Paul Frampton: Off-topic sad news: he gets 4.67 years in prison for drug smuggling; Google News. He may be moved to the U.S. in 2014 if he applies. They found a note, "1grm/200U$S. 2000grms/400000 U$S", handwritten by Frampton himself on which he calculated the price of the stuff. He admitted he wrote it but only after the stuff was found, weighed, and he was accused. So the curious Paul calculated what he was supposed to earn, too. When cops accuse you of anything in Argentina, shut up and don't calculate anything! Even more seriously, however, Paul's e-mail to (fake) Ms Denisa Krajíčková a day before he was caught allegedly "worried about sniffer dogs looking after the small suitcase".This blog entry elaborates upon a simple point made by Nima Arkani-Hamed in a recent talk (and by others). First, look at this IQ test. What is missing in the box?\[
\begin{array}{|c|c|c|c|c|c|}
\hline
j& 0 & 1/2 & 1 & ??? & 2\\
\hline
\end{array}
\] Those of you who have figured out that \(???=3/2\) earned a ticket and they may continue to read. ;-)
The numbers in the table are the spins of the known elementary particles. The Higgs boson became the first discovered spinless elementary particle, one with \(j=0\). Leptons and quarks have \(j=1/2\). The photon, gluon, W-boson, Z-boson – gauge bosons – carry \(j=1\). And a \(j=2\) graviton has to exist because we know that there exist gravitational waves (see e.g. 1993 Physics Nobel Prize) and because all energy at the frequency \(\omega\) is inevitably packaged into quanta of energy \(E=\hbar\omega\), because of the most universal laws of quantum mechanics. Why? If all expectation values etc. \[
\bra\psi L \ket\psi
\] are demanded to be periodic with period \(2\pi / \omega\), it follows that \(\ket\psi\) must be periodic with this period, up to an overall phase. But if \(\ket\psi\) is a linear superposition of various energy eigenstate terms whose time dependence is \(\exp(Et/ i\hbar)\), it follows that between \(t=0\) and \(t=2\pi/\omega\), the relative phases must return to the original value which means that \(E_i-E_j=N\hbar\omega\) for any pair of allowed eigenvalues \(E_i,E_j\). If the two states included in the superposition differ by an addition of a particle or particles, the particle(s) must have \(E=N\hbar\omega\) for \(N\in\ZZ\).
Again, if you don't understand the argument above sufficiently clearly so that you have eradicated all doubts about the existence of gravitons, I kindly ask you to stop reading because you're not qualified to study or discuss the allowed spins of elementary particles.
Why gauge symmetries are needed
Now, when we have hopefully gotten rid of 99.9% of would-be readers :-), we may continue with gauge symmetries. This topic has been mentioned many times on this blog but let's make the point once again. If you deal with fields that manifestly obey the Lorentz symmetry and whose spin is \(j\geq 1\), you critically need a gauge symmetry, otherwise the probabilities would have both signs. But you don't want the probability of your winning in a lottery to be negative. It's worse than just to lose the money: you would be a ghost, a bad ghost, if your probabilities were negative. And that's worse than to be a zombie, believe me.
It's not too hard to see why a gauge symmetry is required. Take a creation operator for a particle, \(a^\dagger(k)\), and produce a one-particle state out of the vacuum \(\ket 0\):\[
a^\dagger \ket 0
\] That's nice. We want this state to have a positive (or at least non-negative) squared norm because this squared norm gives you the probability of the projection operator \(1\), the truth, and it should be 100 percent. For one creation operator, the equality between the norms\[
\braket{0}{0} = \bra 0 a(k)\cdot a^\dagger(k) \ket 0
\] is guaranteed by the commutator below and by the annihilation skill of the annihilation operator:\[
[a(k),a^\dagger(k)]=1,\quad a(k)\ket 0 = 0.
\] Of course, the usual normalization of the creation and annihilation operators involves a \(\delta\)-function but I don't want to go into messy details. Now, the description here is OK for scalars but what happens when the creation operator has some extra indices related to the spacetime?
