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Is spinor the sum of scalar, vector, bi-vector, …, pseudo-scalar?
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Is spinor $psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?
Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $omega$ in Einstein-Cartan gravity
$$
omega = frac14omega^ab_mugamma_agamma_bdx^mu.
$$
The spin connection 1-form $omega$ is a
- Vector in spacetime manifold characterized by differential vector base $dx^mu$.
- Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $gamma_agamma_b$, where $gamma_a$ are Dirac vector basis.
Now back to our initial assertion: a spinor is a
Scalar in spacetime manifold characterized by differential vector base $dx^mu$. The spinor $psi$ is a $0$-form.
Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $gamma_a$: $psi = xi_s + xi_v^agamma_a + xi_bv^abgamma_agamma_b + ...$. The individual coefficients like $xi_s$ and $xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as a whole) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(barpsipsi)$, vector bi-linear (current) $tr(barpsigamma_apsi)$, bi-vector bi-linear $tr(barpsigamma_agamma_bpsi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac vector base $gamma_a$. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor.
Accordingly, there are two kinds (external and internal) of Lorentz transformations:
- External Lorentz transformation (global rotation portion of local diffeomorphism) of electromagnetic gauge field 1-form $A = A_mudx^mu$ is performed in the spacetime manifold characterized by differential vector base $dx^mu$.
- Internal Lorentz transformation of a spinor field $psi$ is performed in the internal spacetime characterized by Dirac vector base $gamma_a$.
Most of times, we are sloppy (and get away with it) with going back/forth between these two (external and internal) spaces without explicitly mentioning it, thanks to the delta function-like soldering 1-form $e$ (vielbein/frame/tetrad, a vector in spacetime manifold characterized by differential vector base $dx^mu$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $gamma_a$) in flat spacetime:
$$
e = e^a_mugamma_adx^mu = delta^a_mugamma_adx^mu,
$$
which nicely bridges the two spaces, "soldering" $gamma_a$ and $dx^mu$ via delta function $delta^a_mu$ and producing Minkowskian metric
$$eta_munu sim tr(e^a_mugamma_ae^b_nugamma_b) = tr(gamma_mu gamma_nu).
$$
vectors tensor-calculus spinors dirac-matrices clifford-algebra
asked 5 hours ago
MadMax
207110
 |Â
show 3 more comments
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Is spinor $psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?
Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $omega$ in Einstein-Cartan gravity
$$
omega = frac14omega^ab_mugamma_agamma_bdx^mu.
$$
The spin connection 1-form $omega$ is a
- Vector in spacetime manifold characterized by differential vector base $dx^mu$.
- Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $gamma_agamma_b$, where $gamma_a$ are Dirac vector basis.
Now back to our initial assertion: a spinor is a
Scalar in spacetime manifold characterized by differential vector base $dx^mu$. The spinor $psi$ is a $0$-form.
Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $gamma_a$: $psi = xi_s + xi_v^agamma_a + xi_bv^abgamma_agamma_b + ...$. The individual coefficients like $xi_s$ and $xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as a whole) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(barpsipsi)$, vector bi-linear (current) $tr(barpsigamma_apsi)$, bi-vector bi-linear $tr(barpsigamma_agamma_bpsi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac vector base $gamma_a$. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor.
Accordingly, there are two kinds (external and internal) of Lorentz transformations:
- External Lorentz transformation (global rotation portion of local diffeomorphism) of electromagnetic gauge field 1-form $A = A_mudx^mu$ is performed in the spacetime manifold characterized by differential vector base $dx^mu$.
- Internal Lorentz transformation of a spinor field $psi$ is performed in the internal spacetime characterized by Dirac vector base $gamma_a$.
