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

[dextercioby]

Arkajad, do share with us if you get a reply from the co-author of <Theory of Spinors>. Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

 

[arkajad]

I checked. He is 76 years old professor emeritus, probably busy writing another book after his "Nature Loves to Hide: Quantum Physics and Reality, a Western Perspective". So, I am not going to disturb him any more with such a small detail that we can figure out ourselves. It is a good exercise anyway. Forces you to think.

 

[arkajad]

Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

Oh, yes. These authors are funny! They introduce the sigmas tilded, with an upper index. They are identical to those ones with lower index. And there is an explicit warning the reader that this is not the same as using eta to change the index position.

 

[Bill_K]

Arkajad, do share with us if you get a reply from the co-author of <Theory of Spinors>. Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

Is there still a question remaining on this? My impression is that the OP had been answered completely, several times, beaten to death in fact! But for what it's worth, I'll summarize it all over again.

The Pauli matrices σ^μ_AB' are a conversion between the Lorentz vector and the (1/2, 1/2) representation of SL(2,C). While they can be represented as a fixed set of numbers, they are not limited to these values, and under a change of basis transform appropriately according to all three of their indices. Yes, covariance must be maintained.


A vector V^μ corresponds to a spinor V^AB' by V^μ = σ^μ_AB' V^AB', or conversely V^AB' = V^μ σ^μ_AB'. (I'm ignoring any leading 1/2's, which in the present context serve only as a distraction. Fill them in if you like.)

The norm of the vector can be written either V_μ V^μ or V_AB' V^AB', requiring the identity η_μν = σ_{μ AB'} σ_ν ^AB'.

I'm sure Carmeli covers this adequately in Chapter 5, but in Chapter 3 he suppresses the spinor indices and tries to write it as a matrix equation. Well it can't be written as a matrix equation without pulling some tricks! The indices simply do not agree, and they must. We need to understand that in Chapter 3 the matrix operations (product and trace) are not the usual ones. The best you can do perhaps is

σ_{μ AB'} σ_ν ^AB' = σ_{μ AB'} ε^B'D' σ_{ν CD'} ε^AC = - σ_{μ AB'} ε^B'D' σ^T_{ν D'C} ε^CA = - Tr(σ_μ ε σ^T_ν ε)

(Where T is the transpose. Or since σ is Hermitian the transpose could be replaced by the complex conjugate.)

 

[arkajad]

The indices simply do not agree, and they must

In fact, they don't. Example: Carmeli's 1977 book and other textbooks mentioned. But we must remember their meaning. For instance, if I introduce a fixed tetrad $e_\mu$ of four tangent vectors, this index does not transform under change of coordinates. It is not a tensorial index. Once you remember the meaning - you are safe.

Physicists quite often instead of doing geometry do what I call "indexology". Which is the art of operating on indices without understanding what is geometrical meaning of these operations. Sometimes this leads to results that are hard to get using geometrical pictures. Most of the time the results of such calculations are correct. But other times it may lead to confusion and errors.

 

[samalkhaiat]

This thread shows how easily one gets into troubles if one confuses TENSOR CONTRACTION, i.e., $A_{ a r } B^{ c r }$, with the corresponding MATRIX MULTIPLICATION, i.e., $A_{ a r } B^{ r c }$. The two expressions become identical, if and only if the followings are true:
1) the first and the second index both belong to identical or equivalent representations of the group in the question, and
2) $B^{ c r } = B^{ r c }$.
We will see that none of these conditions is satisfied in $SL( 2 , C )$. But, before doing the tensor notations of $SL( 2 , C )$, I would like to derive the relation between the Lorentz transformation,
$$x^{ \mu } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ x^{ \nu } , \ \ \Lambda \in SO( 1 , 3 ) , \ \ (1)$$
and the corresponding $SL( 2 , C )$ transformation of the Hermitian matrix $x_{ \mu } \sigma^{ \mu } = \eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau }$:
$$\eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau } \rightarrow \eta_{ \nu \tau } \ x^{ \nu } \ g \ \sigma^{ \tau } \ g^{ \dagger } , \ \ ( g , g^{ \dagger } ) \in SL( 2 , C ) \ (2)$$
I choose to define
$$\eta^{ \tau \mu } \ \sigma_{ \mu } = \sigma^{ \tau } \equiv ( I , \vec{ \sigma } ) ,$$
I also define
$$\eta^{ \tau \mu } \ \bar{ \sigma }_{ \mu } = \bar{ \sigma }^{ \tau } \equiv ( I , - \vec{ \sigma } ) .$$
Notice that $\sigma_{ \mu }$ and $\bar{ \sigma }^{ \nu }$ are the same matrices. However, writing $\sigma_{ \mu } = \bar{ \sigma }^{ \mu }$ would mess up your covariant expressions (some thing that you would not want to do). So, to stay out of troubles we will treat $\sigma^{ \mu }$ and $\bar{ \sigma }^{ \nu }$ as two independent objects.
Now, by direct matrix multiplication, we can easily verify that
$$\mbox{ Tr } ( \sigma^{ \tau } \bar{ \sigma }^{ \rho } ) = 2 \ \eta^{ \tau \rho } . \ \ (3)$$
Contracting both side of Eq(1) with $\eta_{ \mu \tau } \sigma^{ \tau }$ gives
$$\eta_{ \mu \tau } \ x^{ \mu }\ \sigma^{ \tau } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ x^{ \nu } \ \sigma^{ \tau } . \ \ \ (4)$$
Comparing the RHS of Eq(4) with that of Eq(2) leads to
$$\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ \sigma^{ \tau } = \eta_{ \nu \tau } \ g \ \sigma^{ \tau } \ g^{ \dagger } .$$
Multiplying both sides by $\bar{ \sigma }^{ \rho }$ and taking the trace, we find (using Eq(3)) that
$$\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } ( 2 \ \eta^{ \tau \rho } ) = \mbox{ Tr } (g \ \eta_{ \nu \tau } \ \sigma^{ \tau } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) ,$$
or
$$2 \Lambda^{ \mu }{}_{ \nu } \ \delta^{ \rho }_{ \mu } = \mbox{ Tr } ( g \ \sigma_{ \nu } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) .$$
Thus
$$\Lambda^{ \rho }{}_{ \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } ( \bar{ \sigma }^{ \rho } \ g \ \sigma_{ \nu } \ g^{ \dagger } ).$$

