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In 2D CFT, the Euclidean two point correlator of a primary operator $\mathcal{O}$ with conformal weights $h$, $\bar{h}$ is given by

$$ \begin{align} \langle\mathcal{O}(z,\bar{z})\mathcal{O}(0,0)\rangle=\frac{1}{z^{2h}\bar{z}^{2\bar{h}}}=\frac{1}{(x+i\tau)^{2h}(x-i\tau)^{2\bar{h}}} \end{align}$$

where I have set $z=x+i\tau$, with $\tau$ being the Euclidean time.

Now, the usual $i\varepsilon$ prescription for analytic continuation to Lorentzian time goes as follows:

$$ \begin{align}\langle\mathcal{O}(t,x)\mathcal{O}(0,0)\rangle=\langle\mathcal{O}(\tau=it+\varepsilon,x)\mathcal{O}(0,0)\rangle=&\frac{1}{(-x^-+i\varepsilon)^{2h}(x^+-i\varepsilon)^{2\bar{h}}}\\=&e^{-2h\log(-x^-+i\varepsilon)-2\bar{h}\log(x^+-i\varepsilon)} \end{align} $$ $$ \begin{align} \langle\mathcal{O}(0,0)\mathcal{O}(t,x)\rangle=\langle\mathcal{O}(\tau=it-\varepsilon,x)\mathcal{O}(0,0)\rangle=\frac{1}{(-x^--i\varepsilon)^{2h}(x^++i\varepsilon)^{2\bar{h}}}\\ =e^{-2h\log(-x^--i\varepsilon)-2\bar{h}\log(x^++i\varepsilon)} \end{align} $$ where $x^\pm=t\pm x$ are the lightcone coordinates.

The $i\varepsilon$ prescription must next be supplemented by some choice of branch cut for the log functions. There seems to be a couple of choices.

Naive branch cut choice:

Naively, one might be tempted to take the standard branch cut for the logs viz. negative real axis. However, that way one finds that whenever $x^->0,x^+<0$ (spacelike separation with $x<0$) arguments of both the logs are across the branch cut. On the other hand whenever $x^-<0,x^+>0$ (spacelike separation with $x>0$), the arguments don't see any branch cut.

enter image description here

In terms of commutator,

$$ \langle\left[\mathcal{O}(t,x),\mathcal{O}(0,0)\right]\rangle=\begin{cases} 0\text{ for}\;\;x^-<0,x^+>0 \;(\text{spacelike},x>0)\\ - \frac{2i\sin 2\pi s}{(x^-)^{2h}(-x^+)^{2\bar{h}}} \text{ for}\;\;x^->0,x^+<0 \;(\text{spacelike}, x<0)\\ -\frac{2i\sin 2\pi h}{(x^-)^{2h}(x^+)^{2\bar{h}}}\text{ for} \;\;x^->0,x^+>0 \text{ (future timelike)}\\ -\frac{2i\sin 2\pi \bar{h}}{(-x^-)^{2h}(-x^+)^{2\bar{h}}}\text{ for} \;\;x^-<0,x^+<0 \text{ (past timelike)} \end{cases} $$ where $s=h-\bar{h}$ is the spin of $\mathcal{O}$.

While both bosonic and fermionic operators (i.e. $s \in \frac{\mathbb{Z}}{2}$) commute at spacelike separation, for parafermions (or anyons) we get different commutation behaviour depending on the spatial direction, which is quite odd!

Branch cut choice dictated by causality constraints:

a. Operators must always commute at spacelike separation

If we demand that at spacelike separation the commutators always vanish, no matter the spin, then we must choose branch cuts that are always directed away from the spacelike region in the complex $\tau$ plane (as considered in https://arxiv.org/abs/1509.00014). Thus the branch cut emanating from $i|x|$ extends in the positive imaginary direction, while the one starting at $-i|x|$ extends in the negative imaginary direction. The branch cut locations in terms of lightcone coordinates are

$$t>x\iff -x^-<0,\; t<-x \iff x^+<0 \text{ for }x>0\\ t<x \iff -x^->0,\; t>-x\iff x^+>0 \text{ for }x<0\\$$

In other words, we must choose the log branch cuts along negative real axis for $x>0$ but along positive real axis for $x<0$.

enter image description here

For this branch cut choice, the commutator turns out to be

$$ \langle\left[\mathcal{O}(t,x),\mathcal{O}(0,0)\right]\rangle=\begin{cases} 0 \text{ for}\;\;t^2<x^2 \text{ (spacelike)}\\ -\frac{2i\sin 2\pi h}{(x^-)^{2h}(x^+)^{2\bar{h}}}\text{ for} \;\;x^+>x^->0 \text{ (future timelike)}\\ -\frac{2ie^{-2\pi i \Delta}\sin 2\pi \bar{h}}{(x^-)^{2h}(x^+)^{2\bar{h}}}\text{ for} \;\;x^->x^+>0 \text{ (future timelike)}\\ \frac{2i\sin 2\pi \bar{h}}{(-x^-)^{2h}(-x^+)^{2\bar{h}}}\text{ for} \;\;x^-<x^+<0 \text{ (past timelike)}\\ \frac{2ie^{-2\pi i \Delta}\sin 2\pi h}{(-x^-)^{2h}(-x^+)^{2\bar{h}}}\text{ for} \;\;x^+<x^-<0 \text{ (past timelike)}\\ \end{cases} $$ where $\Delta=h+\bar{h}$ is the scaling dimension of $\mathcal{O}$.

