I Why are the gamma-matrices invariant?

[dextercioby]

Sam, you definitely know geometry. Perhaps you can help me with the following question whose explicit answer I haven't seen anywhere: Let's take Minkowski flat spacetime M₄ and its de Rham differential complex built on the cotangent bundle by the exterior derivative d. How exactly do you put an electromagnetic field (one form field $A_{\mu}$ judged in terms of the de Rham complex) in/on spacetime by means of the so-called U(1) gauge bundle, or more precisely why is the space-time differentiation (flat Levi-Civita connection) "equal" or "similar" to the connection in the gauge bundle?

 

[samalkhaiat]

Sam, you definitely know geometry. Perhaps you can help me with the following question whose explicit answer I haven't seen anywhere: Let's take Minkowski flat spacetime M₄ and its de Rham differential complex built on the cotangent bundle by the exterior derivative d. How exactly do you put an electromagnetic field (one form field $A_{\mu}$ judged in terms of the de Rham complex) in/on spacetime by means of the so-called U(1) gauge bundle, or more precisely why is the space-time differentiation (flat Levi-Civita connection) "equal" or "similar" to the connection in the gauge bundle?

And you definitely know how to put me in trouble. Well, it is not really difficult, but to put a connection on principal bundle requires many definitions and then few theorems take your structure closer to the Maurer-Cartan form and Cartan structure equation.
Good account with applications to Dirac and ‘t Hooft-Polykov monopoles, and instantons can be found in
[1] M. Gockeler & T. Schucker : “Differential Geometry, gauge theories, and gravity”, Cambridge University Press. 1990. Chapter 9 & 10.
And my favourite
[2] J. A. de Azcarraga & J. M. Izquierdo: “Lie groups, Lie algebras: cohomology and some application in physics”, Cambridge University Press. 1995. Chapter 2.

 

[Demystifier]

He is absolutely correct, but you are wrong. Weinberg did not, does not and will never say that “Dirac spinor is a Lorentz scalar”, not even in his dreams.

Statement 1 (Weinberg): Dirac field is a coordinate scalar.
Statement 2 (me): A coordinate transformation may be a Lorentz transformation, in which case Dirac field is a coordinate Lorentz scalar.

Please explain how can it be that Statement 1 is right and Statement 2 wrong? Are you saying that a coordinate transformation cannot be a Lorentz transformation?

 

[vanhees71]

A bispinor is a bispinor.

I'm referring only to the usual flat Minkowski spacetime, because I'm not very familiar with spinors in GR. As far as I remember there you have to introduce vierbeins (tetrads), and there can be spinors only in spacetimes where you have tetrades.

In Minkowski space a bispinor (Dirac spinor) behaves under a Lorentz transformation as
$$\psi'(x')=S(\Lambda) \psi(\Lambda^{-1} x'),$$
where
$$S=\exp \left (-\frac{\mathrm{i}}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right ) \quad \text{with} \quad \sigma^{\mu \nu} = \frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

 

[PatrickUrania]

The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

Indeed. The left part of your last equation again satisfies the Clifford algebra, hence it can be used as a new set of γ-matrices. Using that set requires a transformation of the wavefunction that recovers the original wavefunction, only this time there was no frame change. Also, by your last equation that new set is just the Lorentz-transformed original set. That is what has been argued here.

 

[samalkhaiat]

The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

Even though you used a quotation mark “Minkowski vector”, the statement is still misleading:
1) As we all know, objects carrying space-time indices need not be space-time tensors. For example the Levi-Civita connection \Gamma^{\mu}_{\nu\rho} is not a type-(1,2) tensor, and certainly the Dirac \gamma^{\mu} is not a vector.
2) The above equation is not a Lorentz (group) transformation (\mbox{LT}) equation: Notice that on the left-hand-side you have an (matrix) action on the spin indices
S^{-1}_{mp}\gamma^{\mu}_{pq}S_{qn}, while on the right-hand-side the action is on the vector index only
\Lambda^{\mu}{}_{\nu}\gamma^{\nu}_{mn}.
So, the correct transformation of \gamma^{\mu}_{mn} under the matrix spin group of Lorentz must be
\mbox{LT}(\gamma^{\mu}_{mn}) = \Lambda^{\mu}{}_{\nu} \ S^{-1}_{mp} \ \gamma^{\nu}_{pq} \ S_{qn} = \gamma^{\mu}_{mn} .
This is exactly what I proved in my first post in this thread: Under Lorentz transformations, the Dirac gamma’s are invariant numerical matrices.

