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?
Is Lorentz symmetry broken if SUSY is broken?
If human space travel is limited by the G force vulnerability, is there a way to counter G forces?
Arrow those variables!
Why does Kotter return in Welcome Back Kotter
intersection of two sorted vectors in C++
Combinations of multiple lists
Neighboring nodes in the network
Is it possible to download Internet Explorer on my Mac running OS X El Capitan?
Fully-Firstable Anagram Sets
What mechanic is there to disable a threat instead of killing it?
Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?
Stopping power of mountain vs road bike
Twin primes whose sum is a cube
I would say: "You are another teacher", but she is a woman and I am a man
Why is the 'in' operator throwing an error with a string literal instead of logging false?
Today is the Center
Brothers & sisters
1960's book about a plague that kills all white people
Is it canonical bit space?
Can a virus destroy the BIOS of a modern computer?
Blender 2.8 I can't see vertices, edges or faces in edit mode
Alternative to sending password over mail?
Assassin's bullet with mercury
Is it legal for company to use my work email to pretend I still work there?
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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
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