Depolarizing channel operator sum representation
In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with
$frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.
What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.
quantum-information quantum-channel
edited Dec 16 at 23:16
Blue♦
5,64521352
asked Dec 16 at 20:35
user1936752
2626
add a comment |
In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with
$frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.
What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.
quantum-information quantum-channel
edited Dec 16 at 23:16
Blue♦
5,64521352
asked Dec 16 at 20:35
user1936752
2626
add a comment |
In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with
$frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.
What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.
quantum-information quantum-channel
edited Dec 16 at 23:16
Blue♦
5,64521352
asked Dec 16 at 20:35
user1936752
2626
In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with
$frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.
What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.
quantum-information quantum-channel
quantum-information quantum-channel
edited Dec 16 at 23:16
Blue♦
5,64521352
asked Dec 16 at 20:35
user1936752
2626
edited Dec 16 at 23:16
Blue♦
5,64521352
asked Dec 16 at 20:35
user1936752
2626
edited Dec 16 at 23:16
Blue♦
5,64521352
edited Dec 16 at 23:16
Blue♦
5,64521352
edited Dec 16 at 23:16
Blue♦
5,64521352
5,64521352
asked Dec 16 at 20:35
user1936752
2626
asked Dec 16 at 20:35
user1936752
2626
asked Dec 16 at 20:35
user1936752
2626
2626
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
$$
sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
$$
with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
$$
sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
$$
and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).
Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).
Let's look at your example in a bit more detail. Here, $D=2$. You have that
$$
mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
$$
for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.
We now have that
begin{align}
sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
\
& = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
end{align}
inserting the definition of $|Omegarangle$ and using linearity.
This yields
$$
sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
(1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
$$
The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
$frac{p}{2D}mathbb Iotimesmathbb I$.
You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
$$
(sigma_kotimes mathbb I)|Omegarangle ,
$$
where $sigma_k$ are the three Pauli matrices or the identity.
Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.
answered Dec 16 at 21:45
Norbert Schuch
1,253211
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
add a comment |
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This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
$$
sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
$$
with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
$$
sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
$$
and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).
Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).
Let's look at your example in a bit more detail. Here, $D=2$. You have that
$$
mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
$$
for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.
We now have that
begin{align}
sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
\
& = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
end{align}
inserting the definition of $|Omegarangle$ and using linearity.
This yields
$$
sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
(1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
$$
The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
$frac{p}{2D}mathbb Iotimesmathbb I$.
You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
$$
(sigma_kotimes mathbb I)|Omegarangle ,
$$
where $sigma_k$ are the three Pauli matrices or the identity.
Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.
answered Dec 16 at 21:45
Norbert Schuch
1,253211
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
add a comment |
This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
$$
sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
$$
with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
$$
sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
$$
and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).
Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).
Let's look at your example in a bit more detail. Here, $D=2$. You have that
$$
mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
$$
for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.
We now have that
begin{align}
sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
\
& = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
end{align}
inserting the definition of $|Omegarangle$ and using linearity.
This yields
$$
sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
(1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
$$
The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
$frac{p}{2D}mathbb Iotimesmathbb I$.
You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
$$
(sigma_kotimes mathbb I)|Omegarangle ,
$$
where $sigma_k$ are the three Pauli matrices or the identity.
Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.
answered Dec 16 at 21:45
Norbert Schuch
1,253211
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
add a comment |
This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
$$
sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
$$
with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
$$
sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
$$
and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).
Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).
Let's look at your example in a bit more detail. Here, $D=2$. You have that
$$
mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
$$
for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.
We now have that
begin{align}
sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
\
& = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
end{align}
inserting the definition of $|Omegarangle$ and using linearity.
This yields
$$
sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
(1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
$$
The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
$frac{p}{2D}mathbb Iotimesmathbb I$.
You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
$$
(sigma_kotimes mathbb I)|Omegarangle ,
$$
where $sigma_k$ are the three Pauli matrices or the identity.
Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.
answered Dec 16 at 21:45
Norbert Schuch
1,253211
This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
$$
sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
$$
with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
$$
sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
$$
and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).
Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).
Let's look at your example in a bit more detail. Here, $D=2$. You have that
$$
mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
$$
for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.
We now have that
begin{align}
sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
\
& = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
end{align}
inserting the definition of $|Omegarangle$ and using linearity.
This yields
$$
sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
(1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
$$
The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
$frac{p}{2D}mathbb Iotimesmathbb I$.
You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
$$
(sigma_kotimes mathbb I)|Omegarangle ,
$$
where $sigma_k$ are the three Pauli matrices or the identity.
Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.
answered Dec 16 at 21:45
Norbert Schuch
1,253211
edited Dec 17 at 0:16
answered Dec 16 at 21:45
Norbert Schuch
1,253211
answered Dec 16 at 21:45
Norbert Schuch
1,253211
answered Dec 16 at 21:45
Norbert Schuch
1,253211
1,253211
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
add a comment |
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
– user1936752
Dec 16 at 22:06
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
@user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
– Norbert Schuch
Dec 16 at 22:08
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
– user1936752
Dec 16 at 22:26
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
@user1936752 Have edited.
– Norbert Schuch
Dec 17 at 0:17
Thank you very much
– user1936752
Dec 17 at 11:41
Thank you very much
– user1936752
Dec 17 at 11:41
add a comment |
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