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!


1 Answer 1


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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