Does orbital angular mometum has no meaning for single photons?

1. 
   

   In the quantization of free electromagnetic field, it is found that the left-circularly polarised photons corrsponds to helicity $\vec{S}\cdot\hat p=+\hbar$ and right-circularly polarised photons corrsponds to $\vec{S}\cdot\hat p=-\hbar$. They respectively corresponds to the states $$a^{\dagger}_{\vec k,+}|0\rangle, \hspace{0.2cm}\text{and}\hspace{0.2cm} a^{\dagger}_{\vec k,-}|0\rangle$$ where $$a^{\dagger}_{\vec k,\pm}=\frac{1}{\sqrt{2}}(a^{\dagger}_{\vec k,1}\pm a^{\dagger}_{\vec k,2}).$$ This little calculation is performed in the QFT book by Maggiore by looking at the action of the spin operator $S^{ij}$ on these states. But nothing is mentioned about the orbital angular momentum of individual photons. My question is whether individual photons also carry orbital angular momentum? If yes, what are the values of orbital angular momentum in one-particle states? Can the superposition of two photons have orbital angular momentum? If yes, how to determine its possible values?

   
   


2. 
   

   In Classical Electrodynamics (Ref. J. D. Jackson, 3rd edition, page 350) or Classical field theory, the angular momentum of the electromagnetic field is defined as $$\vec J=\epsilon_0\int d^3x \vec x\times (\vec E\times \vec B)$$which can be reduced to the form $$\vec J=\epsilon_0\int d^3x [\vec E\times \vec A+\sum\limits_{i=1}^{3}E_j(\vec x\times \vec\nabla)A_j).$$ The fist term can be identified with the spin contribution of the angular momentum of the field which has its origin in the spin angular momentum of individual photons. The second term is identified with orbital angular momentum of the field? Is there a quantum mechanical origin to this orbital angular momentum?

   
   


3. 
   

   If there is no meaning to orbital angular momentum of individual photons? Is it only a property of that emerges only when collection of photons builds up a classical field?