Looking at the history of the weak force from a more inevitable viewpoint
The Atlantic published an article (plus an audio interview) trying to explain why some physicists such as Stephen Hawking made bets against the Higgs boson. I haven't really understood the answer.
The story started in Jáchymov/Joachimstahl, a Sudeten Czech town with lots of pitchblende. Marie Sklodowska ordered some of this stuff. (In the 1950s, the mines in the local uranium mines would host lots of political prisoners of Stalinism. Four centuries earlier, the town minted silver thalers/tolars – which the dollars were later named after.)
Instead, I was offered an explanation why Lawrence Krauss was skeptical about the Higgs. Well, I didn't need any assistance on this puzzle: the reason is that Krauss is an idiot, of course. Instead of trying to clarify the mysterious logic that led to a wrong answer, it may be more useful to analyze the correct logic that led to the correct answer. ;-)
Fine, so why have people like me been certain that there had to be a Higgs boson somewhere in the remaining corners?
In December 2011, it became a sure thing that a Higgs boson of mass \(124-126\GeV\) existed. In fact, one could calculate and I did calculate that it would be 5-sigma-discovered sometimes in Summer 2012 and I have announced this fact in all my popular lectures in Spring 2012. ;-)
However, we've known that there had to be a Higgs boson for decades. How could we know that? Well, you may write the elegant electroweak theory with its electroweak gauge symmetry that is – somewhat elegantly but not too elegantly – spontaneously broken by the Higgs doublet. And you may say that the theory is rather pretty. You don't have competitors so it must be right.
Fine, so let's start.
A century ago, Marie Sklodowska ordered some pitchblende (uraninite) from Jáchymov, Austria-Hungary. This toxic garbage contained lots of yummy ingredients including radium she became famous for. Radium comes in dozens of isotopes. Their decays include alpha-decays, beta-decays, as well as gamma-decays.
Gamma-decays are about the emission of high-energy photons only. Alpha-decays may be modeled as the tunneling of an alpha-particle (helium-4 nucleus) out of the parent nucleus. Because quantum tunneling has a probability that exponentially depends on the thickness times the height of the barrier, you may get a huge hierarchy of possible lifetimes. Alpha-decaying isotopes may have half-lives between tiny fractions of a second and billions of years.
But we're going to be interested in beta-decays. At the elementary level, beta-decay may always be explained as a decay of a neutron (isolated neutron or a neutron inside a nucleus),\[
n\to p+ e^- + \bar \nu_e
\] or similar reactions in which you move particles from one side of the equation to the other side and change them to antiparticles. Once you know quarks and realize that a neutron is \(n=udd\) while a proton is a \(p=uud\), you may also rewrite the true essence of the beta-decay as\[
d\to u+ e^- + \bar \nu_e
\] but it's really the same thing. It doesn't matter much whether you convert protons and neutrons or quarks. In both cases, they may be organized as doublets. Now, in the 1930s, Enrico Fermi understood that this reaction may be derived from quantum field theory if you add a quartic (fourth-order) term in the quantum fields that are able to create or destroy these particles:\[
\LL_{\rm Fermi} \approx G_F \cdot \psi_d \bar\psi_u \cdot \psi_e \bar\psi_{\nu,e}.
\] The product of the four fermionic fields is able to take one of the four particles, destroy it, and create the remaining three particles at the same point. You may have noticed that the objects \(\psi\) are Dirac (or Weyl/Majorana) spinors and I need to contract the indices in the right way – which must be both Lorentz-invariant as well as compatible with other detailed experiments. So the Lagrangian above is schematic if not sloppy.
Just to be sure, I am introducing you to the weak nuclear force right now.
Fermi and others were able to realize that the objects \(\psi_d\bar\psi_u\) were analogous to the "electric currents" \(j^\mu\) in quantum electrodynamics; note that this current \(\bar\psi \gamma^\mu\psi\) is bilinear in the Dirac fields, too. This current should be contracted – summed over \(\mu\) – with the similar current coming from the leptons, i.e. the electron and its neutrino.