Start with the \(j=1/2\) Dirac spinors. The fields have an extra index \(i=1,2,3,4\) and the relevant commutator – well, we need an anticommutator for fermions but it changes nothing about our discussion – is\[
\{ c_i(k),c^\dagger_j(k) \} = \delta_{ij}
\] The Kronecker delta is actually right but you could make an error – one that would prove that you know something but you don't know it quite well – and write \(\gamma_0\) instead of the identity matrix. It's because \(\psi\) and the Dirac conjugate \(\bar\psi\) rather than the Hermitian conjugate \(\psi^\dagger=\bar\psi\gamma^0\) are naturally conjugate to produce Kronecker deltas. For example, \(\bar\psi\psi\) and not \(\psi^\dagger\psi\) is Lorentz-invariant. However, the equal-time commutator has another time index in the game (the time-like vector enters because it's orthogonal to the slice on which we define the equal-time [anti]commutators), one which translates to \(\gamma^0\), so the two factors of \(\gamma^0\) cancel and you actually do get the Kronecker delta on the right hand side.
Now, notice that we're kind of lucky that the right hand side is a positively definite matrix \(\delta_{ij}\) rather than an indefinite matrix such as \(\gamma^0_{ij}\) we have mentioned. When you create particles with \(c^\dagger_i\), they will still have a positive norm. In fact, even when the role of \(c\) and \(c^\dagger\) is interchanged for one-half of the values of the index \(i\), because you have to fill the Dirac sea and introduce antiparticles as the holes in the Dirac sea, you will still create positive-norm states because the relevant positive definite construct above was the anticommutator which was symmetric. It just didn't care which of the factors was \(c\) and which of them was \(c^\dagger\).
The percentage of readers who have given up has jumped from 99.9% to 99.99%. The remaining reader, if any, knows this stuff anyway and she or more likely he may notice that we have just seen that \(j=1/2\) fermionic operators produce nicely positive definite particle states.
But what about \(j=1\)?
The creation and annihilation operators have to carry an extra vector index \(\mu=0,1,2,3\). The commutators or, equivalently, the normalization of one-particle states is\[
\bra 0 a_{\mu}(k)\cdot a_{\nu}^\dagger(k) \ket 0 = C\cdot g_{\mu\nu}.
\] The right hand side has to be proportional to the metric tensor, otherwise the formula would violate the Lorentz symmetry! But the metric tensor has both positive and negative eigenvalues. So some of the states are positive definite while others are negative definite.
The latter states would be a problem because they would lead to negative probabilities once again. (Well, it's actually the first time in our story when they're a genuine threat.) In fact, it is such a big problem that we don't want too much of this problem. It means that the overall sign must be such that the 3 spatial polarizations are positive definite and only the remaining 1 time-like polarization has a negative squared norm. What do we do with it?
It's simple. We must kill it. But we're physicists, not murderers or terrorists, so we must kill it in an elegant way. An elegant way to kill states (yes, Gazan terrorists, we are able to kill whole states!) is to make them unphysical. A reason allowing us to delegitimize a state and declare this state unphysical is that this state isn't invariant under a gauge symmetry.
For each value of \(k^\alpha\), we need to kill exactly one polarization. So we must have enough generators of symmetries, one per each \(k^\alpha\) or, equivalently, one per \(x^\beta\). Clearly, we need a gauge symmetry. We must have a generator \(j^0(x,y,z)\) for each point in the three-dimensional space and we require all the physical states to obey\[
j^0(x,y,z)\ket\psi = 0.
\] Also, states of the form \(j^0(x,y,z)\ket\lambda\) are "pure gauge" and they have a zero norm which is still harmless because probabilities may be zero. The operators \(j^0\) above have to be symmetry generators (time-like components of a conserved current) and not just some arbitrary operators because we want the labeling "\(\ket\psi\) is unphysical for \(t=0\)" to survive at later times \(t\), too. That's why the charge – integrated current – has to commute with the Hamiltonian. It has to be a symmetry and because it's localized in space and locally conserved, it's a local or gauge symmetry. If it were just some random operator, it wouldn't commute with the Hamiltonian and a ghost-free state at \(t=0\) would typically evolve into a state with ghosts at \(t\gt 0\). But we want to get rid of the zombies permanently!
