1. Jun 21, 2014

pellman

2. Jun 21, 2014

Orodruin

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

3. Jun 21, 2014

OCR

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...

Very much so... IMO.

Last edited: Jun 22, 2014
4. Jun 21, 2014

pellman

Ha! I knew I missed something.

5. Jun 21, 2014

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?

Last edited: Jun 21, 2014
6. Jun 21, 2014

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?

7. Jun 21, 2014

Bill_K

Wikipedia is (partially) wrong. The page says,

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 obeys0S = γ0.

γμ' = (SγμS-1) = S-1† γμ† S = γ00 γμ† γ0S-1γ0 = γ00 γμ S-1γ0 = γ0 γμ' γ0

8. Jun 22, 2014

pellman

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.

9. Jun 23, 2014

Bill_K

I think what it says is that γ0 is the metric for Dirac spinors. That is, 0S = γ0 is analogous to ΛηΛT = η, where η is the Minkowski metric.

10. Jun 23, 2014

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.

11. Jun 23, 2014

pellman

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

12. Jun 23, 2014

Avodyne

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

13. Jun 24, 2014

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.

14. Jun 24, 2014

Bill_K

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.

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.

Last edited: Jun 24, 2014
15. Jun 24, 2014

pellman

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 $\Lambda$.

16. Jun 24, 2014

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.

Last edited: Jun 24, 2014
17. Jun 25, 2014

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.

18. Jun 25, 2014

stevendaryl

Staff Emeritus
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.

19. Jun 26, 2014

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?

20. Jun 27, 2014

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.