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.

 

[Demystifier]

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.

You say that (52) is not wrong, but actually the catch is that there IS a trouble with (52). Namely, the expression (S^{ -1 }\psi)' suggests that you need to transform not only \psi, but also S^{ -1 }. The confusing thing here is that S^{ -1 } is itself an (inverse) transformation, so what does it mean to transform the transformation? For that reason, even though (52) seems right, formally it does not have a proper mathematical form.

What does it mean? It means that (52) is valid only in ONE Lorentz frame (the one in which \Psi=\psi), which is why your result above is inconsistent with (52). Nevertheless, all other equations in this Appendix are valid in all Lorentz frames (and have the proper mathematical form), which makes the whole theory consistent because after (53) there is no further reference to Eq. (52).

Eq. (52) is just a more elegant (but perhaps also more confusing) way to write the second equation in (28). The only purpose of the confusing equation (52) is to justify Eq. (53). But unlike (52), equation (53) is valid in all frames so should not be confusing.

 

[stevendaryl]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR. A frame field is a set of 4 vector fields, written \vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3 with \vec{e}_0 timelike and the others spacelike. For any vector \vec{A}, and for any frame field, we can compute a set of 4 numbers:
A_i = \vec{e}_i \cdot \vec{A}. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object \vec{\tilde{A}} = (A_0, A_1, A_2, A_3). But it's not a vector, since its components are scalars. So what is it?

 

[Demystifier]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR.

Indeed, in GR it is impossible to define a spinor transformation under general coordinate transformations. For that reason, in GR the transformation rules are redefined, such that "spinors" transform as scalars and gamma matrices as components of a vector under general coordinate transformations.

If samalkhaiat was right (which fortunately he is not), then it would be impossible to treat spinors in GR.

A frame field is a set of 4 vector fields, written \vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3 with \vec{e}_0 timelike and the others spacelike. For any vector \vec{A}, and for any frame field, we can compute a set of 4 numbers:
A_i = \vec{e}_i \cdot \vec{A}. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object \vec{\tilde{A}} = (A_0, A_1, A_2, A_3). But it's not a vector, since its components are scalars. So what is it?

These are a collection of 4 scalars. Unfortunately, most GR textbooks do not explain these subtleties very well. A good exception is the Wald's General Relativity textbook, where he introduces two kinds of indices thus avoiding the confusion.

 

[stevendaryl]

These are a collection of 4 scalars. Unfortunately, most GR textbooks do not explain these subtleties very well. A good exception is the Wald's General Relativity textbook, where he introduces two kinds of indices thus avoiding the confusion.

Actually, it occurs to me that it's kind of an "internal symmetry" like isospin. The 4-tuple of scalar values can certainly be treated as a vector, but it's a vector that lives in a different space than the tangent vectors of spacetime.

 

[Demystifier]

Actually, it occurs to me that it's kind of an "internal symmetry" like isospin. The 4-tuple of scalar values can certainly be treated as a vector, but it's a vector that lives in a different space than the tangent vectors of spacetime.

Yes, that's also a correct way to think of it. But I have to slightly correct your terminology. The tangent space (at a point on a curved manifold) is a FLAT space. So the internal space is the tangent space, which should be distinguished from the "physical" curved space.

When the physical space is flat (which is the case in special relativity), the things are more confusing because both physical and tangent space are flat, so it is more difficult to distinguish them. Yet, there is a difference because on the physical space you are allowed to use curved (e.g. polar) coordinates, while on the internal tangent space you are only allowed to use flat coordinates because otherwise you cannnot define spinor transformations in the internal space.

 

[atyy]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR. A frame field is a set of 4 vector fields, written \vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3 with \vec{e}_0 timelike and the others spacelike. For any vector \vec{A}, and for any frame field, we can compute a set of 4 numbers:
A_i = \vec{e}_i \cdot \vec{A}. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object \vec{\tilde{A}} = (A_0, A_1, A_2, A_3). But it's not a vector, since its components are scalars. So what is it?

