SL(2,C) to Lorentz in Carmeli's Theory of Spinors

1. Jul 24, 2013

SL(2,C) to Lorentz in Carmeli's "Theory of Spinors"

On page 56 of "Theory of Spinors", Eq. (3.84a), Carmeli gives the formula for the Lorentz matrix in terms of Pauli matrices and an SL(2,C) matrix g:
$$\Lambda^{\alpha\beta}=(1/2)Tr(\sigma^\alpha g \sigma^\beta g^*)$$
His sigma matrices are the standard ones. It seems to me that there should be ${\Lambda^\alpha}{\phantom{\alpha}}_\beta$ on the LHS. For instance, when g is the identity Lorentz transformation, that is
$${\Lambda^\alpha}{\phantom{\alpha}}_\beta=\delta^\alpha_\beta$$
While with his formula we will get the Minkowski metric matrix instead.
Carmeli's conventions about rising and lowering indices are the standard ones.

I am asking, as I do not want to make some trivial mistake myself while writing my own lecture notes.

2. Jul 24, 2013

dextercioby

Last edited by a moderator: May 6, 2017
3. Jul 24, 2013

Thanks. That confirms my suspicions, since, as I have checked, $\bar{\sigma}^{\mu}=\sigma_\mu.$
P.S. It is, in fact, not so much "lecture notes", but a book to be published: "Quantum fractals".

Last edited by a moderator: May 6, 2017
4. Jul 24, 2013

samalkhaiat

I don’t see any thing wrong with that. The identity in $SL( 2 , C )$, i.e., $g = I_{ 2 \times 2 }$, corresponds to the identity in $SO( 1 , 3 )$. The Minkowski metric is the identity in the Lorentz group:
$$\bar{ x }^{ \alpha } = \Lambda^{ \alpha \beta } x_{ \beta } = \eta^{ \alpha \beta } x_{ \beta } = x^{ \alpha } = I_{ SO( 1 , 3 )} x^{ \alpha } .$$

Sam

5. Jul 25, 2013

According to Carmeli and Malin formula, for $g=I_{2\times 2}$:

$$\Lambda^{ \alpha \beta }=\delta^{ \alpha \beta } = \mbox{diag}(1,1,1,1)$$

(i.e. Kronecker's delta)

$$\Lambda^{ \alpha \beta }=\eta^{ \alpha \beta } = \mbox{diag}(1,-1,-1,-1)$$

6. Jul 25, 2013

dextercioby

Yes, the reasoning in 3.83 is wrong, because $\frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right) = \delta^{\alpha\beta}$, but
$$x'^{\alpha} \neq \delta^{\alpha\beta}x'_{\beta}$$

7. Jul 25, 2013

Thank you. That is also where I have detected the mistake in the derivation.
On the other hand I have checked one by one all 16 expression for ${\Lambda^\mu}_{\phantom{\mu}\nu}$ in Exercise 3.5 on p. 62, and they are all OK.

8. Jul 25, 2013

samalkhaiat

There is no such object in Minkowski space. When it comes to the Lorentz group, I very much trust Carameli. Also, I know the book in question and I do not remember any such blunder. I guess you are unfamiliar with their conventions or/and calculating the TRACE in a naïve way. When we write the relation
$$\sigma^{ \mu } \ \sigma^{ \nu } = - \eta^{ \mu \nu } + 2 i \sigma^{ \mu \nu } ,$$
what we really mean is the following $SL( 2 , C )$ realation
$$\left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } \epsilon_{ A C } + 2 i \left( \sigma^{ \mu \nu } \right)_{ A C } , \ \ \ (1)$$
where (and this is important)
$$\left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } , \ \ \ (2)$$
$$\left( \sigma^{ \mu \nu } \right)_{ AC } = \left( \sigma^{ \mu \nu } \right)_{ C A } , \ \ \ (3)$$
and
$$\epsilon_{ A B } = \epsilon^{ A B } = \left( \begin{array} {cc} 0 & -1 \\ 1 & 0 \end{array} \right) . \ \ \ (4)$$
Now, to calculate the trace of $\sigma^{ \mu } \sigma^{ \nu }$, we contract Eq(1) with $\epsilon^{ A C }$ and use Eq(3) and Eq(4)
$$\mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \epsilon^{ A C } \left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } ( - 2 ) + 0 = 2 \eta^{ \mu \nu } . \ \ (5)$$
Now, if you naively take $\mu = \nu = 1$ in the above equation, you find $2 = - 2$. This is because you are not taking Eq(2) in the consideration. What you should do is the following
$$\mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } ,$$
where $\bar{ \sigma }^{ \nu } = ( I , - \sigma^{ i } ) .$
You said the book writes $\Lambda^{ \mu \nu} \in SO( 1 , 3 )$ as
$$\Lambda^{ \mu \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } g \sigma^{ \nu } \bar{ g } \right) .$$
Therefore, for $g = \bar{ g } = I \in SL( 2 , C )$, you get
$$\Lambda^{ \mu \nu } ( I ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) .$$
Now, if you compare this with Eq(5), you find that
$$\Lambda^{ \mu \nu } ( I_{ 2 \times 2 } ) = \eta^{ \mu \nu } = I \in SO( 1 , 3 ) .$$
So there is no such object as $\delta^{ \mu \nu }$.

