5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Prahar Feb 14 '14 at 13:01
  • $\begingroup$ I guess I am just confused about what exactly should $T(z)\partial_z \phi$ be equal to using Wick's theorem. $\endgroup$ – huyichen Feb 14 '14 at 16:41
  • $\begingroup$ @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? $\endgroup$ – user7757 May 27 '14 at 18:22
  • $\begingroup$ 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. $\endgroup$ – Prahar May 27 '14 at 18:26
  • $\begingroup$ @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. $\endgroup$ – user7757 May 27 '14 at 18:33
1
$\begingroup$

Normal-ordered non-singular terms in the OPE are still there in principle, but they are sometimes omitted (and therefore only implicitly implied) in the notation. This is because many important physical quantities only depend on the singular terms of the OPE.

By the way, speaking of implicitly implied things, be aware that many authors don't write the radial ordering symbol $\cal R$ explicitly.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user7757 May 27 '14 at 18:21
  • $\begingroup$ 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: $\mathcal{T}\partial_{z}\phi(z)\partial_{z}\phi(z)\partial_{w}\phi(w)=:\partial_{z}\phi(z)\partial_{z}\phi(z):\partial_{w}\phi(w)$ $\endgroup$ – user7757 May 27 '14 at 18:28
  • $\begingroup$ OP's relation is an operator identity which is not (yet) applied to a ket vector. Normal-ordered operators are in general not zero. $\endgroup$ – Qmechanic May 27 '14 at 19:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.