Barut's Electrodynamics Identity Problem

[facenian]

On page 25 of his book "Electrodynamics and classical theory of fields and particles" he presents this identity
\sigma_\mu\sigma_\nu-\frac{i}{2}\epsilon_{\mu\nu\beta\alpha}\sigma^\beta\sigma^\alpha=\delta_{\mu\nu}
where \sigma^\mu:(\mathbf{I},-\mathbf{\sigma}) and \sigma_\mu:(\mathbf{I},\mathbf{\sigma}) , \sigma=(\sigma_1,\sigma_2,\sigma_3)\,and\,\sigma_i are the Pauli matrices, \sigma_0=\mathbf{I}
It seems that this can't be right because if we put \mu=1\,\sigma=2 then we have \sigma_1\sigma_2=0 while the correct result is \sigma_1\sigma_2=i\sigma_3

 

[BvU]

Did you sum over double indices ?

 

[facenian]

Yes, Einstein convention is assumed

 

[BvU]

This help ?

 

[facenian]

Although this article has many interesting identities it does not have one like the one I asked.

 

[Demystifier]

On page 25 of his book "Electrodynamics and classical theory of fields and particles" he presents this identity
\sigma_\mu\sigma_\nu-\frac{i}{2}\epsilon_{\mu\nu\beta\alpha}\sigma^\beta\sigma^\alpha=\delta_{\mu\nu}
where \sigma^\mu:(\mathbf{I},-\mathbf{\sigma}) and \sigma_\mu:(\mathbf{I},\mathbf{\sigma}) , \sigma=(\sigma_1,\sigma_2,\sigma_3)\,and\,\sigma_i are the Pauli matrices, \sigma_0=\mathbf{I}
It seems that this can't be right because if we put \mu=1\,\sigma=2 then we have \sigma_1\sigma_2=0 while the correct result is \sigma_1\sigma_2=i\sigma_3

Let me check by myself. For \mu=1\,\sigma=2 the identity above gives
\sigma_1\sigma_2-\frac{i}{2}\epsilon_{12\beta\alpha}\sigma^\beta\sigma^\alpha=0
Therefore
\sigma_1\sigma_2=\frac{i}{2}\epsilon_{12\beta\alpha}\sigma^\beta\sigma^\alpha<br /> =\frac{i}{2}[\epsilon_{1230}\sigma^3\sigma^0+\epsilon_{1203}\sigma^0\sigma^3] <br /> =\frac{i}{2}[\sigma^3\sigma^0-\sigma^0\sigma^3]=\frac{i}{2}[\sigma^3,\sigma^0]=0
So you are right, and Barut is wrong.

 

[facenian]

Thank you. I think it is a pitty because it would have been nice have an identity like that. He uses it to prove the relation between the restricted Lorentz group and Sl(2,C)

 

[Demystifier]

It seems that the Barut's identity is correct if $\epsilon_{\mu\nu\beta\alpha}$ is redefined. With such a redefinition it is zero when two or more indices are equal (which is standard), it is antisymmetric under exchange of spatial indices (which is also standard), but it is symmetric under exchange of a spatial index and time index (which is not standard). In particular, in my calculation above one has to take $\epsilon_{1230}=\epsilon_{1203}=-1$, which leads to the correct result.

 

[vanhees71]

Hm, but this doesn't make sense. You want a Levi-Civita Tensor, which should be totally antisymmetric. Barut's book is full of such typos or even mistakes. Don't trust any formula in it but carefully check the calculations. This is a pity, because conceptionally it's a really great book.

 

[Demystifier]

Hm, but this doesn't make sense. You want a Levi-Civita Tensor, which should be totally antisymmetric.

It seems to me that a Barut-like expression above with a true Levi-Civita is possible only if $\sigma$ 2x2 matrices are replaced with $\gamma$ 4x4 matrices. In particular, $\sigma^0$ is a unit matrix while $\gamma^0$ is not, which makes a big difference.

