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Standard Model: Problem with Masses of Elementary Particles?

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?

One Answer

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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