Undergrad What is SR transformarion for bispinor?

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The discussion centers on the transformation properties of bispinors under Lorentz transformations, highlighting their distinction from four-vectors. It emphasizes that SO(3,1) transformations are block diagonal in the Weyl basis, which is a key characteristic of bispinors. The conversation also delves into the representation theory of the Lorentz group, clarifying that a Dirac spinor is a direct sum of two Weyl spinors, enabling parity conservation in electrodynamics and QCD. Terminological confusion arises between physicists and mathematicians regarding direct sums and tensor products, with participants seeking clarity on these concepts. Overall, the thread provides insights into the mathematical framework underpinning bispinor transformations and their representation in physics.
  • #31
samalkhaiat said:
think of the vector potential and the field tensor of Maxwell

The fact that one particular antisymmetric 4-tensor is the exterior derivative of a 4-vector (which I assume is why you're saying that in this case there are really only 4 independent components, not 6) does not mean every antisymmetric 4-tensor is the exterior derivative of a 4-vector.

samalkhaiat said:
the number of complex components is equal to [2(1) + 1][2(0) + 1] = 3.

Which is 6 real components, not 4.
 
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  • #32
PeterDonis said:
Which is 6 real components, not 4.
Why do you need them to be 4?I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).
 
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  • #33
samalkhaiat said:
Why do you need them to be 4?

I don't. I'm trying to understand how spin-1 can "correspond to an antisymmetric Lorentz tensor", as you said, when "spin-1", to me, means 4 independent components, but "antisymmetric Lorentz tensor" means 6. Basically what I understand you to be saying is that my interpretation of "spin-1" is wrong--"spin-1" is a more general term that includes multiple representations, only one of which (the Lorentz vector) has 4 independent real components.

samalkhaiat said:
I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).

So by the rule you gave, I get that (1/2, 1/2) is spin 1 and (1, 0) is spin 1. Would (1, 0) + (0, 1) still be considered spin 1?
 
  • #34
PeterDonis said:
I don't. I'm trying to understand how spin-1 can "correspond to an antisymmetric Lorentz tensor", as you said, when "spin-1", to me, means 4 independent components, but "antisymmetric Lorentz tensor" means 6. Basically what I understand you to be saying is that my interpretation of "spin-1" is wrong--"spin-1" is a more general term that includes multiple representations, only one of which (the Lorentz vector) has 4 independent real components.
An irreducible representation (j,k) has 2, in general, unrelated numbers. For non-zero mass, these are the spin and dimension: \mbox{Spin}(j,k) = j + k ,\mbox{dim}(j,k) = (2j +1)(2k +1). The rule of identification with objects on Minkowski space-time is as follows: The irreducible representation (j,k) corresponds to a massive spin-(j + k) Lorentz tensor with \mbox{dim}(j,k) independent components. For example the irrep (1,1) corresponds to a (massive) spin-2, traceless symmetric Lorentz tensor: G_{ab} = G_{ba}, \ \eta^{ab}G_{ab} = 0. Check the rule: \mbox{dim}(1,1) = 9 and this is exactly the number of independent components in a traceless symmetric tensor in 4 dimensions.

Would (1, 0) + (0, 1) still be considered spin 1?

Yes \mbox{Spin}(j,0) = \mbox{Spin}\left( (j , 0) \oplus (0 , j)\right)
 
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  • #35
It's clear that you can represent fields of a given spin in different ways. E.g., you can represent a (massive) vector field by a four-vector field ##A^{\mu}(x)##. Nothing else said, it has 4 field-degrees of freedom, but from representation theory we know that a spin-1 particle has only 3 spin (polarization) degrees of freedom. That's usually part of the equations of motion to project out the unwanted spin-degrees of freedom (in this case the spin-0 part). In both standard treatments (Proca or Stueckelberg) you end up with the contraint ##\partial_{\mu} A^{\mu}=0##, which projects out the spin-0 part.

Another way is to use an antisymmetric tensor field with the constraint ##\partial_{\mu} ^{\dagger} F^{\mu \nu}=0##.
 
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  • #36
vanhees71 said:
It's clear that you can represent fields of a given spin in different ways. E.g., you can represent a (massive) vector field by a four-vector field ##A^{\mu}(x)##. Nothing else said, it has 4 field-degrees of freedom, but from representation theory we know that a spin-1 particle has only 3 spin (polarization) degrees of freedom. That's usually part of the equations of motion to project out the unwanted spin-degrees of freedom (in this case the spin-0 part). In both standard treatments (Proca or Stueckelberg) you end up with the contraint ##\partial_{\mu} A^{\mu}=0##, which projects out the spin-0 part.
The constraint is often called Lorentz gauge condition (if it is about photons aka electromagnetic field).
What is the constrint for spinn 1/2 particles like electron? What is analog of Dirac equation(for bispinor (field) notation) for four-vector (field) notation?
 
  • #37
For massive fields ##\partial_{\mu} A^{\mu}## is not necessarily a gauge condition. Only in the Stueckelberg approach to massive vector fields you treat them as Abelian gauge fields. In contradistinction to non-Abelian gauge fields such Abelian Stueckelberg fields can have a mass.

For spin-1/2 fields you don't need constraints. For spin-1 fields you have
$$\Box A^{\mu}=-m^2 A^{\mu}, \quad \partial_{\mu} A^{\mu}=0,$$
where the latter equation is the constraint to project out the unwanted spin-0 part of the four-vector field.
 
  • #38
martinbn said:
There is no difference between the direct sum and direct product
Only when considered as a vector space only; as such they are isomorphic. But they carry additional structure, and the difference is in the implied representation of the Lorentz group!
 