However, it wasn't guaranteed that you should insert exactly a single \(\gamma^\mu\) between \(\bar\psi_u\) and \(\psi_d\). You could also insert nothing (I mean the unit matrix) or other possibilities. Depending on whether you insert \(1\), \(\gamma^\mu\), \(\gamma^\mu\gamma_5\), or \(\gamma^{\mu\nu}\) (the latter is an antisymmetrized product of two gamma matrices), you would call the resulting object scalar, vector, axial vector, or tensor, and the resulting four-fermion interaction would be denoted by the acronym, either S or V or A or T. The same gamma matrices with lower indices are inserted to the other, leptonic bilinear structure and the Lorentz indices (if any) are summed over.
It wasn't clear for quite some time which tensor structure was chosen by Nature to run the weak nuclear interactions and beta-decay. Fortunately, the best and most famous experimenters of the 1960s made the experiments and ended up with a consensus: Nature chooses S or T. Feynman and Gell-Mann declared it was bullshit because they already had a theory in which the interaction was V-A – an equal mixture of vector and axial vector terms. Needless to say, Feynman and Gell-Mann were right and the experimenters were wrong. Feynman had remembered that when he saw the wrong paper for the first time, he noticed that the whole S-T conclusion depended on the last point in a chart – and you shouldn't rely on such points too much because if they were reliable, there would be one more behind them. ;-)
So in the mid 1960s, it was established that a more accurate form of the four-fermion interaction looked like this:\[
\LL_{\rm Fermi} \approx G_F \cdot \bar\psi_u(1-\gamma_5)\gamma^\mu\psi_d \cdot \bar\psi_{\nu,e} (1-\gamma_5)\gamma_\mu \psi_e.
\] The mixture \((1-\gamma_5)\) is what prevents you from assigning a clear name – vector or pseudovector – to the interaction term and that's what violates the left-right symmetry (parity) of the underlying theory. In fact, the interaction above only applies to the left-handed (left-spinning) quarks and leptons and the right-handed (right-spinning) antiquarks and antileptons.
The interaction above is problematic when you try to compute loop corrections. It's because it is nonrenormalizable. Each \(\psi\) has units of \({\rm mass}^{3/2}\) as you can see from the Dirac kinetic term. Four of them have units of \({\rm mass}^6\). So you need to subtract two masses to get the usual Lagrangian density whose units are \({\rm mass}^4\). It means that the dimension of \(G_F\), the coefficient, is \({\rm mass}^{-2}={\rm length}^2\). Such negative mass dimensions mean that in higher-loop diagrams, the higher power of \(G_F\) has to be multiplied by positive powers of the loop momenta and that's what is causing high-energy, ultraviolet divergences that are getting worse as you add additional loops.
So the direct four-fermion interaction is non-renormalizable. More generally, it sucks at higher energies. The founding fathers of the electroweak theory, building on approximate visions that were in place since Oskar Klein's work in the 1930s, realized that one may rebuild the four-fermion interaction into a renormalizable theory in a full analogy with the renormalizable theory of electrodynamics.
The interaction between the four fermionic fields i.e. between the two currents isn't direct, they decided, but it is due to an indirect interaction with an intermediate messenger field \(W_\mu\) which is analogous to the electromagnetic potential \(A_\mu\). So the fundamental Lagrangian doesn't have a direct interaction between the two currents. Much like electromagnetism contains \[
e\cdot \bar\psi \gamma^\mu \psi \cdot A_\mu,
\] the weak interaction includes similar terms such as\[
g\cdot \bar\psi_u (1-\gamma_5) \gamma^\mu \psi_d \cdot W_\mu.
\] The two expressions differ in several respects. First, the coupling constant is called \(g\) and not \(e\). It has a different value but this difference is not exactly profound or conceptual. More importantly, the two \(\psi\) fields in the weak interaction case have different subscripts. It's because unlike the electromagnetism, the weak force is changing the identity of the participating particles: don't forget about the beta-decay which is the exclusive "everyday life phenomenon" that experimentally justifies all this research. Third, there is the extra "chiral" \((1-\gamma_5)\) inserted which makes the weak force left-right-asymmetric.
Finally, the messenger field is called \(W_\mu\) rather than \(A_\mu\). In this case, it's not just a different name. The field \(W_\mu\) has some different properties, too. In particular, it is a field associated with massive particles, the W-bosons, and not with massless particles created by the \(A_\mu\) fields, the photons. The massive character of the W-bosons is what turns the weak interaction into a short-range force.
