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


However, if I use $(\ast)$ directly, replacing $\phi$ by $\phi^{\mathrm{P}}$, why shouldn't I also transform each $\partial^{i}$ to $-\partial^{i}$?

  • $\begingroup$ $\uparrow$ You should for the spatial differentiations. $\endgroup$
    – Qmechanic
    Nov 13, 2018 at 5:54

2 Answers 2


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.

  • $\begingroup$ Thank you. But isn't the parity defined as a symmetry of the spacetime itself? If it's the symmetry of the spacetime itself, then it acts on coordinate also. Could you explain more? $\endgroup$
    – Valac
    Nov 13, 2018 at 1:02
  • $\begingroup$ @NewStudent That's a good question. I added two more paragraphs to try to address that. Not sure if I addressed it well, though. Please let me know if it still isn't satisfying. $\endgroup$ Nov 13, 2018 at 1:13
  • $\begingroup$ I will try to digest your answer. Thank you very much. $\endgroup$
    – Valac
    Nov 13, 2018 at 1:18
  • $\begingroup$ Symmetries may act on fields, or on coordinates (or, sometimes, even both!). Either approach is fine; they are equivalent, cf. Group representations and active/passive transformations. $\endgroup$ Nov 15, 2018 at 0:23

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


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

$$\rho:F^{\prime B}[D(\rho(x))]=R^{B}_{\,\,\,A}F^{A}[D(x)],$$



where $R$ is a constant, so that the naturality condition

$$F^{\prime B}[D(\rho(x))]\phi^{\prime}_{B}(\rho(x))=F^{C}[D(x)]\phi_{C}(x)$$


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


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


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

$$\frac{\partial J^{\prime\mu}(y)}{\partial y^{\mu}}=0$$


$$\tilde{\eta}^{\mu\nu}(y)\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial\phi^{\prime\dagger}(y)}{\partial y^{\nu}}\phi^{\prime}(y)-\phi^{\prime\dagger}\frac{\partial\phi^{\prime}(y)}{\partial y^{\nu}}\right)=0$$

Using the chain rule, one has, in $x$-coordinate,

$$\tilde{\eta}^{\mu\nu}(y(x))\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial x^{\beta}}{\partial y^{\nu}}\frac{\partial}{\partial x^{\alpha}}\left(\frac{\partial\phi^{\prime\dagger}(y(x))}{\partial x^{\beta}}\phi^{\prime}(y(x))-\phi^{\prime\dagger}(y(x))\frac{\partial\phi^{\prime}(y(x))}{\partial x^{\beta}}\right)=0$$


$$\tilde{\eta}^{\mu\nu}(y(x))\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial x^{\beta}}{\partial y^{\nu}}=\eta^{\alpha\beta},$$

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


Therefore, the transformation property of $J_{\mu}$ is indeed consistent with the naturality condition.


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