Edit: Physical Set-up: Below, we've basically defined a map from $SU(2)$ to $SU(3)$. However, there's another way to define a map from (a neighborhood of $1 \in$) $SU(2)$ to $SO(3)\subset SU(3)$, which simply takes $exp(i t L_{\frac{1}{2}, i})\to \exp(i t L_{1, i})$, where $L_{\frac{1}{2}, i}$ are the spin $\frac{1}{2}$ generators and $L_{1, i}$ are the spin $1$ generators. I want to compare these two maps.
Consider the following physical situation. I have two spins both initially in the $|+\frac{1}{2}>$ state. I can consider them as either two separate spins or a single spin $|+1>$.
Suppose I think of them as a single spin-$1$. Then I apply a magnetic field in, say, the $+x$ direction. My single spin-1 will precess about the $x$ axis. The time-evolution matrix will be given by $U= \exp L_{x} t $, where $$ L_{x} = \left(\begin{array}{ccc}0 & 1 & 0 \\-1 & 0 & 0 \\0 & 0 & 0\end{array}\right) $$ (In a particular basis!). This defines an element of $SO(3)$ as my time evolution matrix.
On the other hand, I can consider them as two separate spin-$\frac{1}{2}$'s that evolve under the magnetic field. This gives me an element of $SU(2)$ as my time-evolution matrix. From the the mapping we define below, the element of $SU(2)$ immediately defines a $U_{1}\in SO(3)$ that acts as my time-evolution matrix on the combined system.
Which one is correct?
I've always added angular momenta using Clebsh-Gordan coefficients, but today I was trying the following naive approach of literally tensoring spinors together and was surprised to find that it fails.
Below, I explicitly work out the effect of an $SU(2)$ transformation $v\mapsto U_{\frac{1}{2}}v$, $w\mapsto U_{\frac{1}{2}}w$ on the the Kronecker product $v\otimes w$. Of course, $v\otimes w$ decomposes into a three-index spin $1$ object $\phi$ and a one-index spin-0 invariant $\rho$.
We write the induced $SU(2)$ transformation on the spin $1$ field $\phi$ as a $3\times 3$ matrix $\phi\mapsto U_{1}\phi$. However, although $U_{1}$ is special-unitary, it is not real. Because integer spin representations of $SU(2)$ project to representations of $SO(3)$, I was expecting to get a matrix in $SO(3)$. (In fact, I was expecting to get the matrix shortly after eq. (2) in this wikipedia article.) Can you help me understand what I've done wrong (or why my expectation was wrong)?
Most of what follows is fixing notation and explicitly working things out so that anyone can follow along. If you're familiar with this sort of thing, skip to the end.
Suppose that we have two two-component spinors:
$$v = \left(\begin{array}{c}a \\b\end{array}\right) \hspace{1 cm} w = \left(\begin{array}{c}c \\d\end{array}\right)$$
Under a given $SU(2)$ transformation, $$ U_{\frac{1}{2}}\equiv \left(\begin{array}{cc}\alpha & \beta \\-\beta^{*} & \alpha^{*}\end{array}\right)\left(\begin{array}{c}a \\b\end{array}\right) $$ with $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^{2} + |\beta|^{2} = 1$, we assume $v, w$ transform under $SU(2)$ as $v\mapsto Uv$, $w\mapsto Uw$. Specifically, $$ \left(\begin{array}{c}a \\b\end{array}\right) \mapsto \left(\begin{array}{cc}\alpha & \beta \\-\beta^{*} & \alpha^{*}\end{array}\right)\left(\begin{array}{c}a \\b\end{array}\right) \hspace{1 cm} \left(\begin{array}{c}c \\d\end{array}\right) \mapsto \left(\begin{array}{cc}\alpha & \beta \\-\beta^{*} & \alpha^{*}\end{array}\right)\left(\begin{array}{c}c \\d\end{array}\right) $$
When we add angular momenta, we combine these spinors using the tensor product. The tensor product of $v$ and $w$ is a four component object given by the Kronecker product: $$ (vw)^{i, j} = v^{i}w^{j} $$ where I regard $(i, j)$ as a multi-index that takes values $0\equiv (1, 1), 1 \equiv (1, 2), 2 \equiv (2, 1) 3\equiv (2, 2)$.
The $SU(2)$ transformation above separately preserves the symmetric $v^{(i}w^{j)}$ and anti-symmetric components $v^{[i}w^{j]}$. So we can introduce new fields: a spin-1 three-index $\phi^{i}$: $$ \phi^{0} = v^{1} w^{1} = ac\hspace{.5cm} \phi^{1} = \frac{1}{\sqrt{2}}(v^{1}w^{2} + v^{2}w^{1}) = \frac{1}{\sqrt{2}}(ad+bc)\hspace{.5cm} \phi^{2} = v^{2}w^{2} = bd $$ and a scalar $\rho = \frac{1}{\sqrt{2}}(v^{2}w^{1} - v^{1}w^{2})$.
The above $SU(2)$ transformation induces the following transformation on the components of $\phi$ (this is messy--see the matrix form below): $$ \phi^{0} = ac \mapsto \alpha^{2} ac + \alpha\beta (bc+ ad) + \beta^{2}bd = \alpha^{2} \phi^{0} + \sqrt{2}\alpha\beta \phi^{1} + \beta^{2}\phi^{1} $$ $$ \phi^{1} = \frac{1}{\sqrt{2}}(ad+bc) \mapsto \frac{1}{\sqrt{2}}(|\alpha|^{2}-|\beta|^{2} )(bc+ad) -\sqrt{2}\alpha\beta^{*} ac + \sqrt{2}\alpha^{*}\beta bd = (|\alpha|^{2}-|\beta|^{2})\phi^{1} -\sqrt{2}\alpha\beta^{*}\phi^{0} + \sqrt{2}\alpha^{*}\beta \phi^{2} $$ $$ \phi^{2} = bd \mapsto \beta^{*2} ac - \alpha^{*}\beta^{*}(bc+ad) + \alpha^{*2}bd = \beta^{*2}\phi^{0} - \sqrt{2}\alpha^{*}\beta^{*} \phi^{1} +\alpha^{*2}\phi^{2} $$ while $\rho$ is left invariant: $$ \rho = ad - bc \mapsto (|\alpha|^{2} + |\beta|^{2})(ad - bc) = \rho $$ We can summarize the transformation of $\phi$ as : $$ \phi = \left(\begin{array}{c}\phi^{0} \\\phi^{1} \\\phi^{2}\end{array}\right) \mapsto \left(\begin{array}{ccc}\alpha^{2} & \sqrt{2}\alpha \beta & \beta^{2} \\-\sqrt{2}\alpha\beta^{*} & |\alpha|^{2}-|\beta^{2} & \sqrt{2}\alpha^{*}\beta \\\beta^{*2} & -\sqrt{2}\alpha^{*}\beta^{*} & \alpha^{*2}\end{array}\right)\left(\begin{array}{c}\phi^{0} \\\phi^{1} \\\phi^{2}\end{array}\right) \equiv U_{1}\phi $$
Given a $U_{\frac{1}{2}}$, it immediately defines $U_{1}$ as above. Moreover, $U_{1}$ is a special unitary matrix. However, it's not real! But shouldn't $\phi$ transform under $SO(3)$ and be blind to the fact that $SU(2)$ is really calling the shots? Wouldn't this mean than the induced transformation $U_{1}$ should be in $SO(3)$?
