Running of the Higgs coupling

Why is the coupling constant of the Higgs potential not a constant, but a running constant?

In other words, why is $$\lambda (h) \sim \lambda_{0} - \frac{y_{t}}{16\pi^{2}}\text{ln}\left(\frac{h^{2}}{m^{2}}\right) + \frac{\lambda_{0}^{2}}{16\pi^{2}}\text{ln}\left(\frac{h^{2}}{m^{2}}\right)+\cdots,$$ where $\lambda_{0}$ is the bare coupling and $y_{t}={m_{t}}/{v}$, where $m_{t}$ is the mass of the top quark and $v$ is the Higgs vacuum expectation value?

• Hi! In general, in QFT all coupling constants are running. I don't understand whether you are familiar with this concept and you just want to know about the Higgs in particular or if it is the first time that you hear of a running coupling. – DelCrosB Sep 27 '16 at 8:59
• I would like to know about the Higgs specificallly. – nightmarish Sep 27 '16 at 9:32
• @failexam have you learnt about renormalization and qft? – innisfree Sep 27 '16 at 9:54
• Ok, then why do you expect it not to be running? Loop diagrams arising from the Higgs sector diverge even worse than e.g. QED loops, and so you need renormalization and so on.. I don't thing I am able to give you a complete answer, but if you are familiar with the QFT language this is an excellent reference: sciencedirect.com.ezproxy.its.uu.se/science/article/pii/… – DelCrosB Sep 27 '16 at 9:55
• Have you studied phi-4 theory? It's the canonical QFT. You'll see the quartic running. – innisfree Sep 27 '16 at 11:05