(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.
 A: 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.
