On spontaneous symmetry breaking and Higgs’s mechanism of mass production

1. Mar 27, 2008

agkyriak

On spontaneous symmetry breaking and Higgs’s mechanism of mass production

From lectures: L. Peak and K. Varvell. The Physics of the Standard Model.

Full Lagrangian for fermion and photon

Combine the gauge-invariant Lagrangian density describing a fermion field in the presence of an electromagnetic field with that for the EM field itself
$$\begin{array}{l} L=\bar {\psi }\left[ {\gamma ^\mu \left( {i\partial _\mu -qA_\mu } \right)-m} \right]\psi -\frac{1}{4}F_{\mu \nu } F^{\mu \nu }-j^\mu A_\mu = \\ =\bar {\psi }\left[ {\gamma ^\mu i\partial _\mu -m} \right]\psi -\frac{1}{4}F_{\mu \nu } F^{\mu \nu }-\left( {j^\mu +q\bar {\psi }\gamma ^\mu \psi } \right)A_\mu \\ \end{array}$$
Note that the term coupling to the photon field $$A_\mu$$consists of two parts:
1) The external current density $$j^\mu$$
2) A term corresponding to the fermion field itself $$q\bar {\psi }\gamma ^\mu \psi$$. This is called the electromagnetic current (think flow of the fermion charge) and when coupled to $$A_\mu$$ describes the interaction vertex.

Last edited: Mar 27, 2008
2. Mar 27, 2008

agkyriak

Massive photons?

What would a mass term for the photon look like in the Lagrangian density?
We can use an analogy with the Klein-Gordon case
$$L=\frac{1}{2}\left[ {\partial _\mu \phi \partial ^\mu \phi -m^2\phi ^2} \right]$$
where the mass of the scalar field $$\phi$$ comes in the term $$m^2\phi ^2$$.
Perhaps for the (vector) photon field $$A^\mu$$ we could introduce a term $$m^2A_\mu A^\mu$$.

Unfortunately this is not gauge invariant, since

$$\begin{array}{l} m^2A'_\mu A'^\mu =m^2\left( {A+\partial _\mu \chi } \right)\left( {A+\partial ^\mu \chi } \right)= \\ m^2\left( {A_\mu A^\mu +\left( {\partial _\mu \chi } \right)A^\mu +\left( {\partial ^\mu \chi } \right)A_\mu +\partial _\mu \chi \partial ^\mu \chi } \right)\ne m^2A_\mu A^\mu \\ \end{array}$$

The term containing $$\partial _\mu \chi \partial ^\mu \chi$$ is harmless (it does not contribute to the equations of motion) but the terms linear in $$A^\mu$$do. This is not a problem for electromagnet ism since the photon is massless, but it will be a problem for the weak interaction.

3. Mar 27, 2008

agkyriak

Spontaneous symmetry breaking

Suppose we generalise the Klein-Gordon Lagrangian density to the case where the scalar field $$\Phi$$ is complex, or equivalently is a pair of real scalar fields $$\phi _1$$ and $$\phi _2$$ such that

$$\Phi =\frac{1}{\sqrt 2 }\left( {\phi _1 +i\phi _2 } \right) \quad \Phi ^+=\frac{1}{\sqrt 2 }\left( {\phi _1 -i\phi _2 } \right)$$ or
$$\Phi =\left| \Phi \right|e^{i\theta } \quad \Phi ^+=\left| \Phi \right|e^{-i\theta }$$
The Lagrangian density would then be
$$L\left( {\Phi ,\Phi ^+} \right)=\partial _\mu \Phi ^+\partial ^\mu \Phi -m^2\Phi ^+\Phi .$$
We can take $$\Phi$$ and $$\Phi ^+$$ a.s independent, just as easily as $$\phi _1$$ and $$\phi _2$$.

The state of lowest energy of a system is known as the ground state, or in field theory terminology, the vacuum.

In the above Lagrangian the potential energy density $$V\left( {\Phi ,\Phi ^+} \right)=m^2\Phi ^+\Phi$$ is a minimum when $$\phi _1 =\phi _2 =0$$
But suppose we modify it to ($$\phi _0$$ a real constant)
$$\begin{array}{l} V\left( {\Phi ,\Phi ^+} \right)=\frac{m^2}{2\phi _0^2 }\left[ {\Phi ^+\Phi -\phi _0^2 } \right]^2=V\left( {\Phi ,\Phi ^+} \right)= \\ =\frac{m^2}{2\phi _0^2 }\left( {\Phi ^+\Phi } \right)^2-m^2\Phi ^+\Phi +\frac{1}{2}m^2\phi _0^2 \\ \end{array}$$
The vacuum state $$V\left( {\Phi ,\Phi ^+} \right)=0)$$ now occurs when
$$\left| \Phi \right|=\phi _0$$, which defines a circle in $$\left( {\phi _{1,} \phi _2 } \right)$$ space, i.e. there are an infinity of vacuums. The Lagrangian density has a U(1) symmetry.