The reason why the $\Delta$ dependent phases appear is because flipping the sign of $x$ results in $(-x^-+i\varepsilon)^{-2h}\to (-x^++i\varepsilon)^{-2h}=e^{-2\pi ih}(x^+-i\varepsilon)^{-2h}$ and $(x^+-i\varepsilon)^{-2\bar{h}}\to (x^--i\varepsilon)^{-2\bar{h}}=e^{-2\pi i\bar{h}}(-x^-+i\varepsilon)^{-2\bar{h}}$ and thus $\langle\mathcal{O}(t,x)\mathcal{O}(0,0)\rangle \to e^{-2\pi i \Delta}\langle\mathcal{O}(0,0)\mathcal{O}(t,x)\rangle\rvert_{h\leftrightarrow\bar{h}}$.

b. Commutator of operators at spacelike separation is spin dependent

From Non-zero Euclidean commutator in 2D CFT?, we see that for fractional spins Euclidean commutators are non-zero. So one may expect non-zero commutator even for the Lorentzian correlator.

To get non-zero commutator for spacelike separation, we must align the branch cuts to be towards the spacelike region. This is the opposite of the choice made in a. In terms of lightcone coordinates, the locations are

$$t<x\iff -x^->0,\; t>-x \iff x^+>0 \text{ for }x>0\\ t>x \iff -x^-<0,\; t<-x\iff x^+<0 \text{ for }x<0\\$$

In other words, we must choose the log branch cuts along positive real axis for $x>0$ but along negative real axis for $x<0$.

enter image description here

For this branch cut choice, the commutator turns out to be

$$ \langle\left[\mathcal{O}(t,x),\mathcal{O}(0,0)\right]\rangle=\begin{cases} \frac{2ie^{-2\pi i\Delta}\sin 2\pi s}{(-x^-)^{2h}(x^+)^{2\bar{h}}} \text{ for}\;\;x^-<0,x^+>0 \;(\text{spacelike},x>0)\\ -\frac{2i\sin 2\pi s}{(x^-)^{2h}(-x^+)^{2\bar{h}}} \text{ for}\;\;x^->0,x^+<0 \;(\text{spacelike},x<0)\\ -\frac{2i e^{-2\pi i \Delta} \sin 2\pi \bar{h}}{(x^-)^{2h}(x^+)^{2\bar{h}}}\text{ for} \;\;x^+>x^->0 \text{ (future timelike)}\\ -\frac{2i \sin 2\pi h}{(x^-)^{2h}(x^+)^{2\bar{h}}}\text{ for} \;\;x^->x^+>0 \text{ (future timelike)}\\ \frac{2i e^{-2\pi i \Delta}\sin 2\pi h}{(-x^-)^{2h}(-x^+)^{2\bar{h}}}\text{ for} \;\;x^-<x^+<0 \text{ (past timelike)}\\ \frac{2i\sin 2\pi \bar{h}}{(-x^-)^{2h}(-x^+)^{2\bar{h}}}\text{ for} \;\;x^+<x^-<0 \text{ (past timelike)}\\ \end{cases} $$

I think the last one is the correct branch cut choice, but I am not exactly sure.

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    $\begingroup$ in higher dimensions Integer spin operators commute at spacelike distance, and half-integer spins don't (they anti-commute in fact). Now, in 2d you can have anyons and the requirement is not really about commutators or anticommutators but something in between that depends on the spin, e.g. for equal times $\psi^\dagger(0,x)\psi(0,0)-e^{-i\pi k sign(x)}\psi(0,0)\psi^\dagger(0,x)=0$ where $k$ is the parameter that controls the statistics of the operator. $\endgroup$
    – TwoBs
    Mar 7, 2020 at 16:51
  • $\begingroup$ I'm a little confused by the question. By modularity, all local operators have integer spin, and for nonlocal operators this question of the branch cut is resolved by considering the strings attached to these operators. $\endgroup$ Mar 8, 2020 at 16:55
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    $\begingroup$ First of all, you should start with a single-valued correlator in Euclidean signature. This requires the spin to be an integer. Second, the $i\epsilon$-prescription completely determines the analytic continuation, there is no extra choice. Branch cut is not an actual physical object, branch cut is a line that you draw on paper because you can't draw multi-sheeted Riemann surfaces. The epsilons tell you from which side you should pass the branch point when performing analytic continuation, and that's all you need to know. $\endgroup$ Mar 10, 2020 at 1:13
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    $\begingroup$ If you want to consider non-integer spin operators, you need to be very specific about what you mean. Everybody agrees on what a local operator is, and they have integer spins. Now, as soon as you go outside of the world of local operators, you need to be very specific about what your non-integer-spin operators actually are and treat them accordingly. You should not necessarily expect that the usual $i\epsilon$-prescriptions work for them. $\endgroup$ Mar 10, 2020 at 1:16

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