# Some diracology in traces

Suppose I want to evaluate the trace $p_{\alpha} q_{\beta}\text{Tr}(\gamma^{\alpha} \gamma^0 \gamma^{\beta} \gamma^0)$. Using the standard trace formula for four gamma matrices I arrive at $$p_{\alpha} q_{\beta}\text{Tr}(\gamma^{\alpha} \gamma^0 \gamma^{\beta} \gamma^0) = 4p_{\alpha} q_{\beta}(2g^{0 \alpha} g^{0 \beta} - g^{\alpha \beta})$$ Contracting the indices gives me $4(p^0 q^0 + \mathbf p \cdot \mathbf q)$ which is the answer I know to be correct.

But lets proceed in another way: Write $$p_{\alpha} q_{\beta}\text{Tr}(\gamma^{\alpha} \gamma^0 \gamma^{\beta} \gamma^0) = p_{\alpha} q_{\beta} (\text{Tr}(\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 \beta}))) = p_{\alpha}q_{\beta} (-\text{Tr}(\gamma^{\alpha} \gamma^{\beta}) + 2 \text{Tr}(\gamma^{\alpha} \gamma^{\beta})) )$$ using the clifford algebra and the fact $\gamma_0^2 = 1$. Then we have $$\text{Tr} (\not p \not q) = 4 p \cdot q = 4(p^0 q^0 - \mathbf{p} \cdot \mathbf q)$$ So I have a minus error in the second term. Did I miss a term somewhere?

You go from $$\text{Tr}\big[\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 \beta})\big]$$ to $$-\text{Tr}\big[\gamma^{\alpha} \gamma^{\beta}\big] + 2 \text{Tr}\big[\gamma^{\alpha} \gamma^{\beta}\big]$$ but this is wrong. The correct result is
$$\text{Tr}\big[\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 \beta})\big]=-\text{Tr}\big[\gamma^{\alpha}\gamma^{\beta}\big] +2g^{0 \beta}\text{Tr}\big[\gamma^{\alpha} \gamma^0\big]$$
Finally, take $\text{Tr}[\gamma^{\alpha}\gamma^{\beta}]=4g^{\alpha\beta}$ and $\text{Tr}[\gamma^\alpha\gamma^0]=4g^{\alpha 0}$:
$$\text{Tr}\big[\gamma^{\alpha} \gamma^0 (-\gamma^0 \gamma^{\beta} + 2g^{0 \beta})\big]=-4g^{\alpha\beta}+8g^{0 \beta}g^{\alpha0}$$ which is the same result you got first.
• Okay, thanks! The only question I would have is, why is $\gamma_0 g^{0 \beta} = \gamma^{\beta}$ not valid? – CAF Mar 22 '16 at 10:52
• Because in $\gamma_0 g^{0\beta}$ there is not a sum: the index $0$ is not contracted, but is fixed to the single value "$0$". Take for example $\beta=1$. Then the l.h.s. you have $\gamma_0 g^{01}=0$ (because $g^{01}=0$) while on the r.h.s you have $\gamma^1\neq 0$. You can see that l.h.s$\neq$r.h.s. (if I didn't make myself clear please say so and I'll try to explain better). – AccidentalFourierTransform Mar 22 '16 at 10:55
• Yes it's clear! thanks, I just realised as well - $$\gamma_{\alpha} g^{\alpha \beta} = \gamma^{\beta} = \gamma_0 g^{0 \beta} + \gamma_i g^{i \beta}$$ so what I wrote is indeed not true. I guess something perhaps more conceptually though: The $\gamma_{\mu}$ are matrices so why do they obey the same raising/lowering property with the metric as four vectors do? – CAF Mar 22 '16 at 11:00
• nice! the matrices $\gamma_\mu$ (with lower indices) are defined to be $\gamma_\mu\equiv g_{\mu\nu}\gamma^\nu$. This means: they obey the same raising/lowering properties because we define them to. The matrices with upper indices are given by the Clifford algebra. The matrices with lower indices are defined by contracting with the metric. Its just notation. – AccidentalFourierTransform Mar 22 '16 at 11:06