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

[arkajad]

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.

 

[dextercioby]

Hi Arkajad and welcome back to PF.

For the notes you plan to write, use this book:

https://www.amazon.com/dp/9971503557/?tag=pfamazon01-20

where the right equation is stated and proved:

$$\Lambda^{\mu}_{~~\nu} = \frac{1}{2}\mbox{Tr}\left(\bar{\sigma}^{\mu}g \sigma_{\nu} g^{\dagger}\right)$$

$g\in\mbox{SL(2,C)}$.

 

[arkajad]

Hi Arkajad and welcome back to PF.

For the notes you plan to write, use this book:

https://www.amazon.com/dp/9971503557/?tag=pfamazon01-20

where the right equation is stated and proved:

$$\Lambda^{\mu}_{~~\nu} = \frac{1}{2}\mbox{Tr}\left(\bar{\sigma}^{\mu}g \sigma_{\nu} g^{\dagger}\right)$$

$g\in\mbox{SL(2,C)}$.

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".

 

[samalkhaiat]

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.

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

 

[arkajad]

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

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)

instead of (correct)

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

 

[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
$$x'^{\alpha} \neq \delta^{\alpha\beta}x'_{\beta}$$

 

[arkajad]

Yes, the reasoning in 3.83 is wrong ...

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.

 

[samalkhaiat]

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

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

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

 

[samalkhaiat]

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

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

 

[arkajad]

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

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?

 

[dextercioby]

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

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

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).

 

[arkajad]

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).

Yet

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

while

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

Therefore

$$1=-1$$

So, we are getting a contradiction.

 

[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.

 

[arkajad]

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

 

[arkajad]

Yes, but actually the <trace> is a little exotic .

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.

 

[dextercioby]

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

 

[samalkhaiat]

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?

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

 

[samalkhaiat]

Yet

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

while

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

Therefore

$$1=-1$$

So, we are getting a contradiction.

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

 

[arkajad]

You are mistaken.
Sam

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?

 

[arkajad]

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

Sam

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:

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

 

[samalkhaiat]

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

Do you agree with that?

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).

The traces are the same.

Perhaps you should consider the fact that the trace is the trace -

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

 

[samalkhaiat]

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:

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

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

 

[arkajad]

Really?

Sam

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!

 

[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.

 

[arkajad]

Both lack use of SL(2,C) indices.

They are needed for the formulation of this particular formula, as well as for its derivation.

I was not aware of the fact that samalkhaiat did not have access to Carmeli's book. If he did,
probably the whole discussion would go smoother. Also if I would have explained all the details
and all the context. So, I have learned somethings. Next time I will think twice before asking for help.
Thanks.

 

[arkajad]

My question to all partcipants of this discussion is this:

In two Carmeli's books we have two different formulas

In "Theory of Spinors" , 2000, (with Malin) , Ch. 3.4.2, p. 56 (3.84a):
$$\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr }(\sigma^\alpha g\sigma^\beta g\dagger).$$

In "Group Theory and General Relativity", (1997), Ch. 3-1, p. 36 (3-9a)
$$\Lambda^{\alpha}_{\phantom{\alpha}\beta}=\frac{1}{2}\mbox{Tr }(\sigma^\alpha g\sigma^\beta g\dagger).$$

The same context, the same definitions, the same notation: $\alpha,\beta=0,1,2,3$, $g$ is SL(2,C) matrix, $\Lambda$ - the corresponding Lorentz matrix, $\sigma$ - Pauli matrices, ${}^\dagger$ - Hermitian conjugate of a matrix, Tr - trace of the matrix.

The difference is the eta matrix (Minkowski metric)

$$\Lambda^{\alpha}_{\phantom{\alpha}\beta}=\Lambda^{\alpha\rho}\eta_{\rho\beta}.$$

At least one of the formulas must be wrong. My contention is that the second formula is right, the first one is wrong. Probably other readers will be also puzzled, will google, and will come to this thread. So, we should reach a clear consensus - which one is right and which one is wrong.

P.S. I wrote to Carmeli and my email got returned. I checked: Carmeli died in 2007. I wrote to Malin, did not receive any reply.

 

[dextercioby]

Apparently there's no consensus among the ones that do not use SL(2,C) indices when writing the equality, even though they are obviously not the same. Check out Barut's book on electrodynamics and field theory vs. Bogoliubov's et al. text on axiomatical field theory (1975, English).