 

[vanhees71]

I also don't understand what Barut tries to achieve with this identity. The matrices $\sigma^{\mu}$ are related to the Weyl fermions. You introduce two sets of matrices
$$\sigma^{\mu}=(\mathbb{1},\sigma^j), \quad \overline{\sigma}^{\mu}=(1,-\sigma^i).$$
The matrices with the lower indices are defined as usual
$$\sigma_{\mu}=\eta_{\mu \nu} \sigma^{\nu}, \quad \overline{\sigma}_{\mu}=\eta_{\mu \nu} \overline{\sigma}^{\nu}.$$
Then you map any four vector $x^{\mu}$ uniquely to the hermitean 2X2-matrix
$$X=x^{\mu} \sigma_{\mu} \; \Leftrightarrow \; x^{\mu}=\frac{1}{2} \mathrm{tr} (X \overline{\sigma}^{\mu}).$$
The invariant product is
$$x_{\mu} x^{\mu}=\det X.$$
Thus any $\mathrm{SL}(2,\mathbb{C})$ matrix $L$ defines a Lorentz transformation via
$$X'=A X A^{\dagger}.$$
The "induced" proper orthochronous Lorentz transformation is given by
$${\Lambda^{\mu}}_{\nu}=\frac{1}{2} \mathrm{tr}(\overline{\sigma}^{\mu} A \sigma_{\nu} A^{\dagger}).$$
This defines a homomorphism $\mathrm{SL}(2,\mathbb{C}) \rightarrow \mathrm{SO}(1,3)^{\uparrow}$.

One cannot extend this representation to $\mathrm{O}(1,3)^{\uparrow}$, i.e., you cannot discribe spatial reflections. That's why you have to introduce another kind of two-component spinors, which transform according to the conjugate complex transformation. Then you can put both transformations together to the four-spinor (Dirac spinor) representation, which is reducible for the $\mathrm{SO}(3)^{\uparrow}$ but irreducible wrt. $\mathrm{O}(3)^{\uparrow}$. The two kinds of Weyl fermions making up the Dirac representation are interchanged by spatial reflections, thus establishing the Weyl fermions as the states with definite chirality.

For an excellent introduction to all this, see

Sexl, Urbandtke, Relativity, Groups, Particles, Springer (2001).

 

[facenian]

It seems that the Barut's identity is correct if $\epsilon_{\mu\nu\beta\alpha}$ is redefined. With such a redefinition it is zero when two or more indices are equal (which is standard), it is antisymmetric under exchange of spatial indices (which is also standard), but it is symmetric under exchange of a spatial index and time index (which is not standard). In particular, in my calculation above one has to take $\epsilon_{1230}=\epsilon_{1203}=-1$, which leads to the correct result.

Let me check by myself
\sigma_1\sigma_2=\frac{i}{2}\epsilon_{12\alpha\beta}\sigma^\alpha\sigma^\beta=\frac{i}{2}(-\sigma^0\sigma^3-\sigma^0\sigma^3)=-i\sigma^3=i\sigma_3
\sigma_0\sigma_1=\frac{i}{2}\epsilon_{01\alpha\beta}\sigma^\alpha\sigma^\beta=\frac{i}{2}(-\sigma^2\sigma^3+\sigma^3\sigma^2)=-i^2\sigma^1=\sigma^1=-\sigma_1
The case \sigma_1\sigma_2 came out right but the case \sigma_0\sigma_1 did not.

Hm, but this doesn't make sense. You want a Levi-Civita Tensor, which should be totally antisymmetric. .

It seems to me that an expression like this can be useful despite its non tensorial behavior

 

[samalkhaiat]

On page 25 of his book "Electrodynamics and classical theory of fields and particles" he presents this identity
\sigma_\mu\sigma_\nu-\frac{i}{2}\epsilon_{\mu\nu\beta\alpha}\sigma^\beta\sigma^\alpha=\delta_{\mu\nu}
where \sigma^\mu:(\mathbf{I},-\mathbf{\sigma}) and \sigma_\mu:(\mathbf{I},\mathbf{\sigma}) , \sigma=(\sigma_1,\sigma_2,\sigma_3)\,and\,\sigma_i are the Pauli matrices, \sigma_0=\mathbf{I}
It seems that this can't be right because if we put \mu=1\,\sigma=2 then we have \sigma_1\sigma_2=0 while the correct result is \sigma_1\sigma_2=i\sigma_3

