I Why are the gamma-matrices invariant?

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1. Feb 24, 2016

PatrickUrania

Hi,
I've been studying Dirac's theory of fermions. A classic topic therein is the proof that the equation is covariant. Invariably authors state that the gamma-matrices have to be considered constants: they do not change under a Lorentz-transformation. I am looking for the reason behind this. It seems to me that if you consider them a vector and the wave-function a scalar then all works out OK. The scalar, vector, tensor, pseudoscalar and pseudovector constructed from the gamma-matrices, the wave-function and its adjoint all have the same value as in the classical approach and transform in the correct fashion. What am I missing?

2. Feb 24, 2016

dextercioby

Have you asked yourself the question: what does a Lorentz transformation act on?

3. Feb 24, 2016

stevendaryl

Staff Emeritus
An interesting paper discussing this was written (I think) by Physics Forums' own Demystifier:
http://arxiv.org/pdf/1309.7070.pdf

In this paper, the author claims that the physical content of the Dirac equation is unchanged, whether you view spinors as transforming under Lorentz transformations and gamma matrices as being constant, or vice-versa.

4. Feb 25, 2016

PatrickUrania

@Dextercloby: As far as I can see, the Lorentz transformation acts on everything that lives in spacetime, but only on those objects. That implies it obviously acts on the momentum operator. By the same token it clearly should have an action on the gamma-matrices as they, at least partially, live in spacetime as well. Since the wavefunction has a spacetime point as its argument, that event has to be transformed. The action on the wavefunction itself is much less clear. Looking at it as a set of complex scalars does not seem unreasonable. In that view, the wavefunction exists in a ℂ4 space that is separate from spacetime. The gamma-matrices have 2 indices in that space as well.

5. Feb 25, 2016

PatrickUrania

@stevendaryl: I had read the article but was not aware that the author had a link to this forum. It indeed exactly describes what I'm asking here. What bothers me is that, given that there is a much simpler way to use spinors, why would anyone use the traditional way? I have been looking at some literature concerning tetrads (or vierbeins), that indeed seems to go in that direction. There is also the Dirac-Fock-Weyl equation that takes the Dirac equation to curved space and that - as far as I can see - again seems to point in that direction. Caveat: I am by no means an expert in quantum mechanics on curved spaces. However, the very vast majority still use the traditional way. I can only imagine that there is a good reason to do so, but so far I haven't seen it.

6. Feb 25, 2016

Demystifier

One obvious reason why the simple formalism (I developed in the recent paper you have seen) is not widely used, is that this formalism is not widely known. (The related formalism with tetrads is better known, but is not so simple.)

Another reason is that in most practical calculations it doesn't really matter which formalism you use, because in practice one rarely needs to transform a quantity from one Lorentz frame to another. Instead, one simply works in the fixed laboratory frame, in which there is no difference between the two approaches.

My formalism is much simpler when you want to prove Lorentz covariance of the Dirac equation. But once you prove it (in either a simple or a complicated way), you don't longer need to worry much about it.

In fact, the true reason why I developed this formalism in the first place is because I needed some new bilinear covariant combinations mentioned in the paper, which cannot easily be constructed in the standard formalism. But these new bilinear combinations appear in a certain non-standard formulation of QM, not in standard QM/QFT.

7. Feb 25, 2016

stevendaryl

Staff Emeritus
Just curious: is that Hestene's spacetime algebra approach?

8. Feb 25, 2016

Demystifier

It is not, but maybe there is a relation to it. I would need to refresh my knowledge about the Hestenes approach and think about it.

9. Feb 25, 2016

vanhees71

In the usual operator formulation of relativistic QFT the fields are operators, and a Lorentz transformation is realized as a unitary (ray) representation on that fields. The $\gamma$ matrices are complex-number matrices and thus the QFT operator doesn't do anything to them.

10. Feb 25, 2016

PatrickUrania

Your respons is indeed the standard one. However, why exactly do the γ-matrices have to be just complex numbers? As was explained above, considering them as a vector and the wavefunction as a scalar leads to the same physics, only without the need to introduce another representation of the Lorentz transformation. The article mentioned above explains this in detail. Is there a compelling reason why the γ-matrices have to be just complex numbers?

11. Feb 26, 2016

Demystifier

In curved spacetime, it's not only that $\gamma^{\mu}$ is a vector, but is also a function $\gamma^{\mu}(x)$ depending on the spacetime point $x$. So no, the γ-matrices do not have to be just complex numbers.