In line with Demystifier's comment about the tangent space, the Dirac equation in curved spacetime is written with orthonormal tetrads, eg. Eq 5.1 and 5.2 of http://arxiv.org/abs/1108.3896.

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.

Bill_K comments on this and why tetrads are used to write the Dirac equation in curved spacetine in post #2 of https://www.physicsforums.com/showthread.php?t=668131.

A related thread about the Dirac equation under general coordinate transformations was https://www.physicsforums.com/showthread.php?t=653985, which has several nice comments like haushofer's post #10.

 

[samalkhaiat]

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.

May be you need to recall the following fact in geometry.

“Geometrical objects of a given space are classified by the representations of the global symmetry group of that space”.

Poincare’ group is the global symmetry group of Minkowski space-time (prove it if you have time!)
Thus, there exists a one-to-one correspondence between objects in Minkowski space and representations of Poincare’ group. The dictionary is given as follow

1) Trivial Rep.:
( \mbox{ I } , a ) \leftrightarrow \mbox{ scalar } : \phi ( x )
2) Defining Rep.:
( \Lambda , a ) \leftrightarrow \mbox{ vector } : V^{ \mu } ( x )
3) Bispinor Rep.:
( \sqrt{ \Lambda } , a ) \leftrightarrow \mbox{ Dirac field } : \ \psi_{ i } ( x )
4) Spinorvector Rep.:
( \sqrt{ \Lambda } \otimes \Lambda , a ) \leftrightarrow \mbox{ R-S field } : \ \psi^{ \mu}_{ i } ( x )
5) Rank-n Tensor Rep.:
( \Lambda \otimes \Lambda \otimes \cdots \otimes \Lambda , a ) \leftrightarrow \mbox{ Rank-n Tensor field } : \ T^{ \mu_{1} \mu_{2} \cdots \mu_{n} } ( x )
And the list continues. The point now is this: choosing a frame DOES NOT change the geometrical nature of an object. This is because the representation theory of Poincare’ group does not care less about your frame. For example: evaluating vector, say, in the “lab-frame” or in the “bathroom-frame” does not turn it into non-vector. It is as simple as the saying: there exists no frame where an electron becomes a meson.

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

If \psi satisfies the Dirac equation, i.e. Dirac spinor, then
\Psi_{ 1 } = \psi_{ 1 } , \Psi_{ 2 } = \psi_{ 2 } , \cdots \Psi_{ 4 } = \psi_{ 4 } . \ \ (1)
These will be used below.

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

The above mentioned list does not contain this “scalar bi-spinor”. Therefore, it does not exist: No such object exists in Klein’s geometry and I have never heard of it. We have, however, a scalar-(iso)spinor such as the K-meson doublet ( K^{ + } , K^{ 0 } ) and, we also have a scalar-(iso)vector like the \pi meson triplet ( \pi^{ + } , \pi^{ - } , \pi^{ 0 } ). Here, the word “scalar” refers to the behaviour under Poincare’ group which does not mix different members of the doublet or the triplet. And, the words isospinor and isovector refer to the behaviour of the doublet and the triplet under the action of the internal iso-spin group SU(2), i.e., not space-time spinor and vector.

\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.

Substituting (1) in this column, we find
\Psi = \psi
Thus, if \psi transforms as a bi-spinor, then \Psi transforms as bi-spinor too.
So, you went through all that business of constructing rows and columns and ended up writing
\mbox{ DIRAC } = \mbox{ dirac } .
I am impressed. May be, you need to go through the following simple exercise:
You are given the following set of four numbers
\psi_{ 1 } = 1, \ \psi_{ 2,3,4 } = 0 .
Show that
\Psi = \left( \begin{array} { c } 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right)
is not Lorentz scalar. Determine its form in an arbitrary Lorentz fram.
Solving this exercise will help you a lot, so do it.