Sam

9. Jul 25, 2013

samalkhaiat

Is he doing something along the following lines
$$\bar{ x }^{ \mu } = \delta^{ \mu }_{ \rho } \ \bar{ x }^{ \rho } = \delta^{ \mu }_{ \rho } \ \eta^{ \rho \nu } \ \bar{ x }_{ \nu } = \delta^{ \mu }_{ \rho } ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \rho } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } = ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \mu } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } .$$
If so, why do you think there is an $\delta^{ \mu \nu }$ involved?

Sam

10. Jul 26, 2013

Well, their sigma matrices are given explicitly in (3.78):

As you can see - if I am not mistaken:

$$\mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \delta^{ \mu \nu } . \ \ (5)$$

and NOT

$$\mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \eta^{ \mu \nu } . \ \ (5)$$

Right?

11. Jul 26, 2013

dextercioby

Hi Sam,

as usual your argument is impeccable. Indeed, I've overlooked something:

$$x'^{\alpha} = \delta^{\alpha}_{\beta}x'^{\beta} = \delta^{\alpha}_{\beta}\eta^{\beta\lambda}x'_{\lambda} = \eta^{\alpha\lambda}x'_{\lambda}$$

therefore

$$x'^{\alpha} =\eta^{\alpha\beta}x'_{\beta}$$ (as expected, no delta).

and the proof that

$$\eta^{\alpha\beta} = \frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right)$$

is exactly how you wrote it, because of the $\epsilon$ metric tensor being antisymmetric for SL(2,C) as opposed to symmetric for SU(2).

12. Jul 26, 2013

Yet

$$\eta^{11}=-1$$

while

$$\frac{1}{2}\mbox{tr}(\sigma^1\sigma^1)=1$$

Therefore

$$1=-1$$

So, we are getting a contradiction.

13. Jul 26, 2013

dextercioby

Yes, but actually the <trace> is a little exotic and comes from how spinorial (dotted and undotted) indices act in that product. Their order matters, switching one dotted for one undotted is done with an epsilon matrix. The -1 which 'solves' the contradiction comes from the epsilon above, see the argument by Sam in post #8.

14. Jul 26, 2013

Somewhat desperate checked Carmeli's "Group Theory and Relativity". There it is the way I think it should be:

15. Jul 26, 2013

The trace of a matrix is the trace of a matrix. And sigmas are explicitly given matrices. The only "exotic" thing about them is that Carmeli's sign of $\sigma^2$ ("sigma two") is opposite to the "conventional", most popular sign. But it does not matter for our purpose.

Last edited: Jul 26, 2013
16. Jul 26, 2013

dextercioby

It does look different, doesn't it ? Yet it's the same author copy-pasting from Ruehl's book...

