I am currently reading Schwartz on QFT and the Standard Model and arrived now at Chapter 8.2.2, where he derives a Lagrangian for a massive Spin-1 field. The final Lagrangian looks like this: $$ \mathcal{L}=\frac{1}{2} A_{\mu} \square A_{\mu}-\frac{1}{2} A_{\mu} \partial_{\mu} \partial_{\nu} A_{\nu}+\frac{1}{2} m^{2} A_{\mu}^{2}\tag{8.23} $$ The equations of motion are $$ (\square +m^{2}) A_{\mu}=0 $$ with a Lorentz-invariant constraint: $$ \partial_{\mu} A_{\mu}=0. $$

The Lagrangian was constructed in a way, that $A_{\mu}$ is forced to transform like a 4-vector under Lorentz-transformations in order to keep the Lagrangian Lorentz-invariant.

Thus, we have a 4-dimensional representation of the Lorentz-group acting on the vector space, given by all the $A_{\mu}$ solving above equations.

From later in the book (Chapter 10) we know that, if we have a 4-dimensional representation of the Lorentz-group, then the corresponding representation of the Rotation-group is $3 \oplus 1$ -dimensional.

Now here is my question: The constraint $\partial_{\mu} A_{\mu}=0$ reduces the number of degrees of freedom of the vectors $A_{\mu}$ to 3. Schwartz then states:

Since $\partial_{\mu} A_{\mu}=0$ is a Lorentz-invariant condition, it has to remove a complete representation, which with one degree of freedom can only be the spin-0 component.

Can anyone explain to me, what he means by that/ why that is? In other words: Why does Lorentz-invariance of the constraint necessarily remove the Spin-0 d.o.f. and not one of the three Spin-1 degrees of freedom?


1 Answer 1


A Lorentz-invariant constraint is rotation-invariant, among other things. The three spin-1 degrees of freedom are mixed with each other by rotations, so a rotation-invariant constraint cannot remove one of them without removing all of them. But a spin-0 degree of freedom is not mixed with anything else by rotations, so it can be removed by a rotation-invariant constraint. Since $\partial^\mu A_\mu=0$ is only one constraint, it can only remove one degree of freedom, which must therefore be the spin-0 degree of freedom.

By the way, a Lorentz-invariant constraint is also boost-invariant, but this doesn't affect the preceding argument because a boost mixes different momenta, too. The preceding argument only requires considering the case of zero momentum, which is rotation-invariant, so that only the spin degrees of freedom are affected by the rotations.

  • $\begingroup$ Now it's crystal clear, thank you very much! $\endgroup$
    – sika_98
    Commented May 30, 2021 at 10:02

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