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Can the Kraus decomposition always be chosen to be a statistical mixture of unitary evolutions?

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3

$begingroup$

If $mathcal{E}$ is a CPTP map between hermitian operators on two Hilbert spaces, then we can find a set of operators ${K_j}_j$ such that

$$mathcal{E}(rho)=sum_j K_jrho K_j^dagger $$

in the same spirit as any density matrix $rho$ can be decomposed in a pure states ensemble

$$rho=sum_k p_k |psi_kranglelanglepsi_k| $$

with $sum_k p_k=1$ and be intepreted as a classical statistical mixture of pure states.

Can I always find a Kraus decomposition such that $K_j= sqrt{p_j} U_j$ with $U_j U_j^dagger=mathbb{1}$ and $sum_j p_j=1$ and interpret it as a classical statistical mixture of unitary evolutions?

quantum-information quantum-operation quantum-channel

share|improve this question

edited 6 hours ago

glS

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asked 9 hours ago

user2723984user2723984

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$endgroup$

add a comment | 

3

$begingroup$

If $mathcal{E}$ is a CPTP map between hermitian operators on two Hilbert spaces, then we can find a set of operators ${K_j}_j$ such that

$$mathcal{E}(rho)=sum_j K_jrho K_j^dagger $$

in the same spirit as any density matrix $rho$ can be decomposed in a pure states ensemble

$$rho=sum_k p_k |psi_kranglelanglepsi_k| $$

with $sum_k p_k=1$ and be intepreted as a classical statistical mixture of pure states.

Can I always find a Kraus decomposition such that $K_j= sqrt{p_j} U_j$ with $U_j U_j^dagger=mathbb{1}$ and $sum_j p_j=1$ and interpret it as a classical statistical mixture of unitary evolutions?

quantum-information quantum-operation quantum-channel

share|improve this question

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $mathcal{E}$ is a CPTP map between hermitian operators on two Hilbert spaces, then we can find a set of operators ${K_j}_j$ such that

$$mathcal{E}(rho)=sum_j K_jrho K_j^dagger $$

in the same spirit as any density matrix $rho$ can be decomposed in a pure states ensemble

$$rho=sum_k p_k |psi_kranglelanglepsi_k| $$

with $sum_k p_k=1$ and be intepreted as a classical statistical mixture of pure states.

Can I always find a Kraus decomposition such that $K_j= sqrt{p_j} U_j$ with $U_j U_j^dagger=mathbb{1}$ and $sum_j p_j=1$ and interpret it as a classical statistical mixture of unitary evolutions?

quantum-information quantum-operation quantum-channel

share|improve this question

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

$endgroup$

share|improve this question

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

share|improve this question

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

edited 6 hours ago

glS

5,43511 gold badge99 silver badges4545 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

asked 9 hours ago

user2723984user2723984

45299 bronze badges

1 Answer
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3

$begingroup$

You cannot always find such a Kraus decomposition. Notice that any CPTP map $mathcal E$ which does have a decomposition is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state:
$$ mathcal E(tfrac{1}{d} mathbf 1) = tfrac{1}{d} mathbf 1 . $$
This is true because $U cdot mathbf 1 cdot U^dagger = mathbf 1$ for each unitary $U$, and taking a mixture over several unitaries $U$ does not change the result for the operator $mathbf 1$.
But it is easy to find maps that don't have this property — so for such maps, there is no interpretation as a probabilistic mixture of unitary processes.
For instance, the map $mathcal R$ that resets a qubit to $lvert 0 rangle$ does not preserve the maximally mixed state on a qubit:
$$ mathcal R(rho) ,=, mathrm{tr}(rho) cdot lvert 0 rangle!langle 0 rvert . $$
Of course, this map can be expressed using Kraus operators: for instance, you could take $K_0 = lvert 0 rangle!langle 0 rvert$ and $K_1 = lvert 0 rangle!langle 1 rvert$, which would suffice for $mathcal R(rho) = K_0^{phantomdagger} rho K_0^dagger + K_1^{phantomdagger} rho K_1^dagger$.

