Why are circular unit vectors often defined as $\hat{\mathbf e}_\pm = \mp (\hat{\mathbf e}_x \pm i \hat{\mathbf e}_y)/\sqrt{2}$

Question

When dealing with circular polarizations, spherical harmonics, and generally with any vector-valued rotationally-invariant quantity, it is often a requirement to define complex-valued unit vectors of the form $(1,i,0)$ and $(1,-i,0)$, which have the nice property that under a rotation they go to themselves times a complex phase.

However, many resources, especially the ones with a serious and systematic stance, use a different sign convention, and they define $$ \newcommand{\ue}[1]{\hat{\mathbf{e}}_{#1}} \ue\pm= \mp \frac{1}{\sqrt{2}}\big(\ue x \pm i \ue y\big),$$ with a global sign up front. This is relatively counter-intuitive to me, but it sees a fair amount of use [see 1, 2, 3, 4 for examples], so I imagine there must be some reason for this convention.

In what way does this sign convention simplify things? This isn't the kind of thing you'd do at random, by just introducing gratuitous complexity on a cornerstone formula that should be as simple as possible, so I imagine it's there to reduce complexity elsewhere. What exactly is that 'elsewhere'?

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^{For the benefit of sanity on this thread, \ue{x} has been defined to produce $\ue{x}$.}

Answer 1

One answer, connecting he antisymmetry of the wedge product and the (Lie) commutator, is through the Wigner-Eckart theorem. Let $$ \hat T_{10}=\hat L_z\, \qquad \hat T_{1\pm 1}=\mp \frac{1}{\sqrt{2}} \left(\hat L_x\pm i \hat L_y\right)=\mp \frac{1}{\sqrt{2}}\hat L_{\pm} $$ Then the matrix elements $$ \langle jm'\vert \hat T_{1\mu}\vert jm\rangle=\langle jm;1\mu\vert jm'\rangle \sqrt{j(j+1)} $$ where $\langle jm;1\mu\vert jm'\rangle$ is a Clebsch-Gordan coefficient. Basically, the $-$ sign is required to define $\hat T_{11}$ as the correct $+1$ component of the tensor operator.

The connection with the wedge product is so that $$ \newcommand{\ue}[1]{\hat{\mathbf{e}}_{#1}} [\hat T_{1k},\hat T_{1m}]\leftrightarrow \ue{k}\wedge \ue{m} $$ and indeed in some textbooks the commutator $[\hat T_{1k},\hat T_{1m}]$ is written as $\hat T_{1k}\wedge \hat T_{1m}$

Answer 2

$\newcommand{\ue}[1]{\hat{\mathbf{e}}_{#1}}$ If you define $\ue\pm$ as

$$\ue\pm= \frac{1}{\sqrt{2}}\big(\ue x \pm \mathrm{i} \ue y\big)$$

you get

$$\ue+\wedge \ue- = -\mathrm{i}\ue z.$$

If, instead, you define

$$\ue+ = -\frac{1}{\sqrt{2}}\big(\ue x + \mathrm{i} \ue y\big)$$

and

$$\ue- = \frac{1}{\sqrt{2}}\big(\ue x - \mathrm{i} \ue y\big),$$

the cross product becomes

$$\ue+\wedge \ue- = \mathrm{i}\ue z.$$

Therefore, in the former case, $(\ue+,\ue-,\mathrm{i}{\ue z})$ has the same orientation of $(\ue x,\ue y, \ue z)$, whereas in the latter the orientation is preserved by $(\ue -,\ue +, \mathrm{i}\ue z)$.

Is there any reason to prefer one orientation with respect to the other? I would say no, it's probably just a matter of tradition.