# Transformation of a 2x2 matrix with Pauli matrices

1. Apr 11, 2016

### CAF123

1. The problem statement, all variables and given/known data
Suppose the vector $\phi$ transforms under SU(2) as: $$\phi' = (\exp(-i \alpha \cdot t))_{ij}\phi_j,$$ where $(t_j)_{kl} = −i \epsilon_{jkl}$ and $j, k, l \in \left\{1, 2, 3\right\}.$

Based on $\phi,$ we define the $2 \times 2$ matrix $\sigma = \tau \cdot \phi$ where $\tau$ are the Pauli matrices.

1) Assume that the real parameters $\alpha = (\alpha_1, \alpha_2, \alpha_3)$ are small and show, by expanding up to first order in $\alpha$, that the transformation law for the $2 \times 2$ matrix $\sigma$ is given by $\sigma \rightarrow \sigma' = U\sigma U^{\dagger}$ where $U = \exp(i\alpha \cdot \tau/2)$

(b) By construction, $\sigma$ is hermitian and traceless. Verify that $\sigma'$ is also hermitian and traceless, and that det $\sigma'$ = det $\sigma$. Show that $\phi'$ and $\phi$ are related by a rotation.

2. Relevant equations
The $(t_j)_{kl}$ are 3 matrices each of which are 3x3. These correspond to matrices in the adjoint representation of SU(2). I think I can write $\tau_i \rightarrow u\tau_i u^{\dagger}$, where $u \approx 1 + i \beta^a \tau^b$ with $u \in SU(2)$ - i.e can decompose a general group element in terms of the lie algebra and since the taus are 2x2 matrices the corresponding representation of the generators are also pauli matrices. (I think?)

3. The attempt at a solution
$\sigma_{ij} \rightarrow (\tau_k' \phi_k')_{ij}$. Using the transformation of the taus above and the algebra $[\tau_a/2, \tau_b/2] = i/2 \epsilon_{abc} \tau_c$ I get that $$\tau_k' = \tau_k + 2i \beta_a \epsilon_{akm}\tau_m$$

The transformation of $\phi$ is given in the question and, as far as I understand, it is transforming in the adjoint representation of SU(2).

$$\phi_k' = (1-i\alpha^b(t^b)_{kl})\phi_l = (\delta_{kl} - i \alpha^b (-i \epsilon_{bkl}))\phi_l = \phi_k - \alpha^b \epsilon_{bkl} \phi_l$$

Putting these two transformations together (that of $\phi$ and that of the $\tau_k$ ) I don't get something resembling the answer. Did I go wrong somewhere?

Thanks!

2. Apr 16, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?