Why does factoring out $\pm$ from an expression introduce a minus sign?

Question

This is strongly linked to this previous question of mine.

In this question I am trying to show that $$\frac{\hbar}{\sqrt2}\left(i g_z \pm g_y\right)=\pm\frac{\hbar}{\sqrt{2}}\left(g_y \pm ig_z\right)\tag{*}$$

From $(*)$; it seems that factoring $\pm$ out of the bracket did nothing other than switch the sign of the terms in the bracket. If this is true; then it must be the case that $$\pm(\pm 5) = 5\tag{a}$$ & $$\pm\left(\pm \left(\pm 8\right)\right)=-8\,\tag{b}$$ and for my specific case $$\pm\left(\pm \,i g_z \right) = ig_z\tag{c}$$

---

It's not obvious to me why $(\rm a)$, $(\rm b)$, $(\rm c)$ are true; if I had to guess I would say it's because the $$+\times +=+$$ & $$- \times - = +$$

Is there a better way of thinking about this or am I on the right tracks?

Answer 1

The $\pm$ sign stands for $+$ or $-$.

(In my opinion, it should be avoided as much as possible, since it is likely to cause confusions like this one.)

So you can answer your question by considering each time the two cases :

$$\begin{cases} \text{when $\pm$ stands for $+$} \\ \text{when $\pm$ stands for $-$}.\end{cases}$$

So you indeed have

$$\frac {\hbar }{\sqrt 2} (ig_z+g_y)=+\frac {\hbar }{\sqrt 2} (g_y+ig_z)$$

and also

$$-\frac {\hbar }{\sqrt 2} (ig_z+g_y)=-\frac {\hbar }{\sqrt 2} (g_y-ig_z)$$

so your equality is true for $\pm$.

It is the same for all the other examples you gave, and you can see that

$$\pm(\pm 5)=5$$

and

$$\pm(\pm(\pm 8))=\pm 8.$$

Answer 2

I think that the equality in $(*)$ is not very standard, because if you have free choice at every instance of $\pm$ (which is usually the case), then you could get a $-ig_z$ on the right, which you can't on the left. However, if you want to use $\pm$ with the convention that you have to choose the same sign every time there is a $\pm$ in the same expression (which seems to be the case in your question), then your explanation for $(a)$ and $(c)$ is correct, but in $(b)$ you should have $\pm8$. But I personally find this convention confusing.

Answer 3

Whenever several instances of $\pm$ (or its cousin $\mp$) occur in a single equation, they should be considered as working "in sync"; that is, the equation is best viewed as an abbreviation for two different equations: one where we take the upper sign throughout and one where we take the lower sign throughout. Thus $$\frac{\hbar}{\sqrt2}\left(i g_z \pm g_y\right)=\pm\frac{\hbar}{\sqrt{2}}\left(g_y \pm ig_z\right)\tag{*}$$ is a shorthand for $$\frac{\hbar}{\sqrt2}\left(i g_z + g_y\right)=+\frac{\hbar}{\sqrt{2}}\left(g_y + ig_z\right),\qquad \frac{\hbar}{\sqrt2}\left(i g_z - g_y\right)=-\frac{\hbar}{\sqrt{2}}\left(g_y - ig_z\right)$$ (and whether these variants are to be considered as combined with an "and" or "or" may depend on context).

Likewise, $$ \pm(\pm 5) = 5\tag{a}$$ is a shorthand for $$ +(+ 5) = 5,\qquad -(- 5) = 5,$$ which is absolutely correct. But the claim $$ \pm\left(\pm \left(\pm 8\right)\right)=-8\,\tag{b}$$ is a shorthand for $$ +\left(+ \left(+ 8\right)\right)=8\,\qquad -\left(- \left(- 8\right)\right)=-8$$ which is less correct ...

---

An alternative way to work around the $\pm$ issue is to replace each occurrence of $\pm$ with a multiplication with a newly introduced variable $\sigma$, where $\sigma\in\{-1,1\}$. So your main equality could be written as $$\frac{\hbar}{\sqrt2}\left(i g_z +\sigma g_y\right)=\sigma\frac{\hbar}{\sqrt{2}}\left(g_y +\sigma ig_z\right).$$ The correctness then follows from $\sigma^2=1$. One advantage of this notation over $\pm$ notation is that you make it absolutely clear that all sign choices are what I called "in sync" above; and that you are free to introduce several sign variables $\sigma_i$ if you want to allow the sign choices to be more independent. (Indeed, some consider a notation such as $a\pm b\pm c$ to be ambiguous - is it shorthand for two cases or for four cases?)