Sam

 

[samalkhaiat]

You say that (52) is not wrong, but actually the catch is that there IS a trouble with (52). Namely, the expression (S^{ -1 }\psi)' suggests that you need to transform not only \psi, but also S^{ -1 }. The confusing thing here is that S^{ -1 } is itself an (inverse) transformation, so what does it mean to transform the transformation? For that reason, even though (52) seems right, formally it does not have a proper mathematical form.

What does it mean? It means that (52) is valid only in ONE Lorentz frame (the one in which \Psi=\psi), which is why your result above is inconsistent with (52). Nevertheless, all other equations in this Appendix are valid in all Lorentz frames (and have the proper mathematical form), which makes the whole theory consistent because after (53) there is no further reference to Eq. (52).

Eq. (52) is just a more elegant (but perhaps also more confusing) way to write the second equation in (28). The only purpose of the confusing equation (52) is to justify Eq. (53). But unlike (52), equation (53) is valid in all frames so should not be confusing.

If S \in SO(1,3) , \mbox{ and } \ \psi \in \mathbb{ C } ( 2^{ 2 } ), then S^{ - 1 } \in SO(1,3) \ \mbox{ and } S^{ - 1 } \psi \in \mathbb{ C } ( 2^{ 2 } ). In other words, if \psi is a bi-spinor, then \Psi = S^{ - 1 } \psi is also a bi-spinor.

Sam

 

[samalkhaiat]

Indeed, in GR it is impossible to define a spinor transformation under general coordinate transformations. For that reason, in GR the transformation rules are redefined, such that "spinors" transform as scalars and gamma matrices as components of a vector under general coordinate transformations.

If samalkhaiat was right (which fortunately he is not), then it would be impossible to treat spinors in GR.

I don’t think so my friend, you are totally confused about this. The “transformations” of the gammas in curved space come not from the gammas themselves but from the vielbein fields. Seeking compact notations and convenience, some people choose to redefine the gammas by writing
\gamma^{ \mu } ( x ) = e^{ \mu }{}_{ a } ( x ) \ \gamma^{ a },
where, \mu = 0 , 1, .. , 3 is world index and a = 0 , 1 ,.. , 3 is local Lorentz index. You could do every thing without even introducing \gamma^{ \mu } ( x ).
The generally covariant action is
A = \int d^{ 4 } x \ | e | e^{ \mu }{}_{ a } ( x ) \left( i \bar{ \psi } ( x ) \gamma^{ a } \nabla_{ \mu } \psi ( x ) \right) ,
where
\nabla_{ \mu } \psi = \partial_{ \mu } \psi + \frac{ 1 }{ 8 } \Omega_{ \mu a b } ( x ) [ \gamma^{ a } , \gamma^{ b } ] \psi ,
and \Omega^{ \mu }_{ a b } ( x ) is the spin-connection.
You want more, I am able to tell you “almost” any thing you need to know about this subject.


Sam

 

[stevendaryl]

The point now is this: choosing a frame does change the geometrical nature of an object.

Okay, I explained already why you were wrong. If \psi_A, \psi_B, \psi_C, \psi_D are 4 different spinors, and \psi is another spinor, then do you agree that the four numbers below are all Lorentz scalars:

\Psi_1 = \bar{\psi}_A \psi
\Psi_2 = \bar{\psi}_B \psi
\Psi_3 = \bar{\psi}_C \psi
\Psi_4 = \bar{\psi}_D \psi

Do you agree with that claim? If not, why not?

[edit]

Note: the four scalars above can be put into a column vector \Psi, but this \Psi is NOT a spinor in the sense of representations of the Lorentz group. That's because its four numbers are Lorentz SCALARS. \Psi does NOT get transformed by a Lorentz transformation. So your conclusion, that because \psi = \Psi in the lab frame (meaning the components are equal) then \psi = \Psi in every frame, is wrong. If \Psi were a spinor, that would be true. But it's NOT a spinor---it's a collection of 4 Lorentz scalars.