In electromagnetism, the currents would also "effectively interact" via the electromagnetic field and you would get interaction energies such as \(Q_1 Q_2/r\) where \(r\) is the distance between the charges. Note that each \(Q_1,Q_2\) results from the integral of some \(j^\mu\sim \bar\psi \gamma^\mu\psi\). In the case of the weak force, the \(1/r\) dependence is combined with a quickly, exponentially decreasing factor \(\exp(-m_W r)\) so that the function \(1/r\) effectively turns into a multiple of the delta-function. That's why the original Fermi interaction could have been approximated by a "contact interaction" – all the four participating fermions have to sit at the same spacetime point for the interaction to operate.
Great. So the renormalizability of the four-fermion interaction forces you to introduce a new vector field generalizing the electromagnetic field. It has to be massive. One may also show that for related reasons, the vector bosons have to be associated with a gauge symmetry – they have to be gauge bosons. One realizes that \(SU(2)\times U(1)\) is the minimal group for which the electroweak gauge theory becomes viable. Check the book at the top.
By almost purely theoretical advances, we may argue that Marie Sklodowska's observations of the beta-decay demand the existence of the W-bosons and Z-bosons and indeed, they were discovered at CERN in the early 1980s. Several times earlier, the Nobel prize was already given to the theorists who predicted them – Glashow, Salam, Weinberg. By the way, a week ago, Fox News and others ran a story about Pakistani bigots' success in turning their best scientist Abdus Salam into a taboo heretic. Another reason to thermonuke these breathtaking barbarian fanatics and would-be allies.
Higgs needed for a healthy WW scattering
While I omit the detailed derivation of the existence of the W-bosons and Z-bosons (because we've known that W-bosons and Z-bosons exist from the experiments for 30 years anyway), let me say a few more details about very similar arguments that show that a Higgs boson is necessary as well.
Because the W-bosons exist and are massive, they must have 3 polarizations. That's clear if you go into the W-bosons' rest frame. That's different than for photons that only have 2 transverse polarizations. A photon moving in the \(z\) direction may have its electric field pointing in the \(x\) or \(y\) direction or their combination (or it may be circularly polarized which is still a complex combination of the linear ones) but nothing else. A W-boson may also be longitudinally polarized, i.e. with the field parallel to the direction of motion.
Now, the electroweak theory contains the W-bosons so it must also have a meaningful answer to the following question: What happens if you collide two longitudinally polarized \(W^+\) bosons at high energies? In the leading order in the expansion over the small coupling constants \(g\), the answer is given by the tree-level Feynman diagrams (those without any loops).
(It doesn't matter whether you collide two W-bosons with the same charge or the opposite charge. All the amplitudes are related by the crossing symmetry etc.)
What are the tree-level diagrams contributing to the \(2\to 2\) scattering of \(W^+_L\) bosons? Well, there are three of them. You may connect the 4 external W-boson lines by a direct vertex because there are quartic interactions in Yang-Mills theory. Or you may obtain this interaction between the 4 external lines by inserting an intermediate photon or Z-boson in two possible channels – in \(t\) or \(u\) channel, in my particular formulation of the problem (with two equally charged W-bosons in the initial state).
The resulting scattering amplitude from the three tree-level diagrams is\[
{\mathcal M}_{\rm gauge} = -\frac{g^2}{4 m_W^2} \zav{4-\frac{3}{\rho}} s + {\mathcal O}(u^0).
\] For different assignment of the positive/negative charges, you would get various permutations of \(s,t,u\). Now, \(\rho=m_W^2/m_Z^2\cos^2\theta_W\) is equal to one at the leading (tree-level approximation) so \(4-3/\rho\sim 1\). Nevertheless, you see that a tree-level scattering amplitude increases as \(s\sim E_{c.m.}^2\) at high enough energies. Because the coupling constants are rather small – and run slowly – the higher-order diagrams can't beat this growth.
This growth of the scattering amplitude with energy really means that the probability of scattering exceeds 100 percent for W-boson energies that are comparable to \(1\TeV\). That's too bad because probabilities shouldn't ever surpass 100 percent, regardless of any uncertainty we may have about the detailed laws of physics.