***********

Tensor Contraction & Trace in $SL( 2 , C)$

The matrices $\sigma^{ \mu }$ behave like mixed spinor-tensors indexed by $A = 1,2$ and $\dot{B} = \dot{1}, \dot{2}$ which label the fundamental and anti-fundamental (conjugate) representations respectively,
$$( \sigma^{ \mu } )_{ A \dot{B} } = ( I , \vec{ \sigma } )_{ A \dot{B} } .$$
Complex conjugation turns undotted indices into dotted ones and vice versa. For example, the complex conjugate of the spinor-tensor $\Psi_{ A B \dot{C}}$ is given by
$$\overline{ \Psi_{ A B \dot{C} } } = \bar{ \Psi }_{ \dot{A} \dot{B} C } .$$
Therefore, the hermiticity of $\sigma^{ \mu }$ means, in the $SL( 2 , C )$ index notations, that
$$\overline{ ( \sigma^{ \mu } )_{ A \dot{B} } } \equiv ( \bar{ \sigma }^{ \mu } )_{ \dot{A} B } = ( \sigma^{ \mu } )_{ B \dot{A} } . \ \ (5)$$
Raising and lowering of the indices is done here by the invariant (metric) spinors
$$\epsilon_{ A B } = \epsilon^{ A B } = \epsilon_{ \dot{A} \dot{B} } = \epsilon^{ \dot{A} \dot{B} } , \ \ (6a)$$
which are the 2-dimensional Levi-Civita symbol, i.e. ,
$$\epsilon_{ A B } = - \epsilon_{ B A } = \left( \begin{array} {rr} 0 & 1 \\ -1 & 0 \end{array} \right) . \ \ (6b)$$
They satisfy the relations
$$\epsilon_{ A B } \epsilon^{ C B } = \epsilon^{ C }{}_{ A } = - \epsilon_{ A }{}^{ C } = \delta^{ C }_{ A } . \ \ (6c)$$
The invariance of the metric spinor $\epsilon_{ A B }$ under $SL( 2 , C)$, which is a necessary condition for the consistency of the relations (6), follows from the definition of a determinant
$$\det | g | \epsilon_{ A B } = g_{ A }{}^{ C } \ g_{ B }{}^{ D } \ \epsilon_{ C D } ,$$
and the fact that, for all $g \in SL( 2 , C )$, $\det | g | = 1$.
The invariant trace of tensors is, therefore, obtained by contracting all indices with the appropriate spinor metric. For example, the trace of the tensor $T^{ A B }$ is given by
$$\mbox{Tr}_{ sl(2,C) } ( T ) = \epsilon_{ A B } T^{ A B } . \ \ \ (7)$$
It is clear from this that a traceless matrix, if interpreted as tensor, needs not be traceless tensor. For instance, as an anti-symmetric matrix, the Levi-Civita symbol is traceless, i.e. ,
$$\mbox{Tr}_{ M } ( \epsilon ) = 0 ,$$
while the tensor $\epsilon^{ A B }$, as seen from Eq(7) and Eq(6c), has a non-vanishing trace,
$$\mbox{Tr}_{ sl(2,C) } ( \epsilon ) = \epsilon^{ A }{}_{ A } = - \epsilon_{ A }{}^{ A } = 2 .$$
Another example is the Pauli matrices $\sigma^{ i }$ themselves. While $\mbox{Tr}_{ M } ( \sigma^{ i } ) = 0$, $\mbox{Tr}_{ sl(2,C) } ( \sigma^{ i } )$ does not exist (not even defined). However, the tensor $( \sigma^{ \mu } \sigma^{ \nu } )_{ A B }$ has a trace given by
$$\mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \sigma^{ \mu } )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ A \dot{ B } } = 2 \eta^{ \mu \nu } . \ \ \ (8)$$
This follows directly from the Dirac algebra (upon contracting the indices $A$ and $C$):
$$\{ \sigma^{ \mu } , \sigma^{ \nu } \}^{ C }_{ A } \equiv ( \sigma^{ \mu} )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ C \dot{ B } } + ( \sigma^{ \nu } )_{ A \dot{ B } } ( \sigma^{ \mu } )^{ C \dot{ B } } = 2 \ \delta^{ C }_{ A } \eta^{ \mu \nu } .$$
The hermiticity condition Eq(5) allows us to rewrite the trace in Eq(8) as that of matrix multiplication
$$\mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \bar{ \sigma }^{ \mu } )_{ \dot{ B } A } ( \sigma^{ \nu } )^{ A \dot{ B } } = \mbox{Tr}_{ M } ( \bar{ \sigma }^{ \mu } \sigma^{ \nu } ) .$$
Let us now show that the complex conjugation defined in Eq(5) is consistent with our (index free) definition of $\bar{ \sigma }^{ \mu } = ( I , - \vec{ \sigma } )$. Consider
$$( \sigma^{ \mu } )^{ A \dot{ B } } = \epsilon^{ A C } \ \epsilon^{ \dot{ B } \dot{ D } } \ ( \sigma^{ \mu } )_{ C \dot{ D } } ,$$
now take the complex conjugate of both sides,
$$\overline{ ( \sigma^{ \mu } )^{ A \dot{ B } } } = \overline{ \epsilon^{ A C } } \ \overline{ \epsilon^{ \dot{ B } \dot{ D } } } \ \overline{ ( \sigma^{ \mu } )_{ C \dot{ D } } } .$$
Using the rule of complex conjugation and Eq(5), we find
$$( \bar{ \sigma }^{ \mu } )^{ \dot{ A } B } = - \epsilon^{ B D } \ ( \sigma^{ \mu } )_{ D \dot{ C } } \ \epsilon^{ \dot{ C } \dot{ A } } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ B \dot{ A } } .$$
Therefore
$$\bar{ \sigma }^{ \mu } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ t } .$$
Since $\epsilon^{ t } = \epsilon^{ -1 } = - \epsilon$, then
$$\bar{ \sigma }^{ \mu } = - \epsilon \ ( \sigma^{ \mu } )^{ t } \ \epsilon = ( I , - \vec{ \sigma } ) .$$
I finish this post by writing the completeness relations with respect to the trace
$$\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )^{ C \dot{ D } } = 2 \ \delta^{ C }_{ A } \ \delta^{ \dot{ D } }_{ \dot{ B } } ,$$
which follow from
$$\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )_{ C \dot{ D } } = 2 \ \epsilon_{ A C } \ \epsilon_{ \dot{ B } \dot{ D } } .$$

