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All unitary errors are correctable
Why do error correction protocols only work when the error rates are already significantly low to begin with?Why does the surface (quantum error correction) code have such a high threshold for errors?Are all $[[n, k, d]]$ quantum codes equivalent to additive self-orthogonal $GF(4)^n$ classical codes?Quantum error correction: necessary and sufficient conditionWhat are reliable references on analytical and/or numerical studies of threshold theorems under faulty quantum error correction?Compact way of describing the set of all stabilizer groups for fixed number of physical qubits and encoded logical qubitsIn what situation are three rounds of syndrome measurement required for fault-tolerance in the surface code?Shor 9 qubit code — how are the observables measured and eigenvalues obtained during syndrome measurement?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The Knill-Laflamme condition for a stabilizer $mathcalM$ is
An error with Kraus operators $E_k$ is correctable if either $$E^dagger_kE_linmathcalMquadforall, k,l $$ or there exists $MinmathcalM$ such that $$M,E_k^dagger E_k=0quadforall ,k $$
But consider a unitary error $U$, then $U^dagger U=Iin mathcalM$. Does this mean that all unitary errors are always correctable by any stabilizer? It shouldn't, because for example Shor's code doesn't correct double bit flips. What am I missing?
error-correction stabilizer-code
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
$endgroup$
add a comment |
$begingroup$
The Knill-Laflamme condition for a stabilizer $mathcalM$ is
An error with Kraus operators $E_k$ is correctable if either $$E^dagger_kE_linmathcalMquadforall, k,l $$ or there exists $MinmathcalM$ such that $$M,E_k^dagger E_k=0quadforall ,k $$
But consider a unitary error $U$, then $U^dagger U=Iin mathcalM$. Does this mean that all unitary errors are always correctable by any stabilizer? It shouldn't, because for example Shor's code doesn't correct double bit flips. What am I missing?
error-correction stabilizer-code
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
$endgroup$
$begingroup$
Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
$endgroup$
– Norbert Schuch
Jul 28 at 20:21
$begingroup$
$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
$endgroup$
– user2723984
Jul 28 at 20:26
1
$begingroup$
But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
$endgroup$
– Norbert Schuch
Jul 28 at 20:29
$begingroup$
I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
$endgroup$
– user2723984
Jul 28 at 20:30
1
$begingroup$
$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
$endgroup$
– Norbert Schuch
Jul 28 at 20:32
add a comment |
$begingroup$
The Knill-Laflamme condition for a stabilizer $mathcalM$ is
An error with Kraus operators $E_k$ is correctable if either $$E^dagger_kE_linmathcalMquadforall, k,l $$ or there exists $MinmathcalM$ such that $$M,E_k^dagger E_k=0quadforall ,k $$
But consider a unitary error $U$, then $U^dagger U=Iin mathcalM$. Does this mean that all unitary errors are always correctable by any stabilizer? It shouldn't, because for example Shor's code doesn't correct double bit flips. What am I missing?
error-correction stabilizer-code
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
$endgroup$
The Knill-Laflamme condition for a stabilizer $mathcalM$ is
An error with Kraus operators $E_k$ is correctable if either $$E^dagger_kE_linmathcalMquadforall, k,l $$ or there exists $MinmathcalM$ such that $$M,E_k^dagger E_k=0quadforall ,k $$
But consider a unitary error $U$, then $U^dagger U=Iin mathcalM$. Does this mean that all unitary errors are always correctable by any stabilizer? It shouldn't, because for example Shor's code doesn't correct double bit flips. What am I missing?
error-correction stabilizer-code
error-correction stabilizer-code
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
edited Jul 28 at 20:25
user2723984
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
asked Jul 28 at 19:39
user2723984user2723984
4579 bronze badges
4579 bronze badges
$begingroup$
Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
$endgroup$
– Norbert Schuch
Jul 28 at 20:21
$begingroup$
$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
$endgroup$
– user2723984
Jul 28 at 20:26
1
$begingroup$
But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
$endgroup$
– Norbert Schuch
Jul 28 at 20:29
$begingroup$
I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
$endgroup$
– user2723984
Jul 28 at 20:30
1
$begingroup$
$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
$endgroup$
– Norbert Schuch
Jul 28 at 20:32
add a comment |
$begingroup$
Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
$endgroup$
– Norbert Schuch
Jul 28 at 20:21
$begingroup$
$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
$endgroup$
– user2723984
Jul 28 at 20:26
1
$begingroup$
But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
$endgroup$
– Norbert Schuch
Jul 28 at 20:29
$begingroup$
I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
$endgroup$
– user2723984
Jul 28 at 20:30
1
$begingroup$
$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
$endgroup$
– Norbert Schuch
Jul 28 at 20:32
$begingroup$
Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
$endgroup$
– Norbert Schuch
Jul 28 at 20:21
$begingroup$
Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
$endgroup$
– Norbert Schuch
Jul 28 at 20:21
$begingroup$
$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
$endgroup$
– user2723984
Jul 28 at 20:26
$begingroup$
$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
$endgroup$
– user2723984
Jul 28 at 20:26
1
1
$begingroup$
But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
$endgroup$
– Norbert Schuch
Jul 28 at 20:29
$begingroup$
But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
$endgroup$
– Norbert Schuch
Jul 28 at 20:29
$begingroup$
I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
$endgroup$
– user2723984
Jul 28 at 20:30
$begingroup$
I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
$endgroup$
– user2723984
Jul 28 at 20:30
1
1
$begingroup$
$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
$endgroup$
– Norbert Schuch
Jul 28 at 20:32
$begingroup$
$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
$endgroup$
– Norbert Schuch
Jul 28 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.
