2
$\begingroup$

Take $\Lambda$ to be a Lorentz matrix, it satisfying $\Lambda^T \eta \Lambda=\eta$. By writing $\Lambda=\exp[-\frac{i}{2}\omega_{\mu\nu}\mathcal J^{\mu\nu}]$, we find that the generators satisfy $$ [\mathcal{J}^{\mu\nu},\mathcal{J}^{\rho\sigma}]=i(\eta^{\mu\rho}\mathcal J^{\nu\sigma}+\cdots) \qquad \qquad(1) $$

Now we can find any $n$ dimensional representation just by finding $n\times n$ matrices satisfying $(1)$ and calculating the matrix exponential. For example, the Dirac representation is given by exponentiating the generators $\frac{i}{4}[\gamma^\mu,\gamma^\nu]$, where $\gamma^\mu$ are the Dirac gamma matrices.

Now, let $S^{\mu\nu}$ be the generators of a $n$ dimensional representation, and write $M=\exp[-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}]$. Is it true that this matrix satisfies $M^T \eta M=\eta$? Should perhaps the $T$ symbol be $\dagger$ instead? What should the $\eta$ matrix be for an $n$ dimensional vector space? perhaps it is $\eta=\text{diag}(1,-1,-1,\cdots,-1)$, with $(n-1)$ negative ones?

The thing is, I tried to check $M^T \eta M=\eta$ with several representations, and this is sometimes right, and sometimes not, which made me think that I was doing something wrong. I checked my math and couldn't find any mistake, so now I think that $M^T \eta M=\eta$ is not true in general, but then what is the point of finding such an $M$ matrix? Why we care for more representations of the Lorentz Group, if the resulting matrices don't possess this nice property? I know that quantum fields transform according to different representations of the LG, but why should this be? Why don't we just take any linear transformation? Why would we care for it being related to the Lorentz Group?

I know that if the field transforms as $\phi\to M\phi$, then we can make covariant lagrangians, but this is always related to some properties $M$ satisfies, irrespective of it being related to $\Lambda$. For example, take any matrix satisfying $M^T M=1$; then $\mathcal L=\phi^T\phi$ is covariant, and we dont need $M$ to come from a representation of the LG.

For example, the covariance of the Dirac Lagrangian can be easly proven by noting that if the field transforms as $\psi\to M\psi$, then it is true that $\bar M \gamma^\mu M=\Lambda^\mu_\nu \gamma^\mu$. But to prove this relation we don't need to use $(1)$ nor anything coming from the Lorentz Group, but just some algebra. Couldn't we just find more matrices satisfying nice algebraic relations that allow as to make covariant lagrangians, matrices that are not related to the Lorentz Group?.

$\endgroup$
  • 1
    $\begingroup$ If we were trying to take an axiomatic approach to QFT, we could say that there must exist certain funcions of space-time which must transform in a covariant way. We would like the transformation to be linear, so $\phi^i(x)\to M^i_j(\Lambda) \phi^j(\Lambda x)$. Now the quetion is, why should $M$ sit in a representation of the LG? Why can't it be a more general linear transformation? $\endgroup$ – AccidentalFourierTransform Aug 13 '15 at 14:14
3
$\begingroup$

The defining property of the fundamental representation of the Lorentz group $\mathrm{SO}(1,3)$ $$ M^T\eta M = \eta \quad \forall M\in\mathrm{SO}(1,3)$$ and hence the defining property of the Lorentz group itself does not make sense in representations other than the fundamental, because those are not naturally equipped with a metric "$\eta$" from a physics point of view.

Instead, we know that the Lorentz group is a spacetime symmetry that all proper relativistic laws (quantum or not) should respect in the sense that they should transform "covariantly", i.e. in a proper representation of it. A function of spacetime $\phi(x)$ has several options how to do this - it can be a scalar $\phi$, a vector $\phi^\mu$, a rank-2 tensor $\phi^{\mu\nu}$, and so on, but it has to choose one of these options, so that we know how laws involving it will change when we perform a Lorentz transformation (which corresponds to us changing our frame of reference).

That the concrete realizations of the representations may not preserve some arbitrarily chosen metric on the representation space (which is just the space of the values certain functions take rather than anything like spacetime itself) is wholly irrelevant to this.

$\endgroup$
  • 1
    $\begingroup$ I understand the importance of the fundamental representation (and its tensor products), so I see why we have fields such as scalar, vector, tensor ($\phi,\phi^\mu,\phi^{\mu\nu}$), but why do every field to come from a certain representation? Why can't a field transform according to any other rule? $\endgroup$ – AccidentalFourierTransform Aug 13 '15 at 14:10
  • 1
    $\begingroup$ @qftishard: It can! E.g. the object given by the Christoffel symbols is a connection, and transforms non-linearly under diffeomorphisms of GR. The important thing is that the rule must be such that the physical equations involving the object take the same form after all possible transformations, and in most cases, this is only achieved through a linear transformation rule. Also, in QM, there's Wigner's theorem telling us that every symmetry is either unitarily or anti-unitarily represented. $\endgroup$ – ACuriousMind Aug 13 '15 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.