So the correct amplitude for the longitudinal W-boson scattering has to be different. It can't possibly increase with \(s\) so quickly. Something must slow down the growth. And whatever is able to do so may be interpreted as a new diagram contributing to the scattering process. And any new diagram may be interpreted as the "exchange of some new stuff".
It's important to realize that you can't just modify the contributions of individual Feynman diagrams for high energies by any "fixes" you could find convenient. The Feynman diagrams' contributions to scattering amplitudes are completely determined by the spectrum of particles and their interactions. Even in string theory in which the high-energy behavior of the scattering amplitudes starts to "mysteriously soften", you may still prove that this softening is inseparable from new particle species – in fact, a whole infinite tower of excited string modes.
So there have to be new particles below a \(1\TeV\) or so whose interactions with the W-bosons force you to include new diagrams and those new diagrams erase most of the unwelcome growing term proportional to \(s\). The Standard Model Higgs boson does the job in a very straightforward way. If you add the two diagrams with the exchange of a Higgs boson – in the same two channels in which you exchanged the photon or Z-boson above – you obtain a new contribution to the scattering amplitude,\[
{\mathcal M}_{\rm higgs} = \frac{g^2}{4 m_W^2} s + {\mathcal O}(u^0)
\] That's neat because for \(\rho=1\) as promoted above, this exactly cancels the leading term that grew with \(s\). So the leading high-energy behavior is actually constant and you avoid the conflict with the condition that probabilities should never exceed 100 percent.
You may derive the Feynman rules for the W-bosons, Z-bosons, and the Higgs boson if you start with the "elegant" theory with the electroweak symmetry and you add the symmetry breaking fields, and so on. But as we've seen, you may also start with an ugly theory that describes the dirty experimental data – the four-fermion interaction at the very beginning – and you may keep on refining the theory so that it's well-behaved and cancels an ever greater number of new pathologies. At the end, you end up with a theory that is equivalent to the "pretty" electroweak theory derived from the symmetries – the equivalence boils down to simple field redefinitions. For example, the first approach makes the point \(h=(0,0)\) natural as a starting point (the maximum of the Higgs potential); the second approach is naturally expanded around the physical vacuum \(h=(v,0)\).
One could also try to suppress the high-energy scattering by some more obscure methods, by making the W-bosons composite or by exchanging composite particles. You may get into the muddy waters of compositeness, technicolor, and similar proposals. However, the ability of these theories to cancel the unwanted terms would always look a bit surprising and moreover, all these theories predict some new deviations from the Standard Model physics that would have already been seen. In fact, precision measurements have shown that the compositeness scale couldn't have been below \(10\TeV\) or so.
There are all kinds of reasons why I would have bet that technicolor-like theories couldn't have been right (the natural compatibility of the Higgs mechanism with string theory was a major reason for me) but even if they were right, it would still be true that they must provide us with some new particles whose propagators appear in new diagrams that cancel the pathological growth of the WW scattering amplitudes in the Higgsless Standard Model. I just find a Higgs boson to be the most natural solution – it just "exactly" cancels the term proportional to \(s\).
With a dose of rational thinking and some mathematics, you may "directly see" that there has to exist a Higgs boson just by looking at Marie Curie and her Bohemian pitchblende. Good enough particle physicists have known how to "see" this far for quite some time.
And that's the memo.
The Atlantic published an article (plus an audio interview) trying to explain why some physicists such as Stephen Hawking made bets against the Higgs boson. I haven't really understood the answer.
The story started in Jáchymov/Joachimstahl, a Sudeten Czech town with lots of pitchblende. Marie Sklodowska ordered some of this stuff. (In the 1950s, the mines in the local uranium mines would host lots of political prisoners of Stalinism. Four centuries earlier, the town minted silver thalers/tolars – which the dollars were later named after.)
Instead, I was offered an explanation why Lawrence Krauss was skeptical about the Higgs. Well, I didn't need any assistance on this puzzle: the reason is that Krauss is an idiot, of course. Instead of trying to clarify the mysterious logic that led to a wrong answer, it may be more useful to analyze the correct logic that led to the correct answer. ;-)
Fine, so why have people like me been certain that there had to be a Higgs boson somewhere in the remaining corners?
