Is this section of the wikipedia page for gamma matrices wrong?

[pellman]

http://en.wikipedia.org/wiki/Gamma_matrices#Normalization

See the image below. Which of us is right: me or Wikipedia?

 

[Orodruin]

$\gamma^0$ also transforms ... Wikipedia is usually pretty reliable.

 

[OCR]

http://en.wikipedia.org/wiki/Gamma_matrices#Normalization

Which of us is right: me or Wikipedia?

You probably know all this, but other folks might not...

Wikipedia is not always right, but... if you have a question about the reliability of an article, always look at the View history section. That will show you the edits, vandalism, edit wars, last edit date, and all the other crap that goes on... lol

Probably the best place to look though, is the article Talk page... that tells you more or less how, and by whom the page was developed; you can also add your thoughts... Lol, it's kinda like here on PF...

Wikipedia is usually pretty reliable.

Very much so... IMO.

https://www.google.com/#q=reliability+of+wikipedia

 

[pellman]

$\gamma^0$ also transforms ... Wikipedia is usually pretty reliable.

Ha! I knew I missed something.

 

[pellman]

So am I reading the statement in the article correctly? In any particular frame we are free to choose a basis such that $\gamma^0 (\gamma^\mu)^\dagger \gamma^0 = \gamma^\mu$ but then when we Lorentz transform to another frame, this relation may no longer hold?

 

[pellman]

Also the reason I ask is because I was thinking that the Dirac equation

$$(i\gamma^\mu \partial_\mu -m)\psi=0$$

and its conjugate

$$\bar{\psi}(-i\gamma^\mu \overleftarrow{\partial}_\mu - m)=0$$

are equivalent. However, to derive the conjugate equation from the former requires $\gamma^0(\gamma^\mu)^\dagger \gamma^0 = \gamma^\mu$. So if that relation does not hold in general, what does it mean for the conjugate Dirac equation?

 

[Bill_K]

Wikipedia is (partially) wrong. The page says,

The hermiticity conditions are not invariant under the action $\gamma^\mu \to S(\Lambda) \gamma^\mu {S(\Lambda)}^{-1}$ of a Lorentz transformation $\Lambda$ because $S(\Lambda)$ is not necessarily a unitary transformation due to the noncompactness of the Lorentz group.

It's true that S(Λ) is not unitary. But, in place of that it does obey a condition sufficient to guarantee that the hermiticity conditions are invariant. Instead of SS^† = I, it obeys Sγ⁰S^† = γ⁰.

Start with γ^μ' = Sγ^μS^-1. Then

γ^μ'^† = (Sγ^μS^-1)^† = S^-1† γ^μ† S^† = γ⁰Sγ⁰ γ^μ† γ⁰S^-1γ⁰ = γ⁰Sγ⁰ γ^μ S^-1γ⁰ = γ⁰ γ^μ' γ⁰

 

[pellman]

it obeys Sγ⁰S^† = γ⁰.

I haven't proved this but I checked it for a few cases. Seems to be the case. So this is to say that γ⁰ is invariant under Lorentz transformations? That's a surprise. And now the equivalence of the Dirac equation and its conjugate holds.

I will consider updating the Wikipedia article.

 

[Bill_K]

I haven't proved this but I checked it for a few cases. Seems to be the case. So this is to say that γ⁰ is invariant under Lorentz transformations?

I think what it says is that γ⁰ is the metric for Dirac spinors. That is, Sγ⁰S^† = γ⁰ is analogous to ΛηΛ^T = η, where η is the Minkowski metric.

 

[vanhees71]

The point of the Dirac formalism is that it provides a irreducible representation of the proper Lorentz group, $\mathrm{SO}(1,3)$, i.e., the proper orthochronous Lorentz group augmented with space reflections (parity). It's reducible as a representation of the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$.

The representation can be charcterized by the behavior of the four Dirac-$\gamma$ matrices,
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu},$$
where $\Lambda \in \mathrm{SO}(1,3)$.

