Pauli matrices, Levi-Civita symbol and Einstein notation

Question

It's some time that I notice some sign issues in my calculations of quantum field theory and I started to think that the origin should be addressed to my choice $(\hat{p}_\alpha)\doteq -i\hbar\partial_\alpha$ instead of the $(\hat{p}_\alpha)\doteq i\hbar\partial_\alpha$, but many other things just didn't come right.

So I noticed something that really shook me and is the fact that, attempting to save Einstein convention, I used the Levi-Civita tensor with upper and lower indices, but maybe I just could not do it in such naive way.

I tried to find an answer and I read this and this very interesting one, but I'm dumb and I simply can't grab the meaning: seems like I'm about to understand, but I don't.

I go to the point: consider the well known property of Pauli matrices \begin{equation*} \sigma^i\sigma^j = \delta^{ij}\mathbb{I} +i {\epsilon^{ij}}_k \sigma^k \end{equation*} Here the upper indices are just an "illusion", right? I mean that ${\epsilon^{ij}}_k\equiv\epsilon_{ijk}, {\epsilon^{ij}}_k\neq g^{i\alpha}g^{j\beta}\epsilon_{\alpha\beta k}$ (it's not an $\text{SO}(1,3)$ tensor?); but at the same time that seems to hold for the Kronecker delta! In fact, and I conclude, consider that you have now the expression \begin{equation*} \sigma^i\sigma^j\partial_i\partial_j \end{equation*} where ${\epsilon^{ij}}_k\partial_i\partial_j\equiv{\epsilon^{ji}}_k\partial_j\partial_i=0$. The question is: is this $\partial_j\partial^j=-\nabla$ or, as I suspect, is $\partial_j\partial_j=\nabla$?

In both cases why is that and how I can concile it with the Einstein convention?

This question is really hurting me inside so thanks for help!

Answer 1

The $\sigma^i\sigma^j\partial_i\partial_j$ will lead to $$(\partial_x\sigma^x+\partial_y\sigma^y+\partial_z\sigma^z)(\partial_x\sigma^x+\partial_y\sigma^y+\partial_z\sigma^z)=\partial_x^2 1+\partial_y^21+\partial_z^21$$ The $\sigma^i\sigma^j$ terms with $i\neq j$ vanish as Pauli matrices anticommute and therefore these terms should cancel out. This result is obtained by an Euclidian metric to contract the Pauli Matrices with the partial derivatives (This is in most cases implicitly assumend).

To calculate this with you statement for $\sigma^i\sigma^j$ you can do the following: $$\sigma^i\sigma^j\partial_i\partial_j=\delta^{ij}1\partial_i\partial_j+i\epsilon^{ij}_k\sigma^k\partial_i\partial_j$$ And see directly that the first term ist the same as writing everything out explicitly and the second term vanishes because of symmetry and antisymmetry arguments.

My Advice is if you are unsure write your problem in matrices to see what you are dealing with

The contraction $\partial_i\partial^i$ is in most cases to be used as a short hand notation for the following way:

$$\partial_i\partial^i=g^{ij}\partial_i\partial_j$$ with the metric $g$ and the indices running over all space-time dimensions.

• For the three dimensional case with an Euclidean metric $g^{ij}=\delta^{ij}$ you obtain $$\partial_i\partial^i=g^{ij}\partial_i\partial_j=\delta^{ij}\partial_i\partial_i=\partial_x^2+\partial_y^2+\partial_z^2=\Delta$$ The Laplace operator

• For the four-dimensional case with a minkowski space-time and (-,+++) convention (and $c=1$) you should obtain $$\partial_i\partial^i=g^{ij}\partial_i\partial_j=\eta^{ij}\partial_i\partial_i=-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2=\square$$ Which is called the d‘Alembert operator.

To verify these statements by yourself you can find for the easy cases some matrix representation of your Tensors up to rank two and calculate it with objects you are more familiar with.

I do not fully get the question in your question, this refers to the part with the question mark :)

Answer 2

In Minkowski space $\epsilon^{0123}=-\epsilon_{0123}$ but people differ in which of these two expressions they take to be $+1$, so it's a good idea to always state your conventions. Similarly with $p_\mu$: It is always true that $p^\mu= (E, {\bf p})=m V^\mu =mdx^\mu/d\tau$, but $p_\mu=(-E,{\bf p}) \to -i\hbar \partial_\mu$ only in the $(-,+,+,+)$ metric convention. In the $(+,-,-,-)$ metric we have $p_\mu = (E,-{\bf p}) \to +i\hbar \partial_\mu$. There are similar problems with $A^\mu=(\phi, {\bf A})$ when one wants to use $A_\mu$ in ${\bf p}+e{\bf A}$. Since I've used both metrics at various times, I always work a couple of cases to make sure I have gotten things right.

There is less problem with Pauli matrices as one only uses them in 3-d Euclidean space. Then upstairs and downstairs indices are the same thing.