Depolarizing channel operator sum representation
up vote
1
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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 2 hours ago
Blue♦
5,64011352
asked 4 hours ago
user1936752
2605
add a comment |
up vote
1
down vote
favorite
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 2 hours ago
Blue♦
5,64011352
asked 4 hours ago
user1936752
2605
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 2 hours ago
Blue♦
5,64011352
asked 4 hours ago
user1936752
2605
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 2 hours ago
Blue♦
5,64011352
asked 4 hours ago
user1936752
2605
edited 2 hours ago
Blue♦
5,64011352
asked 4 hours ago
user1936752
2605
edited 2 hours ago
Blue♦
5,64011352
edited 2 hours ago
Blue♦
5,64011352
edited 2 hours ago
Blue♦
5,64011352
5,64011352
asked 4 hours ago
user1936752
2605
asked 4 hours ago
user1936752
2605
asked 4 hours ago
user1936752
2605
2605
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
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 3 hours ago
Norbert Schuch
1,188211
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
3 hours ago
@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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
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 3 hours ago
Norbert Schuch
1,188211
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
3 hours ago
@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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
add a comment |
up vote
3
down vote
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 3 hours ago
Norbert Schuch
1,188211
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
3 hours ago
@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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
add a comment |
up vote
3
down vote
up vote
3
down vote
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 3 hours ago
Norbert Schuch
1,188211
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 3 hours ago
Norbert Schuch
1,188211
edited 1 hour ago
answered 3 hours ago
Norbert Schuch
1,188211
answered 3 hours ago
Norbert Schuch
1,188211
answered 3 hours ago
Norbert Schuch
1,188211
1,188211
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
3 hours ago
@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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
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
3 hours ago
@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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
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
3 hours ago
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
3 hours ago
@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
3 hours ago
@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
3 hours ago
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
3 hours ago
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
3 hours ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
@user1936752 Have edited.
– Norbert Schuch
1 hour ago
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