Sam

 

[arkajad]

This thread shows how easily one gets into troubles if one confuses TENSOR CONTRACTION, i.e., $A_{ a r } B^{ c r }$, with the corresponding MATRIX MULTIPLICATION
...
Sam

What is not clear from your post is: who has confused these two things and exactly where?
I think being absolutely clear about this point will be of value for those who will read this thread.
Otherwise they can be confused.

I also understand that you agree that there is an error in Carmeli-Malin book, while there is no error in the book by Carmeli alone. Even if you did not spell it clearly. Yes?

 

[samalkhaiat]

What is not clear from your post is: who has confused these two things and exactly where?
I think being absolutely clear about this point will be of value for those who will read this thread.
Otherwise they can be confused.

I leave this to the readers. After all these posts, it shouldn’t be hard question to answer: who got “1 = -1” and where, and who said “trace is a trace” and where.

I also understand that you agree that there is an error in Carmeli-Malin book, while there is no error in the book by Carmeli alone. Even if you did not spell it clearly. Yes?

No, I did not agree:
1) I did read the book “Theory of Spinors” 12 years ago and I do not remember seeing any errors in it. However, I do remember that half of the book was about $SL( 2 , C )$ and its representations and the second half was about the spinor formulation of various field theories.
2) You thought that the following equations (from the book) were wrong
$$\mbox{Tr} ( \sigma^{ \alpha } \sigma^{ \beta } ) = 2 \eta^{ \alpha \beta } , \ \ \ \mbox{Tr} ( \sigma^{ \alpha } g \sigma^{ \beta } g^{ \dagger }) = 2 \Lambda^{ \alpha \beta } .$$
I have shown you, several times, that the equations are correct and consistent with each other if the trace is an $SL( 2 , C )$ trace.
3) Moshe Carmeli was a very careful writer and teacher. I have read many of his good published paper and three of his books. I believe that he published more that 30 paper and 5 books. He spent his life working on $SO( 1 , 3 )$ and $SL( 2 ,C )$. In the early 1970’s he formulated his $SL( 2 ,C )$ gauge theory of gravitation and showed that it is equivalent to Einstein’s theory of gravitation. Now days, we speak of the Carameli’s Lagrangian.

Sam

 

[arkajad]

3) Moshe Carmeli was a very careful writer and teacher.

Sam

If so how you explain the fact that in "Theory of Spinors" he has

$$\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)$$

While in "Group Theory and relativity" he has

$$\Lambda^\alpha_{\phantom{\alpha}\beta}= \frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)$$
Do you notice the difference? I mean, not on the right hand side, right hand sides are the same. But on the left hand side there is a very small and subtle difference. If you see this difference, how are you going to explain it? Just address this simple point. I will really appreciate it. Other readers of these pages will appreciate it as well, I am sure.

Right hand sides are the same, left hand sides are different. Not only different, but if you for a moment suppose thath they are equal, you will easily get 1=0. So they are not just different, they contradict one to another.

Remember: alpha and beta spacetime indices. Sigmas are Pauli matrices. They are exactly the same in those two books.

 

[samalkhaiat]

If so how you explain the fact that in "Theory of Spinors" he has

$$\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)$$

This equation is correct, if the trace is that of tensor contraction, i.e., when we treat the sigmas as $SL(2,C)$ mixed tensors. In other words, the trace is what I denoted by $\mbox{Tr}_{ sl(2,C) }$.

While in "Group Theory and relativity" he has

$$\Lambda^\alpha_{\phantom{\alpha}\beta}= \frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)$$
Do you notice the difference?

