What is SR transformarion for bispinor?

[PeterDonis]

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.

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

Which is 6 real components, not 4.

 

[samalkhaiat]

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 dimension 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).

 

[PeterDonis]

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.

I gave you the rule to calculate the spin and dimension 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?

 

[samalkhaiat]

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 $\mbox{Spin}(j,0) = \mbox{Spin}\left( (j , 0) \oplus (0 , j)\right)$

 

[vanhees71]

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 $\partial_{\mu} ^{\dagger} F^{\mu \nu}=0$.

 

[olgerm]

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?

 

[vanhees71]

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.

 

[A. Neumaier]

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!

 

[martinbn]

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.

 

[formodular]

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.

 

[olgerm]

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.

 

[vanhees71]

For spin-1/2 fields you have either the Weyl equation(s) or the Dirac equation.

 

[A. Neumaier]

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.

 

[vanhees71]

In the usual notation: "$(1/2,0) \oplus (0,1/2)$" and $(1/2,1/2)$ are not equivalent".

 

[martinbn]

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.

 

[A. Neumaier]

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.

 

[olgerm]

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.

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?

 

[vanhees71]

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 $s=1$).

 

[samalkhaiat]

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.

 

[olgerm]

So it is absolutely impossible to represent particles whos spin is not 1 with vectorfield?