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In Schwartz's QFT book, he said that the vector representation of the Lorentz group, $V_\mu$ that is four-dimensional, is the direct sum of two irreducible representations of $SO(3)$: a spin-0 representation, which is one-dimensional, and a spin-1 representation, which is three-dimensional. i.e. $4=3\oplus 1$.

Question 1:

Is there an explicit reason why so? I am looking for both a mathematical reason that is also physical. The book just indicated it without stating exactly how.

Is it because the 0th component of the vector is the spin-0 and the other components are spin-1? If so, this doesn't make sense to me since the vector field $A_\mu$ itself is spin-1.

The book then continues to guess a Lagrangian for a massive spin-1 field and came up with $$ \mathcal{L}=-\frac{1}{2}\partial_\nu A_\mu\partial_\mu A_\nu+\frac{1}{2}m^2A^2_\mu $$ Which has the equations of motion $$ (\partial^2+m^2)A_\mu=0 $$ And then says that it has "4 propagating modes" and $A_\mu$ is actually 4 scalar fields and therefore the Lagrangian is not for a vector field. In this case this is $4=1\oplus 1\oplus 1\oplus 1$.

Question 2:

What does he mean by propagating modes? Are the fields just scalar because the equation of motion is for each index of $A_\mu$ or is there a much better meaning of why?

He never actually defined "Propagating modes" at all. Anyways, the book continues with another guess which is the Proca Lagrangian $$ L = \frac{a}{2}A^\mu\partial^2A_\mu+\frac{b}{2}A_\mu\partial_mu\partial_\nu A_\nu +\frac{1}{2}m^2A_\mu A^\mu $$ Where it says that the $\partial_\mu A_\mu$ contraction forces $A_\mu$ to be a 4-vector since if each component $A_\mu$ transforms as scalar then $\partial_\mu A_\mu$ is not Lorentz invariant.

Question 3:

I seem to not understand what the book meant by that. Why is it not Lorentz Invariant?

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1 Answer 1

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Answer 1

Each representation of the Lorentz group is equivalent to a representation of an $SU(2)\otimes SU(2)$. The vector representation in particular is a $(1/2,1/2)$ representation. The generators of the $SU(2)\otimes SU(2)$ group are $J_i^\pm = J_i\pm i K_i$; so for rotations the generators of the two groups are equal and given by the usual $SU(2)$ generators. So if you just add the two $1/2$ with the usual addition rules you find the $0 \oplus 1$ decomposition under rotations.

For a more physical interpretation of this, go in the center of mass frame: you have one degree of freedom $A^0$ that is left unchanged by rotations i.e. it is a scalar, and three degrees of freedom $\vec A$ that form a $SO(3)$ vector.

It is not 100% true that "$A^\mu$ is itself a spin-1". What's true is that it contains a spin-1 representation and that's why you can use this field to describe spin-1 particles. From the Wigner-Eckart theorem this is a necessary condition to have non-zero matrix elements between the particle state and the field operator

$\langle0|A^\mu (x)|p,\lambda\rangle$

Answer 2 and 3

As also explained in the answer linked in the comment, that Lagrangian is equivalent to the sum of 4 independent Lagrangians for four independent scalar fields and one of these scalars appear with the "wrong" sign in the Lagrangian and you have states with negative energies. As it is also written a bit later in the book, just by looking at the equations of motion having four scalars or a 4-vector is exactly the same thing. Nothing is telling you that $A^\mu$ must transform in a particular way under a Lorentz transformation. Third problem is that you have too many degrees of freedom.

The situation is different when you add the other term with the two mixed derivatives

$$A^\mu \partial_\mu \partial_\nu A^\nu$$

Under a Lorentz transformation you get two $\Lambda^{-1}$ matrices from the two partial derivatives. If the field then transforms as $A\to \Lambda A$ then those two matrices cancel out and you end up with what you started: that piece is Lorentz invariant. The Lagrangian that you had before was invariant even if under a Lorentz transformation $x\to \Lambda x$ but $A\to A$.

The point that you need to understand from that chapter is that the free equations of motion for fields are not something you can build by hand randomly or something you can play around with by adding new pieces or modifying old ones. Free equations of motion are 100% fixed by group theory. Even the more complicated looking equations, for example for a massive graviton, are just a sum of projectors on the right Poincaré representation needed for the particle you want to describe. In the vector case for example the equations of motion are equivalent to

$$(\square + m^2)A^\mu = 0 \\ \partial^\mu A_\mu = 0$$

In momentum space and in the center of mass frame ($p^\mu = (p^0,\vec 0)$), the first equation is telling you $p^2=(p^0)^2 = m^2$, while the second is $A^0 = 0$. That is you are projecting your field on-shell and keeping only the spin-1 component.

A fun exercise is to do the same for the graviton. You impose that the two scalar and one vector degrees of freedom must be zero in the center of mass frame, add those condition with the on-shell condition, go in a general frame and in position space and you see that you magically get back the Fierz-Pauli equations of motion.

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  • $\begingroup$ One last question, in the following when he derived the solution to the equation of motion of the massive vector field, the condition $p^\mu\epsilon_\mu^i=0$ implies that there are only 3 of these polarization vectors $\epsilon_\mu^i$. I don't understand how the condition implies that there are only 3. Can you explain to me how so? Thank you. $\endgroup$ Nov 8, 2018 at 19:40
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    $\begingroup$ @Gradient137 Because that equation is telling you that the $\epsilon$ are vector orthogonal to $p^\mu$. In 4D space this means that they form a 3D vector space and thus you have tree basis vectors. $\endgroup$
    – FrodCube
    Nov 8, 2018 at 20:52
  • $\begingroup$ But there are 4 components of $p^\mu$ though. Shouldn't the inner product between $\epsilon$ and $p^\mu$ then goes to a sum of 4 terms? $\endgroup$ Nov 8, 2018 at 21:36
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    $\begingroup$ @Gradient137 I don't see your problem. If I tell you that I'm looking for vectors $w$ orthogonal to a given $v$, the equation is a scalar equation $w\cdot v=0$, but the solution is the whole plane orthogonal to $v$, i.e. you need two vector $w^1$ and $w^2$ to span that space. This is the same but in 4D. $\endgroup$
    – FrodCube
    Nov 9, 2018 at 16:07

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