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How are supersymmetry transformations even defined?
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I am just starting to read about supersymmetry for the first time, and there is something bothering me. Supersymmetry transformations transform between bosonic fields and fermionic fields, but I don't see how this can even be defined. Take the following simple example that appears in the beginning of the notes I'm reading. In four dimensions, say $S$ is a real scalar field, $P$ is a real pseudo scalar field, and $psi$ is a Majorana spinor. Take the Lagrangian to just be
$$mathcalL = - frac12 (partial S)^2 - frac12 (partial P)^2 - frac12 barpsi partial!!!/ psi.$$
Now, $S$ and $P$ are just real fields. (And they truly are classical fields, because Lagrangians are always functions of classical variables, even when we're interested in QFT.) However, $psi$ is made of anti commuting Grassmann variables. So we can see that the $psi$ field is not made of the same "type" of object as the $S$ and $P$ fields. In other words, $S$ is a function
$$ S: mathbbR^4 to mathbbR$$
while I believe $psi$ is a function
$$ psi: mathbbR^4 to textorder 1 elements of Gr(4, mathbbR)$$
where $Gr(4,mathbbR)$ denotes the real Grassmann algebra with 4 generators. (Please correct me if I'm wrong.) By order 1 elements, I mean linear combinations of the four generators of $Gr(4, mathbbR)$. As a vector space, this is equal to $mathbbR^4$.
How can we then consider "supersymmetry" transformations of the following form?
beginalign*
delta_varepsilon S &= barvarepsilon psi \
delta_varepsilon P &= barvarepsilon gamma_5 psi \
delta_varepsilon psi &= partial!!!/ (S + P gamma_5) varepsilon
endalign*
($varepsilon$ is a constant Majorana spinor.) $barvarepsilon psi$ is now a Grassmann number of order 2, which is simply just not the same "type" of thing as $S$, which is just a real number! So how is this "transformation" even defined?
Perhaps I do not understand how a "classical" fermion field is really supposed to work, or how these Grassmann numbers are being used.
quantum-field-theory supersymmetry fermions grassmann-numbers superalgebra
edited 2 days ago
Qmechanic♦
96.8k121631023
asked 2 days ago
user1379857
1,097515
add a comment |Â
up vote
4
down vote
favorite
I am just starting to read about supersymmetry for the first time, and there is something bothering me. Supersymmetry transformations transform between bosonic fields and fermionic fields, but I don't see how this can even be defined. Take the following simple example that appears in the beginning of the notes I'm reading. In four dimensions, say $S$ is a real scalar field, $P$ is a real pseudo scalar field, and $psi$ is a Majorana spinor. Take the Lagrangian to just be
$$mathcalL = - frac12 (partial S)^2 - frac12 (partial P)^2 - frac12 barpsi partial!!!/ psi.$$
Now, $S$ and $P$ are just real fields. (And they truly are classical fields, because Lagrangians are always functions of classical variables, even when we're interested in QFT.) However, $psi$ is made of anti commuting Grassmann variables. So we can see that the $psi$ field is not made of the same "type" of object as the $S$ and $P$ fields. In other words, $S$ is a function
$$ S: mathbbR^4 to mathbbR$$
while I believe $psi$ is a function
$$ psi: mathbbR^4 to textorder 1 elements of Gr(4, mathbbR)$$
where $Gr(4,mathbbR)$ denotes the real Grassmann algebra with 4 generators. (Please correct me if I'm wrong.) By order 1 elements, I mean linear combinations of the four generators of $Gr(4, mathbbR)$. As a vector space, this is equal to $mathbbR^4$.
How can we then consider "supersymmetry" transformations of the following form?
beginalign*
delta_varepsilon S &= barvarepsilon psi \
delta_varepsilon P &= barvarepsilon gamma_5 psi \
delta_varepsilon psi &= partial!!!/ (S + P gamma_5) varepsilon
endalign*
($varepsilon$ is a constant Majorana spinor.) $barvarepsilon psi$ is now a Grassmann number of order 2, which is simply just not the same "type" of thing as $S$, which is just a real number! So how is this "transformation" even defined?