In fact, the γ-matrices satisfy the algebra
$$\{\gamma^{\mu}(x), \gamma^{\nu}(x) \} =2g^{\mu\nu}(x)$$
where $g^{\mu\nu}(x)$ is the spacetime metric. In Minkowski spacetime this can be reduced to
$$\{\gamma^{\mu}, \gamma^{\nu} \} =2\eta^{\mu\nu}$$
Hence γ-matrices can be chosen to be just numbers only when the metric tensor is just numbers.

12. Feb 26, 2016

vanhees71

The $\gamma$ matrices cannot be expressed in terms of field operators and are thus c-number valued matrices.

In GR spinors are, as any fields, local objects, i.e., they have to be defined with respect to comoving vierbeins.

13. Feb 26, 2016

Demystifier

In quantum gravity the vierbein may also be quantized i.e. become a field operator. In this case $\gamma^{\mu}(x)$ also becomes a field operator.

14. Feb 26, 2016

Demystifier

The above gave me a wild idea. Consider a quantity (operator, matrix, or something like that) $I(x)$ satisfying
$$I^2(x)=g(x)$$
where $g(x)$ is the determinant of the metric tensor with signature (-+++). In Minkowski spacetime this reduces to
$$I^2(x)=-1$$
so $I$ can be represented by the imaginary unit $I=i$. But in curved spacetime $I(x)\neq i$.

What if in quantum theory we replace $i$ with $I(x)$? Could it have something to do with merging quantum theory with gravity?

As I said, at this level this is only a wild speculation. But has anybody thought about something like that?

15. Feb 26, 2016

vanhees71

That's a funny idea. I've no clue, if anybody has thought about such a thing, but I'm no expert in attempts to create a consistent QT of gravity.

16. Feb 26, 2016

stevendaryl

Staff Emeritus
I'm not sure whether it is a different way of thinking of the same thing, or not, but in the Hestenes "geometric algebra" approach to the Pauli equation or the Dirac equation, the gamma matrices (or the Pauli matrices) are not treated as matrices, at all, but are just considered vectors. That is, Hestenes considers $\gamma^\mu$ to be a set of 4 basis vectors, rather than 4 components of a single vector.

17. Feb 26, 2016

vanhees71

But $\gamma^{\mu}$ cannot be just a number, because it fulfills the Clifford algebra $[\gamma^{\mu},\gamma^{\nu}]_+=2 g^{\mu \nu}$. Of course you can think of it in terms of abstract objects as well as you can think of the Dirac spinors being just abstract objects. Then the usual physicists' objects are components of spinors and matrix elements of Clifford-algebra elements defined as linear maps on the spinors.

18. Feb 28, 2016

PatrickUrania

I found this article by A. Weldon: "Fermions without vierbeins in curved space-time" (cds.cern.ch/record/466101/files/0009086.pdf or on arxiv)> There is also some explanation of these ideas on: http://www.qgf.uni-jena.de/gk_quantenmedia/downloads/Monitoring2012/talk_lippoldt.pdf. This looks like the correct generalisation of these ideas in curved space-time.
From all the above I think that considering the γ-matrices as a true spacetime vector is not only justified but also greatly simplifies the interpretation of what a spinor actually is.

