Pure states of $M_n(Bbb C)$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 19:33









Martin Argerami

124k1177176




124k1177176










asked Jan 2 at 16:20









mathrookiemathrookie

832512




832512












  • $begingroup$
    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












  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059668%2fpure-states-of-m-n-bbb-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 17:42









Martin ArgeramiMartin Argerami

124k1177176




124k1177176












  • $begingroup$
    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




$begingroup$
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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059668%2fpure-states-of-m-n-bbb-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more