In December 2011, it became a sure thing that a Higgs boson of mass \(124-126\GeV\) existed. In fact, one could calculate and I did calculate that it would be 5-sigma-discovered sometimes in Summer 2012 and I have announced this fact in all my popular lectures in Spring 2012. ;-)
However, we've known that there had to be a Higgs boson for decades. How could we know that? Well, you may write the elegant electroweak theory with its electroweak gauge symmetry that is – somewhat elegantly but not too elegantly – spontaneously broken by the Higgs doublet. And you may say that the theory is rather pretty. You don't have competitors so it must be right.
Off-topic: The Pioneer anomaly has been explained away as an artifact of heat radiating in the direction of motion. No new physics.While this argument isn't as weak as you might think, you may rightfully object that it could have loopholes. A completely different theory could be valid. Is there some more waterproof argument showing that a Higgs boson was needed? Yes, there is. A rather detailed presentation of this logic is given in Dr Jiřà HorejÅ¡Ã's book "Standard Model From Tree Unitarity" that you may buy via the amazon.com link on the left side; he was an undergraduate quantum field theory instructor of mine. (His name means George Up-quark-type in Czech; there is another guy in the group who's called George Down-quark-type.)
Fine, so let's start.
A century ago, Marie Sklodowska ordered some pitchblende (uraninite) from Jáchymov, Austria-Hungary. This toxic garbage contained lots of yummy ingredients including radium she became famous for. Radium comes in dozens of isotopes. Their decays include alpha-decays, beta-decays, as well as gamma-decays.
Gamma-decays are about the emission of high-energy photons only. Alpha-decays may be modeled as the tunneling of an alpha-particle (helium-4 nucleus) out of the parent nucleus. Because quantum tunneling has a probability that exponentially depends on the thickness times the height of the barrier, you may get a huge hierarchy of possible lifetimes. Alpha-decaying isotopes may have half-lives between tiny fractions of a second and billions of years.
But we're going to be interested in beta-decays. At the elementary level, beta-decay may always be explained as a decay of a neutron (isolated neutron or a neutron inside a nucleus),\[
n\to p+ e^- + \bar \nu_e
\] or similar reactions in which you move particles from one side of the equation to the other side and change them to antiparticles. Once you know quarks and realize that a neutron is \(n=udd\) while a proton is a \(p=uud\), you may also rewrite the true essence of the beta-decay as\[
d\to u+ e^- + \bar \nu_e
\] but it's really the same thing. It doesn't matter much whether you convert protons and neutrons or quarks. In both cases, they may be organized as doublets. Now, in the 1930s, Enrico Fermi understood that this reaction may be derived from quantum field theory if you add a quartic (fourth-order) term in the quantum fields that are able to create or destroy these particles:\[
\LL_{\rm Fermi} \approx G_F \cdot \psi_d \bar\psi_u \cdot \psi_e \bar\psi_{\nu,e}.
\] The product of the four fermionic fields is able to take one of the four particles, destroy it, and create the remaining three particles at the same point. You may have noticed that the objects \(\psi\) are Dirac (or Weyl/Majorana) spinors and I need to contract the indices in the right way – which must be both Lorentz-invariant as well as compatible with other detailed experiments. So the Lagrangian above is schematic if not sloppy.
Just to be sure, I am introducing you to the weak nuclear force right now.
Fermi and others were able to realize that the objects \(\psi_d\bar\psi_u\) were analogous to the "electric currents" \(j^\mu\) in quantum electrodynamics; note that this current \(\bar\psi \gamma^\mu\psi\) is bilinear in the Dirac fields, too. This current should be contracted – summed over \(\mu\) – with the similar current coming from the leptons, i.e. the electron and its neutrino.
However, it wasn't guaranteed that you should insert exactly a single \(\gamma^\mu\) between \(\bar\psi_u\) and \(\psi_d\). You could also insert nothing (I mean the unit matrix) or other possibilities. Depending on whether you insert \(1\), \(\gamma^\mu\), \(\gamma^\mu\gamma_5\), or \(\gamma^{\mu\nu}\) (the latter is an antisymmetrized product of two gamma matrices), you would call the resulting object scalar, vector, axial vector, or tensor, and the resulting four-fermion interaction would be denoted by the acronym, either S or V or A or T. The same gamma matrices with lower indices are inserted to the other, leptonic bilinear structure and the Lorentz indices (if any) are summed over.
