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.

  • 1
    $\begingroup$ 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, 2014 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, 2014 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, 2014 at 15:15
  • 1
    $\begingroup$ 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, 2014 at 11:33

2 Answers 2


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.

  • $\begingroup$ 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, 2014 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, 2014 at 8:32

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.