $hat{c}$ are creation and annihilation operators belong to system. $X^pm_j$ are operators belong to fermionic environment. The order of $X^pm_j$, $hat{c}^dagger$ and $hat{c}$ is fixed (no commutation or anti-commutation relation exist between $X^pm_j$ and $hat{c}^dagger$, $hat{c}$ ). $nu_j$ is white noise, $omega_{1t}$ and $omega_{2t}$ are defined as
$$ omega_1(t) = int_{t_0}^{t} C^+(t-tau)nu_{1tau} dtau$$
and
$$ omega_2(t) = int_{t_0}^{t} C^-(t-tau)nu_{2tau} dtau $$
Where $C^pm$ is defined as
$$C^sigma(t)=int_{-infty}^{infty} domega e^{isigmaomega t} f(omega) J(omega)$$
My question here is that how can we decrease the number of noises by combining white noises $nu_j$ and color noises $omega_j$ while they are connected to different operators. I know how to decrease the number of noises if they are connected to same operators. Any idea about different operator case, any article, book ? Your help will be appreciated.










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$endgroup$












  • $begingroup$
    Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
    $endgroup$
    – user617446
    Dec 26 '18 at 13:26










  • $begingroup$
    For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
    $endgroup$
    – Skeptical Khan
    Dec 27 '18 at 8:37


















0












$begingroup$


I have a Quantum SDE containing both white and color noises (open quantum system).
$$ dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{X}^+_1 hat{c})rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$
Here A is a constant, $hat{c}^dagger$ and $hat{c}$ are creation and annihilation operators belong to system. $X^pm_j$ are operators belong to fermionic environment. The order of $X^pm_j$, $hat{c}^dagger$ and $hat{c}$ is fixed (no commutation or anti-commutation relation exist between $X^pm_j$ and $hat{c}^dagger$, $hat{c}$ ). $nu_j$ is white noise, $omega_{1t}$ and $omega_{2t}$ are defined as
$$ omega_1(t) = int_{t_0}^{t} C^+(t-tau)nu_{1tau} dtau$$
and
$$ omega_2(t) = int_{t_0}^{t} C^-(t-tau)nu_{2tau} dtau $$
Where $C^pm$ is defined as
$$C^sigma(t)=int_{-infty}^{infty} domega e^{isigmaomega t} f(omega) J(omega)$$
My question here is that how can we decrease the number of noises by combining white noises $nu_j$ and color noises $omega_j$ while they are connected to different operators. I know how to decrease the number of noises if they are connected to same operators. Any idea about different operator case, any article, book ? Your help will be appreciated.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
    $endgroup$
    – user617446
    Dec 26 '18 at 13:26










  • $begingroup$
    For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
    $endgroup$
    – Skeptical Khan
    Dec 27 '18 at 8:37
















0












0








0





$begingroup$


I have a Quantum SDE containing both white and color noises (open quantum system).
$$ dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{X}^+_1 hat{c})rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$
Here A is a constant, $hat{c}^dagger$ and $hat{c}$ are creation and annihilation operators belong to system. $X^pm_j$ are operators belong to fermionic environment. The order of $X^pm_j$, $hat{c}^dagger$ and $hat{c}$ is fixed (no commutation or anti-commutation relation exist between $X^pm_j$ and $hat{c}^dagger$, $hat{c}$ ). $nu_j$ is white noise, $omega_{1t}$ and $omega_{2t}$ are defined as
$$ omega_1(t) = int_{t_0}^{t} C^+(t-tau)nu_{1tau} dtau$$
and
$$ omega_2(t) = int_{t_0}^{t} C^-(t-tau)nu_{2tau} dtau $$
Where $C^pm$ is defined as
$$C^sigma(t)=int_{-infty}^{infty} domega e^{isigmaomega t} f(omega) J(omega)$$
My question here is that how can we decrease the number of noises by combining white noises $nu_j$ and color noises $omega_j$ while they are connected to different operators. I know how to decrease the number of noises if they are connected to same operators. Any idea about different operator case, any article, book ? Your help will be appreciated.










share|cite|improve this question











$endgroup$




I have a Quantum SDE containing both white and color noises (open quantum system).
$$ dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{X}^+_1 hat{c})rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$
Here A is a constant, $hat{c}^dagger$ and $hat{c}$ are creation and annihilation operators belong to system. $X^pm_j$ are operators belong to fermionic environment. The order of $X^pm_j$, $hat{c}^dagger$ and $hat{c}$ is fixed (no commutation or anti-commutation relation exist between $X^pm_j$ and $hat{c}^dagger$, $hat{c}$ ). $nu_j$ is white noise, $omega_{1t}$ and $omega_{2t}$ are defined as
$$ omega_1(t) = int_{t_0}^{t} C^+(t-tau)nu_{1tau} dtau$$
and
$$ omega_2(t) = int_{t_0}^{t} C^-(t-tau)nu_{2tau} dtau $$
Where $C^pm$ is defined as
$$C^sigma(t)=int_{-infty}^{infty} domega e^{isigmaomega t} f(omega) J(omega)$$
My question here is that how can we decrease the number of noises by combining white noises $nu_j$ and color noises $omega_j$ while they are connected to different operators. I know how to decrease the number of noises if they are connected to same operators. Any idea about different operator case, any article, book ? Your help will be appreciated.







