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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!

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    $\begingroup$ The Levi-Civita symbol is a pseudotensor rather than a tensor. Index notation rules are sometimes broken in group theory texts like this. It is understood that repeated index implies summation but strict adherence to tensor index notation is not always done $\endgroup$
    – Charlie
    Commented Oct 1, 2020 at 10:50
  • $\begingroup$ So it is true that ${\epsilon^{ij}}_k\equiv\epsilon_{ijk}$? And in this situation also $\delta^{ij}\equiv\delta_{ij}$? What that does mean? Thanks $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 11:08
  • $\begingroup$ I don't know about the first equation you've written, I would avoid using a definition sign though. Regarding $\delta^{ij}\equiv \delta_{ij}$ this is certainly true for the Euclidean metric since it is just the identity in matrix form, but not in general. $\endgroup$
    – Charlie
    Commented Oct 1, 2020 at 11:20
  • $\begingroup$ In fact $\delta^{ij}=\delta_{ij}$ in the Euclidean case, but now there is Minkowski metric and this become really confusing to me $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 11:36

2 Answers 2

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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 :)

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  • $\begingroup$ I agree with what you wrote, but sorry this doesn't answer my question. And I'm not so sure that, that $\delta^{ij}$ behaves like expected and may be that $\sigma^i\sigma^j\partial_i\partial_j$ is actually $\partial_j\partial_j$ and not $\partial_j\partial^j$; that's one of the questions! $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 12:22
  • $\begingroup$ The contraction with the Pauli matrices is done with use an Euclidian metric so the index position does not matter $\endgroup$
    – jan0155
    Commented Oct 1, 2020 at 12:32
  • $\begingroup$ Yes I understand that, but the problem arises in four dimensions in which I am. Because this calculation in particular has to do with the hamiltonian of Dirac field interacting with the electromagnetic one in the usual Minkowski metric $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 12:37
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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.

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  • $\begingroup$ Are you sure that momentum has a sign defined by the metric, because I think the sign is "free". It seems to me just a definition, given by the fact that a unitary transformation of four-translation (with derivative as generator) $\boldsymbol{a}$ can be defined as $\exp{i a^\alpha p_\alpha/\hbar}$ such as $\exp{-ia^\alpha p_\alpha/\hbar}$ and nothing changes. $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 13:59
  • $\begingroup$ Maybe between two cases I should choose that giving me $[\hat{r}^i,\hat{p}_j]=i\hbar{\delta^i}_j$? This would arise from the classical $\{r^i,p_j\}={\delta^i}_j$.In that case why am I wrong? Because I should have $\hat{r}^i\hat{p}_j-\hat{p}_j(\hat{r}^i\circ)=\hat{r}^i\hat{p}_j-\hat{p}_j\hat{r}^i-\hat{r}^i\hat{p}_j=-\hat{p}^j\hat{r}_i\equiv +i\hbar{\delta^j}_i$ that is exactly the commutation relation! $\endgroup$
    – Rob Tan
    Commented Oct 1, 2020 at 14:19
  • $\begingroup$ @Rob Surely there's no meaning to raising and lowering indices in the equal time commutator? --- as it is strictly euclidean 3-d But maybe I miss your point? I would have thought that whichever metric we use the operator corresponding to 3-d $m{\bf v}$ should be $-i\hbar {\boldsymbol \partial}$ so as to get NR quantum mechanics $\endgroup$
    – mike stone
    Commented Oct 1, 2020 at 15:30

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