Most of times, we are sloppy (and get away with it) with going back/forth between these two (external and internal) spaces without explicitly mentioning it, thanks to the delta function-like soldering 1-form $e$ (vielbein/frame/tetrad, a vector in spacetime manifold characterized by differential vector base $dx^mu$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $gamma_a$) in flat spacetime:
$$
e = e^a_mugamma_adx^mu = delta^a_mugamma_adx^mu,
$$
which nicely bridges the two spaces, "soldering" $gamma_a$ and $dx^mu$ via delta function $delta^a_mu$ and producing Minkowskian metric
$$eta_munu sim tr(e^a_mugamma_ae^b_nugamma_b) = tr(gamma_mu gamma_nu).
$$
vectors tensor-calculus spinors dirac-matrices clifford-algebra
asked 5 hours ago
MadMax
207110
2
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
1
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
1
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago
 |Â
show 3 more comments
up vote
1
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up vote
1
down vote
favorite
Is spinor $psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?
Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $omega$ in Einstein-Cartan gravity
$$
omega = frac14omega^ab_mugamma_agamma_bdx^mu.
$$
The spin connection 1-form $omega$ is a
- Vector in spacetime manifold characterized by differential vector base $dx^mu$.
- Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $gamma_agamma_b$, where $gamma_a$ are Dirac vector basis.
Now back to our initial assertion: a spinor is a
Scalar in spacetime manifold characterized by differential vector base $dx^mu$. The spinor $psi$ is a $0$-form.
Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $gamma_a$: $psi = xi_s + xi_v^agamma_a + xi_bv^abgamma_agamma_b + ...$. The individual coefficients like $xi_s$ and $xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as a whole) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(barpsipsi)$, vector bi-linear (current) $tr(barpsigamma_apsi)$, bi-vector bi-linear $tr(barpsigamma_agamma_bpsi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac vector base $gamma_a$. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor.
Accordingly, there are two kinds (external and internal) of Lorentz transformations:
- External Lorentz transformation (global rotation portion of local diffeomorphism) of electromagnetic gauge field 1-form $A = A_mudx^mu$ is performed in the spacetime manifold characterized by differential vector base $dx^mu$.
- Internal Lorentz transformation of a spinor field $psi$ is performed in the internal spacetime characterized by Dirac vector base $gamma_a$.
Most of times, we are sloppy (and get away with it) with going back/forth between these two (external and internal) spaces without explicitly mentioning it, thanks to the delta function-like soldering 1-form $e$ (vielbein/frame/tetrad, a vector in spacetime manifold characterized by differential vector base $dx^mu$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $gamma_a$) in flat spacetime:
$$
e = e^a_mugamma_adx^mu = delta^a_mugamma_adx^mu,
$$
which nicely bridges the two spaces, "soldering" $gamma_a$ and $dx^mu$ via delta function $delta^a_mu$ and producing Minkowskian metric
$$eta_munu sim tr(e^a_mugamma_ae^b_nugamma_b) = tr(gamma_mu gamma_nu).
$$
vectors tensor-calculus spinors dirac-matrices clifford-algebra
asked 5 hours ago
MadMax
207110
Is spinor $psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?
Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $omega$ in Einstein-Cartan gravity
$$
omega = frac14omega^ab_mugamma_agamma_bdx^mu.
$$
The spin connection 1-form $omega$ is a
- Vector in spacetime manifold characterized by differential vector base $dx^mu$.
- Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $gamma_agamma_b$, where $gamma_a$ are Dirac vector basis.
Now back to our initial assertion: a spinor is a
Scalar in spacetime manifold characterized by differential vector base $dx^mu$. The spinor $psi$ is a $0$-form.
Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $gamma_a$: $psi = xi_s + xi_v^agamma_a + xi_bv^abgamma_agamma_b + ...$. The individual coefficients like $xi_s$ and $xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as a whole) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(barpsipsi)$, vector bi-linear (current) $tr(barpsigamma_apsi)$, bi-vector bi-linear $tr(barpsigamma_agamma_bpsi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac vector base $gamma_a$. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor.
Accordingly, there are two kinds (external and internal) of Lorentz transformations:
- External Lorentz transformation (global rotation portion of local diffeomorphism) of electromagnetic gauge field 1-form $A = A_mudx^mu$ is performed in the spacetime manifold characterized by differential vector base $dx^mu$.
- Internal Lorentz transformation of a spinor field $psi$ is performed in the internal spacetime characterized by Dirac vector base $gamma_a$.
