# How is the parity transformation defined? Especially for vector or tensor fields?

If I have a scalar field \begin{align} f: \mathbb{R}^3 &\rightarrow \mathbb{C}\\ (x, y, z) &\mapsto f(x, y, z) \end{align}

We can define an operator $$P$$ that takes a function like $$f$$ and gives a new function $$Pf$$. It is clear that for scalar fields like $$f$$ we have

$$Pf(x, y, z) = f(-x, -y, -z)$$

But how is $$P$$ defined for vector fields? Suppose we have

\begin{align} \textbf{F}: \mathbb{R}^3 &\rightarrow \mathbb{R}^3\\ (x, y, z) &\mapsto F_x(x, y, z)\hat{\textbf{x}} + F_y(x, y, z)\hat{\textbf{y}} + F_z(x, y, z)\hat{\textbf{z}} \end{align}

Where each of $$F_x, F_y$$, and $$F_z$$ are scalar fields How is $$P\mathbf{F}$$ defined in this case?

$$P\mathbf{F}(x, y, z) = \mathbf{F}(-x, -y, -z) = F_x(-x, -y, -z)\hat{\mathbb{x}} + F_y(-x, -y, -z)\hat{\mathbb{y}} + F_z(-x, -y, -z)\hat{\mathbb{z}}$$

Or do the basis vectors get transformed as well?

$$P\mathbf{F}(x, y, z) = -\mathbf{F}(-x, -y, -z) = -F_x(-x, -y, -z)\hat{\mathbb{x}} - F_y(-x, -y, -z)\hat{\mathbb{y}} - F_z(-x, -y, -z)\hat{\mathbb{z}}$$

Or maybe just the vectors change and not the components:

$$P\mathbf{F}(x, y, z) = -\mathbf{F}(x, y, z) = -F_x(x, y, z)\hat{\mathbb{x}} - F_y(x, y, z)\hat{\mathbb{y}} - F_z(x, y, z)\hat{\mathbb{z}}$$

I feel like the answer is going to depend on whether $$\textbf{F}$$ is a "axial" or "polar" vector like $$\textbf{B}$$ or $$\textbf{E}$$. But then I feel like the statement "$$\textbf{B}$$ is an axial vector" is more of a statement about how we choose to define some parity operator rather than something intrinsic about the magnetic field.

• The first one (I think) Apr 1, 2022 at 0:28

In a naive sense, if we think of parity inversion as similar to "reflecting an image through mirrors" then when we flip the coordinates we should expect the coordinates flip. So I would say

$$P\mathbf{F}(x, y, z) = - \mathbf{F}(-x, -y, -z)$$

Suppose we have a car with positive $$x$$ position and moving with velocity $$\textbf{v}$$ to the $$+x$$ direction. If we flip this scene then now the car will have negative $$x$$ position and be moving in the $$-x$$ direction. This is the second option in the original question.

However, in physics terminology, some vectors are pseudovectors. An example would be angular momentum

$$\textbf{L} = \textbf{r}\times\textbf{p}$$

If we act the parity operator $$P$$ on $$\textbf{r}$$ and $$\textbf{p}$$ then both vectors flip signs, so their cross product must maintain the same sign.

The frustration, expressed in the original question, is that $$\textbf{L}$$ is an element of the same vector space as $$\textbf{r}$$ and $$\textbf{p}$$ so it makes no sense that the action of the parity operator on $$\textbf{L}$$ would differ from its action on $$\textbf{r}$$ and $$\textbf{p}$$

That is, there seems to be an arbitrary choice made that

$$P\textbf{L} = (P\textbf{r}) \times (P\textbf{p})$$

This, of course, disagrees with the definition of the parity operator I gave at the top of this answer, hence the apparent paradox.

When things are approached in this manner there is no "principled" way to determine whether any given vector is a (polar) vector or a pseudovector. To determine how the parity operator acts on any vector you must ADDITIONALLY be aware of the constitutive relations for the original vector (e.g. you must know that $$\textbf{L}=\textbf{r}\times \textbf{p}$$ and that $$\textbf{r}$$ and $$\textbf{p}$$ are polar vectors).

One mathematically rigorous way to resolve this is to introduce $$\mathbf{L}$$ as the exterior product of $$\mathbf{r}$$ and $$\mathbf{p}$$:

$$\mathbf{L} = \mathbf{r} \wedge \mathbf{p}$$

Now $$\mathbf{L}$$ is an object in a different vector space than $$\mathbf{r}$$ and $$\mathbf{p}$$ so it is not absurd for the parity operator to have a different action on it. Here $$\mathbf{L}$$ is a bivector or 2-blade, but there are also ways were these objects can instead be expressed as differential forms.