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We know that two SU(2) fundamentals have multiplication decompositions, such that $$ 2 \otimes 2= 1 \oplus 3.$$ In particular, the 3 has a vector representation of SO(3). While the 1 is the trivial representation of SU(2).

I hope to see the precise SO(3) rotation from the two SU(2) fundamental rotations.

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So let us first write two SU(2) fundamental objects in terms of an SO(3) object. In particular, we can consider the following three:

$$ |1,1\rangle= \begin{pmatrix} 1\\ 0 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}= | \uparrow \uparrow \rangle,$$ $$|1,0\rangle ={1 \over \sqrt{2} } (\begin{pmatrix} 1\\ 0 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} + \begin{pmatrix} 0\\ 1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix})={1 \over \sqrt{2} }(| \uparrow \downarrow \rangle+ | \downarrow \uparrow \rangle) ,$$ $$|1,-1\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix}= | \downarrow \downarrow \rangle. $$

where the $| \uparrow \rangle$ and $ \downarrow \rangle$ are in SU(2) fundamentals. And we shothand $| \uparrow \uparrow \rangle \equiv | \uparrow \rangle |\uparrow \rangle $ and so on.

question: How do we rotate between $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$, via two SU(2) rotations acting on two SU(2) fundamentals? Namely, that is, construct an SO(3) rotation inside the two SU(2) fundamental rotations? The SU(2) has three generators, parametrized by $m_x,m_y,m_z$; how do we write down the generic SO(3) rotations from two SU(2) rotations?

Let us consider an example, an SU(2) rotation $U$ acting on the SU(2) fundamental $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ give rise to $$ U \begin{pmatrix} 1\\ 0 \end{pmatrix}= \begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2}) & (i m_x -m_y) \sin(\frac{\theta}{2}) \\ (i m_x +m_y) \sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2})-{i m_z} \sin(\frac{\theta}{2}) \\ \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}= \begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2})\\ (i m_x +m_y) \sin(\frac{\theta}{2}) \end{pmatrix} \equiv\cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2}) \begin{pmatrix} 1\\ 0 \end{pmatrix} + (i m_x +m_y) \sin(\frac{\theta}{2}) \begin{pmatrix} 0\\ 1 \end{pmatrix} $$

In other words, the SU(2) rotation $U$ (with the $|\vec{m}|^2=1$) rotates SU(2) fundamentals. Two SU(2) rotations act as $$ UU|1,1\rangle = U \begin{pmatrix} 1\\ 0 \end{pmatrix}U \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2})\\ (i m_x +m_y) \sin(\frac{\theta}{2}) \end{pmatrix}\begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2})\\ (i m_x +m_y) \sin(\frac{\theta}{2}) \end{pmatrix} $$

Hint: Naively, we like to construct $$ L_{\pm} =L_{x} \pm i L_y, $$ such that $L_{\pm}$ is an operator out of two SU(2) rotations acting on two SU(2) fundamentals, such that it raises/lowers between $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$.

question 2: Is it plausible that two SU(2) are impossible to perform such SO(3) rotations, but we need two U(2) rotations?

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The following solution originates from the theory of geometric quantization. I'll not explain the full theory behind it, but I'll give here the solution, then briefly discuss how to check that this is the required solution.

A general $SU(2)$ group element in the fundamental representation can be written as: $$ g = \begin{bmatrix} \alpha & \beta\\ -\bar{\beta}& \bar{\alpha} \end{bmatrix} $$ with $$|\alpha|^2+\beta|^2=1$$ The three dimensional Hilbert space of the three dimensional representation can be parametrized by: $$ \psi (z) = a + b z + cz^2 \quad (1) $$

where $x$ is an indeterminate

The action of $SU(2)$ on this vector space is given by:

$$ (g\cdot \psi)(z) = (-\bar{\beta} z + \bar{\alpha})^2 \psi( \frac{\alpha z + \beta}{-\bar{\beta} z + \bar{\alpha}}) \quad (2)$$

