In particle physics we deal with parity transformations and in particular when we regard weak interactions. While learning this subject I found something that needs more clarification for me. Namely, suppose that a particle has an electric dipole moment $\vec{d}$ and some magnetic moment $\vec{\mu}$. Every source I see states that a permanent dipole moment 'violates' parity symmetry, but I do not see why.

I tried the show that the energy is not invariant under parity symmetry but I am not sure whether this is a rigorous theoretical way of proving this.

$$E = -\vec{\mu}\cdot \vec{B} - \vec{d}\cdot \vec{E}$$

is the energy, suppose I apply the parity operator on this:

$$\mathcal{P}(E) = -\vec{\mu}\cdot \vec{B} + \vec{d}\cdot \vec{E}.$$

So we see that $\mathcal{P}(E)\neq E$ so I would say that parity is 'violated' in order to let the energy $E$ be invariant under parity transformation ($E$ is a scalar so it should be mapped to itself). I used that $\vec{d} \propto \vec{S}$ and that $\vec{E}$ is a vector and $\vec{\mu},\vec{B},\vec{S}$ pseudovectors. $\vec{S}$ is the spin vector. My conclusion is therefore: $\vec{d}=\vec{0}$ or that parity is violated.

Is this 'proof' of showing this theoretically rigorous or am I missing someting?


The usual answer is something along these lines: The only thing specifying a "direction" of an elementary particle is the direction of its spin---so the dipole moment is either parallel to, or antiparallel to, the spin. Suppose that the electric dipole moment of the particle is ${\bf d} = \lambda {\bf S}$ for some scalar $\lambda$. Now look at the system in a mirror. The mirror image of the particle appears to spin in the opposite direction but the electric charge, and hence the electric dipole moment, is unchanged. Thus, in the mirror world, ${\bf d} = -\lambda {\bf S}$.

  • $\begingroup$ But $\vec{s}$ is invariant under a parity transformation right, so where does the minus sign comes from in the second equation? $\endgroup$ – Dani Nov 7 '18 at 15:51
  • $\begingroup$ @Dani There are different definitions of the parity transformation. Theorists like to send ${\bf r} \to -{\bf r}$. Then ${\bf d}\to -{\bf d} $ while ${\bf S}\to {\bf S}$, so the sign of $\lambda$ still changes. I prefer the mirror image definition as it is easier to explain to laypeople but has the same physics content. (The determinant of the map is still $-1$) $\endgroup$ – mike stone Nov 7 '18 at 21:10

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