Is Lorentz symmetry broken if SUSY is broken?
$begingroup$
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
$$P^{mu}|Omegarangle=p^{0}delta^{mu}_{0}|Omegarangle$$
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:
$$P^{mu}U(Lambda)|Omegarangle=(Lambda^{-1})^{mu}_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^{mu}U^{-1}(Lambda)=Lambda^{mu}_{nu}P^{nu}$.
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 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
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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
$$P^{mu}|Omegarangle=p^{0}delta^{mu}_{0}|Omegarangle$$
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:
$$P^{mu}U(Lambda)|Omegarangle=(Lambda^{-1})^{mu}_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^{mu}U^{-1}(Lambda)=Lambda^{mu}_{nu}P^{nu}$.
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 4 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
$$P^{mu}|Omegarangle=p^{0}delta^{mu}_{0}|Omegarangle$$
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:
$$P^{mu}U(Lambda)|Omegarangle=(Lambda^{-1})^{mu}_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^{mu}U^{-1}(Lambda)=Lambda^{mu}_{nu}P^{nu}$.
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 4 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
$$P^{mu}|Omegarangle=p^{0}delta^{mu}_{0}|Omegarangle$$
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:
$$P^{mu}U(Lambda)|Omegarangle=(Lambda^{-1})^{mu}_0p^0U(Lambda)|Omegarangle$$
Since $U(Lambda)P^{mu}U^{-1}(Lambda)=Lambda^{mu}_{nu}P^{nu}$.
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 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
asked 4 hours ago
LucashWindowWasherLucashWindowWasher
1819
1819
add a comment |
add a comment |
1 Answer
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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 4 hours ago
knzhouknzhou
46.2k11124222
$endgroup$
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That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours ago
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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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 4 hours ago
knzhouknzhou
46.2k11124222
$endgroup$
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours 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 4 hours ago
knzhouknzhou
46.2k11124222
$endgroup$
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours 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 4 hours ago
knzhouknzhou
46.2k11124222
$endgroup$
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 4 hours ago
knzhouknzhou
46.2k11124222
answered 4 hours ago
knzhouknzhou
46.2k11124222
answered 4 hours ago
knzhouknzhou
46.2k11124222
answered 4 hours ago
knzhouknzhou
46.2k11124222
46.2k11124222
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours ago
add a comment |
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours ago
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours ago
$begingroup$
That makes so much sense!
$endgroup$
– LucashWindowWasher
2 hours ago
add a comment |
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