Nature chooses one of these as the physical vacuum and "breaks'" this symmetry. This phenomenon is known as spontaneous symmetry breaking.

How does spontaneous symmetry breaking help? Suppose we expand the field $$\Phi$$ around the chosen vacuum state, by writing
$$\Phi =\phi _0 +\frac{1}{\sqrt 2 }\left( {\chi +i\psi } \right)$$
Substituting into
$$L\left( {\Phi ,\Phi ^+} \right)=\partial _\mu \Phi ^+\partial ^\mu \Phi -\frac{m^2}{2\phi _0^2 }\left[ {\Phi ^+\Phi -\phi _0^2 } \right]^2$$
and doing the algebra, the Lagrangian density now becomes
$$L\left( {\Phi ,\Phi ^+} \right)=\partial _\mu \chi ^+\partial ^\mu \chi +\partial _\mu \psi ^+\partial ^\mu \psi -\frac{m^2}{2\phi _0^2 }\left[ {\sqrt 2 \phi _0 \chi +\frac{\chi ^2}{2}+\frac{\psi ^2}{2}} \right]^2$$
Pick out the "free particle" pieces by writing
$$L=L_{free} +L_{int}$$
we have
$$\begin{array}{l} L_{free} =\frac{1}{2}\partial _\mu \chi ^+\partial ^\mu \chi +\frac{1}{2}\partial _\mu \psi ^+\partial ^\mu \psi -m^2\chi ^2 \\ L_{int} =-\frac{\sqrt 2 \chi }{\phi _0 }\left( {\frac{\chi ^2}{2}+\frac{\psi ^2}{2}} \right)-\frac{m^2}{2\phi _0^2 }\left( {\frac{\chi ^2}{2}+\frac{\psi ^2}{2}} \right)^2 \\ \end{array}$$
$$L_{int} ,$$ is a complicated ''self" interaction amongst the fields, which we will leave aside.

We can interpret
$$L_{free} =\frac{1}{2}\partial _\mu \chi ^+\partial ^\mu \chi +\frac{1}{2}\partial _\mu \psi ^+\partial ^\mu \psi -m^2\chi ^2$$
by comparing with the Klein-Gordon Lagrangian density
$$L=\frac{1}{2}\left[ {\partial _\mu \phi \partial ^\mu \phi -m^2\phi ^2} \right]$$
We can see that we have a massive, spinless scalar boson field $$\chi$$ of mass $$\sqrt 2 m$$. This is called a Higgs boson.

A massless, spinless scalar boson field $$\psi$$. This is called a Goldstone boson.

The Higgs boson is like a fluctuation around the vacuum point in the direction in which the potential density increases. The Goldstone boson is like a fluctuation in the direction in winch the potential density is flat.

At this point, we seem to have introduced new fields into our toy theory and not gained a lot. However, the full theory mast be locally gauge invariant, which is not yet the case.

For local gauge invariance we require invariance under
$$\Phi \left( x \right)\to \Phi ^'\left( x \right)=e^{-iq\theta \left( x \right)}\Phi \left( x \right)$$
and the introduction of a gauge field $$A_\mu$$, transforming as
$$A_\mu \left( x \right)\to A_\mu ^' \left( x \right)=A_\mu \left( x \right)+\partial _\mu \theta \left( x \right)$$
with the Lagrangian looking like

$$\begin{array}{l} L\left( {\Phi ,\Phi ^+} \right)=\left[ {\left( {\partial _\mu -iqA_\mu } \right)\Phi ^+} \right]\left[ {\left( {\partial ^\mu +iqA^\mu } \right)\Phi } \right]-\frac{1}{4}F_{\mu \nu } F^{\mu \nu }- \\ -\frac{m^2}{2\phi _0^2 }\left[ {\Phi ^+\Phi -\phi _0^2 } \right]^2 \\ \end{array}$$

where as before
$$F^{\mu \nu }=\partial ^\mu A^\nu -\partial ^\nu A^\mu$$

Again the vacuum state is when $$\left| {\Phi \left( x \right)} \right|=\phi _0$$, and since $$\theta \left( x \right)$$ is arbitrary, we can choose it so that $$\Phi \left( x \right)$$ is real, breaking the symmetry.