Perhaps I do not understand how a "classical" fermion field is really supposed to work, or how these Grassmann numbers are being used.
quantum-field-theory supersymmetry fermions grassmann-numbers superalgebra
edited 2 days ago
Qmechanic♦
96.8k121631023
asked 2 days ago
user1379857
1,097515
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am just starting to read about supersymmetry for the first time, and there is something bothering me. Supersymmetry transformations transform between bosonic fields and fermionic fields, but I don't see how this can even be defined. Take the following simple example that appears in the beginning of the notes I'm reading. In four dimensions, say $S$ is a real scalar field, $P$ is a real pseudo scalar field, and $psi$ is a Majorana spinor. Take the Lagrangian to just be
$$mathcalL = - frac12 (partial S)^2 - frac12 (partial P)^2 - frac12 barpsi partial!!!/ psi.$$
Now, $S$ and $P$ are just real fields. (And they truly are classical fields, because Lagrangians are always functions of classical variables, even when we're interested in QFT.) However, $psi$ is made of anti commuting Grassmann variables. So we can see that the $psi$ field is not made of the same "type" of object as the $S$ and $P$ fields. In other words, $S$ is a function
$$ S: mathbbR^4 to mathbbR$$
while I believe $psi$ is a function
$$ psi: mathbbR^4 to textorder 1 elements of Gr(4, mathbbR)$$
where $Gr(4,mathbbR)$ denotes the real Grassmann algebra with 4 generators. (Please correct me if I'm wrong.) By order 1 elements, I mean linear combinations of the four generators of $Gr(4, mathbbR)$. As a vector space, this is equal to $mathbbR^4$.
How can we then consider "supersymmetry" transformations of the following form?
beginalign*
delta_varepsilon S &= barvarepsilon psi \
delta_varepsilon P &= barvarepsilon gamma_5 psi \
delta_varepsilon psi &= partial!!!/ (S + P gamma_5) varepsilon
endalign*
($varepsilon$ is a constant Majorana spinor.) $barvarepsilon psi$ is now a Grassmann number of order 2, which is simply just not the same "type" of thing as $S$, which is just a real number! So how is this "transformation" even defined?
Perhaps I do not understand how a "classical" fermion field is really supposed to work, or how these Grassmann numbers are being used.
quantum-field-theory supersymmetry fermions grassmann-numbers superalgebra
edited 2 days ago
Qmechanic♦
96.8k121631023
asked 2 days ago
user1379857
1,097515
I am just starting to read about supersymmetry for the first time, and there is something bothering me. Supersymmetry transformations transform between bosonic fields and fermionic fields, but I don't see how this can even be defined. Take the following simple example that appears in the beginning of the notes I'm reading. In four dimensions, say $S$ is a real scalar field, $P$ is a real pseudo scalar field, and $psi$ is a Majorana spinor. Take the Lagrangian to just be
$$mathcalL = - frac12 (partial S)^2 - frac12 (partial P)^2 - frac12 barpsi partial!!!/ psi.$$
Now, $S$ and $P$ are just real fields. (And they truly are classical fields, because Lagrangians are always functions of classical variables, even when we're interested in QFT.) However, $psi$ is made of anti commuting Grassmann variables. So we can see that the $psi$ field is not made of the same "type" of object as the $S$ and $P$ fields. In other words, $S$ is a function
$$ S: mathbbR^4 to mathbbR$$
while I believe $psi$ is a function
$$ psi: mathbbR^4 to textorder 1 elements of Gr(4, mathbbR)$$
where $Gr(4,mathbbR)$ denotes the real Grassmann algebra with 4 generators. (Please correct me if I'm wrong.) By order 1 elements, I mean linear combinations of the four generators of $Gr(4, mathbbR)$. As a vector space, this is equal to $mathbbR^4$.