There is no finite-dimensional unitary ray representation of the special orthochronous Lorentz group, and that's why one necessarily needs a QFT formulation. Physically that's clear from the fact that at relativistic energies collisions can lead to the creation and destruction of particles, i.e., one usually has to use a many-body approach including such processes, and for this the ansatz of a local QFT has been the only successful idea so far (Standard Model of the elementary particles). It can be shown that any ray representation can be induced by a unitary representation of the covering group of the Poincare group. Contrary to the non-relativistic case there do not exist non-trivial central extensions (in the non-relativistic case it's crucial to use a non-trivial central extension of the covering group of the inhomogeneous Galilei group with the mass as a central charge, because the unitary representations of the non-extended Galilei group does not lead to physically sensible quantum theories; in the case of the Poincare group the mass is a Casimir operator characterizing together with the spin the possible unitary representations of the proper orthochronous Poincare group), i.e., the quantum fields behave under Lorentz transformations as
$$U(\Lambda) \psi(x) U^{\dagger}(\Lambda)=S(\Lambda) \psi(\Lambda^{-1} x),$$
where indeed $U(\Lambda)$ is a unitary representation in Fock space.

 

[pellman]

Thank you for the post vanhees71 but it isn't clear to me how it addresses the question of the thread.

 

[Avodyne]

The confusing point is that the γ⁰ in Sγ⁰S^† = γ⁰ is actually not the μ=0 component of γ^μ. It's a different matrix that is given a different name by more careful authors; for example, Weinberg and Srednicki both call it β.

 

[vanhees71]

Of course, $\gamma^0$ is the 0-Component of the matrix-valued four vector given by the Dirac matrices.

The relation $\gamma^{0} (\gamma^{\mu})^{\dagger} \gamma^{0}=\gamma^{\mu}$ is, however, indeed only valid for a special class of realizations of the $\gamma$ matrices, which btw. generate the Clifford algebra of the Minkowski vector space. It's only true for the Dirac standard representation and all representations that are derived from it by a unitary transformation. The $S(\Lambda)$ are not unitary for any Lorentz transformation that is not a pure rotation. Only the rotation group is realized as a unitary representation, not the boosts!

Of course, for a fixed representation you have
$$\gamma^0 (\gamma'{}^{\mu})^{\dagger} \gamma^0=\gamma'{}^{\mu},$$
as has been correctly derived in posting #1, but generally you cannot put $\gamma'{}^0$ instead of $\gamma^0$ in this formula.

 

[Bill_K]

The confusing point is that the γ⁰ in Sγ⁰S^† = γ⁰ is actually not the μ=0 component of γ^μ.

This happens so often on PF. Clear, simple answers are hard to come by, because of a desire to make things as complicated and general as possible. Yes, it's true that γ⁰ is not the only matrix that can be used in Sγ⁰S^† = γ⁰, there's an arbitrary phase that can be thrown in. But the obvious choice is to use γ⁰ itself.

Use something else if you like: SAS^† = A. But this same matrix A must also be used to define the adjoint spinor, $\bar{\psi} \equiv \psi A$, so if you don't go with A = γ⁰, you will also need to use a nonstandard definition for that. The equation SAS^† = A guarantees that $\bar{\psi} \psi$ will transform like a scalar.

It's a different matrix that is given a different name by more careful authors; for example, Weinberg and Srednicki both call it β.

Actually Weinberg Eq(5.4.13) calls β = i γ⁰.

As vanhees71 points out also, the discussion is further obscured in many texts because the author has already chosen a particular representation for the γ's.

 

[pellman]

I think what it says is that γ⁰ is the metric for Dirac spinors. That is, Sγ⁰S^† = γ⁰ is analogous to ΛηΛ^T = η, where η is the Minkowski metric.

That's an interesting way of looking at it. It fits with $\psi^\dagger \gamma^0 \psi$ is the scalar product instead of $\psi^\dagger \psi$

This is the kernel of the question of this thread: If $\psi^\dagger \gamma^0 \psi$ transforms under $\Lambda$ as a scalar, then it must be the case that $S(\Lambda)^{-1}\gamma^0 S(\Lambda) = \gamma^0$. From that it follows that the relation $\gamma^0 (\gamma^\mu)^\dagger \gamma^0 = \gamma^\mu$ is invariant under $\Lambda$.