In this way, not one but two polarizations out of the 4 polarizations of the photon become unphysical and only the transverse two polarizations, \(\ket x\) and \(\ket y\) or, in a different basis, \(\ket R\) and \(\ket L\) remain physical. Yes, the macro \(\backslash{\rm ket}\) is my most useful \(\rm\LaTeX\) macro I have ever created. It produces lots of cheap yet elegant fun.
Just a comment I won't need: the \(j=1\) field may be given an extra internal index \(j\), an adjoint index, e.g. \(j=1,2,\dots 8\) in QCD, and we will need to get rid of a greater number of ghosts (8 times in the case of QCD) so the conserved quantities have to possess this extra index, too: the conserved charges become generators of a non-Abelian group, e.g. \(SU(3)\) in QCD. Also, the gauge bosons may be made massive via the Higgs mechanism. If it is so, they have 3 and not 2 physical polarizations: the meat for the third one is obtained by eating a Goldstone boson.
Because we needed to get rid of one polarization of a photon for each \(k^\alpha\), we needed a current whose total charge is a scalar. What about \(j=2\) fields? If fields have two indices, the inner products roughly look like\[
\bra 0 a_{\kappa\lambda}(k)\cdot a^\dagger_{\mu\nu}(k)\ket 0 \sim g_{\kappa\mu} g_{\lambda\nu}.
\] There could also be permutations of the indices on the right hand side and/or symmetries, antisymmetries, or other constraints on the tensor. At any rate, we have two copies of the metric tensor. The one-particle state created by \(a^\dagger_{\mu\nu}\) will have a negative squared norm if exactly one of the two indices \(\mu,\nu\) is timelike. (We avoid the scary scenario in which all the doubly spatial components have the wrong sign: we couldn't get enough symmetries to kill the beasts.)
Fine, in this case, we need to get rid of the wrong "mixed" components \(a_{i0}\) so roughly speaking (ignoring the single \(a_{00}\) component which actually seems to have the right sign), we need a whole Lorentz vector of conserved charges. We know what such a vector of conserved charges could be: it could be the energy-momentum vector, associated with the spacetime translations via Noether's theorem!
In fact, if your theory is made out of local or nearly point-like particles, the energy-momentum vector is the only conserved vector you may have as long as your theory admits interesting interactions. For extended objects, you could also consider some kind of a stringy "winding number" but that would require a compactification or infinitely long strings so let's ignore this option.
To summarize, if we want to get rid of the negative-norm polarizations of symmetric \(j=2\) fields, such as the metric tensor, we need a gauge symmetry whose integrated charges transform as spacetime vectors. We need a conserved energy-momentum generating spacetime translations and we need to make its components local, i.e. make the generators independent for each point \((x,y,z)\) in the space. In other words, we need the diffeomorphism symmetry!
It's bizarre and cute. Our argument critically depended both on the Lorentz symmetry and quantum mechanics and we derived the existence of the diffeomorphism symmetry in a consistent theory – with non-negative probabilities – that allows you to create spin-two quanta! Isn't it cool? Gravity and principles of general relativity, perhaps broken in surprising ways, are inevitable parts of any consistent theory with fields whose spin is \(j=2\).
Why spins higher than two are not fine for elementary particles
What about \(j=3\) or higher? In that case, we would produce an even larger number of wrong-sign polarizations of the one-particle states created by the creation operators transforming as \(j\geq 3\) tensors. The corresponding conserved charges would have to transform as \(j\geq 2\) tensors. And they have too many components in \(d\geq 4\). In fact, if this high number of tensor components were conserved, one could prove that interactions are so constrained that they de facto vanish. Any momentum exchange between the lowest-mass scalar particles would violate the conservation laws.
This is the essence of the Coleman-Mandula theorem. There can't be conserved charges with spin greater than one. It follows – through our negative-norm-based arguments – that there can't be any fundamental fields with \(j\geq 3\) in your theory.
Well, string theory – and also its currenly fashionable "toy model", the Vasiliev higher-spin theory – circumvents this ban but the ability of these theories to avoid the conclusion critically depends on their having an infinite number of excitations with arbitrarily high spins \(j\) and their subtle interplay.
