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Jagerber48
Jagerber48
02:20
@LPZ ok, after some time I've boiled my question down. The answer to the original question asked here relies on:
$$
\text{ad}(\hat{n}\cdot \sigma)(\sigma) = -2i(\hat{n}\cdot J)\sigma
$$

To prove it we use $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k$ and $(J_i)_{jk} = -\epsilon_{ijk}$ and a bit of algebra elbow grease.

But I'm wondering if it's possible to know that this formula is true more abstractly, just using representation theory and the resulting relationships between $\sigma$ and $J$.
In other words: Is there a "deeper" reason that $\text{ad}(\hat{n}\cdot \sigma)(\sigma) = -2i(\hat{n}\cdot J)\sigma$ other than just working through the algebra..?
 
5 hours later…
LPZ
LPZ
07:35
It comes from the definition of $J$. In a general Lie algebra, given a basis $e_i$, you have the structure coefficients:$$[e_i,e_j] = f_{ij}^ke_k$$ This is used to define the the adjoint action:$$ad(e_i)_j^k = f_{ij}^k$$ $f$ is the generalisation of $J$, and it is immediate that $$ad(x)y = [x,y]$$
Disregard my previous message. Things are usually done in reverse. The relation that you want to prove is assumed to define $J$. In a general Lie algebra, the adjoint action is: $$ad(x)y = [x,y]$$. Given a basis $e_i$, you have the structure coefficients:$$[e_i,e_j] = f_{ij}^ke_k$$. This imposes the coefficients of the matrix:$$ad(e_i)_j^k = f_{ij}^k$$
 
6 hours later…
Jagerber48
Jagerber48
14:09
This seems so close to answering my question. What I take away from the structure coefficients $f_{ijk}^k$ is that the commutation relations (and the adjoint actions) will look the same for any representation of $\mathfrak{so}(3)$. Like $\sigma_i$ are related to a 2D representation and $J_i$ are related to a 3D representation, but their commutation relations are similar:
$$
[\sigma_i, \sigma_j] = 2i \epsilon_{ijk} \sigma_k
$$
$$
[J_i, J_j] = \epsilon_{ijk} J_k
$$
But the expression I'm trying to understand
$$
\text{ad}(\hat{n}\cdot \sigma)(\sigma) = -2i(\hat{n}\cdot J)\sigma
$$
mixes/relates actions from the 2D and 3D representations, $\sigma$ AND $J$. It's this relation between the 2D and 3D representations that is... confusing/surprising/interesting to me.
In components the expression is
$$
\text{ad}(\sigma_i)(\sigma_j) = -2i (J_i)_{jk} \sigma_k
$$
So the coincidence is that both
$$
[J_i, J_j] = f_{ijk} J_k
$$
AND
$$
(J_i)_{jk} = -f_{ijk}
$$
Like both the commutation relations amongst the $J$ is given by $f_{ijk}$ AND the matrix components of $J$ are given by $f_{ijk}$. This seems surprising/special in some way..
 
1 hour later…
Jagerber48
Jagerber48
15:38
@LPZ asked in more detail here: physics.stackexchange.com/questions/819825/…
LPZ
LPZ
16:10
I think my previous comment answers your question. To make things even more explicit, it’s best to write:$$[J_i,J_j]=\epsilon_{ijk}(J_i)_{jk}=\epsilon_{ijk}$$ Compare that to the general formula:$$[e_i,e_j]=f_{ij}^ke_k\\ad(e_i)_j^k=f_{ij}^k$$
I think my previous comment answers your question. To make things even more explicit, it’s best to write:$$[J_i,J_j]=\epsilon_{ijk} \\ (J_i)_{jk}=\epsilon_{ijk}$$ Compare that to the general formula:$$[e_i,e_j]=f_{ij}^ke_k\\ad(e_i)_j^k=f_{ij}^k$$
Perhaps you are mixing up the adjoint representation and this specific representation where the Lie bracket is implemented as commutators of matrices. Once you know the Lie structure, the underlying implementation is irrelevant, it is purely a question of $\mathfrak{so}(3)$
 
6 hours later…
Jagerber48
Jagerber48
21:55
@LPZ I'm almost positive I'm "mixing up the adjoint representation and this specific representation"

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