stochastic-calculus stochastic-analysis sde






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edited Mar 3 at 22:09









Andrews

1,3112423




1,3112423










asked Dec 24 '18 at 10:17









Skeptical KhanSkeptical Khan

193




193












  • $begingroup$
    Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
    $endgroup$
    – user617446
    Dec 26 '18 at 13:26










  • $begingroup$
    For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
    $endgroup$
    – Skeptical Khan
    Dec 27 '18 at 8:37




















  • $begingroup$
    Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
    $endgroup$
    – user617446
    Dec 26 '18 at 13:26










  • $begingroup$
    For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
    $endgroup$
    – Skeptical Khan
    Dec 27 '18 at 8:37


















$begingroup$
Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
$endgroup$
– user617446
Dec 26 '18 at 13:26




$begingroup$
Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question.
$endgroup$
– user617446
Dec 26 '18 at 13:26












$begingroup$
For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
$endgroup$
– Skeptical Khan
Dec 27 '18 at 8:37






$begingroup$
For example, For a case where $nu_{1t}$ and $omega_{1t}$ are connected to same operators, $nu_{1t}$ and $omega_{1t}$ can be replaced by one noise, i.e , $$dotrho(t) = Arho_s + (nu_{1t}hat{c}^dagger hat{X}^-_1 + omega_{1t} hat{c}^dagger hat{X}^-_1 )rho_s +(nu_{2t} hat{X}^+_2hat{c} -omega_{2t}hat{c}^dagger hat{X}^-_2)rho_s $$ In above equation the 2nd term on R.H.S can be written as $(nu_{1t}+ omega_{1t} ) hat{c}^dagger hat{X}^-_1 rho_s$ . Now $(nu_{1t}+ omega_{1t} )$ can be replaced by another noise $alpha_{t}$ whose auto and cross-correlation can be found.
$endgroup$
– Skeptical Khan
Dec 27 '18 at 8:37












1 Answer
1






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oldest

votes


















0












$begingroup$

Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]Qleft[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
, where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms.
as you see, there are 9 entries, one for each possible operator combination.
in your example case in the comments, the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1}+omega_1 & -omega_{2}\
0 & 0 & 0\
nu_2 & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
but in the question the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1} & -omega_{2}\
omega_{1} & 0 & 0\
nu_{2} & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your response, will this similarity transformation decrease the number of noises ?
    $endgroup$
    – Skeptical Khan
    Dec 28 '18 at 8:01












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

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0












$begingroup$

Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]Qleft[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
, where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms.
as you see, there are 9 entries, one for each possible operator combination.
in your example case in the comments, the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1}+omega_1 & -omega_{2}\
0 & 0 & 0\
nu_2 & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
but in the question the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1} & -omega_{2}\
omega_{1} & 0 & 0\
nu_{2} & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your response, will this similarity transformation decrease the number of noises ?
    $endgroup$
    – Skeptical Khan
    Dec 28 '18 at 8:01
















0












$begingroup$

Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]Qleft[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
, where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms.
as you see, there are 9 entries, one for each possible operator combination.
in your example case in the comments, the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1}+omega_1 & -omega_{2}\
0 & 0 & 0\
nu_2 & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
but in the question the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1} & -omega_{2}\
omega_{1} & 0 & 0\
nu_{2} & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your response, will this similarity transformation decrease the number of noises ?
    $endgroup$
    – Skeptical Khan
    Dec 28 '18 at 8:01














0












0








0





$begingroup$

Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]Qleft[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
, where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms.
as you see, there are 9 entries, one for each possible operator combination.
in your example case in the comments, the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1}+omega_1 & -omega_{2}\
0 & 0 & 0\
nu_2 & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
but in the question the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1} & -omega_{2}\
omega_{1} & 0 & 0\
nu_{2} & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.






share|cite|improve this answer









$endgroup$



Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]Qleft[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
, where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms.
as you see, there are 9 entries, one for each possible operator combination.
in your example case in the comments, the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1}+omega_1 & -omega_{2}\
0 & 0 & 0\
nu_2 & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
but in the question the form is
$$left[begin{array}{ccc}
c^{dagger} & X_{1}^{+} & X_{2}^{+}end{array}right]left[begin{array}{ccc}
0 & nu_{1} & -omega_{2}\
omega_{1} & 0 & 0\
nu_{2} & 0 & 0
end{array}right]left[begin{array}{c}
c\
X_{1}^{-}\
X_{2}^{-}
end{array}right]rho_{s}$$
and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 9:17









user617446user617446

4543




4543












  • $begingroup$
    Thanks for your response, will this similarity transformation decrease the number of noises ?
    $endgroup$
    – Skeptical Khan
    Dec 28 '18 at 8:01


















  • $begingroup$
    Thanks for your response, will this similarity transformation decrease the number of noises ?
    $endgroup$
    – Skeptical Khan
    Dec 28 '18 at 8:01
















$begingroup$
Thanks for your response, will this similarity transformation decrease the number of noises ?
$endgroup$
– Skeptical Khan
Dec 28 '18 at 8:01




$begingroup$
Thanks for your response, will this similarity transformation decrease the number of noises ?
$endgroup$
– Skeptical Khan
Dec 28 '18 at 8:01


















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