# How are these Covariant Derivative Identities found?

In David Tong's Gauge Theory notes on page 137 near eq. (3.30) he makes use of the following expressions for the covariant derivative $$D_{\mu}$$

$$\frac{1}{2}[\gamma^{\mu},\gamma^{\nu}]D_{\mu}D_{\nu}=\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}]\tag{1}$$

and

$$e^{-ikx}e^{D^2}e^{ikx}=e^{(D_{\mu}+ik_{\mu})^2}\tag{2}$$

I'm guessing the first is just a change of dummy indices in the second term of the commutator, but I don't see how the indices are dummy.

The second expression I'm more confused about. It looks like $$x^{\mu}$$ is acting like a generator of translation in momentum space, but I'm not sure.

• Repeated indices are summed over so they are always dummy indices. Jun 8 at 9:46
• For the first write $D_\mu D_\nu=\frac12([D_\mu,\,D_\nu]+\{D_\mu,\,D_\nu\})$. Only the antisymmetric part survives contraction with $[\gamma^\mu,\,\gamma^\nu]$, which is antisymmetric. For the second use this so $e^{ikx}e^{D^2}e^{ikX}=e^{D^2}+[e^{D^2},\,ikx]$.
– J.G.
Jun 8 at 10:57
• @J.G. Why is $[ikx,e^{D^{2}}]$ central? Jun 8 at 11:26
• @J.G. also I don't see how $[e^{D^2},ikx]$ gives $e^{(D_{\mu}+ik_{\mu})^2}$ Jun 8 at 11:36
• Ah, I hadn't spotted the centrality requirement.
– J.G.
Jun 8 at 11:40

Hint for eq. (2):

$$e^{-ik\cdot x} f(D) e^{ik\cdot x}~=~f\left(e^{-ik\cdot x} D e^{ik\cdot x}\right)$$

and

\begin{align} e^{-ik\cdot x} D_{\mu} e^{ik\cdot x}~\stackrel{\text{Hadamard}}{=}&~e^{-ik_{\nu} [x^{\nu},\cdot]} D_{\mu}\cr ~=~~~&D_{\mu}+ik_{\nu} [D_{\mu},x^{\nu}]\cr ~=~~~&D_{\mu}+ik_{\mu},\end{align} where we used Hadamard's formula.

• Does the second line follow from some form of BCH formula where we evaluate $[D_{\mu},ikx]$? I'm not entirely sure what the expression in the middle means and how we get it. Jun 8 at 13:40
• I updated the answer. Jun 8 at 13:46

By Leibnitz' rule $$e^{-ikx} \partial_x \{e^{ikx} f(x)\} = e^{-ikx}\{f(x)(\partial_x e^{ikx})+ e^{ikx}(\partial_x f)\}\\ = e^{-ikx}\{ f(x) (ik e^{ikx})+ e^{ikx}(\partial_x f)\}\\ = ik f(x) + \partial_x f(x) =(\partial_x +ik)f(x).$$ As $$f(x)$$ can be anything, we have
$$e^{-ikx}\partial_x e^{ikx}= \partial_x+ik.$$