 

[vanhees71]

Ok, finally I checked it myself, and indeed Wikipedia is correct!

For all four $\gamma$ matrices the representation $S=S(\Lambda)$ of the proper Lorentz group MUST fulfill
$$S \gamma^{\mu} S^{-1} = {\Lambda^{\mu}}_{\nu} \gamma^{\nu}$$
written in the convention of Peskin and Schroeder. The way Wikipedia writes it, is ok too, they just interchange $\Lambda$ and $\Lambda^{-1}$ in their argumentation.

Now, for the class of representations of the Dirac matrices, mentioned in my previous posting, you have
$$\gamma^0 (\gamma^{\mu})^{\dagger} \gamma^0=\gamma^0. \qquad (1)$$
Further all representation matrices $S$ can be generated by representation matrices of the form
$$S=\exp \left (-\frac{\mathrm{i}}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right ),$$
where
$$\sigma^{\mu \nu} = \frac{\mathrm{i}}{4} [\gamma^{\mu},\gamma^{\nu}], \quad \omega_{\mu \nu}=-\omega_{\nu \mu} \in \mathbb{R}.$$
Because of (1) and $(\gamma^{0})^2=1$ it follows that
$$\gamma^0 (\sigma^{\mu \nu})^{\dagger} \gamma^0=\sigma^{\mu \nu},$$
which implies
$$S^{\dagger}=\exp \left (+\frac{\mathrm{i}}{4} \omega_{\mu \nu} \gamma^0 \sigma^{\mu \nu} \gamma^0 \right )=\gamma^0 \exp \left (+\frac{\mathrm{i}}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right ) \gamma^0 = \gamma^0 S^{-1} \gamma^0.$$
Multiplying this from left and right with $\gamma^0$ gives
$$S^{-1}=\gamma^0 S^{\dagger} \gamma^0.$$
Now from
$$\psi'(x')=S \psi(x)$$
it follows
$$\overline{\psi'(x')}=[\psi'(x')]^{\dagger} \gamma^0=\psi^{\dagger}(x) S^{\dagger} \gamma^0=\overline{\psi(x)}\gamma^0 S^{\dagger} \gamma^0=\overline{\psi}S^{-1}.$$
From this it follows that indeed $\overline{\psi(x)}\psi(x)$ is a scalar field as claimed.

But the relation, $\gamma^0 (\gamma^{mu})^{\dagger} \gamma^0 = \gamma^{\mu}$ is not invariant under the rep. of Lorentz transformations, because $S$ is NOT unitary. To see this explicitly let's play a bit:

What holds true is
$$(\gamma'{}^{\mu})^{\dagger}={\Lambda^{\mu}}_{\nu} (\gamma^{\nu})^{\dagger} \; \Rightarrow \; \gamma^0 (\gamma'{}^{\mu})^{\dagger} \gamma^0=\gamma'{}^{\mu}.$$

Now the left-hand side can be rewritten as follows:
$$\gamma^0 (\gamma'{}^{\mu})^{\dagger} \gamma^0=\gamma^0 (S^{-1})^{\dagger} (\gamma^{\mu})^{\dagger} S^{\dagger} \gamma^0 = \underbrace{\gamma^0 (S^{-1}) \dagger \gamma^0}_{S} \gamma^0 \gamma^{\mu} \gamma^0 \underbrace{\gamma^0 S^{\dagger} \gamma^0}_{S^{-1}}=S \gamma^0 \gamma^{\mu} \gamma^0 S^{-1} = S \gamma^0 S^{-1} S \gamma^{\mu} S^{-1} S \gamma^0 S^{-1}=\gamma'{}^0 \gamma'{}^{\mu} \gamma'{}^0 \neq \gamma'{}^{\mu}.$$
QED.

 

[kim george]

Physics is a huge and vast field, I think I might be not knowing the best pick for the asked query but as far as my knowledge is concerned, Sγ0S† = γ0 is analogous to ΛηΛT = η, will be the step included.