Most of times, we are sloppy (and get away with it) with going back/forth between these two (external and internal) spaces without explicitly mentioning it, thanks to the delta function-like soldering 1-form $e$ (vielbein/frame/tetrad, a vector in spacetime manifold characterized by differential vector base $dx^mu$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $gamma_a$) in flat spacetime:
$$
e = e^a_mugamma_adx^mu = delta^a_mugamma_adx^mu,
$$
which nicely bridges the two spaces, "soldering" $gamma_a$ and $dx^mu$ via delta function $delta^a_mu$ and producing Minkowskian metric
$$eta_munu sim tr(e^a_mugamma_ae^b_nugamma_b) = tr(gamma_mu gamma_nu).
$$
vectors tensor-calculus spinors dirac-matrices clifford-algebra
vectors tensor-calculus spinors dirac-matrices clifford-algebra
asked 5 hours ago
MadMax
207110
asked 5 hours ago
MadMax
207110
edited 8 mins ago
asked 5 hours ago
MadMax
207110
asked 5 hours ago
MadMax
207110
asked 5 hours ago
MadMax
207110
207110
2
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
1
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
1
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago
 |Â
show 3 more comments
2
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
1
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
1
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago
2
2
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
1
1
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
1
1
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago
 |Â
show 3 more comments
2 Answers
2
active
oldest
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up vote
4
down vote
There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.
The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.
A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.
However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.
The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.
answered 4 hours ago
ACuriousMind♦
68.8k17118295
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
add a comment |Â
up vote
1
down vote
I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".
The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.
If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.
The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.
answered 30 mins ago
doetoe
3,75021634
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
add a comment |Â
2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
4
down vote
There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.
The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.
A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.
However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.
The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.
answered 4 hours ago
ACuriousMind♦
68.8k17118295
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
add a comment |Â
up vote
4
down vote
There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.
The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.
A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.
However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.
The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.
answered 4 hours ago
ACuriousMind♦
68.8k17118295
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.
The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.
A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.
However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.
The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.
answered 4 hours ago
ACuriousMind♦
68.8k17118295
There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.
The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.
A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.
However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.
The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.
answered 4 hours ago
ACuriousMind♦
68.8k17118295
answered 4 hours ago
ACuriousMind♦
68.8k17118295
answered 4 hours ago
ACuriousMind♦
68.8k17118295
answered 4 hours ago
ACuriousMind♦
68.8k17118295
68.8k17118295
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
add a comment |Â
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor.
– MadMax
3 hours ago
2
2
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
@MadMax That is a non-standard claim. Why do you think so?
– ACuriousMind♦
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
See the book: books.google.com/…
– MadMax
3 hours ago
add a comment |Â
up vote
1
down vote
I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".
The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.
If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.
The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.
answered 30 mins ago
doetoe
3,75021634
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
add a comment |Â
up vote
1
down vote
I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".
The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.
If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.
The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.
answered 30 mins ago
doetoe
3,75021634
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".
The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.
If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.
The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.
answered 30 mins ago
doetoe
3,75021634
I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".
The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.
If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.
The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.
answered 30 mins ago
doetoe
3,75021634
answered 30 mins ago
doetoe
3,75021634
answered 30 mins ago
doetoe
3,75021634
answered 30 mins ago
doetoe
3,75021634
3,75021634
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
add a comment |Â
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in.
– MadMax
22 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Maybe you mean this: the spinor representation of $textSpin(mathbb R^1,3)$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors.
– doetoe
4 mins ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
Actually there are 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do real v.s complex Clifford algebra.
– MadMax
32 secs ago
add a comment |Â
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2
I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is.
– ACuriousMind♦
5 hours ago
Thank you @ACuriousMind! Updated the wording.
– MadMax
5 hours ago
Spinor $psi$ is a sum??
– Qmechanic♦
4 hours ago
1
Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor.
– Qmechanic♦
4 hours ago
1
A Dirac spinor $psi$ is a $4times 1$ column vector.
– Qmechanic♦
4 hours ago