1
$\begingroup$

Maybe I still have trouble understanding what physicists mean when something is a vector but here is how I see it. I use the Einstein summation convention throughout.

Given a set of basis vectors $B = \{\vec{x}_i\}$ and a set of coordinates $C = \{c_i\}$, we represent a physical quantity $\vec{v} = c_i \vec{x}_i$. Now I am free to change my basis to some $\vec{x}'_i = R_{ij}\vec{x}_j$. In this case, I must have $c'_i = R^{-1}_{ji}c_j$ so that I still refer to the same physical quantity after the transformation.

With this in mind, I'm trying to understand what it means when people say that one can make a vector out of the Pauli matrices as explained here https://en.wikipedia.org/wiki/Pauli_matrices#Pauli_vector.

What is special about the Pauli matrices such that when I apply the transformation $R$ on basis of Pauli matrices, it "works"? What exactly wouldn't work if I took some other arbitrary set of 2x2 matrices?

$\endgroup$
  • $\begingroup$ This is e.g. explained in my Phys.SE answer here. $\endgroup$ – Qmechanic Apr 2 '18 at 17:16
2
$\begingroup$

Take $$ R=\left( \begin{array}{cc} e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) & -e^{-\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \\ e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) & e^{\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) \\ \end{array} \right)=R_z(\alpha)R_y(\beta)R_z(\gamma)\, . $$ Then, for instance, \begin{align} R\cdot\sigma_z\cdot R^{-1} = \sigma_z\cos\beta + \sigma_x\cos\alpha\sin\beta +\sigma_y\sin\alpha\sin\beta \end{align} which is the same as the rotation $R$ applied to $\hat z$. The other Pauli matrices also transform into linear combinations of themselves as the corresponding basis vectors do.

$\endgroup$
  • $\begingroup$ Sorry, could you explain why the $R_i(\theta)$ are 2x2 matrices since they rotate vectors in three dimensional space? And if they're 3x3, then how do you compute $R^{-1}\sigma_i R$? $\endgroup$ – user1936752 Apr 3 '18 at 6:17
  • $\begingroup$ @user1936752 The dimension of the vector space is $3$ because each Pauli matrix is a basis vector in that space (of $2\times 2$ hermitian matrices). If you want you can map $\sigma_x\mapsto \hat x= \left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)$ etc and then make $\bar R$ into a $3\times 3$ matrix representation of $R$ with $R\cdot \sigma_k \cdot R^{-1}\mapsto \bar R\hat k$. $\endgroup$ – ZeroTheHero Apr 3 '18 at 9:41
  • $\begingroup$ Is your point then that such a map exists for Pauli matrices but not arbitrary 2x2 matrices? If yes, can I trouble you to explicitly show how one gets the mapping between $\sigma_x$ and $\hat{x}$ mapping? $\endgroup$ – user1936752 Apr 3 '18 at 10:04
  • $\begingroup$ @user1936752 any $2\times 2$ hermitian traceless matrix can be expressed as a linear combination of the $3$ Pauli matrices, just like any vector in $3D$ can be expanded as a linear combination of $3$ linearly independent vectors. The Pauli matrices are trace orthogonal $\hbox{Tr}[(\sigma_k)^\dagger \sigma_m]=2\delta_{km}$ so function as basis vectors for this vector space. $\endgroup$ – ZeroTheHero Apr 3 '18 at 12:32
1
$\begingroup$

Suppose you start out looking for a vector $\vec{ \sigma}$, that you want to satisfy:

$$ {\vec{\sigma} \times \vec{\sigma}} = 2i \vec{ \sigma} $$

So it's a priori a vector with a non-zero cross product with itself. Expanding the cross product:

$$ \sigma_a\sigma_b - \sigma_b\sigma_a = 2i\epsilon_{abc}\sigma_c $$

You can solve that with 2 indices ($\alpha, \beta \in (+\frac 1 2, -\frac 1 2)$):

$$ [\sigma_a]_{\alpha\beta} = (\alpha + \beta)\delta_{a3} +|\alpha-\beta|\delta_{a1} + i(\beta-\alpha)\delta_{a2}$$

which are the Pauli matrices, forming a vector. The challenge now is convincing yourself that they describe an internal degree of freedom that is spin 1/2.

$\endgroup$
0
$\begingroup$

The formulae you see in textbooks for the spin matrices are just one representation of an algebra satisfying $\sigma_a\sigma_b =\delta_{ab}I_2+i \sum_c \epsilon_{abc}\sigma_c$. This condition is preserved under the usual rotational transformation, which obtains a different representation of this algebra. The same one-among-many disclaimer applies to other matrix-valued constants in physics (although the preserved rule will be different), such as Gell-Mann matrices & gamma matrices.

$\endgroup$

Not the answer you're looking for?Browse other questions tagged or ask your own question.