How can we then consider "supersymmetry" transformations of the following form?
beginalign*
delta_varepsilon S &= barvarepsilon psi \
delta_varepsilon P &= barvarepsilon gamma_5 psi \
delta_varepsilon psi &= partial!!!/ (S + P gamma_5) varepsilon
endalign*
($varepsilon$ is a constant Majorana spinor.) $barvarepsilon psi$ is now a Grassmann number of order 2, which is simply just not the same "type" of thing as $S$, which is just a real number! So how is this "transformation" even defined?
Perhaps I do not understand how a "classical" fermion field is really supposed to work, or how these Grassmann numbers are being used.
quantum-field-theory supersymmetry fermions grassmann-numbers superalgebra
quantum-field-theory supersymmetry fermions grassmann-numbers superalgebra
edited 2 days ago
Qmechanic♦
96.8k121631023
asked 2 days ago
user1379857
1,097515
edited 2 days ago
Qmechanic♦
96.8k121631023
asked 2 days ago
user1379857
1,097515
edited 2 days ago
Qmechanic♦
96.8k121631023
edited 2 days ago
Qmechanic♦
96.8k121631023
edited 2 days ago
Qmechanic♦
96.8k121631023
96.8k121631023
asked 2 days ago
user1379857
1,097515
asked 2 days ago
user1379857
1,097515
asked 2 days ago
user1379857
1,097515
1,097515
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.
answered 2 days ago
marmot
53116
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
add a comment |Â
up vote
3
down vote
The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.
answered 2 days ago
Qmechanic♦
96.8k121631023
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.
answered 2 days ago
marmot
53116
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
add a comment |Â
up vote
3
down vote
accepted
This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.
answered 2 days ago
marmot
53116
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.
answered 2 days ago
marmot
53116
This is actually a rather subtle question, which does not really get explained in too many text books. As Qmechanic says, Grassmann variables, i.e. elements of the infinite dimensional Grassmann algebra $Lambda_infty$, have in general body and soul. Now, of course, you may say: wait, isn't the action just
$$ S=int!mathrmd^4x,mathcalL;,$$
and how can $S$ have a soul? The answer is that in this context the action here just something that enters the path integral via
$$ textcorrelator=int mathcalDpsi,mathcalDbarpsi,mathcalDphi,mathcalDP, dots exp(mathrmi,S);,$$
where I renamed your field $S$ to $phi$ in order to distinguish it from the action $S$. The important point to notice is that the correlator on the left-hand side is just an ordinary complex number, and not an element of $Lambda_infty$, because in the path integral we got rid of all algebraic objects. This is because $intmathrmdtheta,theta=1$ for a-numbers. So the upshot is that in this context $mathcalL$ and all the fields are really elements of $Lambda_infty$, and hence $barvarepsilonpsi$ is something which one can shift a scalar field by.
answered 2 days ago
marmot
53116
answered 2 days ago
marmot
53116
answered 2 days ago
marmot
53116
answered 2 days ago
marmot
53116
53116
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
add a comment |Â
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
I see. So the total path integral is both a regular path integral and a Berezin path integral. The action $S$ takes supernumbers as its input, and our transformation is taking place within the inputs of $S$, and so the total path integral is still defined.
– user1379857
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
@user1379857 Yes, that's roughly how it is. In this context, he action is just some object that allows us to compute correlators.
– marmot
2 days ago
add a comment |Â
up vote
3
down vote
The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.
answered 2 days ago
Qmechanic♦
96.8k121631023
add a comment |Â
up vote
3
down vote
The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.
answered 2 days ago
Qmechanic♦
96.8k121631023
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.
answered 2 days ago
Qmechanic♦
96.8k121631023
The Grassmann-even variables $S$ and $P$ are supernumbers, which can have both body and soul, not just body. See e.g. this related Phys.SE post.
answered 2 days ago
Qmechanic♦
96.8k121631023
edited 2 days ago
answered 2 days ago
Qmechanic♦
96.8k121631023
answered 2 days ago
Qmechanic♦
96.8k121631023
answered 2 days ago
Qmechanic♦
96.8k121631023
96.8k121631023
add a comment |Â
add a comment |Â
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