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