Parity Transformation on Classical Fields I've been confused by this parity transformation in classical field theory for a long time. 
Let $\phi(t,\vec{x})$ be a scalar field. Then, up to some constant phase factor, it transforms to $\phi^{\mathrm{P}}(t,\vec{x})=\phi(t,-\vec{x})$. Now, consider the current 
$$j^{\mu}(t,\vec{x})=i\left\{\phi^{\dagger}(t,\vec{x})\partial^{\mu}\phi(t,\vec{x})-(\partial^{\mu}\phi^{\dagger}(t,\vec{x}))\phi(t,\vec{x})\right\}. \quad\quad\quad\quad\quad\quad(\ast)$$
Under a parity transformation $\mathrm{P}:(t,\vec{x})\rightarrow(t,-\vec{x})$, I expect to have 
$$(j^{\mathrm{P}})^{0}(t,\vec{x})=j^{0}(t,-\vec{x}),\quad\vec{j^{\mathrm{P}}}(t,\vec{x})=-\vec{j}(t,-\vec{x})$$
However, if I use $(\ast)$ directly, replacing $\phi$ by $\phi^{\mathrm{P}}$, why shouldn't I also transform each $\partial^{i}$ to $-\partial^{i}$?
 A: The parity transformation doesn't act on the coordinates. It acts on the fields. In other words, it acts on the whole function $\phi$. Given a function $\phi$ whose value at $(t,\vec x)$ is $\phi(t,\vec x)$, the parity transform gives a new function $\phi_P$ whose value at $(t,\vec x)$ is
$$
\phi_P(t,\vec x)=\phi(t,-\vec x).
\tag{1}
$$
Sometimes this is abbreviated by writing 
$$
\vec x\mapsto-\vec x,
\tag{2}
$$
 but that's only a convenient abbreviation. Parity acts on the fields, not on the coordinates.
When we say that a model has parity symmetry, we mean that if a given function $\phi$ satisfies the equations of motion, then so does $\phi_P$. We call this a symmetry of spacetime, again as a way of being concise, but it isn't really a symmetry of spacetime itself; it's a symmetry of the behavior of the scalar fields that occupy spacetime. 
If the transformation $\vec x\mapsto -\vec x$ preserves the spacetime metric (say, the Minkowski metric), then that is a symmetry of spacetime itself, at least if we consider the metric to be a property of spacetime itself. But we don't usually consider the scalar fields to be a property of spacetime itself. Instead, the scalar fields are something that occupies the spacetime. We can contrive a model (a set of equations of motion for the scalar fields) that is not parity-symmetric even if the spacetime itself is parity-symmetric.
A: Let $M$ be a manifold in which a theory $\mathfrak{T}$ is define. Let $\rho:x\rightarrow y$ be a coordinate transformation on $M$. Then $\rho$ is a classical symmetry of $\mathfrak{T}$ iff it preserves the equation of motion. In general, one can write down the EOM of $\mathfrak{T}$ in a generic form
$$F^{A}[D(x)]\phi_{A}(x)=0,$$
where $\phi_{A}$ are the fundamental fields of theory $\mathfrak{T}$, $D(x)$ is a local differential operator (e.g $i\frac{\partial}{\partial t}$, $\partial_{\mu}\partial^{\mu}$, and $\partial\!\!\!/$, etc.) and $F$ is a generic function.
If $\rho$ is a symmetry of $\mathfrak{T}$, then denoting objects in new coordinates by $F^{\prime A}$ and $\phi^{\prime}_{A}$, one has
$$F^{\prime B}[D(y)]\phi^{\prime}_{B}(y)=0.$$
Then, one must have the transformation properties
$$\rho:F^{\prime B}[D(\rho(x))]=R^{B}_{\,\,\,A}F^{A}[D(x)],$$
and 
$$\rho:\phi^{\prime}_{B}(y)=(R^{-1})^{C}_{\,\,\,B}\,\phi_{C}(x),$$
where $R$ is a constant, so that the naturality condition 
$$F^{\prime B}[D(\rho(x))]\phi^{\prime}_{B}(\rho(x))=F^{C}[D(x)]\phi_{C}(x)$$
holds.
