Appendix A Gauge invariance from first principle

In Section 3.1 we wrote down the QED Lagrangian and noted that it follows a certain local U⁢(1)U(1) symmetry. In fact, the logic goes the other way around.

In this appendix, we will derive the complete Lagrangian, including the photon field, from the fact that

ℒ=ψ¯⁢(i⁢γ⋅∂−m)⁢ψ\displaystyle\mathcal{L}=\bar{\psi}(i\gamma\cdot\partial-m)\psi (212)

should be invariant under local gauge transformation

ψ⁢(x)→ei⁢α⁢(x)⁢ψ⁢(x).\displaystyle\psi(x)\to{\rm e}^{i\alpha(x)}\psi(x)\,. (213)

The mass term is obviously invariant but what about the derivative? Due to the gauge symmetry, the derivative ∂μ\partial_{\mu} no longer has any geometric meaning since the phase α⁢(x)\alpha(x) could mess things up. Let us therefore define a new derivate DμD_{\mu} which compares two nearby points along a direction n^μ\hat{n}^{\mu}

n^μ⁢Dμ⁢ψ=limϵ→0ψ⁢(x+ϵ⁢n^)−U⁢(x+ϵ⁢n^,x)⁢ψ⁢(x)ϵ.\displaystyle\hat{n}^{\mu}D_{\mu}\psi=\lim_{\epsilon\to 0}\frac{\psi(x+% \epsilon\hat{n})-U(x+\epsilon\hat{n},x)\psi(x)}{\epsilon}\,. (214)

Here we had to introduce a new object, UU, which for the normal derivative ∂μ\partial_{\mu} is just U=1U=1 but accounts for the change α\alpha. For ℒ\mathcal{L} to be invariant, we need this new derivative to transform like the field itself, i.e.

n^μ⁢Dμ⁢ψ→limϵ→0ei⁢α⁢(x+ϵ⁢n^)⁢ψ⁢(x+ϵ⁢n^)−U′⁢(x+ϵ⁢n^,x)⁢ei⁢α⁢(x)⁢ψ⁢(x)ϵ=!ei⁢α⁢(x)⁢n^μ⁢Dμ⁢ψ.\displaystyle\hat{n}^{\mu}D_{\mu}\psi\to\lim_{\epsilon\to 0}\frac{{\rm e}^{i% \alpha(x+\epsilon\hat{n})}\psi(x+\epsilon\hat{n})-U^{\prime}(x+\epsilon\hat{n}% ,x){\rm e}^{i\alpha(x)}\psi(x)}{\epsilon}\stackrel{{\scriptstyle!}}{{=}}{\rm e% }^{i\alpha(x)}\hat{n}^{\mu}D_{\mu}\psi\,. (215)

The only way to more this work generally is if UU transforms as

U⁢(y,x)→U′⁢(y,x)=ei⁢α⁢(y)⁢U⁢(y,x)⁢e−i⁢α⁢(x).\displaystyle U(y,x)\to U^{\prime}(y,x)={\rm e}^{i\alpha(y)}U(y,x){\rm e}^{-i% \alpha(x)}\,. (216)

Then we have

n^μ⁢Dμ⁢ψ→limϵ→0ei⁢α⁢(x+ϵ⁢n^)⁢ψ⁢(x+ϵ⁢n^)−U⁢(x+ϵ⁢n^,x)⁢ψ⁢(x)ϵ=ei⁢α⁢(x)⁢n^μ⁢Dμ⁢ψ.\displaystyle\hat{n}^{\mu}D_{\mu}\psi\to\lim_{\epsilon\to 0}{\rm e}^{i\alpha(x% +\epsilon\hat{n})}\frac{\psi(x+\epsilon\hat{n})-U(x+\epsilon\hat{n},x)\psi(x)}% {\epsilon}={\rm e}^{i\alpha(x)}\hat{n}^{\mu}D_{\mu}\psi\,. (217)

Taylor-expanding UU gives us with U⁢(x,x)=1U(x,x)=1

U⁢(x+ϵ⁢n^,x)=1−i⁢ϵ⁢n^μ⁢(e⁢Aμ⁢(x))+𝒪⁢(ϵ2).\displaystyle U(x+\epsilon\hat{n},x)=1-i\epsilon\hat{n}^{\mu}(eA_{\mu}(x))+% \mathcal{O}(\epsilon^{2})\,. (218)

Here we had to introduce a field AμA_{\mu} that is the derivative of UU as well as an arbitrary constant ee. It is easy to see that AμA_{\mu} transforms as required and it is no surprise that it will turn into the photon field.