It wasn't clear for quite some time which tensor structure was chosen by Nature to run the weak nuclear interactions and beta-decay. Fortunately, the best and most famous experimenters of the 1960s made the experiments and ended up with a consensus: Nature chooses S or T. Feynman and Gell-Mann declared it was bullshit because they already had a theory in which the interaction was V-A – an equal mixture of vector and axial vector terms. Needless to say, Feynman and Gell-Mann were right and the experimenters were wrong. Feynman had remembered that when he saw the wrong paper for the first time, he noticed that the whole S-T conclusion depended on the last point in a chart – and you shouldn't rely on such points too much because if they were reliable, there would be one more behind them. ;-)
So in the mid 1960s, it was established that a more accurate form of the four-fermion interaction looked like this:\[
\LL_{\rm Fermi} \approx G_F \cdot \bar\psi_u(1-\gamma_5)\gamma^\mu\psi_d \cdot \bar\psi_{\nu,e} (1-\gamma_5)\gamma_\mu \psi_e.
\] The mixture \((1-\gamma_5)\) is what prevents you from assigning a clear name – vector or pseudovector – to the interaction term and that's what violates the left-right symmetry (parity) of the underlying theory. In fact, the interaction above only applies to the left-handed (left-spinning) quarks and leptons and the right-handed (right-spinning) antiquarks and antileptons.
The interaction above is problematic when you try to compute loop corrections. It's because it is nonrenormalizable. Each \(\psi\) has units of \({\rm mass}^{3/2}\) as you can see from the Dirac kinetic term. Four of them have units of \({\rm mass}^6\). So you need to subtract two masses to get the usual Lagrangian density whose units are \({\rm mass}^4\). It means that the dimension of \(G_F\), the coefficient, is \({\rm mass}^{-2}={\rm length}^2\). Such negative mass dimensions mean that in higher-loop diagrams, the higher power of \(G_F\) has to be multiplied by positive powers of the loop momenta and that's what is causing high-energy, ultraviolet divergences that are getting worse as you add additional loops.
So the direct four-fermion interaction is non-renormalizable. More generally, it sucks at higher energies. The founding fathers of the electroweak theory, building on approximate visions that were in place since Oskar Klein's work in the 1930s, realized that one may rebuild the four-fermion interaction into a renormalizable theory in a full analogy with the renormalizable theory of electrodynamics.
The interaction between the four fermionic fields i.e. between the two currents isn't direct, they decided, but it is due to an indirect interaction with an intermediate messenger field \(W_\mu\) which is analogous to the electromagnetic potential \(A_\mu\). So the fundamental Lagrangian doesn't have a direct interaction between the two currents. Much like electromagnetism contains \[
e\cdot \bar\psi \gamma^\mu \psi \cdot A_\mu,
\] the weak interaction includes similar terms such as\[
g\cdot \bar\psi_u (1-\gamma_5) \gamma^\mu \psi_d \cdot W_\mu.
\] The two expressions differ in several respects. First, the coupling constant is called \(g\) and not \(e\). It has a different value but this difference is not exactly profound or conceptual. More importantly, the two \(\psi\) fields in the weak interaction case have different subscripts. It's because unlike the electromagnetism, the weak force is changing the identity of the participating particles: don't forget about the beta-decay which is the exclusive "everyday life phenomenon" that experimentally justifies all this research. Third, there is the extra "chiral" \((1-\gamma_5)\) inserted which makes the weak force left-right-asymmetric.
Finally, the messenger field is called \(W_\mu\) rather than \(A_\mu\). In this case, it's not just a different name. The field \(W_\mu\) has some different properties, too. In particular, it is a field associated with massive particles, the W-bosons, and not with massless particles created by the \(A_\mu\) fields, the photons. The massive character of the W-bosons is what turns the weak interaction into a short-range force.
