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Understanding the classification of quantum states based on partial transposition: representations of the bipartite density matrix

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4

1

$begingroup$

I'm going through some slides on the PPT/NPT criteria along with Horodecki's paper, and I'm kind of stuck. Let's take this slide:

Firstly, why can we write a bipartite density matrix as $sum_ijklrho_kl^ij|iranglelangle j|otimes |kranglelangle l|$? What states are the index labels $i, j, k, l$ referring to?

Secondly, Horodecki says on page 21 that if a $rho_AB$ is separable then the new matrix $rho_AB^T_B$ with matrix elements can be defined in some product basis as $langle m|langle mu |rho_AB^T_B|nrangle|nurangle = langle m|langle nu|rho_AB|nrangle|murangle$. I don't quite understand where they're getting this form from. What are $|mrangle, |nrangle, |murangle, |nurangle$?

entanglement density-matrix

share|improve this question

edited Jun 17 at 15:29

Sanchayan Dutta

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What notation are you familiar with for a bipartite density matrix?
  $endgroup$
  – Mahathi Vempati
  Jun 17 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I kinda thought about it and understand the first notation now. It's basically a tensor product of the outer product representation of density matrices.
  $endgroup$
  – Sanchayan Dutta
  Jun 17 at 15:26
  

  
  
  
  

add a comment | 

4

1

$begingroup$

I'm going through some slides on the PPT/NPT criteria along with Horodecki's paper, and I'm kind of stuck. Let's take this slide:

Firstly, why can we write a bipartite density matrix as $sum_ijklrho_kl^ij|iranglelangle j|otimes |kranglelangle l|$? What states are the index labels $i, j, k, l$ referring to?

Secondly, Horodecki says on page 21 that if a $rho_AB$ is separable then the new matrix $rho_AB^T_B$ with matrix elements can be defined in some product basis as $langle m|langle mu |rho_AB^T_B|nrangle|nurangle = langle m|langle nu|rho_AB|nrangle|murangle$. I don't quite understand where they're getting this form from. What are $|mrangle, |nrangle, |murangle, |nurangle$?

entanglement density-matrix

share|improve this question

edited Jun 17 at 15:29

Sanchayan Dutta

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What notation are you familiar with for a bipartite density matrix?
  $endgroup$
  – Mahathi Vempati
  Jun 17 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I kinda thought about it and understand the first notation now. It's basically a tensor product of the outer product representation of density matrices.
  $endgroup$
  – Sanchayan Dutta
  Jun 17 at 15:26
  

  
  
  
  

add a comment | 

$begingroup$

I'm going through some slides on the PPT/NPT criteria along with Horodecki's paper, and I'm kind of stuck. Let's take this slide:

Firstly, why can we write a bipartite density matrix as $sum_ijklrho_kl^ij|iranglelangle j|otimes |kranglelangle l|$? What states are the index labels $i, j, k, l$ referring to?

Secondly, Horodecki says on page 21 that if a $rho_AB$ is separable then the new matrix $rho_AB^T_B$ with matrix elements can be defined in some product basis as $langle m|langle mu |rho_AB^T_B|nrangle|nurangle = langle m|langle nu|rho_AB|nrangle|murangle$. I don't quite understand where they're getting this form from. What are $|mrangle, |nrangle, |murangle, |nurangle$?

entanglement density-matrix

share|improve this question

edited Jun 17 at 15:29

Sanchayan Dutta

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

$endgroup$

share|improve this question

edited Jun 17 at 15:29

Sanchayan Dutta

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

share|improve this question

edited Jun 17 at 15:29

Sanchayan Dutta

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

asked Jun 17 at 10:06

Sanchayan DuttaSanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

1 Answer
1

active

oldest

votes

5

$begingroup$

For any orthonormal basis that you pick, call it $|e_irangle$, you can write a matrix in terms of that basis as
$$
rho=sum_i,jrho_i,j|e_iranglelangle e_j|.
$$
When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write
$$
|e_irangle=|nrangle|nurangle,
$$
splitting the sum over $i$ into two sums, over $n$ and $nu$. Hence, you can write
$$
rho=sum_n,nu,m,murho_n,nu,m,mu|nranglelangle m|otimes|nuranglelanglemu|.
$$
It just happens that they've chosen to change where the different indices are put.

As for the partial transpose, that is simply the definition of the partial transpose. It may not seem so familiar, but recall with the transpose does,
$$
langle e_i|rho|e_jrangle=langle e_j|rho^T|e_irangle.
$$
All the partial transpose does is that it only switches the left and right indices for the second subsystem, not the first.

