Trace norm inequality involving partial trace

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

While reading Holevo's book on Quantum information, I saw the following result: For any bounded operator $T in mathcalB(H_A otimes H_B)$, where $H_A$ and $H_B$ are Hilbert spaces with finite dimension, we have
$$|Tr_B(T)|_1 leq |T|_1$$
where $Tr_B(T)$ is the partial trace (see this for the definition in Holevo) of $T$ with respect to $H_B$. The norm here is the Trace norm defined as $|T|_1 = Tr(|T|) = Tr(sqrtT^*T)$.

My approach was to first show that for any operator $T$, $Tr(T) = Tr(Tr_B(T))$ and then to show $Tr_B(|T|)geq |Tr_B(T)|$ (operator inequality). From definition, I was able to show the first part easily. But the second part, which is like a Jensen inequality was elusive. Also I'm not sure if it is true but it does hold for regular trace i.e., $Tr(|T|)geq |Tr(T)|$.

Starting with the definition does not seem to help as I do not know how to work with $|Tr_B(T)|$. Could someone throw some light on this? (I appreciate hints as well as a proof if possible) Also is there an alternate, more useful way to define partial trace?

operator-theory hilbert-spaces

share|cite|improve this question

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

add a comment | 

up vote
1
down vote

favorite

While reading Holevo's book on Quantum information, I saw the following result: For any bounded operator $T in mathcalB(H_A otimes H_B)$, where $H_A$ and $H_B$ are Hilbert spaces with finite dimension, we have
$$|Tr_B(T)|_1 leq |T|_1$$
where $Tr_B(T)$ is the partial trace (see this for the definition in Holevo) of $T$ with respect to $H_B$. The norm here is the Trace norm defined as $|T|_1 = Tr(|T|) = Tr(sqrtT^*T)$.

My approach was to first show that for any operator $T$, $Tr(T) = Tr(Tr_B(T))$ and then to show $Tr_B(|T|)geq |Tr_B(T)|$ (operator inequality). From definition, I was able to show the first part easily. But the second part, which is like a Jensen inequality was elusive. Also I'm not sure if it is true but it does hold for regular trace i.e., $Tr(|T|)geq |Tr(T)|$.

Starting with the definition does not seem to help as I do not know how to work with $|Tr_B(T)|$. Could someone throw some light on this? (I appreciate hints as well as a proof if possible) Also is there an alternate, more useful way to define partial trace?

operator-theory hilbert-spaces

share|cite|improve this question

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

While reading Holevo's book on Quantum information, I saw the following result: For any bounded operator $T in mathcalB(H_A otimes H_B)$, where $H_A$ and $H_B$ are Hilbert spaces with finite dimension, we have
$$|Tr_B(T)|_1 leq |T|_1$$
where $Tr_B(T)$ is the partial trace (see this for the definition in Holevo) of $T$ with respect to $H_B$. The norm here is the Trace norm defined as $|T|_1 = Tr(|T|) = Tr(sqrtT^*T)$.

My approach was to first show that for any operator $T$, $Tr(T) = Tr(Tr_B(T))$ and then to show $Tr_B(|T|)geq |Tr_B(T)|$ (operator inequality). From definition, I was able to show the first part easily. But the second part, which is like a Jensen inequality was elusive. Also I'm not sure if it is true but it does hold for regular trace i.e., $Tr(|T|)geq |Tr(T)|$.

Starting with the definition does not seem to help as I do not know how to work with $|Tr_B(T)|$. Could someone throw some light on this? (I appreciate hints as well as a proof if possible) Also is there an alternate, more useful way to define partial trace?

operator-theory hilbert-spaces

share|cite|improve this question

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

While reading Holevo's book on Quantum information, I saw the following result: For any bounded operator $T in mathcalB(H_A otimes H_B)$, where $H_A$ and $H_B$ are Hilbert spaces with finite dimension, we have
$$|Tr_B(T)|_1 leq |T|_1$$
where $Tr_B(T)$ is the partial trace (see this for the definition in Holevo) of $T$ with respect to $H_B$. The norm here is the Trace norm defined as $|T|_1 = Tr(|T|) = Tr(sqrtT^*T)$.