In electromagnetism, the currents would also "effectively interact" via the electromagnetic field and you would get interaction energies such as \(Q_1 Q_2/r\) where \(r\) is the distance between the charges. Note that each \(Q_1,Q_2\) results from the integral of some \(j^\mu\sim \bar\psi \gamma^\mu\psi\). In the case of the weak force, the \(1/r\) dependence is combined with a quickly, exponentially decreasing factor \(\exp(-m_W r)\) so that the function \(1/r\) effectively turns into a multiple of the delta-function. That's why the original Fermi interaction could have been approximated by a "contact interaction" – all the four participating fermions have to sit at the same spacetime point for the interaction to operate.
Great. So the renormalizability of the four-fermion interaction forces you to introduce a new vector field generalizing the electromagnetic field. It has to be massive. One may also show that for related reasons, the vector bosons have to be associated with a gauge symmetry – they have to be gauge bosons. One realizes that \(SU(2)\times U(1)\) is the minimal group for which the electroweak gauge theory becomes viable. Check the book at the top.
By almost purely theoretical advances, we may argue that Marie Sklodowska's observations of the beta-decay demand the existence of the W-bosons and Z-bosons and indeed, they were discovered at CERN in the early 1980s. Several times earlier, the Nobel prize was already given to the theorists who predicted them – Glashow, Salam, Weinberg. By the way, a week ago, Fox News and others ran a story about Pakistani bigots' success in turning their best scientist Abdus Salam into a taboo heretic. Another reason to thermonuke these breathtaking barbarian fanatics and would-be allies.
Higgs needed for a healthy WW scattering
While I omit the detailed derivation of the existence of the W-bosons and Z-bosons (because we've known that W-bosons and Z-bosons exist from the experiments for 30 years anyway), let me say a few more details about very similar arguments that show that a Higgs boson is necessary as well.
Because the W-bosons exist and are massive, they must have 3 polarizations. That's clear if you go into the W-bosons' rest frame. That's different than for photons that only have 2 transverse polarizations. A photon moving in the \(z\) direction may have its electric field pointing in the \(x\) or \(y\) direction or their combination (or it may be circularly polarized which is still a complex combination of the linear ones) but nothing else. A W-boson may also be longitudinally polarized, i.e. with the field parallel to the direction of motion.
Now, the electroweak theory contains the W-bosons so it must also have a meaningful answer to the following question: What happens if you collide two longitudinally polarized \(W^+\) bosons at high energies? In the leading order in the expansion over the small coupling constants \(g\), the answer is given by the tree-level Feynman diagrams (those without any loops).
(It doesn't matter whether you collide two W-bosons with the same charge or the opposite charge. All the amplitudes are related by the crossing symmetry etc.)
What are the tree-level diagrams contributing to the \(2\to 2\) scattering of \(W^+_L\) bosons? Well, there are three of them. You may connect the 4 external W-boson lines by a direct vertex because there are quartic interactions in Yang-Mills theory. Or you may obtain this interaction between the 4 external lines by inserting an intermediate photon or Z-boson in two possible channels – in \(t\) or \(u\) channel, in my particular formulation of the problem (with two equally charged W-bosons in the initial state).
The resulting scattering amplitude from the three tree-level diagrams is\[
{\mathcal M}_{\rm gauge} = -\frac{g^2}{4 m_W^2} \zav{4-\frac{3}{\rho}} s + {\mathcal O}(u^0).
\] For different assignment of the positive/negative charges, you would get various permutations of \(s,t,u\). Now, \(\rho=m_W^2/m_Z^2\cos^2\theta_W\) is equal to one at the leading (tree-level approximation) so \(4-3/\rho\sim 1\). Nevertheless, you see that a tree-level scattering amplitude increases as \(s\sim E_{c.m.}^2\) at high enough energies. Because the coupling constants are rather small – and run slowly – the higher-order diagrams can't beat this growth.
This growth of the scattering amplitude with energy really means that the probability of scattering exceeds 100 percent for W-boson energies that are comparable to \(1\TeV\). That's too bad because probabilities shouldn't ever surpass 100 percent, regardless of any uncertainty we may have about the detailed laws of physics.
