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From the commutation relations for the conformal Lie algebra, we may infer that the dilation operator plays the same role as the Hamiltonian in CFTs. The appropriate commutation relations are

$[D,P_{\mu}] = iP_{\mu}$ and $[D,K_{\mu}] = -iK_{\mu}$,

so that $P_{\mu}$ and $K_{\mu}$ are raising and lowering operators, respectively, for the operator $D$.

This is analogous to the operators $\hat a$ and $\hat a^{\dagger}$ being creation and annihilation operators for $\hat H$ when discussing the energy spectra of the $n$ dimensional harmonic oscillator.

My question is, while $\hat a$ and $\hat a^{\dagger}$ raise and lower the energy by one unit $( \pm \hbar \omega)$ for each application of the operator onto eigenstates of $\hat H$, what is being raised and lowered when we apply $P_{\mu}$ and $K_{\mu}$ onto the eigenvectors of $D$?

Secondly, what exactly do we mean by the eigenvectors of $D$? Are they fields in space-time?

Using the notation of Di Francesco in his book 'Conformal Field Theory', the fields transform under a dilation like $F(\Phi(x)) = \lambda^{-\Delta}\Phi(x)$, where $\lambda$ is the scale of the coordinates and $\Delta$ is the scaling dimension of the fields.

Can I write $F(\Phi(x)) = D\Phi(x) = \lambda^{-\Delta}\Phi(x)$ to make the eigenvalue equation manifest?

Thanks for clarity.

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  • $\begingroup$ 2D CFT or general CFT? Also, are $P_\mu$,$D$ and $K_\mu$ what one would usually call $L_{-1},L_0,L_1$ in the Virasoro algebra? $\endgroup$ – ACuriousMind Jul 26 '14 at 17:58
  • $\begingroup$ Hi ACuriousMind, hmm, I am yet to study 2D CFT (but I know it is a special dimension as far as CFT's go) or the Virasoro algebra. I am using $P_{\mu}, D$ and $K_{\mu}$ to mean, respectively, the translation, dilation and special conformal generators of the infinitesimal transformations. $\endgroup$ – CAF Jul 26 '14 at 18:02
  • $\begingroup$ @ACuriousMind yes you're right, the $L_{-1},L_0,L_1$ in the Virasoro algebra are the corresponding conformal generators in 2D CFT. $\endgroup$ – zzz Jul 27 '14 at 15:15
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    $\begingroup$ As exlained here, $\bar{L}_{-1}$ and $\bar{L}_{0}$ and $\bar{L}_{1}$ have to be included to obtain the full 2D conformal group. $\endgroup$ – Dilaton Jul 29 '14 at 11:33
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The commutation relations $$ [D,P_{\mu}] = +i P_{\mu} , \qquad [D,K_{\mu}] = -i K_{\mu} $$ show that $P_{\mu}$ and $K_{\mu}$ raise and lower the conformal dimension of a state. In other words, if you have a state $|\phi\rangle$ of dimensions $\Delta$, so that $D\, |\phi\rangle = i\Delta |\phi\rangle$, then $$ D \, P_{\mu} \, |\phi\rangle = [D,P_{\mu}]\, |\phi\rangle + P_{\mu}\,D\,|\phi\rangle = i(\Delta + 1) \, P_{\mu} \, |\phi\rangle . \tag{1} $$ While the generators of the conformal group act on the fields (after all they generate a symmetry), I find it easier to think about the action on a state, like the $|\phi\rangle>$ above. According to the state operator correspondence (see for example this question), such states can be obtained by acting by local operators on the vacuum. $P_{\mu}$ is the generator of translations, and hence acts on a local operator $\phi(x)$ as a derivative $$ [ P_{\mu} , \phi(x) ] = i \partial_{\mu} \phi(x) . $$ Equation $(1)$ then tells you that that the derivative carries conformal dimension $1$.

The notation you propose at the end of your question seems a bit dangerous. $D$ is frequently used for denoting the infinitesimal generator of dilatations, but the function $F$ gives the action of the corresponding group element. That being said, you are of course free to introduce whatever notation you find convenient, as long as you make it clear -- both to yourself and to others -- what you mean.

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  • $\begingroup$ Hi Olof, many thanks for reply. Should the eqn (1) not be $$D(P_{\mu}|\phi\rangle) = [D,P_{\mu}]|\phi\rangle + P_{\mu}D|\phi\rangle = i(\Delta + 1)(P_{\mu}|\phi\rangle)?$$ Do you mean to say that $|\phi\rangle$ is the ket associated with the local operator/field $\phi(x)$ in the same way that $|n\rangle \equiv \langle x | u_n\rangle = u_n(x)$ when discussing the harmonic oscillator? $\endgroup$ – CAF Jul 27 '14 at 8:25
  • $\begingroup$ @CAF: Yes you are right, there was a typo in (1). The state $|\phi\rangle$ is any state of dimension $\Delta$, but the easiest case to think about is a state created by acting with the local operator $\phi(x)$. $\endgroup$ – Olof Jul 27 '14 at 8:32
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The answer to both questions is that D act on Hilbert space states. I'll answer them in reverse order.

what exactly do we mean by the eigenvectors of D? Are they fields in space-time?

No, in this context, eigenvectors of D are states living in the Hilbert space of the field theory. Because it is only in this sense that the commutation relations between $D$,$P_\mu$ and $K_\mu$ tell us that $P_\mu$ and $K_\mu$ raise and lower eigenstates of D.

To see this, consider, to the contrary, that the eigenvectors of D are fields. When we write down $D\phi$ where $\phi$ is a field in the CFT, $D$ and $\phi$ are both linear operators acting on the space of states. If we insist that $\phi$ is an eigenvector of $D$ in the sense that $D\phi=E\phi$ where E is some scalar, D has to be a multiple of the identity, and hence will not have a discrete eigenvalue spectrum that can be lowered or raised.

what is being raised and lowered when we apply Pμ and Kμ onto the eigenvectors of D

$P_\mu$ and $K_\mu$ are lower and raise eigenstates in the exact same sense that $a$ and $a^\dagger$ lower and raise eigenstates in say, the harmonic oscillator.

Assuming that $D$ has a discrete spectrum, we can define the state $|E\rangle$ to be the state with eigenvalue $E$: $D |E\rangle=E|E\rangle$. $P_\mu$ is a raising operator in the sense that $P_\mu |E\rangle= |E+\hbar\omega \rangle$. These follow directly from the commutation relations.

These notes by Jared Kaplan gives a good discussion of these issues, in particular look at the discussion leading up to eq. 3.31.

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