# What happens to the charge density under parity?

A question came to me when I tried to think about the parity prperties of the Maxwell's equations.

The charge density $$\rho(\vec{r})$$ actually stands for a scalar quantity $$\rho(x,y,z)$$. Since the total charge of a system is $$Q=\int dV \rho(x,y,z)$$ is independent of the choice of a coordinate system and since $$dV\to -dV$$ under parity, and since another minus will be contributed by the limits of integration, the transformation property of $$\rho(\vec{r})$$ under parity must be $$\rho(\vec{r})\xrightarrow{\rm parity}\rho(-\vec{r})=+\rho(\vec{r}).$$ Now, I can think of two different scalar functions representing the charge density. First $$\rho_1(x,y,z)=ax+by+cz$$ and second $$\rho_2(x,y,z)=ax^2+by^2+cz^2$$. Evidently $$\rho_1(\vec{r})$$ is a function that flips sign under parity. However, the sign of $$\rho_2(\vec{r})$$ does not flip under parity (and satisfies the parity property derived above).

The question is this.

Physically, why is $$\rho_1(\vec{r})$$ not a permissible charge density? What stops one to create such a distribution of charge density which is odd (or even mixed) under parity transformation?

• @G.Smith I did not say $\rho$ has radial symmetry. Neither of my examples, $\rho_1$ or $\rho_2$ is radially symmetric. One however has even parity and the other has odd. – mithusengupta123 Aug 27 '19 at 16:43
• Sorry, I meant to say inversion symmetry, not radial symmetry. – G. Smith Aug 27 '19 at 18:41
• As Wikipedia explains, charge density is even (i.e., unchanged) under spatial inversion. This has nothing to do with how $\rho$ depends on $x$, $y$, and $z$. It has to do with what happens to the charge density at a point when the coordinate system is inverted and that same point now has new, negative coordinates. – G. Smith Aug 27 '19 at 18:49

Let $$Q(\Omega)$$ denote the charge inside the volume $$\Omega\subset\mathbb R^3$$. We have $$Q(\Omega)=\int_\Omega\rho(\vec x)\,\mathrm d\vec x\tag1$$ where $$\vec x\in\mathbb R^3$$. If you change variables $$\vec x\to-\vec x$$ you get $$Q(\Omega)=\int_{\Omega'}\rho(-\vec x)\,\mathrm d\vec x\tag2$$ where $$\Omega'$$ is the image of $$\Omega$$ under $$\vec x\to-\vec x$$ (i.e., it is the mirror image of $$\Omega$$ with respect to the origin).

Note that in general $$\Omega\neq \Omega'$$. We indeed have the equality of the integrals $$\int_\Omega\rho(\vec x)\,\mathrm d\vec x\equiv \int_{\Omega'}\rho(-\vec x)\,\mathrm d\vec x\tag3$$ but this does not mean that the integrands are themselves equal. In other words, in general $$\rho(\vec x)\neq\rho(-\vec x)\tag4$$

Indeed, consider the following trivial 1d example: $$\int_1^2 (x^2+x)\,\mathrm dx=\frac{23}{6}\tag5$$ If you perform the same change of variables as before, you get $$\int_{-2}^{-1}(x^2-x)\,\mathrm dx=\frac{23}{6}\tag6$$ The two integrals are identical – we just performed a change of variables. But the integrands are $$x^2+x$$ and $$x^2-x$$, which are different.

The same thing happens in the 3d case. While it is true that the integral of $$\rho(\vec x)$$ over a region $$\Omega$$ is equal to the integral of $$\rho(-\vec x)$$ over the region $$\Omega'$$, the functions $$\rho(\vec x)$$ and $$\rho(-\vec x)$$ need not be equal. In general, they are not, as the example in the OP demonstrates. The charge density, or any other source for that matter, need not be invariant under parity, or any other preconceived transformation.

More generally, the charge $$Q(\Omega)$$ could also be computed by using any other system of coordinates $$\vec x\to\vec y(\vec x)$$, but one must remember that the integration region gets transformed too, $$\Omega\to y^*\Omega$$, see e.g. wikipedia.

But even if the integration regions were the same, i.e., if for a particular $$\Omega$$ and change of variable $$\vec y(\vec x)$$ we had $$\Omega=y^*\Omega$$, equality of integrals does not imply equality of integrands. So you cannot conclude from $$Q=Q$$ that $$\rho=\rho\circ \vec y$$.

• What is the take-home message of your answer? $\rho(\vec{r})$ is neither a true scalar nor a pseudoscalar? – mithusengupta123 Dec 7 '19 at 16:59
• @mithusengupta123 Scalar with respect to what group? The Lorentz group? Then the answer is no: $\rho$ is the 0 component of a vector $j^\mu$. Of the Galileo group? Then the answer is yes: $\rho$ is a scalar. Some other group? – AccidentalFourierTransform Dec 7 '19 at 18:28
• Scalar under rotation. If it is a scalar under rotation, is it a true scalar or pseudoscalar? – mithusengupta123 Dec 7 '19 at 18:32
• @mithusengupta123 a parity transformation $\vec x\to-\vec x$ is not a rotation (it has determinant $-1$ instead of $+1$). In any case, I'm not sure I understand your question. If it is a scalar, it is a scalar. How could it be a scalar and a pseudo-scalar at the same time? But anyway, regardless of that: the charge density is a scalar. Period. – AccidentalFourierTransform Dec 7 '19 at 18:44
• @mithusengupta123 Yeah that's fine. The charge density is arbitrary: you get to choose it. You can pick a charge density that is a scalar, one that is a pseudo-scalar, or one that is neither. For example, the Dirac density $\rho=\psi^\dagger\psi$ is a scalar under parity, while if you introduce a pion field $\rho'=\pi \psi^\dagger\psi$ this becomes a pseudo-scalar. Finally, if you add these up, $\rho+\rho'$ is neither a scalar nor a pseudo-scalar. In general, you can choose $\rho$ to be anything you want, it need not be parity-even nor parity-odd. – AccidentalFourierTransform Dec 7 '19 at 20:58

I believe you are confused as to what a parity transformation actually is. Parity transformations are changes of coordinate system, in which we simply modify how we label points in space.

