How to decrease number of noises in stochastic differential equation
$begingroup$
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.
stochastic-calculus stochastic-analysis sde
$endgroup$
add a comment |
$begingroup$
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.
stochastic-calculus stochastic-analysis sde
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
stochastic-calculus stochastic-analysis sde
$endgroup$
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
stochastic-calculus stochastic-analysis sde
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
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.
$endgroup$
$begingroup$
Thanks for your response, will this similarity transformation decrease the number of noises ?
$endgroup$
– Skeptical Khan
Dec 28 '18 at 8:01
add a comment |
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$begingroup$
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.
$endgroup$
$begingroup$
Thanks for your response, will this similarity transformation decrease the number of noises ?
$endgroup$
– Skeptical Khan
Dec 28 '18 at 8:01
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
Thanks for your response, will this similarity transformation decrease the number of noises ?
$endgroup$
– Skeptical Khan
Dec 28 '18 at 8:01
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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