# (L&L vol. 4, sec. 14) Wave equation of a particle of spin one

Having trouble fully understanding a number of simple-looking statements while reading section 14 of Landau & Lifshitz volume 4 (second edition).

The section starts discussing that a spin one particle is described by a three-component wave function (in the rest frame). Extending this viewpoint to the relativistic domain, it is said that such a vector can be due to a 4-vector $$\psi^\mu$$ or an antisymmetric 4-tensor $$\psi^{\mu\nu}$$, for which $$\psi^0$$ and the space components $$\psi^{ik}$$ are zero in the rest frame. First question: why does $$\psi^{\mu\nu}$$ have to be antisymmetric?

These quantities are then combined in a relativistic fashion to produce some simple first-order equations, which are $$i\psi_{\mu\nu} = \hat{p}_\mu \psi_\nu - \hat{p}_\nu \psi_\mu,\\ im^2\psi_\mu = \hat{p}^\nu \psi_{\mu\nu},$$ where $$\hat{p}=i\partial$$. The above two relations can be massaged to give $$\hat{p}^\mu \psi_\mu=0,\\ (\hat{p}^2-m^2)\psi_\mu=0.$$

Later on, the conserved current density 4-vector for the spin-one field is given as $$j^\mu=i\left((\psi^{\mu\nu})^*\psi_{\nu}-\psi^{\mu\nu}\psi_\nu^*\right)$$ I fail to verify (using the relations shown above) that indeed this 4-vector is conserved (i.e. $$\partial_\mu j^\mu=0$$). Second question: could someone provide a hint as to how to do that, perhaps I'm overlooking a simple step?

Finally, it is mentioned how the spin-one wavefunctions behave under inversion ($$\hat{P}$$). It is said that if $$\psi^\mu$$ is a pseudovector, inversion affects it according to $$\hat{P}\psi^\mu=(-\psi^0,\psi^i).$$ Third question: why does inversion (an operation on the spatial coordinates) change the sign of the time-component of $$\psi^\mu$$?

Hope someone can help out, and thank you in advance.

1: In the rest frame, $$\psi^{\mu}$$ becomes $$\psi^{\mu} = (0,\psi^i)$$ which is a three-vector in the $$i$$ index under rotations. Similarly if $$\psi^{\mu \nu}$$ is anti-symmetric it can be denoted as $$\psi^{\mu \nu} = (\psi^{0i},\psi^{ij})$$ where there are $$3+3=6$$ degrees of freedom there (note the diagonal entries $$\psi^{\mu \mu}$$ are zero automatically if $$\psi^{\mu \nu}$$ is anti-symmetric). If $$\psi^{ij} = 0$$ then this reduces to the three components $$\psi^{0i}$$ and so it too transforms as a vector in the $$i$$ index under rotations.
2: To show $$\partial_{\mu} j^{\mu} = 0$$ you need to use the equations of motion for $$\psi_{\mu \nu}$$ and $$\psi_{\mu \nu}^*$$, and you also need to use the fact that $$\psi_{\mu \nu}$$ and $$\psi_{\mu \nu}^*$$ are anti-symmetric, the proof is just two lines I encourage you to write it out.
3: In the previous section they discussed internal parity of a scalar as being related to reflections of the coordinate axes, and the overall sign of the wave function being $$\pm$$ depending on whether the wave function is a scalar or a pseudo-scalar, $$P \psi(t,r) = \pm \psi(t,-r)$$. This discussion is just the vector version of that relation, $$P \psi^{\mu} = \pm (\psi^0,-\psi^i)$$, and in the $$-$$ case it gives your question.