Fdtxjr

One verb to replace 'be a member of' a club

Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?

How can saying a song's name be a copyright violation?

How to show a landlord what we have in savings?

My ex-girlfriend uses my Apple ID to log in to her iPad. Do I have to give her my Apple ID password to reset it?

Why is the sentence "Das ist eine Nase" correct?

Is there a hemisphere-neutral way of specifying a season?

What is an equivalently powerful replacement spell for the Yuan-Ti's Suggestion spell?

What is a Samsaran Word™?

Forgetting the musical notes while performing in concert

How exploitable/balanced is this homebrew spell: Spell Permanency?

Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?

Was the old ablative pronoun "med" or "mēd"?

Is there an expression that means doing something right before you will need it rather than doing it in case you might need it?

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

How dangerous is XSS

How to compactly explain secondary and tertiary characters without resorting to stereotypes?

In the UK, is it possible to get a referendum by a court decision?

Car headlights in a world without electricity

What exactly is ineptocracy?

"the same as" in a sentence

Venezuelan girlfriend wants to travel the USA to be with me. What is the process?

ssTTsSTtRrriinInnnnNNNIiinngg

How do conventional missiles fly?

An operator exponential/commutator question

Conjugate exponential integral formula for Lie algebraA linear operator between Banach spaces is weakly continuous iff norm continuous?Lie bracket as defining element for transformationsMatrix exponential converse. Baker-Campbell-HausdorffBaker Campbell Hausdorff formula for unbounded operatorsMatrix of linear operator in different bases [PMA Rudin]Why does $fracddte^X+tY |_t=0$ depend linearly on $Y$ with $X$ fixed?Exponential of unbounded normal operatorExplicit form of this Hermitian operatorUniversal property of the Baker-Campbell-Hausdorff formula

1

$begingroup$

There is "an important lemma" related to the Baker-Campbell-Haussdorff theorem which says that

$$
e^XYe^-X = Y + [X,Y] + frac12![X,[X,Y]]+ldots
$$

Clearly if $[X,Y]=0$ we get (noting that $e^-Xe^X = mathbf1$) that

$$
e^X Y-Ye^X = [e^X,Y]=0
$$

Thus $[X,Y]=0$ implies $[e^X,Y]=0$. Is the converse of this true? If $[e^X,Y]=0$ then does that imply $[X,Y]=0$? By the above formula this would imply that if

$$
[X,Y] + frac12![X,[X,Y]]+ldots =0
$$

then $[X,Y]=0$. It is not immediately obvious to me why this might be the case.

operator-theory lie-algebras matrix-exponential

share|cite|improve this question

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

$endgroup$

add a comment | 

1

$begingroup$

There is "an important lemma" related to the Baker-Campbell-Haussdorff theorem which says that

$$
e^XYe^-X = Y + [X,Y] + frac12![X,[X,Y]]+ldots
$$

Clearly if $[X,Y]=0$ we get (noting that $e^-Xe^X = mathbf1$) that

$$
e^X Y-Ye^X = [e^X,Y]=0
$$

Thus $[X,Y]=0$ implies $[e^X,Y]=0$. Is the converse of this true? If $[e^X,Y]=0$ then does that imply $[X,Y]=0$? By the above formula this would imply that if

$$
[X,Y] + frac12![X,[X,Y]]+ldots =0
$$

then $[X,Y]=0$. It is not immediately obvious to me why this might be the case.

operator-theory lie-algebras matrix-exponential

share|cite|improve this question

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

$endgroup$

add a comment | 

$begingroup$

There is "an important lemma" related to the Baker-Campbell-Haussdorff theorem which says that

$$
e^XYe^-X = Y + [X,Y] + frac12![X,[X,Y]]+ldots
$$

Clearly if $[X,Y]=0$ we get (noting that $e^-Xe^X = mathbf1$) that

$$
e^X Y-Ye^X = [e^X,Y]=0
$$

Thus $[X,Y]=0$ implies $[e^X,Y]=0$. Is the converse of this true? If $[e^X,Y]=0$ then does that imply $[X,Y]=0$? By the above formula this would imply that if

$$
[X,Y] + frac12![X,[X,Y]]+ldots =0
$$

then $[X,Y]=0$. It is not immediately obvious to me why this might be the case.

