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The 1D Ising model at criticality is given by the Hamiltonian $H=-\mathcal{N} \sum_i (\sigma^x_i \sigma^x_{i+1} + \sigma_i^z)$ in terms of Pauli operators and a normalization $\mathcal{N}$. In CFT language, the operator content of this theory is supposed to be described by the primary fields usually called $\mathbb{1}, \epsilon,\psi$ and $\bar{\psi}$ with scaling dimensions $\Delta_\mathbb{1}=0$, $\Delta_\epsilon = 1$ and $\Delta_\psi =\Delta_\bar{\psi}=1/2$. In addition, plenty of papers (such as this one) mention the fields $\sigma$ and $\mu$ with $\Delta_\sigma = \Delta_\mu = 1/8$.

Question: What physical (local) operators/observables do these fields correspond to? My understanding so far:

  • The identity field $\mathbb{1}$ seems to be equivalent to a local identity operator.

  • Is the "energy" $\epsilon$ at site $i$ simply $\sigma^x_i \sigma^x_{i+1} + \sigma_i^z$? The Ising CFT Wikipedia article also mentions a field called $\epsilon^\prime$ with $\Delta_{\epsilon^\prime}=4$, whose meaning is unexplained.

  • The fields $\psi$ and $\bar{\psi}$ seem to correspond to the Majorana modes $c_{2i-1}$ and $c_{2i}$ after Jordan-Wignerizing the Hamiltonian.

  • The "order" $\sigma$ appears to correspond to the Pauli operator $\sigma^x_ i$ at site $i$.

  • I have no idea what the "disorder" $\mu$ corresponds to, but it seems to be an order parameter on the dual lattice.

The only proper textbook I have currently available (Christe/Henkel's "Introduction to Conformal Invariance and Its Application to Critical Phenomena") talks about these fields but doesn't explain $\sigma$ and $\mu$, only mentioning that they can't be written locally in terms of $\epsilon$ and the $\psi,\bar{\psi}$.

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    $\begingroup$ There's a whole chapter about this in the big yellow CFT book. $\endgroup$ – Ryan Thorngren Mar 10 '18 at 22:53
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Lattice-continuum correspondence for the primary fields of the Ising CFT:

  1. Indeed, the identity field $1$ simply corresponds to the identity operator on the lattice. More generally, any lattice operator with a finite expectation value in the ground state has support on the identity field in the continuum.
  2. The energy field $\varepsilon$ corresponds to $\sigma^x_{n} \sigma^x_{n+1} - \sigma^z_n$. Notice the relative minus sign which was not in your formula! This is essential: without the minus sign, the operator would have a non-zero expectation value (of course, since it is the Hamiltonian density), meaning it would (partially) have support on $1$ in the continuum. But with the minus sign it is now odd under the Kramers-Wannier duality. Firstly, this means it is zero at the critical point, but moreover if one adds $g \; \varepsilon$ to the action/Hamiltonian, then depending on the sign of $g$, one lands into the symmetry-broken or paramagnetic phase.
  3. The order field $\sigma$: you are correct that it corresponds to the order parameter $\sigma^x_n$ on the lattice.
  4. The disorder field $\mu$ corresponds to $\sigma^z_1 \sigma^z_2 \cdots \sigma^z_n$ on the lattice. There are a few ways of interpreting this: (a) note that this is the operator that maps to $\sigma^x_n$ under the Kramers-Wannier duality, (b) more physically, this operator creates a domain wall between sites $n$ and $n+1$ (and also near site $1$, but it is convenient to imagine having open boundary conditions there). Hence the fact that the paramagnetic phase has $\langle \mu \rangle = \langle \sigma^z_1 \sigma^z_2 \cdots \sigma^z_n \rangle \neq 0$ is exactly telling us that domain walls have condensed into the ground state. Lastly, (c) from a more general point of view, this non-local operator arises as the string order parameter for the trivial symmetry-protected topological phase. (Hence, if you want to learn more about this latter perspective, just google for `string order parameter'.)
  5. With regards to the fermionic fields $\psi,\bar \psi$, from what I remember they don't exactly correspond to the lattice Majorana operators $\gamma_n$ and $\tilde \gamma_n$, but rather their linear combinations $\gamma_n \pm \tilde \gamma_n$. (One way of seeing this has to be so, is that in the continuum, one gets expressions like $\psi \partial \psi$, yet on the lattice $\gamma$ can only couple to $\tilde \gamma$.)
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  • $\begingroup$ Since you identify $\epsilon$ with $\sigma^x_n \sigma^x_{n+1} - \sigma^z_n = i \tilde{\gamma}_n (\gamma_{n+1} + \gamma_n)$, does that mean we should identify $\psi$ with $\tilde{\gamma}_n$ and $\bar{\psi}$ with $\gamma_{n+1} + \gamma_n$, so $i \psi \bar{\psi}$ gives us a "mass term"? $\endgroup$ – Scytale May 10 '18 at 5:41
  • $\begingroup$ Following this script (page 28), one can transform from physical fermionic operators $c_i$ on a lattice of width $a$ to physical fields $\Psi(x_n) = c_n / \sqrt{a}$. Then the Hamiltonian term proportional to the gap is $c^\dagger_n c_n = 1 + i \gamma_n \tilde{\gamma}_n$, suggesting the identification $\epsilon \sim \gamma_n \tilde{\gamma}_n$ and $\psi, \bar{\psi} \sim c_n, c^\dagger_n$. This seems inconsistent with your reasoning. $\endgroup$ – Scytale May 10 '18 at 14:05
  • $\begingroup$ @Scytale I might have been wrong about the fermionic fields (my point (5)), that's the part I haven't thought about too carefully. But with regards to $\varepsilon$, the reason to prefer $XX-Z$ is that it is odd under Kramers-Wannier, which ensures that its expectation value is zero at the critical point. The suggestion $\varepsilon \sim Z$ (which is what you suggest) is not good because $\langle Z \rangle \neq 0$ at the critical point, which implies that $Z$ has support in the identity field $1$. $\endgroup$ – Ruben Verresen May 10 '18 at 14:22
  • $\begingroup$ In understand your point, but it seems to violate the usual definition $\epsilon = i \psi \bar{\psi}$ where $\psi$ and $\bar{\psi}$ are the continuous version of the fermionic operators on the lattice. Perhaps I am thinking of the wrong continuum limit? $\endgroup$ – Scytale May 10 '18 at 15:46
  • $\begingroup$ @Scytale I don't know why you say that that is the usual definition of $\varepsilon$. Are you confusing the $\varepsilon$-field with the Hamiltonian energy density? $\endgroup$ – Ruben Verresen May 10 '18 at 16:55

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