What happened to the basis of 1⊗1/2 when we write 1/2⊗1?

Question

I am doing an exercise of calculating the Clebsch-Gordan (CG) coefficient of $\frac{1}{2}\otimes 1 =\frac{3}{2} \oplus \frac{1}{2}$. Without realizing that the general convention is to have $j_1 > j_2$, I chose $|\frac{1}{2},1\rangle$ instead of $|1,\frac{1}{2}\rangle$.

Evaluating $\bf{|J,M\rangle}$ for $\bf{|\frac{3}{2},M\rangle}$ wasn't difficult, but the problem arise when I was trying to evaluate $\bf{|\frac{1}{2},M\rangle}$ by using the orthogonality properties and taking the inner product with $\bf{|\frac{3}{2},M\rangle}$.

While evaluating ${\bf{|\frac{1}{2},\frac{1}{2}\rangle}} = \alpha|\frac{1}{2},0\rangle +\beta|-\frac{1}{2},1\rangle$, after taking inner product with ${\bf{|\frac{3}{2},\frac{1}{2}\rangle}} = \sqrt\frac{2}{3}|\frac{1}{2},0\rangle +\sqrt\frac{1}{3}|-\frac{1}{2},1\rangle$, and solving for the unknowns yield $\alpha = \pm\sqrt\frac{1}{3}$ and $\beta = \mp\sqrt\frac{2}{3}$.

Following the phase convention of $\langle j_1,j_2;j_1,(j-j_1)|j,j\rangle $, I found that to obey such convention, I need to have

${\bf{|\frac{1}{2},\frac{1}{2}\rangle}} = \sqrt\frac{1}{3}|\frac{1}{2},0\rangle -\sqrt\frac{2}{3}|-\frac{1}{2},1\rangle$ (my answer)

but the standard CG table reads:

${\bf{|\frac{1}{2},\frac{1}{2}\rangle}} = -\sqrt\frac{1}{3}|0,\frac{1}{2}\rangle +\sqrt\frac{2}{3}|-1,\frac{1}{2}\rangle$

which is just $(-1)$ off the given answer. So it makes me wonder if the observation of:

$|0, \frac{1}{2}\rangle = -|\frac{1}{2},0\rangle$

or more generally,

$|j_1m_1\rangle \otimes |j_2m_2\rangle = - |j_2m_2\rangle \otimes |j_1m_1\rangle $

holds every time? I saw a statement of $ V\otimes W \cong W\otimes V$ but not sure how the basis in these two spaces change.

Answer 1

First, correct your mistake, to $\frac{1}{2}\otimes 1 =\frac{3}{2} \oplus \frac{1}{2}$. In my comment, I suggested it is impossible to make it, if only you use the dimensionality of the multiplets, instead of the spin, to label them, in boldface, so ${\mathbf 2} \otimes {\mathbf 3} ={\mathbf 4} \oplus {\mathbf 2}$. The tensor factors themselves do not have an intrinsic order. It's what you choose to do with them!

Specifically, let's inspect this, and in particular the orthogonal matrix for the ${\mathbf 2}$ you are inspecting. Note the off-diagonal terms must have minus signs for the matrix to be orthogonal, and the C-S condition $\langle j_1,j_1~~j_2,(J-j_1)|J,J\rangle >0 $ in the standard convention, not your language, heavily discriminates for the first factor. So changing the order of rows will have to flip relative signs in the two orderings, according to the identity,
$$ \langle j_1~m_1~~ j_2~m_2|JM\rangle = (-)^{J-j_1-j_2} \langle j_2~m_2~~ j_1~m_1|JM\rangle, $$ with the exponent being -1 for your Clebsches of interest.

Specifically, $$ \langle 1~0~~ 1/2~1/2 |\tfrac{1}{2} ~ \tfrac{1}{2} \rangle = -\sqrt{1/3}, \\ \langle 1~1 ~~1/2 ~ -\!1/2 |\tfrac{1}{2} ~ \tfrac{1}{2} \rangle = \sqrt{2/3} , ~~~~~* \\ \langle 1/2~1/2 ~~ 1~0 |\tfrac{1}{2} ~ \tfrac{1}{2} \rangle = \sqrt{1/3} ,~~~~~* \\ \langle 1/2 ~ -\!1/2 ~~ 1~1 |\tfrac{1}{2} ~ \tfrac{1}{2} \rangle=-\sqrt{2/3} , $$ where the * s indicate the maximal states forced to be positive by the C-S condition. The point is the symmetry condition relates them all properly. Inspection of the entire Clebsch matrix makes it inevitable.