Fdtxjr

how to create a data type and make it available in all Databases?

Why has Russell's definition of numbers using equivalence classes been finally abandoned? ( If it has actually been abandoned).

The use of multiple foreign keys on same column in SQL Server

cryptic clue: mammal sounds like relative consumer (8)

How do you conduct xenoanthropology after first contact?

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

Copenhagen passport control - US citizen

Is there a familial term for apples and pears?

Email Account under attack (really) - anything I can do?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

Need help identifying/translating a plaque in Tangier, Morocco

Why Is Death Allowed In the Matrix?

Can I interfere when another PC is about to be attacked?

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

Why is the design of haulage companies so “special”?

Prevent a directory in /tmp from being deleted

What is the offset in a seaplane's hull?

Do airline pilots ever risk not hearing communication directed to them specifically, from traffic controllers?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

What is the command to reset a PC without deleting any files

How can I fix this gap between bookcases I made?

How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative?

Where to refill my bottle in India?

Copycat chess is back

Spectral decomposition of the resolvent map

Spectral measure associated to eigenvector of self-adjoint operatorProjection valued measure of bounded self-adjoint operator.Cauchy integral type formula for self-adjoint operatorShowing an element is in the resolventInverse spectral theorem, clarification from a video.Construction of exponential for an unbounded operator.$intfrac1mu-lambdadP(mu) (A-lambda I)=I$ & Projection Valued MeasuresCompactness of the resolvent - referencesWhy denote the spectral decomposition of a bounded operator as an integral?Relation between the support of spectral measure and proportionality of the identity operator

2

$begingroup$

Let $P_Omega$ be a projection valued measure and let $R_A(z)=(A-z)^-1$ be the resolvent map.
It can be shown that $$R_A(z)=-sum_j=0fracA^jz^j+1$$

whenever this series is defined.

My question is, given a self adjoint operator $A$, and the fact that we know that $A$ = $int lambda P(dlambda)$,
how to proof that
$$ R_A(z) = intfrac1lambda-zdP(dlambda)$$

This is assumed in Frederic Schuller lecture 11

functional-analysis quantum-mechanics

share|cite|improve this question

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

$endgroup$

add a comment | 

2

$begingroup$

Let $P_Omega$ be a projection valued measure and let $R_A(z)=(A-z)^-1$ be the resolvent map.
It can be shown that $$R_A(z)=-sum_j=0fracA^jz^j+1$$

whenever this series is defined.

My question is, given a self adjoint operator $A$, and the fact that we know that $A$ = $int lambda P(dlambda)$,
how to proof that
$$ R_A(z) = intfrac1lambda-zdP(dlambda)$$

This is assumed in Frederic Schuller lecture 11

functional-analysis quantum-mechanics

share|cite|improve this question

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

$endgroup$

add a comment | 

$begingroup$

Let $P_Omega$ be a projection valued measure and let $R_A(z)=(A-z)^-1$ be the resolvent map.
It can be shown that $$R_A(z)=-sum_j=0fracA^jz^j+1$$

whenever this series is defined.

My question is, given a self adjoint operator $A$, and the fact that we know that $A$ = $int lambda P(dlambda)$,
how to proof that
$$ R_A(z) = intfrac1lambda-zdP(dlambda)$$

This is assumed in Frederic Schuller lecture 11

functional-analysis quantum-mechanics

share|cite|improve this question

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

$endgroup$

share|cite|improve this question

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

share|cite|improve this question

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

asked Mar 22 at 11:22

amilton moreiraamilton moreira

1179

1 Answer
1

active

oldest

votes

0

$begingroup$

This follows from a general property of integrals with respect to projection-valued measures associated to self-adjoint operators. Namely, if $P$ is the unique projection-valued measure such that
$$A=intlambda ,P(dlambda) $$
then for all $f:mathbbRto mathbbC$ Borel measurable and bounded on $sigma(A)$, in symbols $fin B(sigma(A))$, we have
$$f(A)=int f(lambda),P(dlambda) qquad (star)$$
Where $f(A)$ is defined in the sense of the (Borel) functional calculus (see e.g. these notes, pag. 178). Even if you do not know the functional calculus, it suffices for you to know that if $znotin sigma(A)$ (otherwise $R_A(z)$ is not defined) and $f(lambda)=f_z(lambda)=(lambda-z)^-1$, then we have $f_z(A)=(A-z)^-1$, as one would obtain by formally replacing $lambda$ with $A$ in the expression of $f_z$.

