5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Repeated indices are summed over so they are always dummy indices. $\endgroup$
    – Prahar
    Jun 8 at 9:46
  • 2
    $\begingroup$ 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 $e^{ikx}e^{D^2}e^{ikX}=e^{D^2}+[e^{D^2},\,ikx]$. $\endgroup$
    – J.G.
    Jun 8 at 10:57
  • $\begingroup$ @J.G. Why is $[ikx,e^{D^{2}}]$ central? $\endgroup$ Jun 8 at 11:26
  • $\begingroup$ @J.G. also I don't see how $[e^{D^2},ikx]$ gives $e^{(D_{\mu}+ik_{\mu})^2}$ $\endgroup$ Jun 8 at 11:36
  • $\begingroup$ Ah, I hadn't spotted the centrality requirement. $\endgroup$
    – J.G.
    Jun 8 at 11:40

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Jun 8 at 13:40
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 8 at 13:46
2
$\begingroup$

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
$$ e^{-ikx}\partial_x e^{ikx}= \partial_x+ik. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.