Spin and scale dimension of canonical spin-1/2 fields in (1+1)d I am reading the book "Non-perturbative methods in 2 dimensional quantum field theory" by Abdalla, Abdalla and Rothe and have some questions about the Chapter 2.4 "Bosonization of Massless Fermions".
Earlier in the book (2.20 & 2.21) they show that the zero mass two-point functions look like
$$\begin{align}
D^+(x)&=\langle \phi(x)\phi(y) \rangle=-\frac{1}{4\pi}\ln[(i\mu x^++\epsilon)(i\mu x^-+\epsilon)] \tag{2.20}\\
\tilde{D}^+(x)&=\langle \phi(x)\tilde{\phi}(y) \rangle=\frac{1}{4\pi}\ln\frac{i\mu x^++\epsilon}{i\mu x^-+\epsilon} \qquad x^\pm=x^0\pm x^1.
\tag{2.21}\end{align}$$
These two-point functions act under a Lorentz transformations like $D^+(\xi')=D^+(\xi)$ and
$\tilde{D}^+(\xi')=\tilde{D}^+(\xi)+\frac{\chi}{2\pi}$ (see 2.24).
Then they compute an arbitrary correlator (see 1) of bosonized Fermi fields $$\psi^{\alpha,\beta}(x)=:\exp(i\alpha\tilde{\phi}+i\beta\phi(x)):$$
$F(x,y)$ acts under Lorentz transformations like $$F(\Lambda x,\Lambda y)=F(x,y)+n\frac{\alpha\beta}{\pi}\chi.$$ So far so good.

Now my questions:

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*They conclude that the field $\psi^{\alpha,\beta}(x)$ transforms as
\begin{align}
U(\Lambda)\psi^{\alpha,\beta}(x)U^{-1}(\Lambda)=e^{\frac{\alpha\beta}{2\pi}\chi}\psi^{\alpha,\beta}(\Lambda x)
\end{align}
Why is that the case? What is $U(\Lambda)$? Some kind of  unitary operator?
Then they go on and call $\frac{\alpha\beta}{2\pi}$ in the exponent the spin $s$. What has this to do with the spin? (I know that in (1+1)d there is no real spin because there are no rotations and spin is a matter of convention.)


*Then they do this same business with dilatations $x \to e^\lambda x$ and find the transformation law
\begin{align}
U(\lambda)\psi^{\alpha,\beta}(x)U^{-1}(\lambda)=e^{\frac{\lambda}{4\pi}(\alpha^2+\beta^2)}\psi^{\alpha,\beta}(e^\lambda x).
\end{align}
and call $\frac{\alpha^2+\beta^2}{4\pi}$ the scale dimension $\Delta_{\alpha,\beta}$. Same questions as above.


*What is the scale dimension $\Delta_{\alpha,\beta}$ of the canonical spin-$\frac{1}{2}$ field? Is it also $\frac{1}{2}$ like the spin? They say that the case $\alpha=\beta=\pm\sqrt{\pi}$ corresponds to a canonical spin $\frac{1}{2}$ field. (I have no prior experience in conformal field thories.)
 A: Okay, let me try to go through the questions and answer what I can:


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*


What is $U(\Lambda)$? Some kind of unitary operator?

Yes! A Lorentz transformation on scalar fields only transforms the coordinate $U(\Lambda) \phi(x) U(\Lambda)^{-1} = \phi(\Lambda x)$. However, for higher-spin fields it also mixes the components of the field. After rotating 180$^\circ$, a spin-up particle might be spin-down after all (see for example Lorentz transformation of the spinor fields).
In this case, there is only one compenent. However, the field can still be scaled by Lorentz boosts and it scales differently for different parameters $\alpha,\beta$. They call $\frac{\alpha\beta}{2\pi}$ the spin of the field, because it specifies the behavior of the field under boosts (as in 3+1 dimensions).

They conclude that the field $\psi^{\alpha,\beta}(x)$ transforms as ... Why?

As I understand it, they deduce this from the transformation behavior of $F(x,y)$ as follows:
Because $U(\Lambda) |0\rangle = |0\rangle$, it holds that
$$
  \langle0| \psi^{\alpha,\beta}(x_1) \cdots \psi^{\alpha,\beta}(x_n) \psi^{\alpha,\beta}(y_1) \cdots \psi^{\alpha,\beta}(y_n)|0\rangle = \langle0| U(\Lambda) \psi^{\alpha,\beta}(x_1) U(\Lambda)^{-1} \cdots U(\Lambda) \psi^{\alpha,\beta}(x_n) U(\Lambda)^{-1} U(\Lambda) \psi^{\alpha,\beta}(y_1) U(\Lambda)^{-1} \cdots U(\Lambda) \psi^{\alpha,\beta}(y_n) U(\Lambda)^{-1} |0\rangle
$$
Assuming that $\psi^{\alpha,\beta}$ scales a certain way under Lorentz transformation, we make the Ansatz
$$
  U(\Lambda) \psi^{\alpha,\beta}(x) U(\Lambda)^{-1} = e^{f_{\alpha,\beta}} \psi^{\alpha,\beta}(\Lambda x).
$$
We want to find the number $f_{\alpha,\beta}$. From the above equation, we have
$$
  e^{F(x_1, .., x_n, y_1, .., y_n)} = e^{2n f_{\alpha,\beta}} e^{F(\Lambda x_1, .., \Lambda x_n, \Lambda y_1, .., \Lambda y_n)} = e^{2n f_{\alpha,\beta}} e^{F(x_1, .., x_n, y_1, .., y_n)} e^{n \frac{\alpha\beta}{\pi} \chi},
$$
where we used the transformation behavior of $F$. From this equality, we can conclude $f_{\alpha,\beta} = -\frac{\alpha\beta}{2\pi}$. I'm not sure where the wrong sign comes from, if you have an idea please let me know.



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Then they do this same business with dilatations $x \mapsto e^\lambda x$ and find the transformation law ...

I'm sure this works similar to before. Again, the transformation of $F$ will be the crucial indicator here. Are they computing it in the book?

What is $U(\lambda)$?

As before, operators that scale the space can act differently on different fields (they might grow stronger). But I am also not well-versed with such things.



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I'll let you know when I find some good introduction conformal scaling for fields of different spin. But I can't say much from the top of my hat.