Let me mention that fields with spin \(j=5/2\) would have to come with conserved charges with spin \(j=3/2\) which is already too high and prohibits interesting interactions. So \(j=2\) is indeed the highest spin of "ordinary" fundamental fields.
Why the gauge symmetry for \(j=3/2\) fields is inevitably local supersymmetry
But as we mentioned at the beginning, there actually exists an integer or half-integer non-negative number – an allowed value of the spin – that is smaller than \(j=2\) and that we have omitted: \(j=3/2\).
Why have we omitted it? Because we were assholes. Some people would like to omit it even when they're reminded about their mistake. Those people are assholes even right now. I won't name all these Shmoits because I would have to throw up.
So how do our arguments work for \(j=3/2\) fields and what do they tell us about the required symmetries? Well, \(j=3/2\) fields look like \(\psi_{\mu k}\). One index is the Lorentz vector index, another index is a spinor index. They may be constrained by some extra conditions but those conditions don't kill too many components. Such fields are sometimes called Rarita-Schwinger fields and the particles they create are gravitinos but this sentence isn't assumed to hold in the following paragraphs.
Combining (and tensoring) our formulae for \(j=1/2\) and \(j=1\), we may determine that the wrong-sign polarizations of the field are created by the components with \(\mu=0\), i.e. \(\psi_{0k}\). Because the spinor index \(k\) may still take on any value, there is roughly one spinor of the required gauge symmetry generators at each point.
The conserved charges must transform as \(j=1/2\) spinors to allow us to kill the negative-norm states in an elegant way.
Are \(j=1/2\) conserved spinors allowed? Well, they have a smaller spin than the \(j=1\) energy-momentum vector so the answer seems to be Yes and indeed, below 12 spacetime dimensions, the answer is Yes. There can be conserved spinors. We call them \(Q_k\) where the index is a spinor index. And yes, let's say the word: they're generators of supersymmetry, the supercharges!
They're constructed out of fields that respect the spin-statistics relationship and because they're \(j=1/2\) spinors, they have to be fermionic, Grassmann-valued operators. We may ask what their anticommutators are. The anticommutator must involve a \(j=1\) conserved object:\[
\{ Q_{m},\bar Q^{n} \} = (\gamma^\mu)_m^n P_\mu+\dots
\] Ignore the bars and the difference between upper and lower indices if you're not experienced in the algebra involving spinors. The anticommutator of two \(j=1/2\) objects may contain a \(j=1\) object which is conserved as well. And I have already mentioned that the only sensible \(j=1\) "set of charges" is the energy-momentum vector.
(If the anticommutator were zero, the theory would be too trivial once again. Let me avoid this discussion.)
Moreover, we need to make the charges and currents local in spacetime, so we have just derived that a consistent theory with \(j=3/2\) fields has to contain both \(j=3/2\) currents for local supersymmetry as well as the \(j=2\) current for local coordinate transformations (diffeomorphisms) i.e. the stress-energy tensor. A consistent theory with \(j=3/2\) fields inevitably contains supergravity – because it contains both Einstein's gravity as well as (local) supersymmetry. Another article (one that has been written on this blog, in fact) may be written to prove that string/M-theory is the only consistent completion of supergravity.
Despite the extra restrictions, symmetries, and limitations, \(\NNN=1\) supersymmetric theories (supergravity coupled to super-Yang-Mills etc.) are compatible with all the known experimental constraints and conditions, assuming a viable choice of parameters. Nature could "discriminate against" the number \(j=3/2\) and omit this number from the list of "low spins" of allowed fundamental fields. But it would be bizarre, wouldn't it?
There is of course a lot of other circumstantial evidence why supersymmetry exists in Nature: dark matter, hierarchy problem, gauge coupling unification, its existence implied by string theory, and so on.
Jesus
Off-topic but funny: The Catholic Church has found a new heretic. Most scholars believe that Jesus was born between 6 years Before Himself and 4 years Before Himself. The new heretic wrote a book in which he claims that He was actually born several years earlier.
The name of the heretic is Ratzinger and he is employed as the Holy Father in a branch of the church somewhere in Rome, Italy.
This heresy is progress but if Jesus is supposed to have any relationship with Creation whatsoever, there are still 13.73 billion years of adjustments waiting in the pipeline.
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