 

[stevendaryl]

At the risk of making things even more complicated, I would like to know what the Dirac equation looks like in generalized coordinates. The defining relation for the gamma matrices is:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$

where $\eta^{\mu \nu}$ is the Minkowski metric for a cartesian basis. What I'm wondering is whether you can do the Dirac equation in an arbitrary basis by letting the defining relation be:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 g^{\mu \nu}$

where $g^{\mu \nu}$ is the metric in this basis. However, the way that people write the Dirac wave function is:

$\left( \begin{array}\\ \Psi_1(x^\mu)\\ \Psi_2(x^\mu)\\ \Psi_3(x^\mu) \\ \Psi_4(x^\mu)\end{array} \right)$

and there seems to be no connection between the basis $x^\mu$ used to describe the components $\Psi_j(x^\mu)$ and the basis used to describe the gamma matrices. So even if spherical coordinates (for example) are used to describe the components, people continue to use a cartesian basis for the gamma matrices.

That seems kind of inconsistent, because a Lorentz transformation is just a special case of a coordinate transformation, and in that particular special case, people definitely make sure to transform the coordinates and the gamma matrices together.

 

[pellman]

I'm having trouble following your post, vanhees71. Let me go over it a little. If we have

$\psi'(x')=S\psi(x)$ and $\gamma'^0=S\gamma^0 S^{-1}$

Then

$$\bar{\psi'}(x')\psi'(x')=\psi'^{\dagger}(x') \gamma'^0 \psi'(x')$$
$$=\left(S\psi(x)\right)^{\dagger} \left(S\gamma^0 S^{-1}\right) S \psi(x)$$
$$=\psi^{\dagger}(x) S^{\dagger} S \gamma^0 \psi(x)$$

which won't equal $\psi^{\dagger}(x) \gamma^0 \psi(x)$ unless S is unitary. Or did I go wrong somewhere?

 

[vanhees71]

You always use the fixed $\gamma$ matrices to build the covariant bilinear forms, as detailed in my posting for the example of the scalar density, $\overline{\psi} \psi$, i.e., the Dirac symbols are fixed quantities in the Dirac-spinor representation, like $\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)$ for the Minkowski pseudo-metric in the four-vector (fundamental) representation of the Lorentz group.

 

[pellman]

I get it now! If $\psi$ and $\psi'$ are related by a Lorentz transformation, then the matrices that appear in their respective equations are identical. The numerical form of the Dirac equation in the two frames is covariant.

What was confusing me was the relation $S^{-1}\gamma^{\mu} S = {\Lambda^{\mu}}_{\nu} \gamma^\nu$. But what appears in the Dirac equation of the boosted frame is

$$S \left({\Lambda^{\mu}}_{\nu} \gamma^\nu \right) S^{-1} = \gamma^\mu$$

I was thinking that the gamma matrices in the boosted frame were
$S^{-1}\gamma^{\mu} S$. I think this was problem all along. Big thanks!

 

[Demystifier]

An alternative view of Lorentz covariance of the Dirac equation:
http://lanl.arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]

 

[samalkhaiat]

An alternative view of Lorentz covariance of the Dirac equation:
http://lanl.arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]

Regarding Eq(28): Do you know that choosing a "preferred" Lorentz frame does not (and should not) change the tensorial character of any geometrical object? i.e., Do you think that the representations of $SO(1,3)$ are frame dependent?
Regarding Eq(52): why should this equation allow you to conclude that $S^{ -1 } \psi$ is Lorentz scalar? Is your $\psi$ a bispinor (taking values in Minkowski space) or not? It has to be, otherwise you can not identify it with Dirac bispinor. In this case, eq(52) should read
$$\left( S^{ -1 } \psi ( x ) \right)^{ ' }= \psi ( x^{ ' } ) \neq \psi ( x )$$

Sam

 

[samalkhaiat]

I was thinking that the gamma matrices in the boosted frame were
$S^{-1}\gamma^{\mu} S$. I think this was problem all along.