We can now concatenate four comparison operations into a small square

𝒰⁢(x)=U⁢(x,x+ϵ⁢n^)⁢U⁢(x+ϵ⁢n^,x+ϵ⁢n^+ϵ⁢m^)⁢U⁢(x+ϵ⁢n^+ϵ⁢m^,x+ϵ⁢m^)⁢U⁢(x+ϵ⁢m^,x).\displaystyle\mathcal{U}(x)=U(x,x+\epsilon\hat{n})U(x+\epsilon\hat{n},x+% \epsilon\hat{n}+\epsilon\hat{m})U(x+\epsilon\hat{n}+\epsilon\hat{m},x+\epsilon% \hat{m})U(x+\epsilon\hat{m},x)\,. (219)

It is easy to see that 𝒰⁢(x)\mathcal{U}(x) is invariant under the transformation. Starting from

U⁢(x,y)=exp⁡(−i⁢e⁢A⁢(x+y2)⋅(x−y)+𝒪⁢((x−y)3)),\displaystyle U(x,y)=\exp\Big{(}-ieA\big{(}\frac{x+y}{2}\big{)}\cdot(x-y)+% \mathcal{O}\big{(}(x-y)^{3}\big{)}\Big{)}\,, (220)

we find

𝒰⁢(x)\displaystyle\mathcal{U}(x) =exp⁡[−i⁢e⁢ϵ⁢(−n^⋅A⁢(x+n^⁢ϵ2)−m^⋅A⁢(x+n^⁢ϵ+m^⁢ϵ2)+n^⋅A⁢(x+m^⁢ϵ+n^⁢ϵ2)+m^⋅A⁢(x+m^⁢ϵ2))]\displaystyle=\exp\bigg{[}-ie\epsilon\Big{(}-\hat{n}\cdot A\big{(}x+\hat{n}% \tfrac{\epsilon}{2}\big{)}-\hat{m}\cdot A\big{(}x+\hat{n}\epsilon+\hat{m}% \tfrac{\epsilon}{2}\big{)}+\hat{n}\cdot A\big{(}x+\hat{m}\epsilon+\hat{n}% \tfrac{\epsilon}{2}\big{)}+\hat{m}\cdot A\big{(}x+\hat{m}\tfrac{\epsilon}{2}% \big{)}\Big{)}\Bigg{]}
=exp[−ieϵ(∂m^(n^⋅A)−∂n^(m^⋅A))).\displaystyle=\exp\Big{[}-ie\epsilon\big{(}\partial_{\hat{m}}(\hat{n}\cdot A)-% \partial_{\hat{n}}(\hat{m}\cdot A)\big{)}\Big{)}\,. (221)

This proofs that Fμ⁢νF_{\mu\nu} and any functions that depend on Fμ⁢νF_{\mu\nu} are invariant. However, AμA_{\mu} itself is not invariant meaning that a mass term like mγ⁢Aμ⁢Aμm_{\gamma}\ A_{\mu}A^{\mu} would not be allowed. This is the reason that the photon is massless.

You may now wonder about the ZZ and WW bosons. The same argument still applies and we cannot write down a mass for them. In the Standard Model, their masses are dynamically generated through the Higgs mechanism. Basically, the theory contains a scalar field ϕ\phi whose kinetic term includes a covariant deriviative that couples it dynamically to the WW and ZZ bosons. Uniquely among all particles, this field has a non-zero vacuum expectation value (vev) meaning that WW and ZZ get a dynamically generated mass.

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