I'm a little confused about Hermitian conjugation in a radially quantized CFT. Now, in the Minkowski theory, Hermitian conjugation leaves the coordinates invariant, i.e. $t^\dagger = t$ and $x^\dagger = x$. We then Wick rotate the theory to Euclidean time $t \to - i \tau$ and therefore $\tau^\dagger = - \tau$. Next, we define the complex coordinates on the Euclidean plane $w = x + i \tau$ and ${\bar w} = x - i \tau$. The Hermitian conjugate of acts on this coordinate as $w^\dagger = w$ and ${\bar w}^\dagger = {\bar w}$. On the radial plane with $z = e^{- i w}$ we get $$ z^\dagger = e^{i w^\dagger} = e^{i w} = \frac{1}{z} $$

However, Francesco explicitly says (pg. 152, above equation (6.4)), that $z^\dagger = \frac{1}{z^*}$.

Where am I going wrong? Can anyone explain this?

  • $\begingroup$ In the Google books version at books.google.at/books?id=keUrdME5rhIC&dq, there is no such statement. Maybe it is a different version? I checked myself and end up at the same relation as you. $\endgroup$ Nov 25, 2013 at 8:18
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    $\begingroup$ @FredericBrünner There is. Go to pg. 152 ("The Operator Formalism). Above eq. (6.4) it says "In radial quantization, this corresponds to a mapping $z \to 1/z^*$..." I don't think its a typo because he uses this mapping quite a bit in his calculations after that. $\endgroup$
    – Prahar
    Nov 25, 2013 at 14:21

1 Answer 1


The same result is in Kiritsis, so I don't think it is a typo. Here is what I think is going on. Hermitian conjugation is an operation defined for the operators on the hilbert space. As you said, in Minkowski space, this leaves the coordinates invariant, it does not touch t and x. In other words, if I were instead to use the coordinate $z=t+ix$ in the Minkowski theory, when I hermitian conjugate operators I still do not touch z (it does not go to $t-ix$). As you said, when you Wick rotate $\tau=it$ Hermitian conjugation picks up a geometrical meaning as well: it is time inversion. So in the Euclidean theory, when you Hermitian conjugate, you do a time reversal as well. You do not conjugate the whole coordinate z, you just send $\tau \to -\tau$. Of course in radial quantization $r=e^\tau$ to the time reversal amounts to inversion through the circle, which in complex coordinates is $z\to \frac{z}{zz*}=\frac{1}{z*}$. So I think your step $w^\dagger=w$ is wrong, you want $\dagger:\tau-ix \to -\tau-ix$. This will give you the correct relation.

This is basically related to the fact that in radial quantization, hermitian conjugation is equivalent to inversion. For example $(P^\mu)^\dagger=IP^\mu I=K^\mu$. This is why you need conformal (not just scale invariance) to go to the plane from the cylinder: If you do not have $K^\mu$ you cannot construct a sensible adjoint operation on the plane.

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    $\begingroup$ $(P^\mu)^\dagger=IP^\mu$ I, you say. How come adjoint operation looks like a continuous evolution. Is there a infinitesimal version of this? How to see this intuitively? $\endgroup$
    – levitt
    Mar 16, 2017 at 5:42
  • $\begingroup$ @levitt: $I$ is an inversion, which is a discrete transformation. $\endgroup$
    – Styg
    Jun 7, 2019 at 12:08

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