17. Jul 26, 2013

samalkhaiat

You are mistaken. From your first post, I knew that you were calculating the trace in a naïve way. The whole story of post #8 was to point out to you how one should calculate the trace in $SL( 2 , C )$. You go wrong if you treat $( \sigma^{ \mu } \sigma^{ \nu } )_{ ab }$ as an $SU( 2 )$ tensor and calculate the trace using the $SU( 2 )$ symmetric metric $\delta^{ a b }$. In $SU( 2 )$, the trace of the tensor $T$ is
$$\mbox{ Tr }_{ su( 2 )} ( T ) = \delta^{ a b } T_{ a b } ,$$
while in $SL( 2 , C)$, you calculate the trace using the antisymmetric metric $\epsilon^{ A B }$:
$$\mbox{ Tr }_{ sl( 2 , c ) } ( T ) = \epsilon^{ A B } T_{ A B } .$$
So, you should not expect the two expressions to give the same number if you don't apply the correct rule for the given group, because the trace is an INVARIANT GROUP OPERATION. Let me give you relevant examples for the above two equations: In SU(2), we have the following (tensor) relation
$$\left( \sigma^{ i } \sigma^{ j } \right)_{ a b } = \delta^{ i j } \delta_{ a b } + i \epsilon^{ i j k } \left( \sigma^{ k } \right)_{ a b } .$$
Therefore the trace is
$$\mbox{ Tr } \left( \sigma^{ i } \sigma^{ j } \right) = \delta^{ a b } \left( \sigma^{ i } \sigma^{ j } \right)_{ a b } = 2 \delta^{ i j } .$$
In SL(2,C), the corresponding relations reads
$$\left( \sigma^{ i } \sigma^{ j } \right)_{ A B } = - \eta^{ i j } \epsilon_{ A B } + 2 i \left( \sigma^{ i j } \right)_{ A B } ,$$
and
$$\mbox{ Tr } ( \sigma^{ i } \sigma^{ j } ) = \epsilon^{ A B } \left( \sigma^{ i } \sigma^{ j } \right)_{ A B } = 2 \eta^{ i j } .$$

Sam

Last edited: Jul 26, 2013
18. Jul 26, 2013

samalkhaiat

Did you read the paragraph and equation after Eq(5) in post#8? There is no contradiction:
$$\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) = \frac{ 1 }{ 2 } \left( \sigma^{ 1 } \bar{ \sigma }^{ 1 } \right)^{ A }{}_{ A } = - \frac{ 1 }{ 2 } \left( \sigma^{ 1 } \sigma^{ 1 } \right)_{ A A } = - \frac{ 1 }{ 2 } ( 2 ) = - 1.$$

Sam

19. Jul 26, 2013

Or Carmeli is mistaken. His formula from one book is different than his formula from the other book.

Do you agree with that?

The traces are the same.

Perhaps you should consider the fact that the trace is the trace - the same in both books.

If Carmeli would have some other trace in mind, he would certainly DEFINE it BEFORE using it, don't you think so?

20. Jul 27, 2013

You see, under the trace there is a product of two matrices: product of $\sigma^1$ with $\sigma^1,$ Not a product of $\sigma^1$ with $\bar{\sigma}^1$.

Therefore your LHS is not equal to your RHS:

$$\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) \neq\frac{1}{2}\left(\sigma^1\bar{\sigma}^1\right)^A_{\phantom{A}A}$$

Last edited: Jul 27, 2013
21. Jul 27, 2013

samalkhaiat

I don’t have any of Carmeli’s books on me right now. Therefore, I cannot help you with that.
However, you said that Carmeli writes, for Lorentz transformation, the expression
$$\Lambda^{ \mu \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ \mu } g \sigma^{ \nu } g^{ \dagger } ) . \ \ \ (1)$$
This expression is correct and unique for any $\pm g \in SL( 2 , C )$. Even undergraduates should have no problem deriving Eq(1).

Really? So which one of the following would you pick for the trace of the TENSOR $T_{ a b }$?
Is it $T_{ a a }, T^{ a a }, T^{ a }{}_{ a }$ or $T_{ a }{}^{ a }$?
Since $T$ is a TENSOR, the answer will depend on the GROUP under which $T_{ a b }$ transforms as a covariant tensor. For some groups, for example $SO( n )$ and $SU( 2 )$ upper and lower indices are equivalent (representations), therefore any of the above expression gives the correct answer for the trace. But, if the groups in question were $SL( 2 , C )$, $SO( m , n - m )$ or $GL( n )$ then the answer will be (depending on your convention) either $T^{ a }{}_{ a }$ or $T_{ a }{}^{ a }$.

In $SL( 2 , C )$, the $\sigma$’s are mixed spinor-tensor with (ingeneral)
$$\left( \sigma^{ \mu } \right)_{ A \dot{ B } } \neq \left( \sigma^{ \mu } \right)^{ A \dot{ B } } .$$

Indeed,
$$( \sigma^{ 0 } )_{ A \dot{ B } } = ( \sigma^{ 0 } )^{ A \dot{ B } } , ( \sigma^{ 1 } )_{ A \dot{ B } } = - ( \sigma^{ 1 } )^{ A \dot{ B } }, ( \sigma^{ 2 } )_{ A \dot{ B } } = ( \sigma^{ 2 } )^{ A \dot{ B } }, \mbox{ and } \ ( \sigma^{ 3 } )_{ A \dot{ B } } = - ( \sigma^{ 3 } )^{ A \dot{ B } } .$$
And the SL(2,C) INVARIANT trace is given by
$$\mbox{ Tr }_{ sl(2,c) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \sigma^{ \mu } )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ A \dot{ B } } \equiv ( \sigma^{ \mu } )_{ A \dot{ B } } ( \bar{ \sigma }^{ \nu } )^{ \dot{ B } A } \equiv ( \sigma^{ \mu } \sigma^{ \nu } )_{ A }{}^{ A } = 2 \eta^{ \mu \nu }$$
Notice the POSITIONS of the SL(2, C) indices, in particular, when the BARED sigma appears.