This current $J[\phi^{\dagger},\phi]$ can be understood as a one form $d\phi^{\dagger}\phi-\phi^{\dagger}d\phi$. In local coordinate, it is given by 
$$J[\phi^{\dagger},\phi](x)=J[\phi^{\dagger},\phi]_{\mu}(x)dx^{\mu},$$
where $J[\phi^{\dagger},\phi]_{\mu}(x)=(\partial_{\mu}\phi^{\dagger})\phi(x)-\phi^{\dagger}(\partial_{\mu}\phi)(x)$. Then it is clear how $J_{\mu}(x)$ should transform under parity. The transformation is given by a pull-back. 
The minus sign for $\vec{J}(x)$ does not come from replacing $\vec{x}$ by $-\vec{x}$ in $\phi(x)$. In this picture, the pull-back does not change $\phi$ to $\phi_{\mathrm{P}}$. The minus sign comes from the pull-back 
$$dy^{\mu}=\frac{\partial y^{\mu}}{\partial x^{\nu}}dx^{\nu}.$$
To relate this to the above "naturality condition", one considers the conservation equation 
$$d\ast J=0$$ 
This equation should hold universally in all coordinates. 
For simplicity, Let $M$ be a Minkowski spacetime with metric $\eta=(+1,-,\cdots,-1)$. Denoting $\eta^{\mu\nu}$ as the metric on the cotangent bundle $T^{\ast}M$, and $\eta_{\mu\nu}$ as the metric on the tangent bundle $TM$. Then, in local coordinate, one has to show that the equation
$$\eta^{\mu\nu}\partial_{\mu}J_{\nu}=0$$
should be independent of the choice of coordinate. 
Let $\rho:x\rightarrow y$ be a coordinate transformation. Then in $y$-coordinate, one has fields $\phi^{\prime\dagger}$, $\phi^{\prime}$, $J^{\prime\mu}$, and $\tilde{\eta}^{\mu\nu}$ satisfying
$$\frac{\partial J^{\prime\mu}(y)}{\partial y^{\mu}}=0$$
or
$$\tilde{\eta}^{\mu\nu}(y)\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial\phi^{\prime\dagger}(y)}{\partial y^{\nu}}\phi^{\prime}(y)-\phi^{\prime\dagger}\frac{\partial\phi^{\prime}(y)}{\partial y^{\nu}}\right)=0$$
Using the chain rule, one has, in $x$-coordinate,
$$\tilde{\eta}^{\mu\nu}(y(x))\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial x^{\beta}}{\partial y^{\nu}}\frac{\partial}{\partial x^{\alpha}}\left(\frac{\partial\phi^{\prime\dagger}(y(x))}{\partial x^{\beta}}\phi^{\prime}(y(x))-\phi^{\prime\dagger}(y(x))\frac{\partial\phi^{\prime}(y(x))}{\partial x^{\beta}}\right)=0$$
But 
$$\tilde{\eta}^{\mu\nu}(y(x))\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial x^{\beta}}{\partial y^{\nu}}=\eta^{\alpha\beta},$$ 
one has, in $x$-coordinate
$$\eta^{\alpha\beta}\frac{\partial}{\partial x^{\alpha}}\left(\frac{\partial\phi^{\prime\dagger}(y(x))}{\partial x^{\beta}}\phi^{\prime}(y(x))-\phi^{\prime\dagger}(y(x))\frac{\partial\phi^{\prime}(y(x))}{\partial x^{\beta}}\right)=0$$
Since $\phi$ and $\phi^{\dagger}$ are scalar fields, one has
$$\phi^{\prime}(y(x))=\phi(x),\quad\phi^{\prime\dagger}(y(x))=\phi^{\dagger}(x)$$
Therefore, the transformation property of $J_{\mu}$ is indeed consistent with the naturality condition. 
