Is Lorentz symmetry broken if SUSY is broken?Multiple vacua vs. vev's in qftIs broken supersymmetry compatible with a small cosmological constant?Why must SUSY be broken?Lorentz transformation of the vacuum stateSupersymmetric background and fermion variationsVacuum energy and supersymmetryCan Poincare representations be embedded in non-standard Lorentz representations?What does soft symmetry breaking physically mean?SUSY vacuum has 0 energy?What does Lorentz index structure say about a full-fledged correlator?
90's TV series where a boy goes to another dimension through portal near power lines
How to prevent "they're falling in love" trope
How to model explosives?
I Accidentally Deleted a Stock Terminal Theme
Arrow those variables!
AES: Why is it a good practice to use only the first 16bytes of a hash for encryption?
Assassin's bullet with mercury
prove that the matrix A is diagonalizable
How can I make my BBEG immortal short of making them a Lich or Vampire?
Is it possible to run Internet Explorer on OS X El Capitan?
Why does Kotter return in Welcome Back Kotter
Emailing HOD to enhance faculty application
Why are electrically insulating heatsinks so rare? Is it just cost?
Is it unprofessional to ask if a job posting on GlassDoor is real?
What exploit are these user agents trying to use?
How can saying a song's name be a copyright violation?
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
What's the difference between 'rename' and 'mv'?
Took a trip to a parallel universe, need help deciphering
In a Spin are Both Wings Stalled?
Watching something be written to a file live with tail
Is Lorentz symmetry broken if SUSY is broken?
If a Gelatinous Cube takes up the entire space of a Pit Trap, what happens when a creature falls into the trap but succeeds on the saving throw?
Where does SFDX store details about scratch orgs?
Is Lorentz symmetry broken if SUSY is broken?
Multiple vacua vs. vev's in qftIs broken supersymmetry compatible with a small cosmological constant?Why must SUSY be broken?Lorentz transformation of the vacuum stateSupersymmetric background and fermion variationsVacuum energy and supersymmetryCan Poincare representations be embedded in non-standard Lorentz representations?What does soft symmetry breaking physically mean?SUSY vacuum has 0 energy?What does Lorentz index structure say about a full-fledged correlator?
$begingroup$
I have seemingly convinced myself that the entire Poincare group is spontaneously broken if one of the supersymmetric charges is spontaneously broken.
We know that if one of the supersymmetric charges is spontaneously broken, then a vacuum with zero three-momentum MUST have a nonzero energy. There is no way to re-scale the Hamiltonian since the supersymmetry algebra provides an absolute scale. Let's suppose the vacuum is an eigenstate of $P^mu$, then we have
$$P^mu|Omegarangle=p^0delta^mu_0|Omegarangle$$
If we lorentz transform this equation with the unitary operator $U(Lambda)$, we find that a new state $U(Lambda)|Omegarangle$ solves the eigenvalue equation:
$$P^muU(Lambda)|Omegarangle=(Lambda^-1)^mu_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^muU^-1(Lambda)=Lambda^mu_nuP^nu$.
Therefore we have a whole family of vacua which are orthogonal and related by a lorentz transformation.
Is there something I am missing here? Is this even a bad thing?
quantum-field-theory special-relativity supersymmetry lorentz-symmetry symmetry-breaking
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
$endgroup$
add a comment |
$begingroup$
I have seemingly convinced myself that the entire Poincare group is spontaneously broken if one of the supersymmetric charges is spontaneously broken.
We know that if one of the supersymmetric charges is spontaneously broken, then a vacuum with zero three-momentum MUST have a nonzero energy. There is no way to re-scale the Hamiltonian since the supersymmetry algebra provides an absolute scale. Let's suppose the vacuum is an eigenstate of $P^mu$, then we have
$$P^mu|Omegarangle=p^0delta^mu_0|Omegarangle$$
If we lorentz transform this equation with the unitary operator $U(Lambda)$, we find that a new state $U(Lambda)|Omegarangle$ solves the eigenvalue equation:
$$P^muU(Lambda)|Omegarangle=(Lambda^-1)^mu_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^muU^-1(Lambda)=Lambda^mu_nuP^nu$.
