Wick's theorem for calculating OPE

I am trying to understand a calculation using Wick's theorem. Let $T(z)$ be the analytic part of a stress-energy tensor, and $\phi(z)$ a free boson field.

Now, $$T(z)\partial_{w}\phi(w)=-2\pi:\partial_{z}\phi(z)\partial_{z}\phi(z):\partial_{w}\phi(w).$$ Using Wick's theorem, we know that $$:\partial_{z}\phi(z)\partial_{z}\phi(z):\partial_{w}\phi(w)=:\partial_{z}\phi(z)\partial_{z}\phi(z)\partial_{w}\phi(w):+2\langle \partial_{z}\phi(z)\partial_{w}\phi(w)\rangle :\partial_{z}\phi(z):$$. Then why is this just equal to $$2\langle \partial_{z}\phi(z)\partial_{w}\phi(w)\rangle \partial_{z}\phi(z)?$$ as stated in many CFT books?

• Just to be clear, the CFT books do not quite say that. What they say is $T(z) \partial_z \phi \sim - 2 \pi \langle \partial_z \phi \partial_z \phi \rangle \partial_z \phi$. The crucial thing is $\sim$. $\sim$ is different from $=$ upto non-singular terms as @Qmechanic has explained. – Prahar Feb 14 '14 at 13:01
• I guess I am just confused about what exactly should $T(z)\partial_z \phi$ be equal to using Wick's theorem. – huyichen Feb 14 '14 at 16:41
• @Prahar: Isn't the normal ordered term zero, because in OPE's we implicitly assume it is the vacuum expectation value which is zero, or is it because the term is non-singular. How can you say it is non singular? – user7757 May 27 '14 at 18:22
• The conformal normal-ordered term is finite only under vacuum expectation value. As an operator, it is not zero. It is non-singular by definition - conformal ordering is defined by taking an operator product and subtracting out all the singularities. – Prahar May 27 '14 at 18:26
• @Prahar: I think you misunderstood me. I am talking about the $:\partial_{z}\phi(z)\partial_{z}\phi(z)\partial_{w}\phi(w):$ which we has been omitted in the expression. Isn't this 0? By 'conformal ordering' I guess you are talking about the expression above that. – user7757 May 27 '14 at 18:33

By the way, speaking of implicitly implied things, be aware that many authors don't write the radial ordering symbol $\cal R$ explicitly.
• Also, correct me if I am wrong. The LHS of wick's theorem has time ordering, or radial ordering whatever is the case, and in the above expression: $\mathcal{T}\partial_{z}\phi(z)\partial_{z}\phi(z)\partial_{w}\phi(w)=:\partial_{z}\phi(z)\partial_{z}\phi(z):\partial_{w}\phi(w)$ – user7757 May 27 '14 at 18:28