However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.
On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=sqrtp_kU_k$, and additionally $E_0=sqrtqI$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
$endgroup$
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
$endgroup$
– Norbert Schuch
Jul 28 at 22:01
$begingroup$
thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
$endgroup$
– user2723984
Jul 28 at 22:05
add a comment |
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$begingroup$
If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.
However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.
On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=sqrtp_kU_k$, and additionally $E_0=sqrtqI$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
$endgroup$
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
$endgroup$
– Norbert Schuch
Jul 28 at 22:01
$begingroup$
thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
$endgroup$
– user2723984
Jul 28 at 22:05
add a comment |
$begingroup$
If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.
However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.
On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=sqrtp_kU_k$, and additionally $E_0=sqrtqI$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
$endgroup$
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
$endgroup$
– Norbert Schuch
Jul 28 at 22:01
$begingroup$
thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
$endgroup$
– user2723984
Jul 28 at 22:05
add a comment |
$begingroup$
If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.
However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.
On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=sqrtp_kU_k$, and additionally $E_0=sqrtqI$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
$endgroup$
If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected.
However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, because you know which error has been applied.
On the other hand, if you have several unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. Thus, we have $E_k=sqrtp_kU_k$, and additionally $E_0=sqrtqI$. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
answered Jul 28 at 20:30
Norbert SchuchNorbert Schuch
1,6443 silver badges11 bronze badges
1,6443 silver badges11 bronze badges
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
$endgroup$
– Norbert Schuch
Jul 28 at 22:01
$begingroup$
thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
$endgroup$
– user2723984
Jul 28 at 22:05
add a comment |
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
$endgroup$
– Norbert Schuch
Jul 28 at 22:01
$begingroup$
thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
$endgroup$
– user2723984
Jul 28 at 22:05
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
Then the condition is worded a bit weirdly I think. To me a double bit flip is "an error" and following this condition it is correctable. Should I take the condition to mean that an ensemble of errors is correctable only if the map induced by the most general linear combination of them has Kraus operators that satisfy the Knill-Laflamme condition?
$endgroup$
– user2723984
Jul 28 at 21:43
$begingroup$
@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
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– Norbert Schuch
Jul 28 at 22:01
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@user2723984 In QECC, you look at the map which corresponds to an average error (including no error) during a given evolution. There is nothing weird about that. Indeed, the right formulation for that is a CPTP map with Kraus operators $E_k$, which are not unique -- an unknown error is nothing but a noisy evolution of the quantum system, nothing more.
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– Norbert Schuch
Jul 28 at 22:01
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thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
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– user2723984
Jul 28 at 22:05
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thank you, I misinterpreted the condition as I was considering each possible flip an individual error that could either be correctable or not, rather than a general error. So I expected the condition to be fulfilled individually by single flip errors and not by double flips in the bit flip code for example.
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– user2723984
Jul 28 at 22:05
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Why would it mean that? The second condition is never satisfied. Why would the first be? (And what is $mathcal M$?)
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– Norbert Schuch
Jul 28 at 20:21
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$mathcalM$ is a stabilizer (I think they're called like that, an abelian subgroup of the Pauli group, and we use as codewords eigenstates of matrices in $mathcalM$ with eigenvalue $1$) the first condition, since there is only one Kraus operator, is always satisfied, since $U^dagger U=IinmathcalM$
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– user2723984
Jul 28 at 20:26
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But is has to be satisfied for all error pairs $E_k^dagger E_l$. Unless your error is always a fixed unitary. In that case, it can obviously be corrected, because you know what happened to your system.
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– Norbert Schuch
Jul 28 at 20:29
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I don't understand, $E_k$ are the Kraus operators of one error, not of all possible errors that can happen to the system, or have I misunderstood?
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– user2723984
Jul 28 at 20:30
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$E_k$ are the Kraus operators of the channel which describes the error - for instance, for unitary errors $U_k$ which occur with probability $p_k$, the channel would be something like $rhomapsto p_k U_krho U_k^dagger + qrho$, where $q$ is the probability that no error occurs. So if there are several unitary errors which can occur (or even just one error or no error), there is more than one $E_k$, and the first condition need not be satisfied for all pairs $E_k$, $E_l$.
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– Norbert Schuch
Jul 28 at 20:32