Physics Asked on June 2, 2021
In his book "Modern Particle Physics", Mark Thomson explains two problems with masses of elementary particles in the SM:
(i) If we take the QED Lagrangian $mathcal L = bar{psi}left( igamma^{mu}D_{mu} – mright)psi – frac{1}{4}F_{munu}F^{munu}$, where $D_{mu} = partial_{mu} + iqA_{mu}$ is the covariant derivative. Introducing a mass term of the form $frac{1}{2}m^2_{gamma}A_{mu}A^{mu}$ would break the required invariance under $U(1)$. This I understand so far.
(ii) On page 469, in Chapter 17.4, he writes:
The problem with particle masses is not restricted to the gauge bosons. Writing the electron spinor field as $psi = e$, the electron mass term in QED Lagrangian can be writtern in terms of the chiral particle states as $$-m_{e}bar{e}e = -m_{e}bar{e}left[ frac{1}{2}left( 1-gamma^{5}right) + frac{1}{2}left( 1+gamma^{5}right) right]e$$
Short comment from my side: On page 142, he wrote that:
[…] any spinor $u$ can be decomposed into left- and right-handed chiral components with
$$ u = frac{1}{2}left( 1+gamma^{5}right)u + frac{1}{2}left( 1-gamma^{5}right)u. $$
Okay, continuing with the calculation on p. 469:
$$-m_{e}bar{e}e = […] = -m_{e}bar{e}left[ frac{1}{2}left( 1-gamma^{5}right)e_{L} + frac{1}{2}left( 1+gamma^{5}right)e_{R} right] = -m_{e}left( bar{e}_{R}e_{L} + bar{e}_{L}e_{R} right) quad (17.16)$$
In the SU(2)$_{L}$ gauge transformation of the weak interaction, left-handed particles transform as weak isospin doublets and right-handed particles as singlets, and therefore the mass term of $(17.16)$ breaks the required gauge invariance.
Question:
How can we understand his last sentence?
If we take the mass term $-m_{e}bar{e}e$ and apply an SU(2)$_{L}$ gauge transformation $U$, we get: $$Uleft( -m_{e}bar{e}eright) = -m_{e}Ubar{e}e overset{(17.16)}{=} -m_{e}Uleft( bar{e}_{R}e_{L} + bar{e}_{L}e_{R} right) = -m_{e}left{ left(Ubar{e}_{R}right)e_{L} + bar{e}_{R}left(Ue_{L}right) + left(Ubar{e}_{L}right)e_{R} + bar{e}_{L}left(Ue_{R}right)right}.$$ Now, we know that the weak charged-current interaction only couples to left-handed chiral particle states and right-handed chiral anti-particle states. Thus, the last two terms have to vanish, and we get: $$-m_{e}Ubar{e}e = -m_{e}left{ left(Ubar{e}_{R}right)e_{L} + bar{e}_{R}left(Ue_{L}right) right} overset{!}{=} -m_{e}bar{e}e overset{(17.16)}{=} -m_{e}left( bar{e}_{R}e_{L} + bar{e}_{L}e_{R} right)$$ Carefully comparing the last equation, we see that gauge invariance is not given, and we thus need the Higgs mechanism.
Answered by user248824 on June 2, 2021
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