I have never seen this book. However, by looking at this non-covariant equation, I guess he used the (non-covariant) matrix equality between $\sigma^{ \beta }$ and $\bar{ \sigma }_{ \beta }$ which I warned about in post#41. So, if you plug
$$\sigma^{ \beta } = \bar{ \sigma }_{ \beta }$$
back into his equation, you get the correct and covariant equation
$$2 \ \Lambda^{ \alpha }{}_{ \beta } = \mbox{Tr}_{M} ( \sigma^{ \alpha } \ g \ \bar{ \sigma }_{ \beta } \ g^{ \dagger } )$$
In this equation, the trace is just that of ordinary matrix multiplication.

I mean, not on the right hand side, right hand sides are the same.

On the contrary, the RHS’s are different. The trace in the first equation is $\mbox{Tr}_{sl(2,C)}$ while the trace in the second equation is $\mbox{Tr}_{M}$.

Sam

 

[arkajad]

On the contrary, the RHS’s are different.
Sam

On the contrary, they are exactly the same. I did copy and paste. And the traces are exactly the same. Though you would have to check the book in order to understand that this is the case. I strongly suggest you do it. Otherwise you will have to guess. Guessing is never a proof of anything.

"Theory of Spinors":

"Group Theory and General Relativity":

 

[samalkhaiat]

On the contrary, they are exactly the same. I did copy and paste. And the traces are exactly the same.

These are meaningless sentences. Denoting the trace by $\mbox{Tr}$ is the same, how one calculates the trace is the difference.

Though you would have to check the book in order to understand that this is the case. I strongly suggest you do it.

I don’t need to see any book, because I am an expert on the subject.

Otherwise you will have to guess. Guessing is never a proof of anything.

Clearly, now you are playing politics with me, because my posts left you with no meaningful questions. Sir, I am a theoretical physicist not politician.

Good Luck

Sam

 

[arkajad]

Sir, I am a theoretical physicist not politician.

Good Luck

Sam

Even theoretical physicists should use logic. My question was about particular formulas in particular books. You did not address my questions. You have addressed your own questions. I understand you were trying to help, and I thank you for your effort. But you did not help. Perhaps one of the reasons was that you have the wrong idea that experts can't make errors. They can and they do. Studying history of science can help to fix this erroneous idea. Usually, when two experts have contradicting ideas about some subject, a third and a fourth expert is being called. But here we do not have too many experts as it appears. Perhaps because it is summer hollidays time.

As another expert on the subject let me state my hypothesis: Carmeli's error in "Theory of Spinors" was due to Malin. Malin was editing, as it appears from the almost identical content of parts of two books, sometimes by simple copy and paste. He was struck but apparent lack consistency in Carmeli's formula. He had a wrong idea that the equation must be covariant. You also seem to share this wrong idea. But these equations do not have to be covariant for the simple idea that they are not dealing with tensors. On the left hand side are the components of the Lorentz matrix. Lorentz matrix is a numerical object - not a tensor. It does not depend on the frame of reference. It operates on frames, but it is not a frame. So, Malin had this wrong idea, and he "fixed" Carmeli's formula without much thinking. Then he had to "fix" the proof. To prove a wrong formula he had to make an error in his proof. Which he did.

That is my hypothesis. When the third expert will come - it will be interesting to know his/her opinion about my hypothesis.

 

[arkajad]

Here is the evidence of the copy and paste job in the two Carmeli's book. Word by word, with just two "little" differences:

Only one of them can be right. Now I know it is Carmeli alone who is right. Carmeli-Malin is wrong. But the proofs are wrong in both. Probably Malin did not notice the mistake in the proof, and wrongly corrected the right formula incorrectly assuming that the proof was correct.

 

[Bill_K]

I have never before seen such a big deal made out of such a trivial point. You've repeatedly been given full explanations, all the way back to dextercioby's #2. So I can't imagine what good it would do to repeat it again, since you are apparently only interested in publicizing Carmeli's "mistake" rather than understanding it.

Nevertheless, :uhh: the point, once again, is that Tr(σ^μσ^ν) is not a covariant quantity unless you do one of two things: a) replace one of the σ's, by defining a barred quantity σ^−μ which reverses the sign of one component, and writing Tr(σ^−μσ^ν), or b) redefine the Tr operation itself to implicitly include this replacement. There is nothing especially strange about redefining Tr. Doing this is similar to what we do by redefining the dot product operation u·v to include the minus sign when dealing with 4-vectors.

 

[arkajad]

So I can't imagine what good it would do to repeat it again, since you are apparently only interested in publicizing Carmeli's "mistake" rather than understanding it.

Wrong conclusion.

Now I understand it. One formula is right, the other is wrong. Both proofs are wrong. Students should be warned. This is not publicizing, this is telling the truth. Physics is about truth, mathematics even more so. Books are known to contain errors and it is a service for the community to point out each such error if it is not in an erratum.

Do check both books, study the relevant parts carefully. Otherwise you will be guessing. Guessing will never replace a rigorous mathematical reasoning.

Once more: thanks for you trying to help. But you did not help. In fact you have introduced an unnecessary confusion into a simple subject. If you have checked both books this could certainly be avoided.

 

[vanhees71]

Well, I've just seen this thread. Perhaps it's good to just have a fresh look. Usually what the book seems to use (I only can look at the quotes given in this threat) the mapping of the four vector is the usual one to a SL(3,C) spinor of second rank with one normal and one dotted index.