So the correct amplitude for the longitudinal W-boson scattering has to be different. It can't possibly increase with \(s\) so quickly. Something must slow down the growth. And whatever is able to do so may be interpreted as a new diagram contributing to the scattering process. And any new diagram may be interpreted as the "exchange of some new stuff".
It's important to realize that you can't just modify the contributions of individual Feynman diagrams for high energies by any "fixes" you could find convenient. The Feynman diagrams' contributions to scattering amplitudes are completely determined by the spectrum of particles and their interactions. Even in string theory in which the high-energy behavior of the scattering amplitudes starts to "mysteriously soften", you may still prove that this softening is inseparable from new particle species – in fact, a whole infinite tower of excited string modes.
So there have to be new particles below a \(1\TeV\) or so whose interactions with the W-bosons force you to include new diagrams and those new diagrams erase most of the unwelcome growing term proportional to \(s\). The Standard Model Higgs boson does the job in a very straightforward way. If you add the two diagrams with the exchange of a Higgs boson – in the same two channels in which you exchanged the photon or Z-boson above – you obtain a new contribution to the scattering amplitude,\[
{\mathcal M}_{\rm higgs} = \frac{g^2}{4 m_W^2} s + {\mathcal O}(u^0)
\] That's neat because for \(\rho=1\) as promoted above, this exactly cancels the leading term that grew with \(s\). So the leading high-energy behavior is actually constant and you avoid the conflict with the condition that probabilities should never exceed 100 percent.
Directly related breaking news: Five minutes ago, ATLAS released a paper that, for the first time, brings experimental evidence in favor of the interaction HWW between the Higgs boson and a pair of W-boson external lines, exactly the cubic vertex that was used (twice) to derive \({\mathcal M}_{\rm higgs}\) above. The significance of the signal has exceeded 3 sigma.A single Higgs boson isn't necessarily the only way how to cancel the pathologically growing terms in the scattering amplitude; it's just the simplest one. However, you may think about "qualitatively different" analytic forms of the canceling contributions and you will find out that for them to play the canceling job, each of them has to look like the exchange of a scalar boson. Whenever it does the job, the couplings of the scalar boson are exactly such that you may derive them from a pretty gauge theory with a spontaneously broken symmetry (although the breaking may be done by several Higgs fields, not just one).
You may derive the Feynman rules for the W-bosons, Z-bosons, and the Higgs boson if you start with the "elegant" theory with the electroweak symmetry and you add the symmetry breaking fields, and so on. But as we've seen, you may also start with an ugly theory that describes the dirty experimental data – the four-fermion interaction at the very beginning – and you may keep on refining the theory so that it's well-behaved and cancels an ever greater number of new pathologies. At the end, you end up with a theory that is equivalent to the "pretty" electroweak theory derived from the symmetries – the equivalence boils down to simple field redefinitions. For example, the first approach makes the point \(h=(0,0)\) natural as a starting point (the maximum of the Higgs potential); the second approach is naturally expanded around the physical vacuum \(h=(v,0)\).
One could also try to suppress the high-energy scattering by some more obscure methods, by making the W-bosons composite or by exchanging composite particles. You may get into the muddy waters of compositeness, technicolor, and similar proposals. However, the ability of these theories to cancel the unwanted terms would always look a bit surprising and moreover, all these theories predict some new deviations from the Standard Model physics that would have already been seen. In fact, precision measurements have shown that the compositeness scale couldn't have been below \(10\TeV\) or so.
There are all kinds of reasons why I would have bet that technicolor-like theories couldn't have been right (the natural compatibility of the Higgs mechanism with string theory was a major reason for me) but even if they were right, it would still be true that they must provide us with some new particles whose propagators appear in new diagrams that cancel the pathological growth of the WW scattering amplitudes in the Higgsless Standard Model. I just find a Higgs boson to be the most natural solution – it just "exactly" cancels the term proportional to \(s\).
With a dose of rational thinking and some mathematics, you may "directly see" that there has to exist a Higgs boson just by looking at Marie Curie and her Bohemian pitchblende. Good enough particle physicists have known how to "see" this far for quite some time.
And that's the memo.
Why there had to be a Higgs boson
Reviewed by DAL
on
July 18, 2012
Rating:
Reviewed by DAL
on
July 18, 2012
Rating:
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