# Regarding the Dirac Hamiltonian's use of summation notation:

Einstein summation notation, as I understand it:
By writing $A_i B^i$ one implicitly means a sum over elements of the rank 1 tensors A and B. The key is the contraction of an "up" and a "down" index. In a formalism where a metric raises/lowers we should see this as an inner product, where the metric encodes the information of how an inner product is taken in such a space.

This notation is convenient tool that I employ on a daily basis. However, the Hamiltonian for the Dirac Equation of QFT fame can be written: $$H = \alpha_i p_i + \beta m$$ Two down indices? Summed together?

Now, in this case we're considering a flat Minkowski space-time with $i$ summing over just the spacial indices. As such, we can raise and lower the indices for "free" with a Euclidean metric.

Is this not an abuse of notation? This is not Einstein's summation convention but instead a bastardised summation notation in which we just do not write summation symbols?

Surely it would be more explicit to write: $$H = \alpha_i p^i + \beta m$$

Am I right here? Am I having some kind of mathematical mental breakdown? Both?

• As the Euclidean metric is just the identity matrix, there is no need to distinguish between up and down indices in this case, – Virgo Nov 14 '16 at 17:18

As long as you understand what you mean, you can use whatever notation you want. For example, when you write $p_i \alpha_i$ you might mean $$p_i\alpha_i\equiv p_1\alpha_1+p_2\alpha_2+p_3\alpha_3$$ or $$p_i\alpha_i\equiv -p_1\alpha_1-p_2\alpha_2-p_3\alpha_3$$ or any other convention you may want to follow. But as soon as you want to share your results with others, you must clearly explain what you mean by some expression.
For example, the combination $$a_0 b_0-a_1b_1-a_2b_2-a_3b_3$$ is very useful. Some people will write $a_\mu b^\mu$ for this particular combination, while some others will write $a_\mu b_\mu$ instead.
• Thanks I assumed this was the case. My main gripe is that I've also seen people write $a_i b_i$ with $i$ as a free index: crazy people! – Vielbein Nov 14 '16 at 23:34
As it was pointed out, we are in Euclidean space, so the metric is the unit matrix $I$, if you are in Minkowski space, similar things hold with $g_{\mu\nu} =\pm(+,-,-,-)I$ (don't care about the sign as long as you stick to one convention). There can be other metrics in general relativity, but as long as you are not in GR, you can ignore the position of the indices.