Therefore we have a whole family of vacua which are orthogonal and related by a lorentz transformation.
Is there something I am missing here? Is this even a bad thing?
quantum-field-theory special-relativity supersymmetry lorentz-symmetry symmetry-breaking
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
$endgroup$
add a comment |
$begingroup$
I have seemingly convinced myself that the entire Poincare group is spontaneously broken if one of the supersymmetric charges is spontaneously broken.
We know that if one of the supersymmetric charges is spontaneously broken, then a vacuum with zero three-momentum MUST have a nonzero energy. There is no way to re-scale the Hamiltonian since the supersymmetry algebra provides an absolute scale. Let's suppose the vacuum is an eigenstate of $P^mu$, then we have
$$P^mu|Omegarangle=p^0delta^mu_0|Omegarangle$$
If we lorentz transform this equation with the unitary operator $U(Lambda)$, we find that a new state $U(Lambda)|Omegarangle$ solves the eigenvalue equation:
$$P^muU(Lambda)|Omegarangle=(Lambda^-1)^mu_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^muU^-1(Lambda)=Lambda^mu_nuP^nu$.
Therefore we have a whole family of vacua which are orthogonal and related by a lorentz transformation.
Is there something I am missing here? Is this even a bad thing?
quantum-field-theory special-relativity supersymmetry lorentz-symmetry symmetry-breaking
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
$endgroup$
I have seemingly convinced myself that the entire Poincare group is spontaneously broken if one of the supersymmetric charges is spontaneously broken.
We know that if one of the supersymmetric charges is spontaneously broken, then a vacuum with zero three-momentum MUST have a nonzero energy. There is no way to re-scale the Hamiltonian since the supersymmetry algebra provides an absolute scale. Let's suppose the vacuum is an eigenstate of $P^mu$, then we have
$$P^mu|Omegarangle=p^0delta^mu_0|Omegarangle$$
If we lorentz transform this equation with the unitary operator $U(Lambda)$, we find that a new state $U(Lambda)|Omegarangle$ solves the eigenvalue equation:
$$P^muU(Lambda)|Omegarangle=(Lambda^-1)^mu_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^muU^-1(Lambda)=Lambda^mu_nuP^nu$.
Therefore we have a whole family of vacua which are orthogonal and related by a lorentz transformation.
Is there something I am missing here? Is this even a bad thing?
quantum-field-theory special-relativity supersymmetry lorentz-symmetry symmetry-breaking
quantum-field-theory special-relativity supersymmetry lorentz-symmetry symmetry-breaking
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 3 hours ago
LucashWindowWasherLucashWindowWasher
1819
1819
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.
I can hear you complaining that this messes up the SUSY algebra since $Q, Q sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.
answered 3 hours ago
knzhouknzhou
46.1k11124222
$endgroup$
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
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: "151"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470609%2fis-lorentz-symmetry-broken-if-susy-is-broken%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
$begingroup$
No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.
I can hear you complaining that this messes up the SUSY algebra since $Q, Q sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.
answered 3 hours ago
knzhouknzhou
46.1k11124222
$endgroup$
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
add a comment |
$begingroup$
No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.
I can hear you complaining that this messes up the SUSY algebra since $Q, Q sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.
answered 3 hours ago
knzhouknzhou
46.1k11124222
$endgroup$
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
add a comment |
$begingroup$
No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.
I can hear you complaining that this messes up the SUSY algebra since $Q, Q sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.
answered 3 hours ago
knzhouknzhou
46.1k11124222
$endgroup$
No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.
I can hear you complaining that this messes up the SUSY algebra since $Q, Q sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.
answered 3 hours ago
knzhouknzhou
46.1k11124222
answered 3 hours ago
knzhouknzhou
46.1k11124222
answered 3 hours ago
knzhouknzhou
46.1k11124222
answered 3 hours ago
knzhouknzhou
46.1k11124222
46.1k11124222
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
add a comment |
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
1 hour ago
add a comment |
Thanks for contributing an answer to Physics 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.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470609%2fis-lorentz-symmetry-broken-if-susy-is-broken%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
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
