Pure states of $M_n(Bbb C)$
$begingroup$
I saw a reference book,there is a statement:the pure states of $M_n(Bbb C)$ are the rank 1 projections of $M_n(Bbb C)$.
By definition of states,they should be the positive linear functional of norm 1.So the pure states should be positive linear functional of $M_n(Bbb C)$.How to interpret the statement.
operator-theory operator-algebras c-star-algebras
edited Jan 2 at 19:33
Martin Argerami
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
$endgroup$
add a comment |
$begingroup$
I saw a reference book,there is a statement:the pure states of $M_n(Bbb C)$ are the rank 1 projections of $M_n(Bbb C)$.
By definition of states,they should be the positive linear functional of norm 1.So the pure states should be positive linear functional of $M_n(Bbb C)$.How to interpret the statement.
operator-theory operator-algebras c-star-algebras
edited Jan 2 at 19:33
Martin Argerami
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
$endgroup$
$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36
add a comment |
$begingroup$
I saw a reference book,there is a statement:the pure states of $M_n(Bbb C)$ are the rank 1 projections of $M_n(Bbb C)$.
By definition of states,they should be the positive linear functional of norm 1.So the pure states should be positive linear functional of $M_n(Bbb C)$.How to interpret the statement.
operator-theory operator-algebras c-star-algebras
edited Jan 2 at 19:33
Martin Argerami
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
$endgroup$
I saw a reference book,there is a statement:the pure states of $M_n(Bbb C)$ are the rank 1 projections of $M_n(Bbb C)$.
By definition of states,they should be the positive linear functional of norm 1.So the pure states should be positive linear functional of $M_n(Bbb C)$.How to interpret the statement.
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
edited Jan 2 at 19:33
Martin Argerami
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
edited Jan 2 at 19:33
Martin Argerami
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
edited Jan 2 at 19:33
Martin Argerami
128k1183183
edited Jan 2 at 19:33
Martin Argerami
128k1183183
edited Jan 2 at 19:33
Martin Argerami
128k1183183
128k1183183
asked Jan 2 at 16:20
mathrookiemathrookie
922512
asked Jan 2 at 16:20
mathrookiemathrookie
922512
asked Jan 2 at 16:20
mathrookiemathrookie
922512
922512
$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36
add a comment |
$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36
$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36
add a comment |
1 Answer
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It is easy to check that any linear functional on $M_n(mathbb C)$ is of the form $Xlongmapsto operatorname{Tr}(AX)$ for some $Ain M_n(mathbb C)$. Under this correspondence positive functional correspond with positive matrices, and so states correspond with positive matrices of trace 1. Among these, one can check that the pure states are precisely those given by the rank-one projections.
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
add a comment |
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$begingroup$
It is easy to check that any linear functional on $M_n(mathbb C)$ is of the form $Xlongmapsto operatorname{Tr}(AX)$ for some $Ain M_n(mathbb C)$. Under this correspondence positive functional correspond with positive matrices, and so states correspond with positive matrices of trace 1. Among these, one can check that the pure states are precisely those given by the rank-one projections.
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
add a comment |
$begingroup$
It is easy to check that any linear functional on $M_n(mathbb C)$ is of the form $Xlongmapsto operatorname{Tr}(AX)$ for some $Ain M_n(mathbb C)$. Under this correspondence positive functional correspond with positive matrices, and so states correspond with positive matrices of trace 1. Among these, one can check that the pure states are precisely those given by the rank-one projections.
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
add a comment |
$begingroup$
It is easy to check that any linear functional on $M_n(mathbb C)$ is of the form $Xlongmapsto operatorname{Tr}(AX)$ for some $Ain M_n(mathbb C)$. Under this correspondence positive functional correspond with positive matrices, and so states correspond with positive matrices of trace 1. Among these, one can check that the pure states are precisely those given by the rank-one projections.
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
$endgroup$
It is easy to check that any linear functional on $M_n(mathbb C)$ is of the form $Xlongmapsto operatorname{Tr}(AX)$ for some $Ain M_n(mathbb C)$. Under this correspondence positive functional correspond with positive matrices, and so states correspond with positive matrices of trace 1. Among these, one can check that the pure states are precisely those given by the rank-one projections.
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
answered Jan 2 at 17:42
Martin ArgeramiMartin Argerami
128k1183183
128k1183183
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
add a comment |
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
$begingroup$
Yes, you use the same definition: you take them to be the extreme points of the state space. The states on an operator system $S$ are exactly the same as the states on $C^*(S)$, so there is nothing new or different.
$endgroup$
– Martin Argerami
Jan 2 at 18:49
add a comment |
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$begingroup$
As every linear functional on $M_n(mathbb C)$ is of the form $Amapstooperatorname{tr}(AB)$ for some matrix $B$, states and density matrices have a one-to-one correspondence (and thus the pure states correspond exactly to rank-1 projections). For a more detailed answer, check out this link.
$endgroup$
– Frederik vom Ende
Jan 2 at 16:45
$begingroup$
That is indeed true. Assume that $B,B'in M_n(mathbb C)$ generate the same state $phi$ (via the trace). Define $A_{xy}:zmapstolangle x,zrangle y$ for arbitrary $x,yinmathbb C^n$, then $langle x,Byrangle=operatorname{tr}(A_{xy}B)=phi(A_{xy})=operatorname{tr}(A_{xy}B')=langle x,B'yrangle$, so all matrix elements of $B,B'$ coincide (and thus $B=B'$).
$endgroup$
– Frederik vom Ende
Jan 2 at 20:36