Collecting the 4 numerical gamma matrices in a “4-vector” form $\gamma^{ \mu } = ( \gamma^{ 0 } , \gamma^{ i } )$ does not mean that $\gamma^{ \mu }$ transforms (or behave) like 4-vector.
Dirac’s gammas or Pauli’s sigmas are FRAME-INDEPENDENT numerical matrices: When it comes to Lorentz transformations/ 3-rotations, $\gamma^{ \mu } / \sigma^{ i }$ are not different from ordinary numbers.
The relation
$$S^{ - 1 } ( \Lambda ) \gamma^{ \mu } S ( \Lambda ) = \Lambda^{ \mu }{}_{ \nu } \gamma^{ \nu } ,$$
IS NOT a transformation law for the gamma’s, rather IT IS a condition on the $SO(1,3)$-representation matrices, $S ( \Lambda_{ 1 } ) S( \Lambda_{ 2 } ) = S ( \Lambda_{ 1 } \Lambda_{ 2 } )$, that allows you to use the same $\gamma^{ \mu }$’s in all Lorentz frames.
This is funny, only few days ago I pointed out to similar misunderstanding regarding the behaviour of Pauli’s sigma matrices under $SO(3)$ rotations. See post#45 in

www.physicsforums.com/showthread.php?t=754419&page=3

Sam

 

[stevendaryl]

The relation
$$S^{ - 1 } ( \Lambda ) \gamma^{ \mu } S ( \Lambda ) = \Lambda^{ \mu }{}_{ \nu } \gamma^{ \nu } ,$$
IS NOT a transformation law for the gamma’s, rather IT IS a condition on the $SO(1,3)$-representation matrices, $S ( \Lambda_{ 1 } ) S( \Lambda_{ 2 } ) = S ( \Lambda_{ 1 } \Lambda_{ 2 } )$, that allows you to use the same $\gamma^{ \mu }$’s in all Lorentz frames.

Well, I think that there are multiple interpretations. There is an algebra for working with vectors called Clifford algebras, which is an associative algebra with the vector multiplication rule:

$e_\mu e_\nu + e_\nu e_\mu = 2 g_{\mu \nu}$

An arbitrary element of this algebra can be written as a linear combination of basis elements, and there are 16 of them: the identity, a pseudoscalar $\rho = e_0 e_1 e_2 e_3$, 4 vectors $e_0, e_1, e_2, e_3$, 6 bivectors $e_0 e_1, e_0 e_2, e_0 e_3, e_1 e_2, e_2 e_3, e_3 e_1$, 4 pseudo-vectors $\rho e_0, \rho e_1, \rho e_2, \rho e_3$.

This is of course the algebra of gamma-matrices. But under the Clifford algebra interpretation, $\gamma_\mu$ is a basis vector in the $\mu$ direction, rather than the $\mu$ component of a vector. So as a vector, $\gamma_\mu$ would transform under a Lorentz transformation according to

$\gamma_\mu' = \Lambda^\nu_\mu \gamma_\nu'$

 

[samalkhaiat]

Well, I think that there are multiple interpretations.

There are no “multiple interpretations”. There is misunderstanding and confusion.
This is rigorous mathematics not QM.

There is an algebra for working with vectors called Clifford algebras, which is an associative algebra with the vector multiplication rule:

$e_\mu e_\nu + e_\nu e_\mu = 2 g_{\mu \nu}$

An arbitrary element of this algebra can be written as a linear combination of basis elements, and there are 16 of them: the identity, a pseudoscalar $\rho = e_0 e_1 e_2 e_3$, 4 vectors $e_0, e_1, e_2, e_3$, 6 bivectors $e_0 e_1, e_0 e_2, e_0 e_3, e_1 e_2, e_2 e_3, e_3 e_1$, 4 pseudo-vectors $\rho e_0, \rho e_1, \rho e_2, \rho e_3$.