Indeed, let us check that $f_zin B(sigma(A))$. Since $znotin sigma(A)$, $f_z$ is continuous on $sigma(A)$ and since $sigma(A)$ is compact, $f_z$ is also bounded. Therefore, substituting in $(star)$ we obtain
$$(A-z)^-1=int frac1lambda-z,P(dlambda) $$
as desired.

share|cite|improve this answer

edited Mar 25 at 15:21

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

$endgroup$

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%2f3158025%2fspectral-decomposition-of-the-resolvent-map%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

This follows from a general property of integrals with respect to projection-valued measures associated to self-adjoint operators. Namely, if $P$ is the unique projection-valued measure such that
$$A=intlambda ,P(dlambda) $$
then for all $f:mathbbRto mathbbC$ Borel measurable and bounded on $sigma(A)$, in symbols $fin B(sigma(A))$, we have
$$f(A)=int f(lambda),P(dlambda) qquad (star)$$
Where $f(A)$ is defined in the sense of the (Borel) functional calculus (see e.g. these notes, pag. 178). Even if you do not know the functional calculus, it suffices for you to know that if $znotin sigma(A)$ (otherwise $R_A(z)$ is not defined) and $f(lambda)=f_z(lambda)=(lambda-z)^-1$, then we have $f_z(A)=(A-z)^-1$, as one would obtain by formally replacing $lambda$ with $A$ in the expression of $f_z$.

Indeed, let us check that $f_zin B(sigma(A))$. Since $znotin sigma(A)$, $f_z$ is continuous on $sigma(A)$ and since $sigma(A)$ is compact, $f_z$ is also bounded. Therefore, substituting in $(star)$ we obtain
$$(A-z)^-1=int frac1lambda-z,P(dlambda) $$
as desired.

share|cite|improve this answer

edited Mar 25 at 15:21

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

$endgroup$

add a comment | 

0

$begingroup$

This follows from a general property of integrals with respect to projection-valued measures associated to self-adjoint operators. Namely, if $P$ is the unique projection-valued measure such that
$$A=intlambda ,P(dlambda) $$
then for all $f:mathbbRto mathbbC$ Borel measurable and bounded on $sigma(A)$, in symbols $fin B(sigma(A))$, we have
$$f(A)=int f(lambda),P(dlambda) qquad (star)$$
Where $f(A)$ is defined in the sense of the (Borel) functional calculus (see e.g. these notes, pag. 178). Even if you do not know the functional calculus, it suffices for you to know that if $znotin sigma(A)$ (otherwise $R_A(z)$ is not defined) and $f(lambda)=f_z(lambda)=(lambda-z)^-1$, then we have $f_z(A)=(A-z)^-1$, as one would obtain by formally replacing $lambda$ with $A$ in the expression of $f_z$.

Indeed, let us check that $f_zin B(sigma(A))$. Since $znotin sigma(A)$, $f_z$ is continuous on $sigma(A)$ and since $sigma(A)$ is compact, $f_z$ is also bounded. Therefore, substituting in $(star)$ we obtain
$$(A-z)^-1=int frac1lambda-z,P(dlambda) $$
as desired.

share|cite|improve this answer

edited Mar 25 at 15:21

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

$endgroup$

add a comment | 

$begingroup$

This follows from a general property of integrals with respect to projection-valued measures associated to self-adjoint operators. Namely, if $P$ is the unique projection-valued measure such that
$$A=intlambda ,P(dlambda) $$
then for all $f:mathbbRto mathbbC$ Borel measurable and bounded on $sigma(A)$, in symbols $fin B(sigma(A))$, we have
$$f(A)=int f(lambda),P(dlambda) qquad (star)$$
Where $f(A)$ is defined in the sense of the (Borel) functional calculus (see e.g. these notes, pag. 178). Even if you do not know the functional calculus, it suffices for you to know that if $znotin sigma(A)$ (otherwise $R_A(z)$ is not defined) and $f(lambda)=f_z(lambda)=(lambda-z)^-1$, then we have $f_z(A)=(A-z)^-1$, as one would obtain by formally replacing $lambda$ with $A$ in the expression of $f_z$.

Indeed, let us check that $f_zin B(sigma(A))$. Since $znotin sigma(A)$, $f_z$ is continuous on $sigma(A)$ and since $sigma(A)$ is compact, $f_z$ is also bounded. Therefore, substituting in $(star)$ we obtain
$$(A-z)^-1=int frac1lambda-z,P(dlambda) $$
as desired.

share|cite|improve this answer

edited Mar 25 at 15:21

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

$endgroup$

share|cite|improve this answer

edited Mar 25 at 15:21

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623

answered Mar 25 at 14:14

Lorenzo QuarisaLorenzo Quarisa

3,765623