operator-theory lie-algebras matrix-exponential

share|cite|improve this question

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

$endgroup$

share|cite|improve this question

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

share|cite|improve this question

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

asked Mar 20 at 20:13

Jagerber48Jagerber48

14810

1 Answer
1

active

oldest

votes

2

$begingroup$

If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.

I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is probably a less technical way of doing this.

For $sin[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0notin D$, and $sigma(e^X)subset D$. Note that
$$
f_s(e^X)=(f_scircexp)(X)=e^sX,
$$
since we have $f_scircexp (z)=e^sz$. Then, since products are continuous, and integrals are limits of sums,
$$
f_s(e^X)Y=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)Y
=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1Y,dzright)
=Yleft(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)=Yf_s(e^X).
$$
(note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^-1Y=Y(z-e^X)^-1$). So we have that
$$
e^sXY=Ye^sX, sin[0,1].
$$
Now differentiate both sides and evaluate at $s=0$, to get
$$
XY=YX.
$$

share|cite|improve this answer

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

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',
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

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

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155960%2fan-operator-exponential-commutator-question%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.

I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is probably a less technical way of doing this.

For $sin[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0notin D$, and $sigma(e^X)subset D$. Note that
$$
f_s(e^X)=(f_scircexp)(X)=e^sX,
$$
since we have $f_scircexp (z)=e^sz$. Then, since products are continuous, and integrals are limits of sums,
$$
f_s(e^X)Y=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)Y
=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1Y,dzright)
=Yleft(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)=Yf_s(e^X).
$$
(note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^-1Y=Y(z-e^X)^-1$). So we have that
$$
e^sXY=Ye^sX, sin[0,1].
$$
Now differentiate both sides and evaluate at $s=0$, to get
$$
XY=YX.
$$

share|cite|improve this answer

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

2

$begingroup$

If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.

I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is probably a less technical way of doing this.

For $sin[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0notin D$, and $sigma(e^X)subset D$. Note that
$$
f_s(e^X)=(f_scircexp)(X)=e^sX,
$$
since we have $f_scircexp (z)=e^sz$. Then, since products are continuous, and integrals are limits of sums,
$$
f_s(e^X)Y=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)Y
=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1Y,dzright)
=Yleft(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)=Yf_s(e^X).
$$
(note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^-1Y=Y(z-e^X)^-1$). So we have that
$$
e^sXY=Ye^sX, sin[0,1].
$$
Now differentiate both sides and evaluate at $s=0$, to get
$$
XY=YX.
$$

share|cite|improve this answer

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

$begingroup$

If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.

I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is probably a less technical way of doing this.

For $sin[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0notin D$, and $sigma(e^X)subset D$. Note that
$$
f_s(e^X)=(f_scircexp)(X)=e^sX,
$$
since we have $f_scircexp (z)=e^sz$. Then, since products are continuous, and integrals are limits of sums,
$$
f_s(e^X)Y=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)Y
=left(frac12pi iint_Gamma f(z),(zI-e^X)^-1Y,dzright)
=Yleft(frac12pi iint_Gamma f(z),(zI-e^X)^-1,dzright)=Yf_s(e^X).
$$
(note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^-1Y=Y(z-e^X)^-1$). So we have that
$$
e^sXY=Ye^sX, sin[0,1].
$$
Now differentiate both sides and evaluate at $s=0$, to get
$$
XY=YX.
$$

share|cite|improve this answer

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

share|cite|improve this answer

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 21 at 3:11

Martin ArgeramiMartin Argerami

129k1184185