This is of course the algebra of gamma-matrices. But under the Clifford algebra interpretation, $\gamma_\mu$ is a basis vector in the $\mu$ direction, rather than the $\mu$ component of a vector. So as a vector, $\gamma_\mu$ would transform under a Lorentz transformation according to

$\gamma_\mu' = \Lambda^\nu_\mu \gamma_\nu'$

You brought this on yoursef, I can use The Clifford group $\mbox{ Pin }( 1 , 3 )$ and its relation to the Lorentz group $\mbox{ SO }( 1 , 3 )$ and show you that the so-called Dirac representation (i.e., the isomorphism from the complexified Clifford algebra onto $\mathbb{ C }( 2^{ 2 } )$)
$$\rho : \mbox{ pin }( 1 , 3 )^{ \mathbb{ C } } \rightarrow \mathbb{ C }( 2^{ 2 } ) ,$$
is given by constant and frame-independent matrices
$$\rho ( e^{ \mu } ) \equiv \gamma^{ \mu }$$

Sam

 

[stevendaryl]

There are no “multiple interpretations”. There is misunderstanding and confusion.

So are you saying that the geometric algebra approach is mistaken? I had the impression that mathematically, it was equivalent to what was done using matrices, but without actually ever introducing matrices. In the geometric algebra approach, the terms $\gamma^\mu$ are interpreted as basis vectors, not matrices at all.

 

[pellman]

I learned something important in my last post but the opening post still eludes me. The original question came from pondering the conjugate of the Dirac equation. Let us start by writing the equation in components as

$$i(\gamma^\mu)_{jk}\partial\psi_k - m\psi_j = 0$$

we take the complex conjugate

$$-i(\gamma^\mu)^*_{jk}\partial\psi^*_k - m\psi^*_j = 0$$

$(\gamma^\mu)^*_{jk}$ is the kjth component of $(\gamma^\mu)^{\dagger}$.

$$-i(\partial\psi^*_k )(\gamma^\mu)^{\dagger}_{kj} - m\psi^*_j = 0$$

Multiplying by $-\gamma^0$ from the right and using the fact $(\gamma^0)^2=1$, we have

$$i(\partial\psi^*_m ) (\gamma^0)_{ml} (\gamma^0)_{lk} (\gamma^\mu)^{\dagger}_{kj}(\gamma^0)_{jh} + m\psi^*_j (\gamma^0)_{jh} = 0$$

which is

$$\bar{\psi}(i\gamma^0(\gamma^\mu)^\dagger \gamma^0 \overleftarrow{\partial}_\mu + m)=0$$

So we need $\gamma^0(\gamma^\mu)^\dagger \gamma^0 =\gamma^\mu$ for this to take the usual form $$\bar{\psi}(i\gamma^\mu \overleftarrow{\partial}_\mu + m)=0$$ . But the claim in the Wikipedia article is that while we can choose a set of gamma matrices such that this is true in one frame, the relation does not in general hold other frames? But this would mean that while the Dirac equation is covariant its conjugate is not?

 

[samalkhaiat]

So are you saying that the geometric algebra approach is mistaken?

No, I did not even mention that unnecessary garbage.

I had the impression that mathematically, it was equivalent to what was done using matrices, but without actually ever introducing matrices.

Yes okay, check your information. Even in that approach, local observables
$$O ( x ) = \bar{ \psi } ( x ) \Gamma \psi ( x ) , \ \ \mbox{ for any Clifford number } \ \Gamma \in \{ I , \gamma_{ 5 } , \gamma^{ \mu } , \gamma_{ 5 } \gamma^{ \mu } , [ \gamma^{ \mu } , \gamma^{ \nu } ] \} ,$$
inherit their transformation properties from $\psi ( x )$ not from $\Gamma$. Otherwise, you run into contradiction. Try it.

What next, forms and differential geometry, if you bring it up I can prove the same results for you. Try me.

In the geometric algebra approach, the terms $\gamma^\mu$ are interpreted as basis vectors, not matrices at all.