share|improve this answer

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be worth noting that even for unital channels, this is not always possible.
  $endgroup$
  – Norbert Schuch
  54 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

You cannot always find such a Kraus decomposition. Notice that any CPTP map $mathcal E$ which does have a decomposition is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state:
$$ mathcal E(tfrac{1}{d} mathbf 1) = tfrac{1}{d} mathbf 1 . $$
This is true because $U cdot mathbf 1 cdot U^dagger = mathbf 1$ for each unitary $U$, and taking a mixture over several unitaries $U$ does not change the result for the operator $mathbf 1$.
But it is easy to find maps that don't have this property — so for such maps, there is no interpretation as a probabilistic mixture of unitary processes.
For instance, the map $mathcal R$ that resets a qubit to $lvert 0 rangle$ does not preserve the maximally mixed state on a qubit:
$$ mathcal R(rho) ,=, mathrm{tr}(rho) cdot lvert 0 rangle!langle 0 rvert . $$
Of course, this map can be expressed using Kraus operators: for instance, you could take $K_0 = lvert 0 rangle!langle 0 rvert$ and $K_1 = lvert 0 rangle!langle 1 rvert$, which would suffice for $mathcal R(rho) = K_0^{phantomdagger} rho K_0^dagger + K_1^{phantomdagger} rho K_1^dagger$.

share|improve this answer

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be worth noting that even for unital channels, this is not always possible.
  $endgroup$
  – Norbert Schuch
  54 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You cannot always find such a Kraus decomposition. Notice that any CPTP map $mathcal E$ which does have a decomposition is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state:
$$ mathcal E(tfrac{1}{d} mathbf 1) = tfrac{1}{d} mathbf 1 . $$
This is true because $U cdot mathbf 1 cdot U^dagger = mathbf 1$ for each unitary $U$, and taking a mixture over several unitaries $U$ does not change the result for the operator $mathbf 1$.
But it is easy to find maps that don't have this property — so for such maps, there is no interpretation as a probabilistic mixture of unitary processes.
For instance, the map $mathcal R$ that resets a qubit to $lvert 0 rangle$ does not preserve the maximally mixed state on a qubit:
$$ mathcal R(rho) ,=, mathrm{tr}(rho) cdot lvert 0 rangle!langle 0 rvert . $$
Of course, this map can be expressed using Kraus operators: for instance, you could take $K_0 = lvert 0 rangle!langle 0 rvert$ and $K_1 = lvert 0 rangle!langle 1 rvert$, which would suffice for $mathcal R(rho) = K_0^{phantomdagger} rho K_0^dagger + K_1^{phantomdagger} rho K_1^dagger$.

share|improve this answer

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be worth noting that even for unital channels, this is not always possible.
  $endgroup$
  – Norbert Schuch
  54 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

You cannot always find such a Kraus decomposition. Notice that any CPTP map $mathcal E$ which does have a decomposition is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state:
$$ mathcal E(tfrac{1}{d} mathbf 1) = tfrac{1}{d} mathbf 1 . $$
This is true because $U cdot mathbf 1 cdot U^dagger = mathbf 1$ for each unitary $U$, and taking a mixture over several unitaries $U$ does not change the result for the operator $mathbf 1$.
But it is easy to find maps that don't have this property — so for such maps, there is no interpretation as a probabilistic mixture of unitary processes.
For instance, the map $mathcal R$ that resets a qubit to $lvert 0 rangle$ does not preserve the maximally mixed state on a qubit:
$$ mathcal R(rho) ,=, mathrm{tr}(rho) cdot lvert 0 rangle!langle 0 rvert . $$
Of course, this map can be expressed using Kraus operators: for instance, you could take $K_0 = lvert 0 rangle!langle 0 rvert$ and $K_1 = lvert 0 rangle!langle 1 rvert$, which would suffice for $mathcal R(rho) = K_0^{phantomdagger} rho K_0^dagger + K_1^{phantomdagger} rho K_1^dagger$.

share|improve this answer

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges

answered 8 hours ago

Niel de BeaudrapNiel de Beaudrap

7,11411 gold badge1212 silver badges4141 bronze badges