To put this in the form with the Pauli and the unit matrix one better should set
$$\sigma_0=\mathbb{1}_2, \quad \sigma_j=\textbf{Pauli matrix}.$$
Then the mapping of the four-vector is
$$x^{\mu} \mapsto Q=x^{\mu} \sigma_{\mu}.$$
Now, indeed the $\sigma_{\mu}$ build the Clifford algebra of the Euclidean $\mathbb{R}^4$, i.e.,
$$\mathrm{tr} (\sigma_{\mu} \sigma_{\nu})=\delta_{\mu \nu}.$$
Thus we have to use the rule
$$\sigma^{\mu}=\sigma_{\mu}$$
for these matrices (it's a bit a pity that there are two rules of index dragging now in this formalism).

On the other hand
$$\det Q=x_{\mu} x^{\mu}=\eta_{\mu \nu} x^{\mu} x^{\nu}.$$
We note that the mapping of the four vector $x$ to $Q$ is a one-to-one correspondence between four vectors and hermitean 2x2 matrices.

Now let $g \in \mathrm{SL}(2,\mathbb{C})$. Then according to the transformation rules for the mixed 2nd-rank spinor one has
$$Q'=g Q g^{\dagger},$$
which is again hermitean and thus uniquely defines a linear map between four vectors $x \mapsto x'$. Since by definition $\mathrm{det} g=\mathrm{det} g^{\dagger}=1$ we have
$$x_{\mu} x^{\mu} = \mathrm{det} Q=\mathrm{det} Q'=x'_{\mu} x'^{\mu},$$
i.e., the map $x \mapsto x'$ is a Lorentz transformation.

Now we have
$$x'^{\mu}=\mathrm{tr}(\sigma^{\mu} Q')=\frac{1}{2} \mathrm{tr}(\sigma^{\mu} g Q g^{\dagger}) = \frac{1}{2} \mathrm{tr}(\sigma_{\mu} g \sigma_{\nu} g^{\dagger}) x^{\nu},$$
which means that the Lorentz transform is given by
$${\Lambda^{\mu}}_{\nu}=frac{1}{2} \mathrm{tr}(\sigma^{\mu} g \sigma_{\nu} g^{\dagger}).$$
This shows that indeed the quote from the one book with the Eq. (3-9a) is correct, while the other book's equation (3.84a) has the index of the Lorentz transformation at the wrong place.

Finally some remarks on this mapping between $\mathrm{SL}(2,\mathbb{C})$:

Taking $(x^{\mu})=(1,0,0,0)$ we get
$Q=\mathbb{1}$ and $Q'=g g^{\dagger}$, which shows that the component $Q'^{1 \dot{1}} \geq 0$, i.e., the Lorentz tranform is orthochronous. Further for $g=\mathbb{1}$ we obviously get $\Lambda=\mathbb{1}$ too. Since the $\mathrm{SL}(2,\mathrm{C})$ is simply connected, this means that the determinant of all so constructed $\Lambda$ is +1 and thus we have a map from $\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}(1,3)^{\uparrow}$, the proper orthochronous Lorentz group. It's also easy to see that this mapping is a group homomorphism. One can also show that this is homomorphism surjective and that to each proper orthocrhonous Lorentz transformation $\Lambda$ there are two $\pm g$ in $\mathrm{SL}(2,\mathbb{C})$ that map into $\Lambda$ under this epimorphism. This means that $\mathrm{SO}(1,3)^{\uparrow}$ is isomorphic to the factor group $\mathrm{SL}(2,\mathbb{C})/\{1,-1\}$.

 

[arkajad]

Well, I've just seen this thread. Perhaps it's good to just have a fresh look.

Thanks. I appreciate your fresh look as well as the additional helpful comments.

 

[samalkhaiat]

Here is the evidence of the copy and paste job in the two Carmeli's book. Word by word, with just two "little" differences:

Only one of them can be right. Now I know it is Carmeli alone who is right. Carmeli-Malin is wrong. But the proofs are wrong in both. Probably Malin did not notice the mistake in the proof, and wrongly corrected the right formula incorrectly assuming that the proof was correct.

Your remarks are utterly trivial. The equations, you labelled by “Mistake!” and “Not OK!” are indeed correct and OK provided you calculate the trace in $SL( 2 , C )$. That is
$$\mbox{Tr}_{ sl(2,C) } ( \sigma^{ \alpha } \sigma_{ \beta } ) = 2 \ \delta^{ \alpha }_{ \beta } . \ \ \ (1)$$
If you think this equation is wrong, then you should show us, by “rigorous mathematical reasoning”, why it is wrong?
Now, let us follow Carmeli:
$$x^{ \prime \alpha } = \delta^{ \alpha }_{ \beta } \ x^{ \prime \beta } = \frac{ 1 }{ 2 } \mbox{Tr}_{ sl(2,C) } ( \sigma^{ \alpha } \sigma_{ \beta } ) \ x^{ \prime \beta } = \frac{ 1 }{ 2 } \mbox{Tr}_{ sl(2,C) } ( \sigma^{ \alpha } \sigma^{ \beta } ) \ x^{ \prime }_{ \beta } .$$
You marked this equation as “Mistake!”. Show us, by “rigorous mathematical reasoning”, where is the mistake?
If we continue with Carmeli’s arguments, we arrive at
$$\Lambda^{ \alpha \beta } = \frac{ 1 }{ 2 } \mbox{Tr}_{ sl(2,C) } ( \sigma^{ \alpha } \ g \ \sigma^{ \beta } \ g^{ \dagger } ) .$$
You marked this equation by “Not OK!”. Show me, by “rigorous mathematical reasoning”, why it is not OK?