My approach was to first show that for any operator $T$, $Tr(T) = Tr(Tr_B(T))$ and then to show $Tr_B(|T|)geq |Tr_B(T)|$ (operator inequality). From definition, I was able to show the first part easily. But the second part, which is like a Jensen inequality was elusive. Also I'm not sure if it is true but it does hold for regular trace i.e., $Tr(|T|)geq |Tr(T)|$.

Starting with the definition does not seem to help as I do not know how to work with $|Tr_B(T)|$. Could someone throw some light on this? (I appreciate hints as well as a proof if possible) Also is there an alternate, more useful way to define partial trace?

operator-theory hilbert-spaces

share|cite|improve this question

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

asked Aug 6 at 13:37

Gautam Shenoy

7,01211444

7,01211444

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

The proof I know uses the partial trace definition via the trace, e.g. for $Tinmathcal B_1(mathcal H_Aotimesmathcal H_B)$ now $operatornametr_B(T)$ is the unique operator such that $$operatornametr(operatornametr_B(T)X)=operatornametr(T(Xotimesmathbb 1_B))$$ for all $Xinmathcal B(mathcal H_A)$. ($mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $mathcal B$)
The key observations are the following:

1. Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $|U|=1,$.




2. For any $Xinmathcal B_1(mathcal H),Yinmathcal B(mathcal H)$ one has $|operatornametr(XY)|leq |X|_1|Y|,$.

The rest is simple as now there exists a unitary matrix $U$, such that

$$
|operatornametr_B(T)|_1=operatornametr(|operatornametr_B(T)|)=operatornametr(Uoperatornametr_B(T))=operatornametr((Uotimesmathbb 1_B)T)leq |T|_1underbrace_mathbb 1_B=|T|_1,.
$$

(Side note that $operatornametr((Uotimesmathbb 1_B)T)=|operatornametr((Uotimesmathbb 1_B)T)|$ (non-negative) as we showed that it equals $|operatornametr_B(T)|_1$, so we can actually use the above trace inequality.)

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
  – Gautam Shenoy
  Aug 7 at 6:19
  

  
  
  
  

add a comment | 

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873878%2ftrace-norm-inequality-involving-partial-trace%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The proof I know uses the partial trace definition via the trace, e.g. for $Tinmathcal B_1(mathcal H_Aotimesmathcal H_B)$ now $operatornametr_B(T)$ is the unique operator such that $$operatornametr(operatornametr_B(T)X)=operatornametr(T(Xotimesmathbb 1_B))$$ for all $Xinmathcal B(mathcal H_A)$. ($mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $mathcal B$)
The key observations are the following:

1. Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $|U|=1,$.




2. For any $Xinmathcal B_1(mathcal H),Yinmathcal B(mathcal H)$ one has $|operatornametr(XY)|leq |X|_1|Y|,$.

The rest is simple as now there exists a unitary matrix $U$, such that

$$
|operatornametr_B(T)|_1=operatornametr(|operatornametr_B(T)|)=operatornametr(Uoperatornametr_B(T))=operatornametr((Uotimesmathbb 1_B)T)leq |T|_1underbrace_mathbb 1_B=|T|_1,.
$$

(Side note that $operatornametr((Uotimesmathbb 1_B)T)=|operatornametr((Uotimesmathbb 1_B)T)|$ (non-negative) as we showed that it equals $|operatornametr_B(T)|_1$, so we can actually use the above trace inequality.)

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
  – Gautam Shenoy
  Aug 7 at 6:19
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The proof I know uses the partial trace definition via the trace, e.g. for $Tinmathcal B_1(mathcal H_Aotimesmathcal H_B)$ now $operatornametr_B(T)$ is the unique operator such that $$operatornametr(operatornametr_B(T)X)=operatornametr(T(Xotimesmathbb 1_B))$$ for all $Xinmathcal B(mathcal H_A)$. ($mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $mathcal B$)
The key observations are the following:

1. Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $|U|=1,$.




