Following the explanation of radial quantization of Slava Rychkov in link the states are defined on the spheres with which we foliate the space-time and the generator that moves the states from one surface to another is the dilations generator $D$ which plays the role of the Hamiltonian.

He says that the states (for simplicity scalar states under $SO(D)$ rotations) living on the spheres are classified according to their scaling dimension, that is

$$ D |\Delta> \ = i \Delta |\Delta>. $$

Then, in section 3.1.3 the author explains how to construct states living on the sphere by inserting operators inside the sphere.

When we insert an operator at the origin, we generate an eigenstate of the dilations generator $D$, namely

$$ D |\Delta> \ = D\ \Phi_\Delta(0)\ |0>= i \Delta |\Delta>. $$

Then, if we insert an operator $ \Phi_\Delta(x)$ with $x\neq 0$ the resulting state $|\Psi> =\Phi_\Delta(x) |0> $ in not an eigenstate of $D$, whereas it is a superposition of eigenstates. Moreover, when he start to discuss the OPE in Sec. 3.3, above the equation (3.72) he says that $|\Psi>$ lives in the surface of the sphere.

The question is: Since $|\Psi>$ is not an eigenstate of $D$, we cannot assign an eigenvalue to this state. But in this picture the states living on the sphere are labelled by their eigenvalue respect to $D$. It seems there is a contradiction, can anyone clarify the problem?

  • If states with certain eigenvalues are "living on the spheres", then so are linear combinations of them. What's the problem here? – ACuriousMind Feb 16 '17 at 19:08
  • A state which lives on a sphere has definite eigenvalue, i.e. $\Delta$. A linear combination of states with different eigenvalues, e.g. $|\Delta_1>+|\Delta_2>$, obviously is not an eigenstate of $D$ – apt45 Feb 16 '17 at 19:12
  • Sure it's not an eigenvector. But given a space of states in which both $\lvert \Delta_1\rangle$ and $\lvert \Delta_2\rangle$ lie, their linear combination also lies in it since it is a vector space. "Radial quantization" does not somehow magically overrule the basic rules of quantum mechanics. – ACuriousMind Feb 16 '17 at 19:15
  • He says that the states living on the spheres are classified according to their scaling dimension, $\Delta$, that is the eigenvector of $D$. So, does not the definition imply that states on the surface of the sphere are labelled by their eigenvalue respect to D? – apt45 Feb 16 '17 at 19:17
up vote 2 down vote accepted

You are misunderstanding what "the states living on the spheres will be classified according to their scaling dimension" is supposed to mean.

The author just means that, usually, you will use the basis for the space of states on a sphere that diagonalizes the dilation operator. In no way is it intended to imply that linear combinations of the eigenstates lie somehow "outside" of this space - they cannot lie outside, if it is to be a vector space.

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