In Dirac’s theory and Dirac representation of Clifford algebra, which we were talking about, the gammas are matrices not numbers.
Lorentz group has the following natural actions:

1) On the index space of fields and their arguments, $SO(1,3)$ acts by finite-dimensional matrix representation:
$$\psi_{ r } ( x ) \rightarrow \psi^{ ' }_{ r } ( x ) = S_{ r }{}^{ s } ( \Lambda ) \psi_{ s } ( \Lambda^{ - 1 } x ) .$$
2) On Hilbert space, it acts by infinite-dimensional unitary representation
$$\psi_{ r } ( x ) \rightarrow \psi^{ ' }_{ r } ( x ) = U^{ \dagger } ( \Lambda ) \psi_{ r } ( x ) U ( \Lambda ) = S_{ r }{}^{ s } ( \Lambda ) \psi_{ s } ( \Lambda^{ - 1 } x ) .$$

$SO(1,3)$ has no natural action on the Dirac representation of the Clifford algebra, $\rho ( e^{ \mu } ) = \gamma^{ \mu }$, or on its unitary equivalent representation $u^{ \dagger } \gamma^{ \mu } u$.

Sam

 

[Demystifier]

Well, I think that there are multiple interpretations.

There are no “multiple interpretations”.

I also think that there ARE different interpretations, and the paper in my post above presents one alternative interpretation in which gamma matrices transform as a vector, while "Dirac" wave function transforms as a scalar.

Of course, Samalkhaiat is right that this is rigorous mathematics, not QM. But rigorous mathematics is always based on some axioms, and there is always some freedom in choosing axioms. So if you change some axioms, you obtain different results (theorems). The paper mentioned above changes some axioms. When stevendaryl and I say "different interpretations" we really mean "different axioms".

In fact, rigorous mathematics does know the concept of different interpretations:
http://en.wikipedia.org/wiki/Interpretation_(model_theory)

The important point to stress is that such a change of axioms does not change physics. The physically measurable quantities are not affected by this change of axioms. As explained in the paper above, this is analogous to a transformation from Schrodinger to Heisenberg picture of QM, which also can be viewed as a change of "axioms" without changing physics.

 

[vanhees71]

I'm not sure what you mean that there are "different interpretations for the Dirac field" (which should be seen as a quantized field, of course). The Dirac field is the local realization of the representation $(1/2,0) \oplus (0,1/2)$ of the Lie algebra of the Lorentz group. In this sense it's unique. For details, see my QFT manuscript on my homepage.

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Nikolic's formulation seems to be a correct alternative way (I just haven't studied the paper in detail, so I cannot say for sure whether it's all mathematically correct, but it looks good in principle) to look at the transformation properties of the "classical" Dirac equation, which is physically equivalent to the usual one. One should note that the observables are necessarily built out of the 16 bilinear forms, $\overline{\psi} \Gamma_J \psi$, with $\Gamma_j \in \{\mathbb{1},\gamma^5, \gamma^{\mu}, \gamma^5 \gamma^{\mu},\sigma^{\mu \nu} \}$, where $\sigma^{\mu \nu}=\mathrm{i}/2 [\gamma^{\mu},\gamma^{\nu}]$.

 

[stevendaryl]

No, I did not even mention that unnecessary garbage.

It's hard for me to have a discussion with someone who gets so emotional about gamma matrices.

 

[stevendaryl]

Of course, Samalkhaiat is right that this is rigorous mathematics, not QM. But rigorous mathematics is always based on some axioms, and there is always some freedom in choosing axioms. So if you change some axioms, you obtain different results (theorems). The paper mentioned above changes some axioms. When stevendaryl and I say "different interpretations" we really mean "different axioms".

Yes. To me, rigorous mathematics proves theorems, which say: In any structure in which X is true, Y is also true. Rigorous mathematics does not prove statements of the form: The correct way to think about X is Y. I think it's a misuse of mathematics to make such a claim.

Now, what you can do is to point out that a particular interpretation of a formalism is inconsistent. Certain collections of axioms together are contradictory. Samalkhaiat seems to be saying that, but he didn't make clear which axioms are contradictory.

 

[samalkhaiat]

I also think that there ARE different interpretations, and the paper in my post above presents one alternative interpretation in which gamma matrices transform as a vector, while "Dirac" wave function transforms as a scalar.