In relativity, non-covariant expressions are necessarily wrong. That “identity” is neither covariant nor algebraically correct:
1) The object “\delta_{\mu\nu}” has no meaning in Minkowski space-time, because it has no well defined transformation law under the Lorentz group, i.e., not a covariant Lorentz tensor.
2) Even if one assumes that the object “\delta_{\mu\nu}” plays the role of a symmetric Kronecker-type delta, one finds that, on one hand, (\sigma_{\mu}\sigma_{\nu} - \delta_{\mu\nu}) has no definite symmetry, while, on the other hand, \epsilon_{\mu\nu\alpha\beta}\sigma^{\alpha}\sigma^{\beta} is certainly antisymmetric in (\mu,\nu). Therefore, the two expressions, (\sigma_{\mu}\sigma_{\nu} - \delta_{\mu\nu}) and \epsilon_{\mu\nu\alpha\beta}\sigma^{\alpha}\sigma^{\beta} cannot be proportional to one another.
In the 4D Minkowski space-time, the correct identity which expresses the Levi-Civita tensor \epsilon in terms of the \sigma-matrices is the following one \sigma_{\mu} \bar{\sigma}_{\nu}\sigma_{\rho} = (\eta_{\mu\rho} \sigma_{\nu} - \eta_{\nu\rho} \sigma_{\mu} - \eta_{\mu\nu} \sigma_{\rho} ) + i \epsilon_{\mu\nu\rho\tau} \sigma^{\tau} .

 

[facenian]

In the 4D Minkowski space-time, the correct identity which expresses the Levi-Civita tensor \epsilon in terms of the \sigma-matrices is the following one \sigma_{\mu} \bar{\sigma}_{\nu}\sigma_{\rho} = (\eta_{\mu\rho} \sigma_{\nu} - \eta_{\nu\rho} \sigma_{\mu} - \eta_{\mu\nu} \sigma_{\rho} ) + i \epsilon_{\mu\nu\rho\tau} \sigma^{\tau} .

Can you please mention a book where this relation is explained or derived?

 

[samalkhaiat]

Can you please mention a book where this relation is explained or derived?

You may find it in textbooks on supersymmetry and/or representation theory of the Lorentz group SL(2,C).

 

[samalkhaiat]

On page 25 of his book "Electrodynamics and classical theory of fields and particles" he presents this identity
\sigma_\mu\sigma_\nu-\frac{i}{2}\epsilon_{\mu\nu\beta\alpha}\sigma^\beta\sigma^\alpha=\delta_{\mu\nu}
where \sigma^\mu:(\mathbf{I},-\mathbf{\sigma}) and \sigma_\mu:(\mathbf{I},\mathbf{\sigma}) , \sigma=(\sigma_1,\sigma_2,\sigma_3)\,and\,\sigma_i are the Pauli matrices, \sigma_0=\mathbf{I}
It seems that this can't be right because if we put \mu=1\,\sigma=2 then we have \sigma_1\sigma_2=0 while the correct result is \sigma_1\sigma_2=i\sigma_3

Also, I should have said that the “identity” becomes correct if you put a bar on one of the sigmas, and relpace \delta with the metric tensor \eta_{\mu\nu} i.e., \sigma_{\mu}\bar{\sigma}_{\nu} + \frac{i}{2} \epsilon_{\mu\nu\alpha\beta} \ \sigma^{\alpha} \ \bar{\sigma}^{\beta} = \eta_{\mu\nu} .
Take, \mu = 3, \ \nu = 0, the identity gives you \sigma_{3} = \frac{i}{2} \epsilon_{0123}(\sigma^{1}\bar{\sigma}^{2} - \sigma^{2}\bar{\sigma}^{1}) . Using the numerical relations: \sigma^{i} = - \sigma_{i}, \bar{\sigma}^{\mu} = \sigma_{\mu} and \epsilon_{0123} = 1, we get the correct algebra [\sigma_{1},\sigma_{2}] = 2i \sigma_{3} . Another check: take \mu=1, \ \nu=2: \sigma_{1}\bar{\sigma}_{2} = \frac{i}{2} \epsilon_{1230}(\bar{\sigma}^{3}- \sigma^{3}) = - \frac{i}{2} \epsilon_{0123}(\sigma_{3} + \sigma_{3}) , So, \sigma_{1}\bar{\sigma}_{2} = -i \sigma_{3}, \ \Rightarrow \ \sigma_{1}\sigma_{2} = i \sigma_{3} .

 

[strangerep]

Barut's book is full of such typos or even mistakes. Don't trust any formula in it but carefully check the calculations.

I, too, have come to grief with some of Barut's material (in his group theory book). I have learned not to trust any of his formulas until I've verified them very carefully myself.

 

[facenian]

Exelent! thank you all guys it's been very helpful