What you're saying is analogous to saying:

In the rest frame of a particle, the 4-momentum P can be represented by the row vector V = (mc, 0, 0, 0). Since P=V in one frame, they must be equal in all frames. Therefore, in every frame, P = (mc, 0, 0, 0).
​

The reasoning is wrong, because V is not a vector in the sense of representations of the Lorentz transforms. V is 4 numbers that happen to be equal to the four numbers in the vector P in the rest frame of the particle.

 

[atyy]

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.

This article puts fermions in curved spacetime with tetrads, and uses tetrads and the "constant gamma matrices" to define "spacetime gamma matrices" which do have the form of your guess \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 g^{\mu \nu}. http://particlephd.wordpress.com/2009/02/07/the-spin-connection/

 

[Demystifier]

You want more, I am able to tell you “almost” any thing you need to know about this subject.

Just one thing. How does the spinor transform under general coordinate transformations?

 

[samalkhaiat]

Just one thing. How does the spinor transform under general coordinate transformations?

How much do you know about the representation theory of GL(n) and its Lie algebra?

 

[Demystifier]

How much do you know about the representation theory of GL(n) and its Lie algebra?

I'm not an expert, but I think I know enough to understand an argument based on it. Just write down what you think to be a good explanation, and if necesary, I can ask you some additional questions.

 

[stevendaryl]

I'm not an expert, but I think I know enough to understand an argument based on it. Just write down what you think to be a good explanation, and if necesary, I can ask you some additional questions.

It seems to me that the details of how spinors transform under various coordinate transformations is orthogonal to Sam's unresolved complaints about the alternative way of doing the Dirac equation. It seems to me that the alternative way is clearly equivalent to the usual way, and there is nothing about the details of transformations that could change that. Those details factor through the argument, and don't affect it in any way.

The confusion is from failing to distinguish between a spinor, which is a geometric object, and a column matrix, which is the representation of that spinor in a particular frame.

Pick two frames, F and F'. Let \psi be a spinor described by the Dirac equation. Let \Psi be the representation of \psi in frame F and let \Psi' be the representation in F'. Let \partial_\mu mean differentiation with respect to F coordinates, and \partial'_\mu mean differentiation with respect to F' coordinates. Then the usual covariance of the Dirac equation tells us the following two equations hold:

1. (-i \gamma^\mu \partial_\mu + m) \Psi = 0
2. (-i \gamma^\mu \partial'_\mu + m) \Psi' = 0

Now, letting \Lambda be the transformation matrix relating the two coordinate systems, we can rewrite: \partial_\mu = \Lambda^\nu_\mu \partial'_\nu. So equation 1 becomes:

(-i \gamma^\mu \Lambda^\nu_\mu \partial'_\nu + m) \Psi = 0

If we define \Gamma^\nu to be \gamma^\mu \Lambda^\nu_\mu, then we have:

1. (-i \Gamma^\nu \partial'_\nu + m) \Psi = 0
2. (-i \gamma^\mu \partial'_\mu + m) \Psi' = 0

In equations (1) and (2), \Psi and \Psi' are matrices, not spinors. In frame F', you can use either equation; they are both valid equations, and they are equivalent, in the sense that the following equations hold:

1. U(\Lambda) \gamma^\mu U^{-1}(\Lambda) = \Lambda^\mu_\nu \gamma^\mu
2. \Psi' = U(\Lambda) \Psi
3. \Gamma^\nu = \Lambda^\mu_\nu \gamma^\mu

(for the appropriate matrix U(\Lambda))

In these equations, you can view (1) as the defining equation for U(\Lambda). Equations (2) and (3) are then just the definitions of what we mean by \Psi' and \Gamma^\mu

So the only aspect of this alternative way of doing the Diract equation that is not true by definition is equation (1) for the matrix U(\Lambda). The details of U are important for relating the two approaches, but the details are unimportant for seeing that the two approaches are equivalent.