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Example of PPT channel that is not entanglement breaking
I was looking at the PPT$^2$ conjecture (that if $Phi$ is a PPT quantum channel, then $Phi^2$ is entanglement breaking) and it said that there are examples of PPT channels that are not entanglement breaking. However, I can neither find any examples nor think of them. What is an example of such a channel?
operator-algebras quantum-computation quantum-information
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
add a comment |
I was looking at the PPT$^2$ conjecture (that if $Phi$ is a PPT quantum channel, then $Phi^2$ is entanglement breaking) and it said that there are examples of PPT channels that are not entanglement breaking. However, I can neither find any examples nor think of them. What is an example of such a channel?
operator-algebras quantum-computation quantum-information
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
add a comment |
I was looking at the PPT$^2$ conjecture (that if $Phi$ is a PPT quantum channel, then $Phi^2$ is entanglement breaking) and it said that there are examples of PPT channels that are not entanglement breaking. However, I can neither find any examples nor think of them. What is an example of such a channel?
operator-algebras quantum-computation quantum-information
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
I was looking at the PPT$^2$ conjecture (that if $Phi$ is a PPT quantum channel, then $Phi^2$ is entanglement breaking) and it said that there are examples of PPT channels that are not entanglement breaking. However, I can neither find any examples nor think of them. What is an example of such a channel?
operator-algebras quantum-computation quantum-information
operator-algebras quantum-computation quantum-information
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
asked May 16 '17 at 17:54
Sam JaquesSam Jaques
870311
870311
add a comment |
add a comment |
1 Answer
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By Choi's theorem, $T:M_nrightarrow M_n$ is completely positive iff $A=(Idotimes T)(vv^t)$ is a state (non-normalized) in $M_notimes M_n$ , where $v=sum_{i=1}^ne_iotimes e_i$ and ${e_1,ldots,e_n}$ is the canonical basis of $mathbb{C}^n$.
Since $vv^tin M_notimes M_n$ is an entangled state (non-normalized) then
if $A=(Idotimes T)(vv^t)$ is entangled then $T$ is not entanglement breaking.
Let $Ain M_notimes M_n$, $n>2$, be any PPT entangled state. Its associated completely positive map $T$ is not entanglement breaking, but $S(cdot)=T(cdot)^t$ still is completely positive $($because $(Idotimes T(cdot)^t)(vv^t)$ is a state).
So $S((cdot)^t)^t=T((cdot)^t)$ is completely positive too.
Let $Rin M_n$ be such that $R^*T^*(Id)R=Id$. Then $RT(cdot)R^*$ is completely positive and its adjoint $R^*T^*(cdot)R$ is unital. So $RT(cdot)R^*$ is a PPT channel, but it is not entanglement breaking.
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
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By Choi's theorem, $T:M_nrightarrow M_n$ is completely positive iff $A=(Idotimes T)(vv^t)$ is a state (non-normalized) in $M_notimes M_n$ , where $v=sum_{i=1}^ne_iotimes e_i$ and ${e_1,ldots,e_n}$ is the canonical basis of $mathbb{C}^n$.
Since $vv^tin M_notimes M_n$ is an entangled state (non-normalized) then
if $A=(Idotimes T)(vv^t)$ is entangled then $T$ is not entanglement breaking.
Let $Ain M_notimes M_n$, $n>2$, be any PPT entangled state. Its associated completely positive map $T$ is not entanglement breaking, but $S(cdot)=T(cdot)^t$ still is completely positive $($because $(Idotimes T(cdot)^t)(vv^t)$ is a state).
So $S((cdot)^t)^t=T((cdot)^t)$ is completely positive too.
Let $Rin M_n$ be such that $R^*T^*(Id)R=Id$. Then $RT(cdot)R^*$ is completely positive and its adjoint $R^*T^*(cdot)R$ is unital. So $RT(cdot)R^*$ is a PPT channel, but it is not entanglement breaking.
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
add a comment |
By Choi's theorem, $T:M_nrightarrow M_n$ is completely positive iff $A=(Idotimes T)(vv^t)$ is a state (non-normalized) in $M_notimes M_n$ , where $v=sum_{i=1}^ne_iotimes e_i$ and ${e_1,ldots,e_n}$ is the canonical basis of $mathbb{C}^n$.
Since $vv^tin M_notimes M_n$ is an entangled state (non-normalized) then
if $A=(Idotimes T)(vv^t)$ is entangled then $T$ is not entanglement breaking.
Let $Ain M_notimes M_n$, $n>2$, be any PPT entangled state. Its associated completely positive map $T$ is not entanglement breaking, but $S(cdot)=T(cdot)^t$ still is completely positive $($because $(Idotimes T(cdot)^t)(vv^t)$ is a state).
So $S((cdot)^t)^t=T((cdot)^t)$ is completely positive too.
Let $Rin M_n$ be such that $R^*T^*(Id)R=Id$. Then $RT(cdot)R^*$ is completely positive and its adjoint $R^*T^*(cdot)R$ is unital. So $RT(cdot)R^*$ is a PPT channel, but it is not entanglement breaking.
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
add a comment |
By Choi's theorem, $T:M_nrightarrow M_n$ is completely positive iff $A=(Idotimes T)(vv^t)$ is a state (non-normalized) in $M_notimes M_n$ , where $v=sum_{i=1}^ne_iotimes e_i$ and ${e_1,ldots,e_n}$ is the canonical basis of $mathbb{C}^n$.
Since $vv^tin M_notimes M_n$ is an entangled state (non-normalized) then
if $A=(Idotimes T)(vv^t)$ is entangled then $T$ is not entanglement breaking.
Let $Ain M_notimes M_n$, $n>2$, be any PPT entangled state. Its associated completely positive map $T$ is not entanglement breaking, but $S(cdot)=T(cdot)^t$ still is completely positive $($because $(Idotimes T(cdot)^t)(vv^t)$ is a state).
So $S((cdot)^t)^t=T((cdot)^t)$ is completely positive too.
Let $Rin M_n$ be such that $R^*T^*(Id)R=Id$. Then $RT(cdot)R^*$ is completely positive and its adjoint $R^*T^*(cdot)R$ is unital. So $RT(cdot)R^*$ is a PPT channel, but it is not entanglement breaking.
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
By Choi's theorem, $T:M_nrightarrow M_n$ is completely positive iff $A=(Idotimes T)(vv^t)$ is a state (non-normalized) in $M_notimes M_n$ , where $v=sum_{i=1}^ne_iotimes e_i$ and ${e_1,ldots,e_n}$ is the canonical basis of $mathbb{C}^n$.
Since $vv^tin M_notimes M_n$ is an entangled state (non-normalized) then
if $A=(Idotimes T)(vv^t)$ is entangled then $T$ is not entanglement breaking.
Let $Ain M_notimes M_n$, $n>2$, be any PPT entangled state. Its associated completely positive map $T$ is not entanglement breaking, but $S(cdot)=T(cdot)^t$ still is completely positive $($because $(Idotimes T(cdot)^t)(vv^t)$ is a state).
So $S((cdot)^t)^t=T((cdot)^t)$ is completely positive too.
Let $Rin M_n$ be such that $R^*T^*(Id)R=Id$. Then $RT(cdot)R^*$ is completely positive and its adjoint $R^*T^*(cdot)R$ is unital. So $RT(cdot)R^*$ is a PPT channel, but it is not entanglement breaking.
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
edited Nov 21 '18 at 23:53
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
answered Nov 4 '18 at 21:10
DanielDaniel
4,18911022
4,18911022
add a comment |
add a comment |
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