3
$\begingroup$

my question regards equation (165) of [1], namely, how to write the conformal Casimir as a differential operator in the "usual" $z,\bar{z}$ coordinates. If one inspects the definition of the Casimir operator as \begin{equation} C = -\frac{1}{2}L^{ab}L_{ab}, \end{equation} where $L_{ab}$ are the embedded $SO(d+1,1)$ generators defined in (38) of [1], it is possible to write the action of the Casimir in the "12" Channel of a 4pt function of identical scalars ($\langle \phi(x_{1})\phi(x_{2})\phi(x_{3})\phi(x_{4}) \rangle$), where \begin{equation} C\phi(x_{1})\phi(x_{2}) = -\frac{1}{2}\left(\mathcal{L}^{ab}_{1}+\mathcal{L}^{ab}_{2}\right)\left(\mathcal{L}_{ab,1}+\mathcal{L}_{ab,2}\right)\phi(x_{1})\phi(x_{2}), \end{equation} where subscript $i$ means the operator acting on coordinates $x_{i}$ (162) of [1]. Given the representation of the scalar field in respect to the conformal group generators (scattered throughout [1]) \begin{align} &[P_{\mu},\phi(x)] = \partial_{\mu}\phi(x),\\ &[D,\phi(x)] = \left( x\cdot\partial + \Delta \right)\phi(x),\\ &[M_{\mu\nu},\phi(x)] = -\left(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu} \right)\phi(x) ,\\ &[K_{\mu},\phi(x)] = \left(2 x_{\mu}(x\cdot\partial) - x^{2}\partial_{\mu} + 2\Delta x_{\mu} \right)\phi(x), \end{align} one can easily read $[L_{ab},\phi(x)]$ from the above (I assumed the SO(d+1,1) metric as $\text{diag}(-1,1,\cdots,1)$, with the "-1" sign in the -1 component). By acting as \begin{equation} L^{ab}L_{ab}\phi(x_{1})\phi(x_{2}) = [L^{ab},[L_{ab},\phi(x_{1})\phi(x_{2})]], \end{equation} I can indeed generate differential operators of first and second order depending on $x_{1},x_{2},\partial_{1},\partial_{2}$. My problem now lies in how to explicitly arrive at (165), and I believe it has to do with the choice of conformal frame. In section 5.1 of [1], the authors define the conformal frame as \begin{align} &x_{1} = (0,0,\cdots,0)\\ &x_{2} = (x,y,0,\cdots,0)\\ &x_{3} = (1,0,\cdots,0)\\ &x_{4}\to\infty. \end{align} Does it mean that I need to take $x_{1}^{\mu} = 0^{\mu}$ and $\frac{\partial}{\partial x_{1}^{\mu}} = 0$ in the Casimir operator in order to reproduce (168)? Which would yield only derivatives in respect to the $x_{2}$ point, which can be transformed into ($z,\bar{z}$) via a complex plane identification? Then I should act in (158) to recover $\mathcal{D}$ (168)?

If someone has a better understanding on how to reproduce the desired equation, I would much appreciate the help.

[1] David Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, https://arxiv.org/abs/1602.07982

$\endgroup$

1 Answer 1

1
$\begingroup$

Here is a "brute force" way of computing the Casimir differential operator:

  1. Act on eq. (158) with the commutators and work you way through the prefactor $1/(x_{12}^{2\Delta_\phi} x_{34}^{2\Delta_\phi})$ to see how the Casimir operator acts on the function $g_{\Delta, \ell}(u,v)$. You should obtain a second-order differential equation in $\partial/\partial x_1$ and $\partial/\partial x_2$.
  2. Use the chain rule to figure out how to write $\partial/\partial x_1$ and $\partial/\partial x_2$ in terms of $\partial/\partial z$ and $\partial/\partial \bar{z}$. This means computing $\partial z/\partial x_1$ from the definition of $z$ (and so on for $\bar{z}$ and $x_2$) in eq. (187) of [1].
  3. Write you result in terms of $z$ and $\bar{z}$. If you don't do mistakes this part should be very simple.

This is not an easy calculation. It is possible that you can simplify it by being clever. But at least it is guaranteed to work.

$\endgroup$
2
  • $\begingroup$ Point 1 can be agreed upon, the problem is point 2. If I am not mistaken, $z,\bar{z}$ coordinates are only defined in the conformal frame, in which the point $x_{1}$ is fixed to the origin. Which means there is no coordinate transformation from $x_{1}$ to $z,\bar{z}$, in fact $z,\bar{z}$ are the only freedom you have ($x_{2}$ coordinates). Having said that, how do we treat derivatives $\frac{\partial}{\partial x^{\mu}_{1}}$ in this frame? $\endgroup$ Jan 26, 2022 at 18:39
  • 1
    $\begingroup$ $z$ and $\bar{z}$ are defined in terms of $x_1$, $x_2$, $x_3$ and $x_4$ in eq. (187). I'm editing my answer to include this point. $\endgroup$
    – M.Jo
    Jan 27, 2022 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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