2. For any $Xinmathcal B_1(mathcal H),Yinmathcal B(mathcal H)$ one has $|operatornametr(XY)|leq |X|_1|Y|,$.

The rest is simple as now there exists a unitary matrix $U$, such that

$$
|operatornametr_B(T)|_1=operatornametr(|operatornametr_B(T)|)=operatornametr(Uoperatornametr_B(T))=operatornametr((Uotimesmathbb 1_B)T)leq |T|_1underbrace_mathbb 1_B=|T|_1,.
$$

(Side note that $operatornametr((Uotimesmathbb 1_B)T)=|operatornametr((Uotimesmathbb 1_B)T)|$ (non-negative) as we showed that it equals $|operatornametr_B(T)|_1$, so we can actually use the above trace inequality.)

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
  – Gautam Shenoy
  Aug 7 at 6:19
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The proof I know uses the partial trace definition via the trace, e.g. for $Tinmathcal B_1(mathcal H_Aotimesmathcal H_B)$ now $operatornametr_B(T)$ is the unique operator such that $$operatornametr(operatornametr_B(T)X)=operatornametr(T(Xotimesmathbb 1_B))$$ for all $Xinmathcal B(mathcal H_A)$. ($mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $mathcal B$)
The key observations are the following:

1. Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $|U|=1,$.




2. For any $Xinmathcal B_1(mathcal H),Yinmathcal B(mathcal H)$ one has $|operatornametr(XY)|leq |X|_1|Y|,$.

The rest is simple as now there exists a unitary matrix $U$, such that

$$
|operatornametr_B(T)|_1=operatornametr(|operatornametr_B(T)|)=operatornametr(Uoperatornametr_B(T))=operatornametr((Uotimesmathbb 1_B)T)leq |T|_1underbrace_mathbb 1_B=|T|_1,.
$$

(Side note that $operatornametr((Uotimesmathbb 1_B)T)=|operatornametr((Uotimesmathbb 1_B)T)|$ (non-negative) as we showed that it equals $|operatornametr_B(T)|_1$, so we can actually use the above trace inequality.)

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

The proof I know uses the partial trace definition via the trace, e.g. for $Tinmathcal B_1(mathcal H_Aotimesmathcal H_B)$ now $operatornametr_B(T)$ is the unique operator such that $$operatornametr(operatornametr_B(T)X)=operatornametr(T(Xotimesmathbb 1_B))$$ for all $Xinmathcal B(mathcal H_A)$. ($mathcal B_1$ is the trace class but no worries, in finite dimensions this is the same as $mathcal B$)
The key observations are the following:

1. Every operator admits a polar decomposition $T=U|T|$ with $U$ unitary (or in infinite dimensions: $U$ partial isometry), so in particular $|U|=1,$.




2. For any $Xinmathcal B_1(mathcal H),Yinmathcal B(mathcal H)$ one has $|operatornametr(XY)|leq |X|_1|Y|,$.

The rest is simple as now there exists a unitary matrix $U$, such that

$$
|operatornametr_B(T)|_1=operatornametr(|operatornametr_B(T)|)=operatornametr(Uoperatornametr_B(T))=operatornametr((Uotimesmathbb 1_B)T)leq |T|_1underbrace_mathbb 1_B=|T|_1,.
$$

(Side note that $operatornametr((Uotimesmathbb 1_B)T)=|operatornametr((Uotimesmathbb 1_B)T)|$ (non-negative) as we showed that it equals $|operatornametr_B(T)|_1$, so we can actually use the above trace inequality.)

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 19:53

Frederik vom Ende

4081119

answered Aug 6 at 19:53

Frederik vom Ende

4081119

answered Aug 6 at 19:53

Frederik vom Ende

4081119

4081119

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
  – Gautam Shenoy
  Aug 7 at 6:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
  – Gautam Shenoy
  Aug 7 at 6:19
  

  
  
  
  

1

1

Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
– Gautam Shenoy
Aug 7 at 6:19

Superb explanation. I had gotten to the polar decomposition part but for some reason i didnt think of applying the trace inequality. Thanks.
– Gautam Shenoy
Aug 7 at 6:19

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873878%2ftrace-norm-inequality-involving-partial-trace%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email