Consider the massless scalar field Hamiltonian, \begin{align} H = \frac{1}{2}\int \Pi^2- (\partial_x\phi)^2 dx \end{align} with $\Pi \sim \partial_t\phi$ the conjugate field of $\phi$. This Hamiltonian is treated in a number of texts, in particular Shankar's text on Bosonization on page 1841 and the Conformal Field Theory book by Di Francesco, Mathieu and Sénéchal (chapter 6 and 9). Furthermore, my question also relates to this text by Von Delft and Schoeller on bosonization.

Shankar defines the vertex operator as the normal ordered operator $:e^{i\alpha\phi}:$. He states (p1843) that the one-point correlator is unity, $\langle :e^{i\alpha\phi}:\rangle =1$, since the exponential can be expanded and only the first term does not vanish.

But just before that he also states that the two-point correlator $$ \langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle $$ vanishes unless $\alpha + \beta = 0$. This is the neutrality condition and it follows from the fact that $\phi\rightarrow \phi + a$ is a symmetry of the Hamiltonian, which needs to be respected by the (two-point) correlators. That sounds convincing enough, but then why does the one-point correlator not vanish according to that same argument?

To add to (my) confusion the normal ordered form is given by $$ :e^{i\alpha\phi}: = e^{i\alpha\phi_+}e^{i\alpha\phi_-} $$ where $\phi_{\pm}$ contain all creation/annihilation operators. So using some rules of exponentiated operators (see appendix C of Von Delft) one can reorder the operators such that $$ :e^{i\alpha\phi(z_1)}: :e^{i\beta\phi(w_1)}: = f(z-w) :e^{i(\alpha+\beta)\phi(w_1)}: + \cdots $$ with $f(z-w)$ a (singular) c-number. This is of course just the OPE. But that would imply that if the right hand side has a non-vanishing correlator, then the left hand side is non-vanishing as well. If the one-point correlator does not vanish, why does the two-point correlator need to obey the neutrality condition?

Di Francesco has some other, more elaborate proofs using Ward identities (Chapter 9) that show that indeed $\alpha+\beta = 0$ in order for the correlator to be non-zero. In particular that means that, according to their conventions, any $N$-point correlator vanishes unless the neutrality condition is satisfied. It could be that the different texts have different conventions that I'm missing and everything is fine. Still, a different question arises: If the exponentials in the normal ordered operators are expanded, and only the first (unit) term is kept, then surely the one-point correlator is just $1$? Why does this approach fail, according to their conventions?

Let me also mention Von Delft and Schoeller. They also state that the one point correlator of the normal ordered operator $:e^{i\alpha\phi}:$ does not vanish. Instead they define the vertex operator (chapter 9) as $$ V_\alpha = \left(\frac{L}{2\pi}\right)^{-\frac{\lambda^2}{2}}:e^{i\alpha\phi}: $$ with $L$ the system size (periodic boundary conditions). They state that "evidently $\langle V_\alpha\rangle =\delta_{alpha,0} $ in the limit of $L\rightarrow\infty$.". Does the neutrality condition then only hold in the $L\rightarrow\infty$ limit?. This would invalidate Di Fransesco's treatment, I presume.

So, in short, I'm quite confused on how to "merge" these different treatments.

  • $\begingroup$ DiFransesco's treatment is incorrect, the correct answer is Von Delft's and Schoeller, but the reason is the infrared problem with a single scalar, and it probably makes no serious problem for the rest of the book. You should always think of the scalar in path-integral in imaginary time, where the answers to these things are obvious, and the normal ordering is just obvious subtractions of propagators between the same vertex. $\endgroup$
    – Ron Maimon
    Commented May 19, 2012 at 22:51
  • $\begingroup$ @RonMaimon this seems like an answer, why post in the comments? $\endgroup$
    – tmac
    Commented May 19, 2012 at 23:44
  • $\begingroup$ @tmac: Because I didn't check it in detail, nor did I work it out enough to give a detailed explanation (also, I could be wrong about Von-Delft and Schoeller, it just "feels" right because it has to have a dependence on the boundaries, but maybe they screwed up a factor somewhere, I didn't check yet. Also, I need to work things out with normal ordered operator formalism just to check, I don't use this often). I just wanted to give a quick heads up to the OP in case he needs a quick answer, and a longer answer when I know what I'm talking about. $\endgroup$
    – Ron Maimon
    Commented May 20, 2012 at 0:42
  • $\begingroup$ Hi Ron, thanks for your comment. I think you are correct -- Di Fransesco is probably too "sloppy", and the neutrality condition only holds in the $L\rightarrow\infty$ limit, provided the vertex operators are properly normalized. The normal ordered exponential $:e^{i\alpha\ph}:$ without normalization should have a unit vacuum expectation value. $\endgroup$
    – Olaf
    Commented May 23, 2012 at 18:40
  • $\begingroup$ I think the same question is asked here, and the answer seems to be correct and complete physics.stackexchange.com/questions/462925/… $\endgroup$ Commented Mar 3, 2019 at 11:00

1 Answer 1


Well, first of all it must be clearly stated the composite operators of which quantum field theory you consider. I can offer you one setup in which everything is perfectly clear: the compactified free boson living on the cylindrical world-sheet (= toroidal compactification of the closed string). It then has to be said what is meant by normal ordering: for me it will mean putting the annihilation operators of the oscillatory string modes right to the creation operators but not ordering the zero modes (i.e. the position and the momentum of the string). Now the vanishing or non vanishing of correlators like $\langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle$ or $\langle :e^{i\alpha\phi}: \rangle$ can be seen just by performing calculations in the zero mode sector and the result is compatible with the neutrality condition: the 2-point correlator vanishes unless $\alpha+\beta=0$ and 1-point correlator vanishes unless $\alpha=0$. How does it come about? Well, the Hilbert space of the theory $H$ is the tensor product $L^2(S^1)\otimes F_L\otimes F_R$ (I am ignoring the winding zero mode sector for simplicity), where $S^1$ is the target circle on which the string is compactified and $F_{L,R}$ are the Fock spaces for the left and right movers, respectively. In particular, the vacuum $\vert 0\rangle\in H$ can be written as the direct product $\vert 0\rangle_0\otimes \vert 0\rangle_L\otimes \vert 0\rangle_R$, where $\vert 0\rangle_0$ stands for the normalized constant function in $L^2(S^1)$ and $\vert 0\rangle_{L,R}$ are the Fock vacua. The vertex operator $V(\alpha)=:e^{i\alpha\phi}:$ has now the structure $V(\alpha)=e^{i\alpha\phi_0+i\alpha pt}: e^{i\alpha\hat\phi}:$ where the $\hat\phi$ is the oscillator part of the field $\phi$. Let us now calculate
$$\langle :e^{i\alpha\phi}: \rangle= \ _0\langle 0\vert e^{i\alpha\phi_0+i\alpha pt}\vert 0\rangle_0 \ _L\langle 0\vert\otimes\ _R\langle 0\vert : e^{i\alpha\hat\phi}:\vert 0\rangle_R\otimes \vert 0\rangle_L= \ _0\langle 0\vert e^{i\alpha\phi_0 +i\alpha pt }\vert 0\rangle_0 =$$ $$=e^{\frac{i\alpha^2t}{2}}\int_{S^1}e^{i\alpha\phi_0}d\phi_0 = \delta_{\alpha,0}.$$ The calculation of $\langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle$ is slightly more involved but the result is indeed proportional to $\delta_{\alpha+\beta,0}$ because of the contribution from the zero mode sector.


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