However, I have an issue with Carmeli’s argument that led to the-silly-looking Eq(3.9a) which you marked as “OK”. He starts the argument using Eq(1) for the $SL( 2 , C )$ trace, then he jumps to conclude Eq(3.9a) which deploys ordinary trace.
In both books the derivation is silly because they have not defined the space under which the trace is taken. They should have (as I have done in the first part of post #41) defined the $\bar{ \sigma }^{ \beta }$ and used the fact that
$$\mbox{Tr} ( \sigma^{ \alpha } \bar{ \sigma }^{ \beta } ) = 2 \ \eta^{ \alpha \beta } .$$

Sam

 

[samalkhaiat]

Wrong conclusion.

Now I understand it. One formula is right, the other is wrong. Both proofs are wrong. Students should be warned. This is not publicizing, this is telling the truth. Physics is about truth, mathematics even more so. Books are known to contain errors and it is a service for the community to point out each such error if it is not in an erratum.

Do check both books, study the relevant parts carefully. Otherwise you will be guessing. Guessing will never replace a rigorous mathematical reasoning.

Once more: thanks for you trying to help. But you did not help. In fact you have introduced an unnecessary confusion into a simple subject. If you have checked both books this could certainly be avoided.

Your arguments in this thread were empty of any mathematical content, let alone “rigorous mathematical reasoning”. The only time you used mathematics, you got the embarrassing result $1 = - 1$. If you want logically consistent and mathematically rigorous statements, then you should read my posts carefully.
I suggest you print out my posts and use them as a companion to help you understand Carmeli’s book.

Sam

 

[vanhees71]

If you look at my posting, you'll see that this equation is really incorrect. Note that in my posting I give the complete calculation to show that the correct equation is
$${\Lambda^{\alpha}}_{\beta}=\frac{1}{2} \mathrm{tr}(\sigma^{\alpha} g \sigma^{\beta} g^{\dagger}).$$

 

[Bill_K]

If you look at my posting, you'll see that this equation is really incorrect. Note that in my posting I give the complete calculation to show that the correct equation is
Λ^α_β=1/2 tr(σ^αgσ^βg^†).

The indices don't match. To get them to match, you have to write it (as you did in your post) as Λ^α_β = ½ Tr(σ ^α g σ_β g^†) where σ_β = η_αγσ^γ. This changes the sign of one of the sigmas, like we have been saying all along.

 

[arkajad]

The indices don't match.

The indices do not have to match. The formula does not concern tensors. It concerns a set of fixed numerical matrices.

Also notice that Tr in the formula is not invariant under SL(2,C) transformations $Q\mapsto AQA^*,\,A\in SL(2,C).$.

 

[vanhees71]

The indices don't match. To get them to match, you have to write it (as you did in your post) as Λ^α_β = ½ Tr(σ ^α g σ_β g^†) where σ_β = η_αγσ^γ. This changes the sign of one of the sigmas, like we have been saying all along.

No that's the point! It is $\sigma_{\mu}=\sigma^{\mu}$. The $\sigma$ build the Clifford algebra of Euclidean $\mathbb{R}^{4}$ not of Minkowskian $\mathbb{R}^{(1,3)}$.


A manifest covariant formalism is provided by Dirac spinors rather than Weyl spinors!

 

[arkajad]

The $\sigma$ build the Clifford algebra of Euclidean $\mathbb{R}^{4}$

In fact don't they build the Clifford algebra of $\mathbb{R}^{3}$ rather than $\mathbb{R}^{4}$?

 

[samalkhaiat]

A manifest covariant formalism is provided by Dirac spinors rather than Weyl spinors!

What is it non covariant in the following equations?
$$( \sigma^{ \mu } )_{ A \dot{ B } } \partial_{ \mu } \chi^{ A } = m \bar{ \eta }_{ \dot{ B } } ,$$
$$( \sigma^{ \mu } )_{ A \dot{ B } } \partial_{ \mu } \eta^{ A } = m \bar{ \chi }_{ \dot{ B } } .$$
Irreducible representations of the Lorentz group know absolutely nothing about Dirac’s bispinors. In fact, Dirac bispinor is a Parity invariant, 4-Dimensional Reducible Representation of $SO( 1 , 3 ) + \mbox{ Parity }$, i.e.,
$$\psi = \left( \begin{array}{c} \chi_{ A } \\ \bar{ \eta }^{ \dot{ A } } \end{array} \right) \sim ( \frac{ 1 }{ 2 } , 0 ) \oplus ( 0 , \frac{ 1 }{ 2 } ) .$$
Here is a question to the readers. What prevents us from representing Dirac bispinor by the $( 1/2 , 0 ) \oplus ( 1/2 , 0 )$ or $( 0 , 1/2 ) \oplus ( 0 , 1/2 )$ Reducible Representations of $SO( 1 , 3 ) + \mbox{ Parity }$?

Sam

 

[samalkhaiat]

... Tr in the formula is not invariant under SL(2,C) transformations...

Can you prove this (ridiculous) claim of yours?

The Tr in the formula, when written in full details, means
$$\Lambda^{ \alpha }{}_{ \beta } ( g ) = \frac{ 1 }{ 2 } \ g^{ A }{}_{ C } \ \bar{ g }^{ \dot{ B } }{}_{ \dot{ D } } \ ( \sigma^{ \alpha } )^{ C \dot{ D } } \ ( \sigma_{ \beta } )_{ A \dot{ B } } .$$
When all the indices of $SL( 2 , C )$ are contracted, we say that the object in question is (scalar) invariant under $SL( 2 , C )$ transformations. So, I insist that you show us how is it that the trace is not invariant?

I also asked you ( in post # 54) to prove few more (false) claims you made in this thread, but you have not replied yet.

