# Why is the relation $M_W=M_Z\cos\theta_W$ true only at tree-level?

In Glashow-Weinberg-Salam electroweak theory, the relation $$M_W=M_Z\cos\theta_W\tag{1}$$ is said to be remain true only at the tree-level; it receives corrections from the loop diagrams. See here. But shouldn't the relation $$(1)$$ be always valid if $$M_W$$ and $$M_Z$$ are defined to be physical masses i.e., bare+loop corrections? Please correct me if I have some erroneous impression.

• Why do you think loop corrections should preserve such a relationship? After all, the $W$ and $Z$ couple very differently to all the other particles. – knzhou May 26 '19 at 15:34
• Edited! @AccidentalFourierTransform – SRS May 26 '19 at 15:39
• Doesn't $M_Z$ and $M_W$ receive corrections? If so, don't we redefine $M_Z+\delta_1$ and $M_Z+\delta_2$ to be $M_Z$ and $M_W$? @knzhou – SRS May 26 '19 at 15:46
• ?? You redefine $M_Z$ and $M_W$ to be the tree value plus corrections, so the 1-loop "physical" masses are different than the tree ones. Veltman and Ross show you the former violate the custodial symmetry present at the tree but not the 1-loop level. What on earth are you asking, given the cottage industry you are invoking? – Cosmas Zachos May 26 '19 at 18:31
• Could you perhaps articulate your "impression" in PDG mainstream contemporary language? – Cosmas Zachos May 26 '19 at 18:50

$$\frac{-ig_{\mu \nu}}{p^2 - M_Z^2 + i\epsilon}, \quad \frac{-ig_{\mu \nu}}{p^2 - (cos\theta_W M_Z)^2 + i\epsilon}$$
• So, if I understand you correctly, the quantities $M_Z$ and $M_W$ are defined by the tree-level relations, once and for all. Also see my comment to knzhou. @Vicky – SRS May 26 '19 at 15:48
• @SRS No, what I'm tealling you is that Proca Lagrangians for $W^{\pm}$ and $Z$ drive you to the relation you are asking for but the masses you see there are 'bare' mass. The real mass of those bosons comes from the complete propagator that is given by corrections such as self-energies. When you compute that corrections, then in the complete propagator you see that to the bare mass is added a term. Now from Källen-Lehmann propagator representation you know that the bare mass plus that addtional term (actually its derivative) (cont.) – Vicky May 26 '19 at 15:55
• gives you the correct the mass. Forget what I said about the energy dependent angle. The masses change, the relation among them too becuse if $M_W$ changes and it is equal to $M_Z cos\theta_W$ and to conserve that relation you would need to assume some relation among $\theta_W$ and the energy: $M_W(p^2) = M_Z(p^2)cos\theta_W(p^2)$ – Vicky May 26 '19 at 16:02