19. Mar 8, 2016

samalkhaiat

See the theorem below.
You are missing a lot. Let me list few for you.
1) The Lorentz group is a matrix Lie group. As such, $SO(1,3)$ sees every index carried by a given object and transforms that index in the appropriate representation matrix. So, for $\Gamma^{\mu}_{ab}$ the index $a$ transforms in the spinor representation space $S_{4}$, $b$ in the conjugate spinor representation $\bar{S}_{4}$ and $\mu$ in the vector representation $D^{(1/2,1/2)}$. This means that $$\Gamma^{\mu}_{ab} \in S_{4}(\Sigma) \oplus \bar{S}_{4}(\bar{\Sigma}) \oplus D^{v}(M) .$$
So, the action of Lorentz algebra (infinitesimal transformation) on $\Gamma^{\mu}$ is, therefore, given by
\begin{align*} \delta^{\mu\nu} \Gamma^{\rho}_{ac} &= \Sigma^{\mu\nu}_{ab}\Gamma^{\rho}_{bc} - \bar{\Sigma}^{\mu\nu}_{cb}\Gamma^{\rho}_{ab} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} \\ &= \Sigma^{\mu\nu}_{ab}\Gamma^{\rho}_{bc} - \Gamma^{\rho}_{ab}\Sigma^{\mu\nu}_{bc} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} \\ &= [ \Sigma^{\mu\nu} , \Gamma^{\rho} ]_{ac} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} . \end{align*}
Now we can prove the theorem:
The matrices $\Sigma^{\mu\nu} = - \frac{i}{4}[\Gamma^{\mu} , \Gamma^{\nu} ]$ generate the spinor representation of $SO(1,3)$ if and only if $\delta \Gamma = 0$.
Proof:
i) Using the Dirac algebra $\big \{ \Gamma^{\mu} , \Gamma^{\nu} \big \} = 2 \eta^{\mu\nu}$, we can calculate the commutator
\begin{align*} [\Sigma^{\mu\nu} , \Gamma^{\rho} ]_{ac} &= i \left( \eta^{\mu\rho} \Gamma^{\nu}_{ac} - \eta^{\nu\rho} \Gamma^{\mu}_{ac} \right) \\ &= i \left( \eta^{\mu\rho}\delta^{\nu}_{\sigma} - \eta^{\nu\rho}\delta^{\mu}_{\sigma} \right) \Gamma^{\sigma}_{ac} \\ &= - (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} . \end{align*}
In the last equality we introduced the generator matrices $M^{\mu\nu}$ of the vector representation
$$(M^{\mu\nu})^{\rho}_{\sigma} = - i \left( \eta^{\mu\rho}\delta^{\nu}_{\sigma} - \eta^{\nu\rho}\delta^{\mu}_{\sigma} \right) .$$
Thus
$$\delta^{\mu\nu}\Gamma^{\rho}_{ac} = - (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} + (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} = 0 .$$
ii) Conversely, starting from $\delta \Gamma = 0$, we arrive at the generator matrices of the spinor representation $\Sigma^{\mu\nu} = (-i/4) [ \Gamma^{\mu} , \Gamma^{\nu}]$ as the Clifford solution of
$$[\Sigma^{\mu\nu}, \Gamma^{\rho}]_{ac} = i \left( \eta^{\mu\rho} \Gamma^{\nu}_{ac} - \eta^{\nu\rho} \Gamma^{\mu}_{ac} \right) .$$
This is nothing but the infinitesimal version of
$$S^{-1}(\Lambda) \Gamma^{\mu}S(\Lambda) = \Lambda^{\mu}{}_{\nu}\Gamma^{\nu} , \ \ \ \ \ \ (1)$$
which follows from the form-invariance of the Dirac equation. Manny textbooks say that (1) is the Lorentz transformation law for Dirac’s gammas. This is wrong, because (as we have just shown) (1) follows from the invariance of the $\Gamma$’s under Lorentz transformations. One should mention that the above theorem can be proved for any matrix Lie group.
2) Quantum theory can be made independent of a frame if the Hilbert space of the theory carries a unitary representation of the Poincare’ group, i.e.,
$$|\psi_{\alpha} \rangle \to U(\Lambda , a) |\psi_{\alpha}\rangle , \ \ \forall |\psi_{\alpha}\rangle \in \mathcal{H} ,$$ or, in terms of multi-component wave-function $$\psi^{'}_{\alpha}(x) = \langle x |U(\Lambda , a)|\psi \rangle = \mathcal{U}(\Lambda , a) \psi_{\alpha}(x) . \ \ \ \ (2)$$ But the Hilbert space of relativistic n-component wave functions is simply the space $$\mathcal{H} \cong \mathcal{F}(\mathbb{R}^{4}) \otimes \mathbb{C}^{n} ,$$ where $\mathcal{F}(\mathbb{R}^{4})$ is the space of functions on Minkowski spacetime. So, the Poincare’ group transforms the argument of the Dirac wavefunction (which comes from $\mathcal{F}(\mathbb{R}^{4})$) and mixes its components (which come from $\mathbb{C}^{4}$) :
$$\psi^{'}_{\alpha} (x) = S_{\alpha \beta}(\Lambda) \psi_{\beta}(\Lambda^{-1}(x-a)) . \ \ \ \ \ (3)$$ So, saying that the Dirac wavefunction transforms like scalar function will simply mean the followings very bad things:
(i) the operator $U(\Lambda , a)$ cannot be a unitary operator within the time independent scalar product of the Hilbert space of solutions of the Dirac equation, and (ii) the infinitesimal generator of $\mathcal{U}(\Lambda, a)$ does not contain the spin-1/2 matrices.

20. Mar 8, 2016

PatrickUrania

Thanks for the response. This is the first time I see an actual reason for this behavior. I don't understand everything though. Could you point me to a reference that explains this in some more detail?