Sam

 

[arkajad]

Can you prove this (ridiculous) claim of yours?

Of course I can. Take as an example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

Then $\det(A)=1$ therefore $A\in SL(2,C).$

We have

$$A A^\dagger =\begin{pmatrix}2&1\\1&1\end{pmatrix}$$

and $\mbox{Tr}( A A^\dagger )=3$

With $Q=I$ (the identity matrix) we have

$$\mbox{Tr}(AQA^\dagger)=3\neq 2=\mbox{Tr}(Q)$$

This means Tr is not SL(2,C)-invariant. It is SU(2) ( and even U(2) )-invariant.

 

[vanhees71]

I think there is so much confusion in this thread that I cannot help further. Just some final remarksfrom my side.

(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$ but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

The correct decomposition of the Dirac equation into coupled Weyl-spinor, leading to the chiral representation of the Dirac matrices reads
$$\mathrm{i} \sigma^{\mu} \partial_{\mu} \eta=m \chi,$$
$$\mathrm{i} \overline{\sigma}^{\mu} \partial_{\mu} \chi=m \eta.$$
Here $\chi$ and $\eta$ are two two-component spinors, called the righthanded and lefthanded components of the four-component Dirac spinor and
$$\sigma^{\mu}=(\mathbb{1}_2,+\vec{\sigma}), \quad \overline{\sigma}^{\mu}=(\mathbb{1}_2,-\vec{\sigma}).$$
Note again that also here the index on the $\overline{\sigma}$ matrices is a contravariant (upper) index!

(ii) In the Weyl-spinor representations $\mathrm{D}(1/2,0)$ and $\mathrm{D}(0,1/2)$ the skew-symmetric bilinear forms $\xi_A \xi^A=\eta_{AB} \xi^A \xi^B$ and $\chi_{\dot{A}} \chi^{\dot{A}}=\eta_{\dot{A} \dot{B}} \chi^{\dot{A}} \chi^{\dot{B}} =$ are invariant under the corresponding $\mathrm{SL}(2,\mathbb{C})$ transformations, where the spinors with the dotted indices transform with the conjugate complex matrix:
$$\xi'=A \xi, \quad \chi'=A^{*} \chi.$$
Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved, because the transformation reads
$$Q'=g Q g^{\dagger}$$
and $g$ is in general not unitary. The trace is only invariant under the subgroup $\mathrm{SU}(2)$, which provides the usual spin-1/2 representation of rotations for Weyl spinors. Also note that
$$\mathrm{tr} Q=2 x^0,$$
which of course changes under general Lorentz transformations and is invariant only under rotations, and thus the representation $g$ of a general Lorentz transformation changes the trace but unitary ones don't. The unimodular transformations $g$, however, keep $\mathrm{det} Q=x_{\mu} x^{\mu}$ invariant as it must be.

(iii) A very good book on this topic is

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

 

[arkajad]

Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved

I guess you wanted to say:

The trace of the $Q=x^{\mu} \sigma_{\mu}$ is not conserved

 

[arkajad]

There is an additional remark that can be useful for the readers of this thread. Namely, if the homomorphism from SL(2,C) to the restricted Lorentz group is defined as in the first Carmaeli's book "Group Theory and General Relativity" (correctly):
$${\Lambda(g)^\alpha}_{\phantom{\alpha}\beta}=\mbox{Tr }(\sigma^\alpha g\sigma^\beta g^\dagger)$$
then
$$\mbox{Tr }\Lambda(g)=|\mbox{Tr }g|^2.$$
This is Eq. (3.85) in "Theory of Spinors" and not numbered equation on p. 36 in Carmeli's "Group theory and General Relativity".

 

[arkajad]

(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$ but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

Perhaps it may be useful here, using simple matrix algebra, to understand why massless two-component spinor equation is SL(2,C) invariant.

So, suppose $\Psi$ is a spinor field transforming according to the contragradient representation of SL(2,C):
$$\Psi\mapsto \tilde{\Psi}=g^{*-1}\Psi.$$
while the coordinates transform in a standard way:
$$\tilde{x}^\mu={\Lambda^\mu}_{\phantom{\mu}\nu}x^\nu$$
so that
$$\tilde{\partial}_\nu={\Lambda^\rho}_{\phantom{\rho}\nu}\partial_\rho$$
Then
$$\sigma^\nu\tilde{\partial}_\nu\tilde{\Psi}=g\tilde{\partial}_\nu g^{-1}\sigma^\nu g^{-1*}\tilde{\partial}_\nu\Psi$$
$$=g{{\Lambda^{-1}}^\nu}_{\phantom{\nu}\mu}\sigma^\mu\tilde{\partial}_\nu\Psi$$
$$=g\sigma^\mu\partial_\mu\Psi$$
Therefore, owing to the fact that $g$ is invertible, $\sigma^\mu\partial_\mu\Psi=0$ if and only if $\sigma^\nu\tilde{\partial}_\nu\tilde{\Psi}=0$

 

[samalkhaiat]

Just some final remarksfrom my side.
(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$

$( 1/2 , 0 ) \otimes ( 0 , 1/2 )$ is the 4-vector irreducible representation. May be you meant to write $\oplus$.

but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

No, I meant exactly what I said: Dirac “spinor” is parity-invariant, Reducible representation of $SO( 1 , 3 ) + P$. By this I meant the piece $\Lambda_{ P } SO^{ \uparrow } ( 1 , 3 )$, sometimes written as $\Lambda_{ P } L_{ + }^{ \uparrow }$, which is the $L_{ - }^{ \uparrow }$ piece of the pseudo-orthognal group $O( 1 , 3 )$.
Dirac bispinor does not exist in the Lorentz group $SO^{ \uparrow } ( 1 , 3 )$. Only when we allow for parity, we can construct the parity-invariant direct sum $( 1/2 , 0 ) \oplus ( 0 , 1/2 )$ with the generator written in block diagonal, i.e., Reducible: Parity action on the generators does not form an in-equivalent class of representations. This is the reason for the question, I asked at the end of post # 61, and nobody cared to answer.