Let's say you write down a coordinate system in which some point $$p$$ has coordinates $$(x,y,z)=(1,0,1)$$, and let the charge density there be $$\rho$$. Now, you decide that you don't like those coordinates. You want to use different coordinates $$(x',y',z')=(-x,-y,-z)$$, so in this coordinate system, $$p$$ has coordinates $$(-1,0,-1)$$.

When I relabel my points, what happens to $$\rho$$ at the point $$p$$? The answer should be obvious - nothing happens to $$\rho$$. It's exactly the same as it was before, because the charge density does not care about how I choose to label points in space.

This can be formalized by saying that when $$\mathbf r \mapsto \mathbf r'$$, $$\rho \mapsto \rho'$$ such that $$\rho'(\mathbf r') = \rho(\mathbf r)$$. If the coordinate transformation is written $$\mathbf r' = T[\mathbf r]$$, then we can say that $$\rho'(\mathbf r') = \rho\left(T^{-1}[\mathbf r']\right)$$

Quantities which transform in this way under all coordinate transformations are called scalars. If we restrict our attention to parity transformations, then such quantities are called parity-even. This immediately implies that scalars must be parity-even; if $$\rho'(\mathbf r') = \rho\left(T^{-1}[\mathbf r']\right)$$ for all coordinate transformations, then it certainly is true in the specific case of parity inversions. The reverse, however, is not necessarily true.

Rather than talking about a scalar quantity, we could talk about a vector quantity. Consider the velocity of a fluid at some point $$p$$, $$\mathbf v = v_x \hat e_x + v_y \hat e_y + v_z \hat e_z$$, and perform the same transformation. What happens?

$$\hat e_{x'} = -\hat e_{x}$$, $$\hat e_{y'}=-\hat e_y$$, and $$\hat e_{z'} = -\hat e_z$$. Therefore, the components of the vector must also transform in the same way, i.e. $$v_{x'} = -v_x$$ $$v_{y'} = -v_y$$ $$v_{z'} = -v_z$$

Therefore, $$\pmatrix{v_x\\v_y\\v_z}\mapsto \pmatrix{v_{x'}\\v_{y'}\\v_{z'}}= -\pmatrix{v_x\\v_y\\v_z}$$ so we say that the velocity is parity-odd. Note that this is a bit of an abuse of terminology - we should really say that the components of the velocity are parity-odd.

From both of the above examples, you should see that "symmetry" under parity transformations revolves completely around changes of coordinate system; in that sense, it is a gauge symmetry, not a real symmetry of a physical system. For some parity transformation $$\mathbf r \mapsto T(\mathbf r)$$, we say that a quantity $$f(\mathbf r)$$ is parity-even if

$$f'(\mathbf r') = f\left(T^{-1}(\mathbf r')\right)$$ and we say it is parity-odd if $$f'(\mathbf r') = - f\left(T^{-1}(\mathbf r')\right)$$

On the other hand, you are talking about actual, honest-to-goodness symmetries of specific charge distributions, i.e. those such that

$$\rho(\mathbf r') = \rho\left(T^{-1}(\mathbf r')\right)$$ or $$\rho(\mathbf r') = -\rho\left(T^{-1}(\mathbf r')\right)$$

The lack of primes on the $$\rho$$'s on the LHS of these equations is crucial. Parity "symmetry" is not a statement about the universe, but rather a statement about how our description of the universe changes when we pick new labels for points.

Your parity-odd charge distribution $$\rho_1$$ has nonzero dipole moment. It's a totally legal charge distribution.

It's frequently convenient to decompose a charge distribution into multipoles. There are parity-even terms in a multipole expansion (monopole, quadrupole, etc.) and parity-odd terms (dipole, octupole, etc.).

The only requirement from parity symmetry is that inverting space twice has to return you to the original state: for any state $$\left|\psi\right>$$, you have to have $$\hat P\hat P \left|\psi\right> = \left|\psi\right>$$. That means that the only allowed eigenvalues of the parity operator are $$\pm 1$$, but it doesn't mean that any particular observable must be in a parity eigenstate.

• Hi @rob Unfortunately this does not answer my question. I have shown that $\rho(\vec{r})$ must be even under parity. This is also state in the Wikipedia article en.m.wikipedia.org/wiki/Parity_(physics) as G. Smith points out. Then I asked that how is a parity-odd charge distribution like $\rho(x,y,x)=ax+by+cz$ is allowed? – mithusengupta123 Oct 17 '19 at 5:12
• Your proof does not apply to your parity-odd charge distribution because its total charge is $Q=0=-Q$. A multipolar expansion of a charge distribution with $Q\neq 0$ will put all of the total charge in the parity-even monopole term. – rob Oct 17 '19 at 14:39