# Problem understanding Barut's book

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1. Oct 14, 2015

### 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 $\sigma_1\sigma_2=i\sigma_3$

2. Oct 14, 2015

### BvU

Did you sum over double indices ?

3. Oct 14, 2015

### facenian

Yes, Einstein convention is assumed

4. Oct 14, 2015

### BvU

5. Oct 14, 2015

### facenian

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

6. Oct 14, 2015

### Demystifier

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
$$\sigma_1\sigma_2=\frac{i}{2}\epsilon_{12\beta\alpha}\sigma^\beta\sigma^\alpha =\frac{i}{2}[\epsilon_{1230}\sigma^3\sigma^0+\epsilon_{1203}\sigma^0\sigma^3] =\frac{i}{2}[\sigma^3\sigma^0-\sigma^0\sigma^3]=\frac{i}{2}[\sigma^3,\sigma^0]=0$$
So you are right, and Barut is wrong.

7. Oct 14, 2015

### 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)

8. Oct 14, 2015

### 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.

9. Oct 14, 2015

### 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.

10. Oct 14, 2015

### Demystifier

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.

11. Oct 14, 2015

### 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).

12. Oct 14, 2015

### facenian

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

13. Oct 14, 2015

### samalkhaiat

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 $$\sigma_{\mu} \bar{\sigma}_{\nu}\sigma_{\rho} = (\eta_{\mu\rho} \sigma_{\nu} - \eta_{\nu\rho} \sigma_{\mu} - \eta_{\mu\nu} \sigma_{\rho} ) + i \epsilon_{\mu\nu\rho\tau} \sigma^{\tau} .$$

14. Oct 14, 2015

### facenian

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

15. Oct 14, 2015

### samalkhaiat

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

16. Oct 15, 2015

### samalkhaiat

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, $$\sigma_{1}\bar{\sigma}_{2} = -i \sigma_{3}, \ \Rightarrow \ \sigma_{1}\sigma_{2} = i \sigma_{3} .$$

17. Oct 15, 2015

### strangerep

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.

18. Oct 15, 2015

### facenian

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