Permanent electric dipole moment parity violation

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, $${\bf d} = -\lambda {\bf S}$$.
• But $\vec{s}$ is invariant under a parity transformation right, so where does the minus sign comes from in the second equation? – Dani Nov 7 '18 at 15:51
• @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 $-1$) – mike stone Nov 7 '18 at 21:10