Sam

22. Jul 27, 2013

samalkhaiat

CLEARLY, you do not know much (if any thing) about the spinor-tensor notations of $SL( 2 , C )$. This is why you are REPEATING and INSISTING on REPEATING your MISTAKES. So, I have to repeat what I have said to you before “READ THE PARAGRAPH AND EQUATION AFTER Eq(5) IN POST #8” or go and read about the tensor notations in $SL( 2 , C )$. Until then, you have to take, what I have given you about the trace, as a definition.
$$\mbox{ Tr }_{ sl(2,c) } ( \sigma^{ i } \sigma^{ j } ) = \left( \sigma^{ i } \bar{ \sigma }^{ j } \right)_{ A }{}^{ A } = - \left( \sigma^{ i } \sigma^{ j } \right)_{ c c } = - \mbox{ Tr }_{ su(2) } ( \sigma^{ i } \sigma^{ j } ) .$$
Notice, that the LHS gives you $\eta^{ i j }$, while the trace on the RHS gives you $\delta^{ i j }$. Thus, we arrive at the identity
$$\eta^{ i j } = - \delta^{ i j } .$$
So, there is no monkey business, such as $1 = - 1$ in here.
I think, I have given you enough information in this thread. If you want me to continue and do proper business, then you should put some efforts and learn more about $SL( 2 , C )$.

Sam

23. Jul 27, 2013

Really. Indices alpha,beta take values 0,1,2,3. There are no spinor indices in this Carmeli's equation.

So, please, do not mix spinor indices. In Carmeli's book they are introduced only in Chapter 5. The equation I am discussing is in chapter 3.

Indices alpha,beta are space-time indices. You rise and lower them with the 4x4 flat Minkowski metric.

Sigma matrices are explicitly given. They are 2x2 matrices, and under the trace is their product.
Square of each of these matrices is the identity matrix. Its trace is 2.

$\Lambda$ is 4x4 Lorentz transformation. Normally it maps vectors into vectors, thus it has one upper and one lower index. The identity transformation has coefficients $\delta^\alpha_\beta$

We can rise the lower index to get [\itex]\eta{\lpha\beta}[/itex]

Let me repeat: This is chapter 3 in Crameli's book. No spinor indices, no bars, no epsilons, no dots over spinor indices. All elementary. Rising, lowering of spinor indices comes only in Ch. 5.

Moreover, almost the same is repeated in the other Carmeli's book. Except that this time his formula is correct. In one of his book there is an evident error. Now that I know THAT, I know that the formula with which I have started this thread is erroneous.

Your confusion probably comes from the fact that you do not have Carmeli's book, and I was not clear enough by introducing all the assumptions, notation, index ranges.

Yea, I know, spinor indices MAY BE tricky owing to the antisymmetry of epsilon metric, somewhat strange notation etc. But HERE we do not need all that. We are dealing only with Lorentz 0,1,2,3
indices and a normal trace of a 2x2 matrix.

Anyway: thanks a lot for trying to help me!!!

24. Jul 27, 2013

dextercioby

Hi Arkajad, this has developed into an argument, which shouldn't have been the case. So earlier on the first page I understood that the <Tr (...)> is not so straightforward, because the sigma's have both SL(2,C) and Lor(1,3) indices and Carmeli's writing in eqn. 3.8a hides something important, namely that the equation has no spinorial indices, which is wrong. His equation only makes sense when all dotted and undotted indices are placed and a clear meaning to what Tr(...) is given. Also quite 'disturbing' is that the $\Lambda$ has the second index raised, which means that his series of equalities should have been written

$$\Lambda^{\alpha}_{\ ~ \sigma} \eta^{\sigma\beta} = \Lambda^{\alpha\beta} = ...$$

Oh, and yes, his treatment in one book is not the same as in the other book. :D Both lack use of SL(2,C) indices. That's why I suggested the book by Mueller-Kirsten and Wiedemann. There's no (from my perspective) better treatment for Lorentz spinors than in the that book.

Last edited: Jul 27, 2013
25. Jul 27, 2013