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  • #39
A. Neumaier said:
Only when considered as a vector space only; as such they are isomorphic. But they carry additional structure, and the difference is in the implied representation of the Lorentz group!
How are they different!? The action is componentwise.
 
  • #40
PeterDonis said:
Is this correct? A 4-vector has 4 components, but an antisymmetric 4-tensor has 6 independent components.

The anti-symmetric tensor is built from the direct product of a dotted spin ##k = \tfrac{1}{2}## and an un-dotted spin ##l=\tfrac{1}{2}## ##\mathrm{SL}(2,\mathbb{C})## spinor, in which case it has ##(2k + 1)(2l + 1) = 4## components - an anti-symmetric tensor built from the direct product of two ##\mathrm{SO}(3,1)## (four)-vectors has ##6## components.
 
  • #41
vanhees71 said:
For spin-1/2 fields you don't need constraints. For spin-1 fields you have
$$\Box A^{\mu}=-m^2 A^{\mu}, \quad \partial_{\mu} A^{\mu}=0,$$
where the latter equation is the constraint to project out the unwanted spin-0 part of the four-vector field.
For spinn 1 field I have equations:
$$\Box A^{\mu}=-m^2 A^{\mu}, \quad \partial_{\mu} A^{\mu}=0$$
, but are the equations for spinn 1/2-spinn particle represented in four-vector field way? If represented with bispinor, the equation is Direac equation.
 
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  • #42
For spin-1/2 fields you have either the Weyl equation(s) or the Dirac equation.
 
  • #43
martinbn said:
How are they different!? The action is componentwise.
The representations ##2\oplus \bar 2## and ##2\otimes \bar2## are both 4-dimensional, but not equivalent.
 
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  • #44
In the usual notation: "##(1/2,0) \oplus (0,1/2)##" and ##(1/2,1/2)## are not equivalent".
 
  • #45
A. Neumaier said:
The representations ##2\oplus \bar 2## and ##2\otimes \bar2## are both 4-dimensional, but not equivalent.
This was already discussed! The example compares direct sum and tensor product, not direct sum and direct product.
 
  • #46
martinbn said:
This was already discussed! The example compares direct sum and tensor product, not direct sum and direct product.
There is only one product of representations; how one calls it is secondary.
 
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  • #47
vanhees71 said:
For spin-1/2 fields you have either the Weyl equation(s) or the Dirac equation.
you said that it is possible to represent 1/2-spinn field with 4-vector instead of spinor-field.
vanhees71 said:
It's clear that you can represent fields of a given spin in different ways. E.g., you can represent a (massive) vector field by a four-vector field ##A^{\mu}(x)##
How to write same condition that Dirac and Weyl equations are about spinors about four-vector field ##A^{\mu}(x)## that represents 1/2-spinn particle?
Or alternatively how to find four-vector-field ##A^{\mu}(x)## if I know bispinor-field?
 
  • #48
No, I never said this. Four-vector components transform differently from Weyl or Dirac fields. You cannot represent spin-1/2 particles with vector fields (that's also easy to see, simply because ##s=1/2## is different from ##s=1##).
 
  • #49
haushofer said:
I'm also used to that last interpretation in the context of supergravity,
Where in supergravity do you find such bi-spinor interpretation?
e.g. in the susy-transfo of vielbeine.

Under local supersymmetry, the frame field transforms as

\delta_{\epsilon}e_{\mu}{}^{a} = \frac{1}{2}\left( \epsilon^{\beta} \ (\sigma^{a})_{\beta \dot{\alpha}} \ \bar{\psi}_{\mu}^{\dot{\alpha}} + \bar{\epsilon}_{\dot{\beta}} \ (\bar{\sigma}^{a})^{\alpha \dot{\beta}} \ \psi_{\alpha \mu} \right) , where \psi_{\alpha \mu} \in (\frac{1}{2} , 1) and \bar{\psi}_{\mu}^{\dot{\alpha}} \in (1 , \frac{1}{2}), are Weyl spinor-vectors, i.e., Lorentz vectors taking values in the 2-dimenstional spin space \mathbb{C}^{2}. Introducing the Majorana bispinor-vector (the superpartner of the gravitational field or the gravitino field) \Psi_{\mu} = \begin{pmatrix} \psi_{\mu \alpha} \\ \bar{\psi}^{\dot{\alpha}}_{\mu} \end{pmatrix} \in \left(\frac{1}{2} , 1 \right) \oplus \left(1 , \frac{1}{2} \right) , the Majorana bispinor \bar{\Upsilon} = \left( \epsilon^{\beta} , \bar{\epsilon}_{\dot{\beta}}\right), and the 4 \times 4 Dirac matrices \gamma^{a} = \begin{pmatrix} 0 & (\sigma^{a})_{\beta \dot{\alpha}} \\ (\bar{\sigma}^{a})^{\dot{\beta}\alpha} & 0 \end{pmatrix} , we can write \delta_{\epsilon}e_{\mu}{}^{a} = \frac{1}{2} \bar{\Upsilon} \gamma^{a} \Psi_{\mu} . The frame field can be used to convert the world index (\mu) on the gravitinos field to a tangent-space index (a): \psi_{\alpha}^{a} = e_{\mu}{}^{a} \psi_{\alpha}^{\mu}, and the local Lorentz index can be converted into a pair of spinor indices: \psi_{\alpha}{}^{\dot{\beta} \beta} = (\bar{\sigma}^{a})^{\dot{\beta}\beta} \psi_{\alpha}^{a}. Thus, the gravitino can be described by a Majorana bispinor-vector field \Psi_{\mu}(x) or, equivalently, by a pair of (mixed) rank-3 spin tensor fields.
 
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  • #50
So it is absolutely impossible to represent particles whos spin is not 1 with vectorfield?
 

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