In fact, this definition of
$$
langle n,nu|rho|m,murangle=langle n,mu|rho^T_B|m,nurangle
$$
is a general definition. It's not about whether $rho$ is separable or not. The point is that if it is separable, then $rho^T_B$ is still a valid quantum state. On the other hand, if $rho$ were not separable, there is no reason why $rho^T_B$ need be a valid quantum state. In particular, it could have negative eigenvalues.

share|improve this answer

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, btw I think there was a small typo...edited it. :)
  $endgroup$
  – Sanchayan Dutta
  Jun 17 at 17:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SanchayanDutta indeed :)
  $endgroup$
  – DaftWullie
  Jun 17 at 17:55
  

  
  
  
  

add a comment | 

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5

$begingroup$

For any orthonormal basis that you pick, call it $|e_irangle$, you can write a matrix in terms of that basis as
$$
rho=sum_i,jrho_i,j|e_iranglelangle e_j|.
$$
When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write
$$
|e_irangle=|nrangle|nurangle,
$$
splitting the sum over $i$ into two sums, over $n$ and $nu$. Hence, you can write
$$
rho=sum_n,nu,m,murho_n,nu,m,mu|nranglelangle m|otimes|nuranglelanglemu|.
$$
It just happens that they've chosen to change where the different indices are put.

As for the partial transpose, that is simply the definition of the partial transpose. It may not seem so familiar, but recall with the transpose does,
$$
langle e_i|rho|e_jrangle=langle e_j|rho^T|e_irangle.
$$
All the partial transpose does is that it only switches the left and right indices for the second subsystem, not the first.

In fact, this definition of
$$
langle n,nu|rho|m,murangle=langle n,mu|rho^T_B|m,nurangle
$$
is a general definition. It's not about whether $rho$ is separable or not. The point is that if it is separable, then $rho^T_B$ is still a valid quantum state. On the other hand, if $rho$ were not separable, there is no reason why $rho^T_B$ need be a valid quantum state. In particular, it could have negative eigenvalues.

share|improve this answer

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, btw I think there was a small typo...edited it. :)
  $endgroup$
  – Sanchayan Dutta
  Jun 17 at 17:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SanchayanDutta indeed :)
  $endgroup$
  – DaftWullie
  Jun 17 at 17:55
  

  
  
  
  

add a comment | 

5

$begingroup$

For any orthonormal basis that you pick, call it $|e_irangle$, you can write a matrix in terms of that basis as
$$
rho=sum_i,jrho_i,j|e_iranglelangle e_j|.
$$
When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write
$$
|e_irangle=|nrangle|nurangle,
$$
splitting the sum over $i$ into two sums, over $n$ and $nu$. Hence, you can write
$$
rho=sum_n,nu,m,murho_n,nu,m,mu|nranglelangle m|otimes|nuranglelanglemu|.
$$
It just happens that they've chosen to change where the different indices are put.

As for the partial transpose, that is simply the definition of the partial transpose. It may not seem so familiar, but recall with the transpose does,
$$
langle e_i|rho|e_jrangle=langle e_j|rho^T|e_irangle.
$$
All the partial transpose does is that it only switches the left and right indices for the second subsystem, not the first.

In fact, this definition of
$$
langle n,nu|rho|m,murangle=langle n,mu|rho^T_B|m,nurangle
$$
is a general definition. It's not about whether $rho$ is separable or not. The point is that if it is separable, then $rho^T_B$ is still a valid quantum state. On the other hand, if $rho$ were not separable, there is no reason why $rho^T_B$ need be a valid quantum state. In particular, it could have negative eigenvalues.

share|improve this answer

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, btw I think there was a small typo...edited it. :)
  $endgroup$
  – Sanchayan Dutta
  Jun 17 at 17:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SanchayanDutta indeed :)
  $endgroup$
  – DaftWullie
  Jun 17 at 17:55
  

  
  
  
  

add a comment | 

$begingroup$

For any orthonormal basis that you pick, call it $|e_irangle$, you can write a matrix in terms of that basis as
$$
rho=sum_i,jrho_i,j|e_iranglelangle e_j|.
$$
When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write
$$
|e_irangle=|nrangle|nurangle,
$$
splitting the sum over $i$ into two sums, over $n$ and $nu$. Hence, you can write
$$
rho=sum_n,nu,m,murho_n,nu,m,mu|nranglelangle m|otimes|nuranglelanglemu|.
$$
It just happens that they've chosen to change where the different indices are put.

As for the partial transpose, that is simply the definition of the partial transpose. It may not seem so familiar, but recall with the transpose does,
$$
langle e_i|rho|e_jrangle=langle e_j|rho^T|e_irangle.
$$
All the partial transpose does is that it only switches the left and right indices for the second subsystem, not the first.

In fact, this definition of
$$
langle n,nu|rho|m,murangle=langle n,mu|rho^T_B|m,nurangle
$$
is a general definition. It's not about whether $rho$ is separable or not. The point is that if it is separable, then $rho^T_B$ is still a valid quantum state. On the other hand, if $rho$ were not separable, there is no reason why $rho^T_B$ need be a valid quantum state. In particular, it could have negative eigenvalues.

share|improve this answer

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

$endgroup$

share|improve this answer

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

edited Jun 17 at 17:23

Sanchayan Dutta

7,87744 gold badges1616 silver badges6262 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges

answered Jun 17 at 15:39

DaftWullieDaftWullie

18.2k11 gold badge77 silver badges4646 bronze badges