# Parity transformation and mirror reflection

I have some trouble understanding what exactly is parity transformation.

The definition of parity transformation is a flip in the sign of all three spatial coordinates, ie $$(x,y,z) \rightarrow (-x,-y,-z).$$

Consider a stationary particle at a position $$(a,b,c)$$ in space described by a coordinate system $$(x,y,z)$$. Does parity transformation mean that the particle is still at the exact point in space but its position is now described by $$(-a,-b,-c)$$?

But often parity is talked about as a mirror reflection and it seems to me that a mirror reflection means physically moving the particle from point $$(a,b,c)$$ to $$(-a,-b,-c)$$ in a coordinate system $$(x,y,z)$$.

Which of the above 2 cases is parity transformation really referring to? If it refers to both cases, why are the two cases the same? In one case a particle is fixed in space while in another case a particle is moved to another point in space.

• The second interpretation is the correct one, you must inverse the position of the particle. Feb 15, 2019 at 16:41

First of all, there are two conventions - 'active' and 'passive' points of view. Within the former, you would say that under the parity transformation the particle has changed its position in space from $$(a,b,c)$$ to $$(-a,-b,-c)$$. Within the latter, you'd say that the particle stays at the same point of space, but the coordinate system has been changed in such a way that the new coordinates of the same particle (staying at the very same location) are $$(-a,-b,-c)$$. Clearly, both conventions are equivalent. Typically, the 'active' one is used in Physics.

You have correctly defined the parity transformation as the change of signs of all the coordinates: it changes all the vectors $$\vec{r}\to-\vec{r}$$, leaving, therefore, only the zero vector invariant.

The reflection, in turn, always happens relative to a certain plane. Say, a reflection relative to the $$x\,y$$ plane works as $$(x,y,z)\to(x,y,-z)$$. Importantly,

1. This operation leaves all the vectors belonging to the $$x\,y$$ plane invariant.

2. It can be turned into parity by applying an additional rotation in the $$x\,y$$ plane.

The second circumstance in the reason why these two operations are often confused. In many cases, we only care about the transformation 'up to a rotation'. Note that if you apply either parity transformation or reflection to a solid body, there's no way of rotating it back into the original positions. In this sense, these two operations are equivalent.

Also, keep in mind that in a different number of dimensions things work slightly differently.

• Is it not true that handedness of a rotation transforms differently in the active vs. the passive view? Nov 28, 2021 at 22:46
• See a good discussion here. Nov 30, 2021 at 4:48

These are more general properties of the parity which I think, they give a better explanation of what this symmetry is really about.

The general definition of parity is an operator $$\mathcal{P}$$ with the properties $$\mathcal{P} = \mathcal{P}^*$$ and $$\mathcal{P}^n=\mathbb{1}$$, $$*$$ denotes complex conjugation. Most of the time, people stick with $$\mathcal{P}^2=\mathbb{1}$$. Perhaps it is also easier to look parity in a discrete Hilbert space in a one-dimensonal system. For this it is enough to look for mirror symmetry. That is, consider a tight-binding chain in which each site is described by the basis

$$$$\lbrace |\psi_1\rangle,\dots,|\psi_N\rangle\rbrace.$$$$

The parity $$\mathcal{P}$$ shoudl act on this basis as $$|\psi_1\rangle\rightarrow|\psi_N\rangle$$, $$|\psi_2\rangle\rightarrow|\psi_{N-1}\rangle$$, and so forth. Fixing the basis, you can write the parity operator as

$$$$\mathcal{P} = \begin{pmatrix} & & & 1 \\ & &1& \\ &\unicode{x22f0}& &\\ 1 & & & \end{pmatrix},$$$$

all the other entries are zero. Furthermore, this representation is also suitable to represent parity in discrete models which are not one-dimensional. For example benzene, ethylene etc.

Finally, from my point of view, parity is explicitly put into action when you integrate relativity in quantum mechanics, i.e. in Quantum Field Theory. There, the position is $$x^\mu = (t,\vec{x})$$ and fixing a metric, the parity operator is given by the matrix

$$$$\mathcal{P}^\mu_{\;\nu} = \begin{pmatrix} 1 & & & \\ &-1 & &\\ & & -1 & \\ & & & -1 \end{pmatrix}.$$$$

Then $$\mathcal{P}: (t,\vec{x}) \rightarrow (t,-\vec{x})$$, which amounts to a matrix multiplication $$\mathcal{P}x$$.

As a bonus, returning to non-relativistic quantum mechanics, people have studied a lot parity-symmetry Hamiltonians for the past 10 years or so.