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 $SO(3)$?

  • $\begingroup$ You must expect $\:U_1\in SU(3)\:$ and this is the case. Why do you expect the special unitary matrix $\:U_1\:$ to be real ? It doesn't represent a rotation in space. The complex 3-vector $\:\phi\:$ belongs in a complex 3-dimensional Hilbert space and not in the real 3-dimensional space $\:\mathbb{R}^3$. In the complex 3-dimensional Hilbert space the "rotations" are special unitary transformations $\:U\in SU(3)$. $\endgroup$ – Frobenius Oct 25 '18 at 18:42
  • $\begingroup$ Take a look in the Example A in my answer here : What is the symmetry of the pion triplet (π−,π0,π+)?. $\endgroup$ – Frobenius Oct 25 '18 at 18:48
  • $\begingroup$ Ah it's great to see that we got the same result I thought I was going crazy (and nice answer there btw!). But there's still a conceptual difficulty I'm facing. I've added a physical set-up above that cuts to the heart of it. The basic question is: should my Hamiltonian contain the generators for $SU(2)$ in the adjoint or $SO(3)$ in the fundamental? Aren't the the same? How can they give rise to Time-evolution operators that live in different groups? (SU(3) vs SO(3)). $\endgroup$ – MDM Oct 25 '18 at 19:32
  • $\begingroup$ Are you cool with everything in here? $\endgroup$ – Cosmas Zachos Oct 25 '18 at 20:56
  • $\begingroup$ Yes! And I think we've worked out the answer. It is broadly related to your answers Cosmas (in our notation, your answer would be considering induced rotations of $v\otimes w^{\dagger}$ + h.c.). But I think the think the simple answer is that matrices of the form $U_{1}$ above are unitarily equivalent to $SO(3)$. That is, there exists a unitary matrix $V$ such that for every $U_{1}$, there is an $O_{1}\in SO(3)$ with $U_{1} = VO_{1}V^{\dagger}$. I won't have the time to work this out for the next couple days, but it should be the matrix that maps between the generators. $\endgroup$ – MDM Oct 25 '18 at 21:09

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