4. Mar 27, 2008

agkyriak

Higgs’s mechanism of mass production

Proceeding as before, we expand about the chosen vacuum, writing
$$\Phi \left( x \right)=\phi _0 +\frac{h\left( x \right)}{\sqrt 2 }$$
with $$h\left( x \right)$$ real.

Substitution into the Lagrangian density now gives
$$\begin{array}{l} L\left( {\Phi ,\Phi ^+} \right)=\left[ {\left( {\partial _\mu -iqA_\mu } \right)\left( {\phi _0 +\frac{h\left( x \right)}{\sqrt 2 }} \right)} \right] \left[ {\left( {\partial ^\mu +iqA^\mu } \right)\left( {\phi _0 +\frac{h\left( x \right)}{\sqrt 2 }} \right)} \right]- \\ -\frac{1}{4}F_{\mu \nu } F^{\mu \nu }-\frac{m^2}{2\phi _0^2 }\left[ {\sqrt 2 \phi _0 h+\frac{1}{2}h^2} \right]^2 \\ \end{array}$$
and we can again write $$L=L_{free} +L_{int}$$

Rearranging terms
$$\begin{array}{l} L_{free} =\frac{1}{2}\partial _\mu h\partial ^\mu h-m^2h^2-\frac{1}{4}F_{\mu \nu } F^{\mu \nu }+q^2\phi _0^2 A_\mu A^\mu \\ L_{int} =q^2A_\mu A^\mu \left( {\sqrt 2 \phi _0 h+\frac{1}{2}h^2} \right)-\frac{m^2h^2}{2\phi _0^2 }\left( {\sqrt 2 \phi _0 h+\frac{1}{4}h^2} \right) \\ \end{array}$$

How to interpret $$L_{free}$$ now?

We still have a Higgs boson (now denoted by $$h)$$ of mass $$\sqrt 2 m$$.

There is now a gauge boson $$A_\mu$$ with a mass! (the term $$q^2\phi _0^2 A_\mu A^\mu$$ represents a field with mass $$\sqrt 2 q\phi _0 )$$.

There is no longer Goldstone boson (it has been "eaten" by the gauge field to give it a mass).

Spontaneous symmetry breaking has introduced a way of giving mass to the gauge boson of the theory, at the expense of introducing a new scalar particle (the Higgs boson).

The theory turns out to remain renormalizable (calculations give sensible results) following spontaneous symmetry breaking.

Last edited: Mar 27, 2008
5. Mar 27, 2008

humanino

Free lecture ?
Can you prove that ?

6. Mar 27, 2008

agkyriak

Proving normalization for the full Weinberg-Salam theory, t'Hooft and Veltman won the 1999 Nobel Prize.

If interested, please, see for example
http://www.slac.stanford.edu/library/nobel/nobel1999.html [Broken]

Last edited by a moderator: May 3, 2017
7. Mar 27, 2008

humanino

Well, since you were lecturing us, I thought maybe you could outline the proof here. That would be interesting.

8. Mar 27, 2008

BenTheMan

Hmmm I suspect a typo... (I thnk you're missing a $$\Phi$$ in the kinetic term).

9. Mar 27, 2008

Haelfix

The proof in full generality is lengthy and wonderfully unlovely. Don't be mean!

10. Mar 27, 2008

humanino

I did not want to be mean, I apologize if I seemed mean . I really would greatly appreciate such an outline, because everytime I attacked myself to this subject I "lost sight of the forest for the trees". There are several general proofs for instance, some of them applying to non-gauge QFTs. If you take Collins' CUP "Renormalization", it is full of examples and extremely useful when you want to learn the technics as a student, but if you want the general proof, it is barely outlined (and not obvious to me that it is straightforward ).

11. Mar 28, 2008

agkyriak

Unfortunately, that would be very complex LaTex text.
I think it would be more interesting, if I told about some interpretation of the string theory, in which the generation of mass does not require renormalization.

12. Mar 28, 2008

agkyriak

You are right. Thanks. I have already corrected.

13. Mar 28, 2008

blechman

Have you checked out Peskin&Schroder's chapter on R$_\xi$ gauges? I thought it was actually pretty well-done.