# 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 $$-\partial^{i}$$?

• $\uparrow$ You should for the spatial differentiations. – Qmechanic Nov 13 '18 at 5:54

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.

• 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? – Libertarian Monarchist Bot Nov 13 '18 at 1:02
• @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. – Chiral Anomaly Nov 13 '18 at 1:13
• I will try to digest your answer. Thank you very much. – Libertarian Monarchist Bot Nov 13 '18 at 1:18
• 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. – AccidentalFourierTransform Nov 15 '18 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

$$F^{A}[D(x)]\phi_{A}(x)=0,$$

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.