The correct decomposition of the Dirac equation into coupled Weyl-spinor, leading to the chiral representation of the Dirac matrices reads

In the representation theory, we do not decompose the Dirac equation. On the contrary, we construct the Dirac equation from the equations satisfied by the two in-equivalent fundamental representations of $SO^{ \uparrow }( 1 , 3 )$, and demanding Parity-Invariance.

Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved, because the transformation reads
$$Q'=g Q g^{\dagger}$$
and $g$ is in general not unitary

I don’t know the meaning of this statement! As a matrix, $Q = x_{ \mu } \sigma^{ \mu }$, has nothing to do with group $SL( 2 , C )$. Therefore, it is meaningless to ask about the invariance of $\mbox{Tr} Q$ with respect to $SL( 2 , C )$ transformations. In fact ( and I said this before), the $SL( 2 , C )$ trace of the mixed spinor-tensor $Q_{ A \dot{ B } } = x_{ \mu } ( \sigma^{ \mu } )_{ A \dot{ B } }$ is not even defined in the group.

Sam

 

[samalkhaiat]

Of course I can. Take as an example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

Then $\det(A)=1$ therefore $A\in SL(2,C).$

We have

$$A A^\dagger =\begin{pmatrix}2&1\\1&1\end{pmatrix}$$

and $\mbox{Tr}( A A^\dagger )=3$

With $Q=I$ (the identity matrix) we have

$$\mbox{Tr}(AQA^\dagger)=3\neq 2=\mbox{Tr}(Q)$$

This means Tr is not SL(2,C)-invariant. It is SU(2) ( and even U(2) )-invariant.

Thank you very much, you made me laugh. Is this your understanding of the invariance of the trace operation? I ground my 12 years old kid for a week, if she reproduces your argument. Who said $\mbox{Tr} ( A Q A^{ \dagger } )$ must equal to $\mbox{Tr} Q$? And why should they be equal?
Look, I don’t know how many different ways I have explained this for you, here is another way. Consider the Matrix $Q = x_{ \mu } \sigma^{ \mu }$, square it by ordinary matrix multiplication and take (1/2) of its trace, you find
$$\frac{ 1 }{ 2 } \mbox{Tr} ( Q^{ 2 } ) = \frac{ 1 }{ 2 } Q_{ a c } Q_{ c a } = x_{ 0 }^{ 2 } + x_{ 1 }^{ 2 } + x_{ 2 }^{ 2 } + x_{ 3 }^{ 2 } .$$
This is the invariant metric of the Orthogonal group $SO( 4 )$. So, why didn’t we get the Invariant Lorentz metric? The reason is this: As a matrix, $Q$ has nothing to do with the group $SL( 2 , C )$ and because we calculated the $SO( 4 )$-invariant trace, we got Euclidean metric.
To avoid this embarrassing situation, people often calculate $\det | Q |$ in order to generate the Lorentz metric with correct signature. Even in this case, we get the correct signature because of the presence of the $SL( 2 , C )$ metric $\epsilon$ in the definition of $\det | Q |$. To see the beauty of $SL( 2 , C )$ tensor calculas, let us repeat the calculation by treating $Q$ as the mixed spinor-tensor $Q_{ A \dot{ B } } = x_{ \mu } ( \sigma^{ \mu } )_{ A \dot{ B } }$ and evaluate the $SL( 2 , C )$-invariant trace
$$\mbox{Tr} ( Q^{ 2 } ) = Q^{ A \dot{ B } } Q_{ A \dot{ B } } = x_{ \mu } x_{ \nu } ( \sigma^{ \mu } )^{ A \dot{ B } } ( \sigma^{ \nu } )_{ A \dot{ B } } = 2 \left( x_{0}^{2} - x_{1}^{2} - x_{2}^{2} - x_{3}^{2}\right) .$$
Here the Lorentz invariant metric with correct signature arises naturally because the invariant trace requires the presence of $SL( 2 , C )$ spinor metric $\epsilon$:
$$( \sigma^{ \mu } )^{ A \dot{ B } } ( \sigma^{ \nu } )_{ A \dot{ B } } = \epsilon^{ A C } \epsilon^{ \dot{ B } \dot{ D } } ( \sigma^{ \mu } )_{ C \dot{ D } } ( \sigma^{ \nu } )_{ A \dot{ B } } .$$
Now, look the two calculations and recall my warning: Do not confuse Matrix Multiplication with Tensor Contraction. Get It?

Sam

 

[arkajad]

As a matrix, $Q$ has nothing to do with the group $SL( 2 , C )$

Sam

It does, as is indicated, for instance, in this part Carmeli's monograph (Eq. (3.80)):

----------------------------------------------------------------------------------
$\det(Q')=\det(Q)$ so "determinant is SL(2,C) invariant".
But $\mbox{Tr }(Q')\neq \mbox{Tr }(Q)$, so "trace is not SL(2,C) invariant" (what is also evident from the fact that $\mbox{Tr }(Q)=2x^0$).