Introduction to Operator Product Expansion and more

  1. 5周前
    4周前Turgon 重新编辑

    The topic of this thread is about Operator Product Expansion (OPE) and related stuffs. I'll briefly introduce OPE with a simple $\phi^4$ theory, emphasize the importance of large momentum expansion and take 4-fermion theory as an example to prove this point, maybe talk a little about complications in CFT, applications in curved spacetime (quantization, etc.) or DIS and QCD sum rule, even quasi PDF (if I'm ever gonna master them, that is). Nowadays people tend to use EFT in place of OPE, but somehow I believe there're subtle diffenerces.

    History of OPE
    Well, Kim Wilson invented it in 1969, shortly afterwards Zimmermann proved it mathematically in a lecture in 1970. HEP physicists adopted it to prove the factorization theorem of DIS experiment, and it was a triumph. Then it became famous. There're a lot of applications ever since, but I won't make a list here.

    What is OPE?
    What is OPE? In short, it's Laurent expansion. We can have a general expression as following:
    $$T\phi(x)\prod_i\phi(x_i)\approx\sum_iC_i(x_i^\mu)[\mathcal{O}_i(x^\mu)]$$
    where $[\mathcal{O}_i(x)]$ is renormalized local/composite operator and $C_i$ is often called Wilson coefficient.

    $\phi^4$ example and tree level contributions
    Let's take $\phi^4$ theory (3+1 dimension) as an example:
    $$\mathcal{L}=(\partial_\mu\phi)^2/2-m^2\phi^2/2-g\phi^4/24+\text{counterterms}.$$
    We can start with OPE btw. two operators and put them into a 4-point Green function so we can actually do some calculation:
    $$\langle 0|T\phi(x)\phi(0)\tilde\phi(p_1)\tilde\phi(p_2)|0\rangle\approx\sum_iC_i(x^\mu)\langle 0|[\mathcal{O}_i(0)\tilde\phi(p_1)\tilde\phi(p_2)]|0\rangle$$
    and we can define its momentum space counterpart
    $$\newcommand{\dd}{\text{d}}T\tilde\phi(q)\phi(0)=\int\dd^4 xe^{iq\cdot x}T\phi(x)\phi(0)\approx\sum_i\tilde C_i(q^\mu)[\mathcal{O}_i(0)].$$
    (We use tilde to denote momentum space quantities and square bracket to denote renormalized quantities. )

    Now we can compute the tree diagrams of the Green function:

    QQ截图20190113182534.png

    we have
    $$\frac{i}{p_1^2-m^2}\frac{i}{p_2^2-m^2}\big[e^{-ip_1\cdot x}+e^{-ip_2\cdot x}\big]$$
    Expand it around $x=0$ we have
    $$\frac{i}{p_1^2-m^2}\frac{i}{p_2^2-m^2}[2-i(p_1+p_2)\cdot x-(p_1\cdot x)^2/2-(p_2\cdot x)^2/2+\cdots]$$
    These are exactly the OPE relation in coordinate space with lowest order Wilson coefficients. This is normally how we derive the basic forms of $\mathcal{O}_i$s.

    The OPE relation is just
    $$T\phi(x)\phi(0)=\phi^2(0)+\frac{1}{2}x^\mu\partial\mu\phi^2+\frac{1}{2}x^\mu x^\nu\phi\partial\mu\partial\nu\phi+\cdots$$
    where all $\phi$ fields appear in the r.h.s. are evaluated at $x=0$. The result is finite, so there's no need to involve renormalization label.

    It's easy to prove that this relation is an operator relation thus independent of external states. One may start with a 6-point Green function ($\phi(x)\phi(0)$ and 4 external legs) and check if this relation is still valid.

    From the look of it, OPE is trivially Taylor expansion. However, one must note that we haven't compute loop corrections yet. So for classical scenarios, Taylor expansion might be enough, but there're quantum fluctuations we must take into account.

  2. 4周前Turgon 重新编辑

    One Loop Corrections

    In last section we derived that
    $$T\phi(x)\phi(0)=\phi^2(0)+\cdots$$
    Now that's assume there're loop corrections to the coefficient of $\phi^2$. We're not sure whether the correction is finite or not, by experience it's more likely to be divergent. To our prior knowledge, $T\phi(x)\phi(0)$ gives finite result when $x$ is not zero (all divergence will be suppressed by the exponential factor). Thus we can define a renormalized operator $[\phi^2]$ and construct
    $$T\phi(x)\phi(0)=C(x)[\phi^2(0)]+\cdots.$$
    This way, $C(x)$ is also well-defined. We already know the leading order of $C$, so
    $$C(x)=1+\frac{g}{16\pi}c(x)$$
    The factor is purely set for convenience.

    Consider the following diagram (I hope I didn't mess up with some signs or factors, nor J. Collins):

    QQ截图20190120015454.png

    $$\frac{i}{p_1^2-m^2}\frac{i}{p_2^2-m^2}ig\int\frac{\dd^4q}{(2\pi)^4}\frac{e^{-iq\cdot x}}{(q^2-m^2)((q-p_1-p_2)^2-m^2)}$$

    T.B.D

 

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