How can a bi-spinor be a scalar? As for your paper, It seems to me that your work rests on inconsistent definition. On page 11 your wrote:

Starting from (11), consider the transformation

$$( S^{ -1 } \psi )^{ ' } = S S^{ -1 } \psi = \psi . \ \ \ \ \ (52)$$

This shows that $S^{ -1 } \psi$ transforms as a scalar, so I define the scalar
$$\Psi = S^{ -1 } \psi . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (53)$$

Before I show the inconsistency between (52) and (53), let us recall the meaning of eq(11) which you started with,

$$\psi^{ ' } ( x^{ ' } ) = S \psi ( x ) . \ \ \ \ \ (11)$$

Implicit in (11) is the fact that the 4 components of Dirac spinor mix under Lorentz transformations, i.e., $\psi$ transforms non-trivially by the $4 \times 4$ matrix representation of $SO(1,3)$:

$$S = \exp ( - \frac{ i }{ 2 } \omega_{ \mu \nu } \Sigma^{ \mu \nu } ) .$$

OK, let us start with eq(53). If $\Psi$ is a SCALAR (as you defined it to be), then it should transforms trivially (by the identity matrix) under the Lorentz group, i.e., Lorentz transformations should not mix the components of $\Psi$:
$$\Psi^{ ' } = ( S^{ - 1 } \psi )^{ ' } = I_{ 4 \times 4 } ( S^{ -1 } \psi ) = S^{ - 1 } \psi . \ \ \ (1)$$
This is clearly inconsistent with (52) unless $S = S^{ -1 } = I_{ 4 \times 4 }$ which we know it is not. Therefore, either eq(52) is wrong (which is not wrong) or eq(53) does not define a scalar. In fact (52) shows that $\Psi$ is not a scalar.
Now, we look at (52):

$$( S^{ -1 } \psi )^{ ' } = S ( S^{ -1 } \psi ) .$$

This means that the object $S^{ - 1 } \psi$ transforms by the non-trivial matrix representation $S$ of $SO(1,3)$. Thus, $S^{ - 1 } \psi$ is not (and does not define) a scalar as you claim in (53).

When stevendaryl and I say "different interpretations" we really mean "different axioms".

No body have problem with "consistent" set of axioms.

Sam

 

[stevendaryl]

How can a bi-spinor be a scalar?

Well, certainly a column vector of 4 complex numbers can be a Lorentz scalar, in the sense that those 4 numbers are not changed under a Lorentz transformation. So if you have a 4-component spinor $\psi$ that transforms in such-and-such a way under Lorentz transformations, and you have a particular frame $F$, then the 4 quantities $\Psi$ which are the components of $\psi$ in frame $F$ are 4 (Lorentz) scalars.

It's the same sort of thing as in the definition of invariant mass. Energy is not a scalar, it is a component of a 4-vector. But if you fix a frame (for example, the rest frame of the particle), then the energy as measured in that frame is a Lorentz scalar.

Possibly a clearer way to think about it: Let $\psi_A, \psi_B, \psi_C, \psi_D$ be four different spinors such that in the lab frame, their conjugates have the representations:

$\bar{\psi}_A = (1,\ 0,\ 0,\ 0)$
$\bar{\psi}_B = (0,\ 1,\ 0,\ 0)$
$\bar{\psi}_C = (0,\ 0,\ 1,\ 0)$
$\bar{\psi}_D = (0,\ 0,\ 0,\ 1)$

Then for any $\psi$, the following four numbers are Lorentz-scalars:

$\Psi_1 = \bar{\psi}_A \psi$
$\Psi_2 = \bar{\psi}_A \psi$
$\Psi_3 = \bar{\psi}_A \psi$
$\Psi_4 = \bar{\psi}_A \psi$

Then you can put the 4 numbers together into a "scalar bi-spinor":

$\left( \begin{array}\\ \Psi_1 \\ \Psi_2 \\ \Psi_3 \\ \Psi_4\end{array} \right)$

This column of scalars is constructed to be equal to the representation of $\psi$ in the lab frame.