spinor representation

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< Lie groups | SO(3)(Redirected from Lie group/SO(3)/spinor representation)

1 Motivation
2 Construction
- 2.1 From SU(2) to SO(3)
  - 2.1.1 Axis-angle form
- 2.2 From SO(3) to SU(2)
3 See also
4 References

Motivation

SO(3) is defined to be the set of linear transformations that preserve the inner product of two vectors in $\mathbb{R}^3\,$ . To that end, consider the vector with components $r^i\,$ and form the $2\times2\,$ matrix

$r = r^i \sigma_i\,$ (summation implicit)

where $\sigma_i\,$ are the three Pauli matrices. Explicitly,

$r= \left( \begin{matrix} r^3 & r^1-i r^2 \\ r^1+i r^2 & -r^3 \end{matrix}\right)$ .

It follows that $\operatorname{tr}\, r = 0\,$ and $\det\, r = -\Vert \mathbf{r}\Vert^2\,$ . Also, $r\,$ is Hermitian. All three these properties are preserved under unitary transformations. Furthermore, since

$\sigma_i \sigma_j = \delta_{ij} \cdot I + i \varepsilon_{ijk} \sigma_k \,$ ,

it is easy to see that

$r^i = \frac{1}{2}\operatorname{tr}\, (r \sigma_i)\,$ ,

which also implies

(1)

$\mathbf{r}\cdot\mathbf{v} = r^i v^i = \frac{1}{2}\operatorname{tr}\, (r^\dagger v)\,$ .

What is the effect of a special orthogonal transformation on $r^i\,$ , i.e., $r^i \to r^{ i\prime} = O^i_j r^j\,$ ? It is a linear function on the elements of $r\,$ that preserves the inner product (1):

$r \to U_L r U_R\,$ ,

so that

$\operatorname{tr} ( r^\dagger v ) = \operatorname{tr} ( U_R^\dagger r U_L^\dagger U_L v U_R) = \operatorname{tr} (r^\dagger v)\qquad \forall \,u,v\,$ ,

so that $U_R U_R^\dagger = U_L^\dagger U_L = 1\,$ , i.e., $U_L, U_R \in U(2)\,$ . However, it must also preserve the Hermiticity of $r\,$ , so that $U_R = U_L^\dagger\,$ ^[1]:

$r \to U r U^\dagger\,$ .

For definiteness, we may require that any $U\,$ be smoothly connected to the identity, so we choose $\det U = 1\,$ and $U \in SU(2)\,$ .

Finally, the bijection should preserve group multiplication, so that

$\mathbf{r} \to \mathbf{r}\,' = O' (O \mathbf{r}) = (O' O) \mathbf{r}\,$ ,

implies

$r \to r' = U' U r U^\dagger U^{'\dagger} = (U'U) r (U'U)^{\dagger}\,$ ,

where $U'U\,$ is the transformation corresponding to $O'O\,$ .

The matrices $U\,$ form a representation of $SO(3)\,$ , known as the spinor representation, and act on a complex vector space of two-component objects called spinors, via $\psi \to \psi^' = U \psi\,$ . Note that the transformation law $r \to r' = U r U^{\dagger}\,$ is also the transformation law obeyed by the outer product of two spinors, namely $\psi \chi^{\dagger} \to U \psi \chi^{\dagger} U^{\dagger}\,$ . We have therefore learned that vectors transform under the direct product of the spinor representation $SU(2)\,$ with itself.

Note that if $U \in SU(2)\,$ corresponds to $O \in SO(3)\,$ , then so does $-U \in SU(2)\,$ . Thus $SU(2)\,$ is the double cover of $SO(3)\,$ .

Construction

From SU(2) to SO(3)

Since

$r \to r' = O^i_j r^j \sigma_i = r^j U \sigma_j U^{\dagger}\,$ ,

we require that

$O^i_j \sigma_i = U \sigma_j U^{\dagger}\,$ ,

from which

$O^i_j = \frac{1}{2} \operatorname{tr}\,\left(U \sigma_j U^{\dagger} \sigma_j\right)\,$ .

Axis-angle form

We can write an arbitrary element of $SU(2)\,$ as

$U = e^{i\theta \hat{n}\cdot \frac{\vec{\sigma}}{2}}\,$ .

Now, recall the identity for Pauli matrices:

$e^{i\theta \hat{n}\cdot \frac{\vec{\sigma}}{2}} = \cos\left(\frac{\theta}{2}\right) + i \hat{n}\cdot\vec{\sigma} \sin\left(\frac{\theta}{2}\right)\,$ .

Writing

$r' = e^{i\theta \hat{n}\cdot \frac{\vec{\sigma}}{2}} r e^{-i\theta \hat{n}\cdot \frac{\vec{\sigma}}{2}}\,$ ,

the brave reader will obtain:

$X = n_x \sin\left(\frac{\theta}{2}\right)$

$Y = n_y \sin\left(\frac{\theta}{2}\right)$

$Z = n_z \sin\left(\frac{\theta}{2}\right)$

$W = \cos\left(\frac{\theta}{2}\right)$

$O = \begin{bmatrix} 1-2(Y^2+Z^2) & 2XY + 2ZW & 2XZ - 2YW\\ 2XY - 2ZW & 1-2(X^2+Z^2) & 2YZ + 2XW\\ 2XZ + 2YW & 2YZ - 2XW & 1-2(X^2+Y^2)\\ \end{bmatrix}$