In my problem I'm looking at a spin-$\frac{1}{2}$ particle witch charge q, which is represented by some Dirac spinor $\Psi$ solving the Dirac equation $$i\partial_t\Psi = \hat{H}\Psi$$ with given Dirac operator $\hat{H} := c\alpha_i(\hat{p} - \frac{q}{c}A^i) + \beta mc^2+\mathbb{1_{4x4}}q\Phi$, where $\alpha_i=\left(\begin{matrix} 0 & \sigma_i \\ \sigma_i & 0 \\ \end{matrix}\right)$ and $\beta=\left(\begin{matrix} \mathbb{1_{4x4}} & 0 \\ 0 & -\mathbb{1_{4x4}} \\ \end{matrix}\right)$ using the Pauli-matrices $\sigma_i$ and a 4x4 unity matrix $\mathbb{1_{4x4}}$.

The original task is to show the commutation relations $\left[\hat{J}^i,\hat{J}^j\right]=i\epsilon^{ijk}\hat{J}^k$, $\left[\hat{J}^i,\hat{J}^2\right]=0$. Showing the first should automatically result in showing the second with some commutation rules, but my problem is with $\hat{J}^i:= \mathbb{1_{4x4}}\hat{L}^i+\hat{S}^i$, I calculated: $$\left[\hat{J}^i,\hat{J}^j\right]=\left[\mathbb{1_{4x4}}\hat{L}^i+\hat{S}^i,\mathbb{1_{4x4}}\hat{L}^j+\hat{S}^j\right]$$ now using the commutation relation $$\left[A+B,C+D\right] = \left[A,C+D\right] +\left[B,C+D\right]= \left[A,C\right]+\left[A,D\right] +\left[B,C\right] +\left[B,D\right]$$ I get four terms: $$\left[\mathbb{1_{4x4}}\hat{L}^i,\mathbb{1_{4x4}}\hat{L}^j\right]+\left[\mathbb{1_{4x4}}\hat{L}^i,\hat{S}^j\right]+\left[\hat{S}^i,\mathbb{1_{4x4}}\hat{L}^j\right]+\left[\hat{S}^i,\hat{S}^j\right]$$ defining $\hat{L}^i:= \epsilon_{ijk}\hat{x}^j\hat{p}^k$ and $\hat{S}^i:=\frac{1}{2}\hbar\left(\begin{matrix} \sigma_i & 0 \\ 0 & \sigma_i \\ \end{matrix}\right)$, it follows: $$\left[\mathbb{1_{4x4}}\hat{L}^i,\mathbb{1_{4x4}}\hat{L}^j\right]=i\epsilon^{ijk}\hat{L}^k\text{ and }\left[\hat{S}^i,\hat{S}^j\right]=\frac{1}{2}i\hbar^2\sum_{k=1}^3\epsilon^{ijk}\sigma^k\mathbb{1_{2x2}}$$

My Problem lies now with the mixed commutations $\left[\mathbb{1_{4x4}}\hat{L}^i,\hat{S}^j\right]$, $\left[\hat{S}^i,\mathbb{1_{4x4}}\hat{L}^j\right]$, which boil down to something like $\left[\epsilon^{ijk}\hat{x}^j\hat{p}^k,\sigma^i\right]$. I thought about saying switching the commutator arguments should give a minus sign and the mixed terms cancel eachother. This could maybe gained by switching indices in the levi-civita symbol ($\epsilon^{ijk}=-\epsilon^{jik}$) after renaming i to j and vice versa. I don't really know, whether I can do that and even when doing so I don't get the results right. Calculating the commutators directly seems difficult to me, for I don't know how to solve $\left[\epsilon^{ijk}\hat{x}^j\hat{p}^k,\sigma^i\right]$. I hope my problem is stated clearly enough and a hint at what I'm missing would be great.

This is my first post so some advice on how to post is always welcome. Thanks!


The mixed commutators are both zero. There is no $x$ or $p$ dependence in the $\sigma_i$'s, and no spin indices in $x$ or $p$.

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