 

[Demystifier]

under the matrix spin group of Lorentz ...
the Dirac gamma’s are invariant numerical matrices.

I think that nobody here doubts that. But it looks as if you fail to realize that there is a thing called the group of Lorentz coordinate transformations, which is not the same thing as matrix spin group of Lorentz. Even though the group is the same, the corresponding transformations are not. (Your mathematics is very sophisticated, in fact much more sophisticated than mine, so I'm sure you know that, in abstract algebra, the concept of abstract group is one thing, while realization of group as a group of transformations of some concrete objects is another. By choosing different objects on which a transformation will act, one obtains different realizations of the same group.)

The Lorentz coordinate transformation is just a special case of a general coordinate transformation (the diffeomorphism group), so what is true for general coordinate transformations must also be true for Lorentz coordinate transformations. So if the Dirac gamma transforms as a vector under general coordinate transformations (and Weinberg says it does), then the same Dirac gamma transforms as a vector under Lorentz coordinate transformations (which is what I repeat over and over again).

And this is not in a conflict with your correct claim that Dirac gamma is invariant under matrix spin group of Lorentz. We are both right, and the conflict is only apparent because
(i) we talk about different realizations of the same Lorentz group, and
(ii) we use a somewhat different language (admittedly, yours being more sophisticated than mine, creating an illusion that your statements sound "more correct" than mine).

 

[vanhees71]

Indeed. The left part of your last equation again satisfies the Clifford algebra, hence it can be used as a new set of γ-matrices. Using that set requires a transformation of the wavefunction that recovers the original wavefunction, only this time there was no frame change. Also, by your last equation that new set is just the Lorentz-transformed original set. That is what has been argued here.

There are no wave functions in relativistic QT, only field operators!

 

[vanhees71]

Even though you used a quotation mark “Minkowski vector”, the statement is still misleading:
1) As we all know, objects carrying space-time indices need not be space-time tensors. For example the Levi-Civita connection \Gamma^{\mu}_{\nu\rho} is not a type-(1,2) tensor, and certainly the Dirac \gamma^{\mu} is not a vector.
2) The above equation is not a Lorentz (group) transformation (\mbox{LT}) equation: Notice that on the left-hand-side you have an (matrix) action on the spin indices
S^{-1}_{mp}\gamma^{\mu}_{pq}S_{qn}, while on the right-hand-side the action is on the vector index only
\Lambda^{\mu}{}_{\nu}\gamma^{\nu}_{mn}.
So, the correct transformation of \gamma^{\mu}_{mn} under the matrix spin group of Lorentz must be
\mbox{LT}(\gamma^{\mu}_{mn}) = \Lambda^{\mu}{}_{\nu} \ S^{-1}_{mp} \ \gamma^{\nu}_{pq} \ S_{qn} = \gamma^{\mu}_{mn} .
This is exactly what I proved in my first post in this thread: Under Lorentz transformations, the Dirac gamma’s are invariant numerical matrices.

Well, I used the usual physicist's slang, according to which a "Lorentz transformation" of a spinor or tensor field is defined by the representation these fields live on. For the Dirac-spinor field this means
$$[\hat{U}(\Lambda) \hat{\psi}(x) \hat{U}^{\dagger}(\Lambda)]^a={S^a}_{b}(\Lambda) \hat{\psi}^b(\Lambda^{-1} x).$$
Of course the Dirac matrices act in spinor space (indices $a$ and $b$ in the formula).

 

[PatrickUrania]

There are no wave functions in relativistic QT, only field operators!

In QFT you are absolutely right. In Dirac theory it's still about a wave function.

 

[Demystifier]

In QFT you are absolutely right. In Dirac theory it's still about a wave function.

Or you can call it classical field, so that everybody is happy.

 

[vanhees71]

Well, but then you need a lot of handwaving, called "hole theory", and you end up with something that's a complicated version of quantum field theory. That's why I prefer to say relativistic QT should be introduced as relativistic QFT right away.