  1. To see that this is a representation, one can check that the composition of the action of two group elements coincides with the action of their product.
  2. To see that this is a faithful $SO(3)$ representation but not a faithful $SU(2)$, we can easily see that for the nontrivial element of the center: $$ g_c = \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix} $$ We have for every $\psi$ $$ (g_c\cdot \psi)(z) = \psi(z)$$
  3. Although, the "spherical" components $a, b, c$ are complex. To see that the representation is real, one can see that the "Cartesian" components $ (a+c), b, i^{-1}(a-c)$ transform by means of only real combinations of $\alpha$ and $\beta$.
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  • $\begingroup$ @ David Bar Moshe, thanks for the answer, +1. Your answer is in a more advanced level. :) $\endgroup$ – annie heart Apr 22 '18 at 16:01
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    $\begingroup$ Then I guess maybe you can also solve this or at least share your opinions: math.stackexchange.com/q/2745276 - thanks. $\endgroup$ – annie heart Apr 22 '18 at 16:03
  • $\begingroup$ @ David, it is easy to see from $$ UU|1,1\rangle = U \begin{pmatrix} 1\\ 0 \end{pmatrix}U \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2})\\ (i m_x +m_y) \sin(\frac{\theta}{2}) \end{pmatrix}\begin{pmatrix} \cos(\frac{\theta}{2})+{i m_z} \sin(\frac{\theta}{2})\\ (i m_x +m_y) \sin(\frac{\theta}{2}) \end{pmatrix} $$ when $m_y=\pm 1, m_x=m_z=1$ and $\theta=\pi$, we have $$ UU|1,1\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix}=|1,-1\rangle$$ $\endgroup$ – annie heart Apr 22 '18 at 16:31
  • $\begingroup$ Do you have any precise answer how to find $U$, such that $$ UU|1,1\rangle = \frac{1}{\sqrt 2} \begin{pmatrix} 1\\ 0 \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} + \frac{1}{\sqrt 2} \begin{pmatrix} 1\\ 0 \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix}=|1,0\rangle ?$$ $\endgroup$ – annie heart Apr 22 '18 at 16:33
  • $\begingroup$ @annie heart there is no such a $U$. Please observe that if you act by some $U \otimes U$ on the highest weight vector: $ \begin{pmatrix} 1\\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1\\ 0 \end{pmatrix}$, then the result will always be a separable vector of the form $ \begin{pmatrix} a\\ b\end{pmatrix} \otimes \begin{pmatrix} a\\ b \end{pmatrix}$. But the vector that you want to reach is entangled, and there is no way to change the entanglement state by a local transformation. $\endgroup$ – David Bar Moshe Apr 23 '18 at 10:23
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Perhaps the following is helpful:

  1. OP's eq. (1) is to be understood as a relation between complex representations of $SU(2)$, i.e complex vector spaces. Recalling that the fundamental $SU(2)$ representation ${\bf 2}\cong \overline{\bf 2}$ is isomorphic to the complex conjugate representation, let us instead consider the isomorphism $$ {\bf 2}\otimes \overline{\bf 2}~\cong~{\bf 1}\oplus {\bf 3}. \tag{A}$$

  2. The left-hand side of eq. (A) can be realized as the real vector space $u(2)$ of $2\times 2$ Hermitian matrices. The group $SU(2)$ acts on $u(2)$ via conjugation. Given a spinor $| \psi\rangle\in {\bf 2}$, then $$ {\bf 1}\oplus su(2)~\cong~ u(2)~\ni~| \psi\rangle \langle\psi | ~=~\frac{1}{2}\sum_{\mu=0}^3x^{\mu} \sigma_{\mu}, \qquad (x^0,x^1,x^2,x^3) ~\in~\mathbb{R}^4. \tag{B}$$ The triplet ${\bf 3}$ corresponds to the traceless part, that is: $su(2)$. Hence the spinor $| \psi\rangle$ represents the 3-vector $\vec{r}=(x^1,x^2,x^3)$. See also this related Phys.SE post.

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  • $\begingroup$ Thanks! +1, the same question as to David, do you have a precise rotation from how to find U, such that UU|1,1⟩=(1/√2) (10)(01)+(1/√2)√(10)(01)=|1,0⟩ by some U? $\endgroup$ – annie heart Apr 22 '18 at 17:53

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