This question is more a request for explanations. I'm reading now the Di Francesco book in attempt to understand how the free field representations of 2d CFTs are constructed. The first steps in constructing the Wakimoto representation seem natural to me:

enter image description here

What authors say later about $\beta_0$ and $\gamma_0$ does not surprise me as well - indeed, it seems plausible to replace them with Laurent polynomials in order for their zero modes to obey the original Lie algebra, while the polynomials themselves - the corresponding vertex algebra.

However, the part about $p$ and $q$ confuses me a lot - I would never guess these steps myself:

enter image description here

Why do we need the vacuum of the Fock space to be the eigenvector of the operator $p$? Why don't we just start with some $|0\rangle$?

And why does $p$ have to take its values in the weight lattice of $su(2)$?? (confuses me most...)

Basically, why do we need the operators / fields $p$ and $q$ in order to construct the representation? Why not just $\beta$ and $\gamma$, while keeping $j$ a number?

  • 1
    $\begingroup$ Perhaps you wish to contrast to the Segal-Bargmann space. You are instructed to embed spin j modules of su(2) in the infinite dimensional Fock space $1,x,x^2,x^3,....$, the module of the infinite-dimensional Heisenberg Lie algebra, so to find a device to pick a highest state $x^{2j+1}$ in it. In so many words, $|p;0\rangle$ specifies the lowest state, 1, of the specific module with highest state $x^{2j+1}$. If you follow the math, why is the language, sheer renaming, bothering you? $\endgroup$ – Cosmas Zachos Jan 21 '17 at 23:54
  • $\begingroup$ A more tastefully symmetric embedding of su(2) modules in Fock space is the original variant of this, the celebrated 3/4 -century Holstein-Primakoff transformation. (Their s is your j.) $\endgroup$ – Cosmas Zachos Jan 22 '17 at 21:09
  • $\begingroup$ You are calling the space of polynomials in $x$ "the module of the Heisenberg Lie algebra". But in this case I would expect the operators $\hat{x}$ and $\hat{p}$ to be defined in the usual way, $\hat{x} = x \cdot$, $~~\hat{p}=-i\dfrac{\operatorname{d}}{\operatorname{d}x}$. $\endgroup$ – mavzolej Jan 22 '17 at 22:07
  • $\begingroup$ Also, why do we need a special 'device' to fit the spin-$j$ $su(2)$ module into the Heisenberg one? The operators (15.268) are designed in such a way that, besides the correct commutation relations, the condition (15.271) is satisfied automatically, and the monimial $x^{2j+1}$ is annihilated by $f_0$... $\endgroup$ – mavzolej Jan 22 '17 at 22:08
  • $\begingroup$ NO, don't expect the hatted new variables to have anything to do with the x variable of γ, β. Baaad language! The hatted variables are a gimmick to read off and project into the weight/spin lattice of the js, is all... Yes, the structure is self-explanatory, except for an ugly nonhermiticity twist, and it is just the Holstein-Primakoff map (above) you probably learned about in college, in pompous mathematese.... $\endgroup$ – Cosmas Zachos Jan 22 '17 at 22:14

Your Answer

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

Browse other questions tagged or ask your own question.