 

[arkajad]

Check out Barut's book on electrodynamics and field theory vs. Bogoliubov's et al. text on axiomatical field theory (1975, English).

Checked both. Both give the second formula. (Their sigmas with lower index are exactly the same as Carmeli's sigmas with upper index - up to an irrelevant here sign of $\sigma^2$).

 

[dextercioby]

The other sources I've checked use sigma bar. They also reduce to the second formula.

 

[arkajad]

In fact the derivation is straighforward.

1) Define $\Lambda$ in terms of an SL(2,C) matrix $A$ by
$$A\sigma_\mu A^\dagger=\sigma_\nu\,\Lambda^\nu_{\phantom{\nu}\mu}$$
2) Check that
$$\mbox{Tr }(\sigma_\mu \sigma_\nu )= 2\delta_{\mu\nu}$$
3) Multiply the first equation by $\sigma_\rho$ and take the trace of both sides. Use the second equation. Rename the indices.

 

[dextercioby]

Yes, but you lack covariance. The indices summed over must be one up, one down, so that 1 should be

$$A \sigma_{\mu} A^{\dagger} = \sigma^{\nu} \eta_{\nu\lambda} \Lambda^{\lambda}_{\ ~ \mu}$$

and $\delta_{\mu\nu} \equiv \eta_{\mu\nu}$.

 

[arkajad]

But we do not demand covariance from sigma matrices. They are just a convenient fixed basis in the space of all Hermitian matrices. Once this is remembered, the rest follows. Carmeli defines his basis having upper indices, others define their basis with lower indices. Does not matter. It is just a basis that is numbered. Perhaps it should be numbered neutrally as $\sigma(\mu)$. Then there would be less reasons for a confusion.

 

[arkajad]

Perhaps I should add a clarification. We have Minkowski space M and its dual M*. Usually a basis in a vector space is written using lower indices for numbering, basis in the duac space with upper indices. Usually we identify M with Hermitian 2x2 matrices. Then we choose a basis in the space of Hermitian matrices - the Pauli sigma matrices - with lower indices for numbering of basis elements. So, we have $\sigma_\mu$ matrices defined. If we want to introduce some $\sigma^\nu$ - we have to define them first. It is up to us how we define them, but first they have to be defined. Then consequences from our definion can be derived.

But, on the other hand, we could also decide to identify the dual space with Hermitian matrices. Again, we need to introduce a basis, this time it can be convenient to use upper indices for numbering the elements of the basis.

Of course M and M* can be identified using eta, If we decide to do this, we must take care of consistency of all our definitions.

In some books the indices numbering a basis (a frame, or "moving frame" in differential geometry context, or "tetrad") the indices numbering the basis element are written differently that indices of components of tensors. This is done for a good reason - to avoid confusion. Physicists tend to think that any object carrying an index is necessarily a tensor. But this not true. The idex may simply number the four vectors of a fixed tetrad. Textbooks written by mathematicians are, I think, in this respect, more reliable than textbooks written by physicists. These later more often rely on the fact that reader will be able to figure out the meaning of the symbols used from the context.

 

[Bill_K]

In some books the indices numbering a basis (a frame, or "moving frame" in differential geometry context, or "tetrad") the indices numbering the basis element are written differently that indices of components of tensors. This is done for a good reason - to avoid confusion. Physicists tend to think that any object carrying an index is necessarily a tensor. But this not true. The idex may simply number the four vectors of a fixed tetrad.

As you say, we often encounter quantities having two different types of indices, such as a tetrad basis e_a^μ. However, rather than regarding the tetrad index as "fixed", a better viewpoint is that both types of index have a transformation group associated with them. In fact there is a local group that transforms one set of basis vectors into another, which may be taken to be the Lorentz group or something more general. The tetrad index is transformed under this group, and may also be raised and lowered using an appropriate metric.

Another common situation in which two types of indices are encountered is the embedding of surfaces in a space of higher dimension, with some indices following the transformations of the spatial coordinates and some following the surface coordinates.

 

[arkajad]

A However, rather than regarding the tetrad index as "fixed", a better viewpoint is that both types of index have a transformation group associated with them.

Indeed, you are right. I would just add to what you just wrote that this point of view becomes especially important in gauge theories, for instnce when we want to consider gravitation as a gauge theory of the Lorentz group (or, even better, of SL(2.C)) .