Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone give a simple expose on Coleman Mandula theorem and what Mandelstam variables are?

Coleman-Mandula is often cited as being the key theorem that leads us to consider Supersymmetry for unification. An overview discussion with sufficient detail is missing in many popular texts. So how does Coleman Mandula theorem actually meet its no-go claim?

share|cite|improve this question
I'd suggest you try asking two separate questions as the Coleman-Mandula theorem (not theory) has nothing to do with Mandelstam variables. – pho Jan 23 '11 at 4:03
I am not totally without reference here, Supersymmetry Demystifed by Patrick Labelle makes reference to Mandelstam variables in the discussion of Coleman Mandula. The reference gives an example of one condition for the proof is that elastic scattering amplitudes are analytic functions of Mandelstam variables. – Humble Jan 23 '11 at 11:59
up vote 5 down vote accepted

This is a bit of a sketch;

The $S$-matrix acts on shift the state or momentum state of a particle. A state with two particle states $|p, p’\rangle$ is acted upon by the $S$ matrix through the $T$ matrix $$ S~=~1~–~i(2\pi)^4 \delta^4(p~–~p’)T $$ So that $T|p, p’\rangle\ne0$. For zero mass plane waves scatter at almost all energy. The Hilbert space is then an infinite product of n-particle subspaces $H~=~\otimes_nH^n$. As with all Hilbert spaces there exists a unitary operator $U$, often $U~=~exp(iHt)$, which transforms the states S acts upon. $U$ transforms n-particle states into n-particle states as tensor products. The unitary operator commutes with the $S$ matrix $$ SUS^{-1}~=~[1 – i(2\pi)^4 \delta^4(p~–~p’)T]U[1~+~i(2π)^4 \delta^4(p~–~p’)T^\dagger] $$$$ = U~+~i(2\pi)^4 \delta^4(p~–~p’)[TU~–~UT^\dagger] ~+~[(2\pi)^4 \delta^4(p~–~p’)]^2(TUT^\dagger). $$ By Hermitian properties and unitarity it is not difficult to show the last two terms are zero and that the S-matrix commutes with the unitary matrix. The Lorentz group then defines operator $p_\mu$ and $M_{\mu\nu}$ for momentum boosts and rotations. The $S$-matrix defines changes in momentum eigenstates, while the unitary operator is generated by a internal symmetries $A_a$, where the index a is within some internal space (the circle in the complex plane for example, and we then have with some $$ [A_a,~p_\mu]~=~[A_a,~M_{\mu\nu}]~=~0. $$ This is a sketch of the infamous “no-go” theorem of Coleman and Mundula. This is what prevents one from being able to place internal and external generators or symmetries on the same footing.

The way around this problem is supersymmetry. The generators of the supergroup, or a graded Lie algebra, have 1/2 commutator group elements $[A_a,~A_b]~=~C_{ab}^cA_c$ ($C_{ab}^c$ = structure constant of some Lie algebra), plus another set of graded operators which obey $$ \{{\bar Q}_a, Q_b\}~=~\gamma^\mu_{ab}p_\mu, $$ which if one develops the SUSY algebra you find this is a loophole which allows for the intertwining of internal symmetries and spacetime generators. One might think of the above anti-commutator as saying the momentum operator, as a boundary operator $p_\mu~= -i\hbar\partial_\mu$ which has a cohomology, where it results from the application of a Fermi-Dirac operator $Q_a$. Fermi-Dirac states are such that only one particle can occupy a state, which has the topological content of $d^2~=~0$. This cohomology is the basis for BRST quantization.

share|cite|improve this answer
I did not discuss Mandelstam variable, but I will lead that to the next contributor. – Lawrence B. Crowell Jan 23 '11 at 1:56
Very awesome, thanks... Just a note for future readers, Mandelstam variables are discussed in answer provided by Lubos Motl. – Humble Jan 23 '11 at 12:17

Mandelstam variables $s,t,u$ are quantities with units of squared momentum (or mass) that describe the Lorentz-invariant part of the information about the momenta and energy in $2\to 2$ scattering processes: $$ s= (p_1+p_2)^2, \quad t=(p_1-p_3)^2,\quad u=(p_1-p_4)^2$$ If you Lorentz transform the incoming momenta $p_1,p_2$ as well as outgoing ones $p_3,p_4$, the quantities above don't change. So by relativity, all the nontrivial information about the collision is encoded in functions of $s,t,u$. Moreover, $s+t+u=4m^2$ if all four particles' masses are equal to $m$; this is generalized easily if they're different. The Mandelstam variables may also be generalized easily to the case of more than 4 external lines - there are more variables in that case. So the Mandelstam variables are a simple thing, given by the formulae above.

Coleman-Mandula theorem

The Coleman-Mandula theorem shows that theories with (bosonic) symmetry groups that mix the spatial (geometric) and internal parts, i.e. they don't have the form $G_{spacetime}\times G_{internal}$, the interactions essentially vanish, so such theories are not usable for any realistic modeling. So for example, if someone hypothesized that an $E_8$ symmetry may include both rotations in space as well as the Standard Model gauge group, his hypothesis would be instantly ruled out by the theorem.

Concerning the proof, I think that they take the lightest scalar excitation in the given theory, two or several of them, and scatter them. The extra symmetries - besides energy and momentum etc. - will inevitably mean that some tensors have to be conserved in a collision, too. Because these tensors may only depend on the momenta of the particles we scatter and they have too many components that must remain unchanged, they can show that the momenta after the scattering have to be essentially equal to those before the scattering. This already means that the interactions kind of vanish universally.

The theorem assumes that the conserved quantities can't be spinors - with half-integral spin. When fermionic conserved quantities with half-integer spin - supersymmetries - are allowed, one finds out that it is possible to circumvent the original theorem, in theories with supersymmetry, because the conserved spinors are not too constraining. One gets supersymmetric theories but their possibilities are still restricted by a supersymmetric extension of the Coleman-Mandula theorem, the Haag-Lopuszanski-Sohnius theorem.

Cheers LM

share|cite|improve this answer

For a textbook reference on the Coleman-Mandula theorem, the third opus of Weinberg's QFT is what you need. The whole proof, with all the details (even the ones left to the reader in the original article) are in appendix B (page 12 to 22). The proof is based only on very general principle of relativistic quantum mechanics as stated in his chapter 2 and 3 (in the first opus), without any need of a local QFT.

But that is not all. Before his full proof, Weinberg gives a simpler (but only partial) proof in section 24.1 (page 1 to 4), which is enough to see clearly the place where relativity makes all the difference.

EDIT (after kyle's fair comment): let's sketch weinberg's kinematical proof of one piece of the theorem.

let all symmetry generators that commute with the 4-momentum $p_m$ form a lie algebra spanned by the generators $B_a$. let a lorentz transformation ($L$), represented on the hilbert space by the unitary operator $U(L)$, act on the $B_a$.

$U(L)B_aU^{-1}(L)$ commutes with ${L_m}^n p_n$ and so with $p_m$. therefore it can be written has a linear combinaison of $B_a$:

$$U(L)B_aU^{-1}(L) = \sum {D_a}^b(L)B_b$$

which can be shown to satisfy the same commutation relations as $B_a$. using this and the commutation of all the generators with $p_m$, and under the assumption that the symmetry generators other than $p_m$ span a compact semi-simple lie algebra (noted as $B_A$), you can construct out of the coefficients ${D_a}^b$, a unitary finite-dimensional representation of the lorentz group. but, because the lorentz group in non-compact, the only such representation is the trivial one. therefore $B_A$ commutes with $U(L)$.

finally, because the $B_A$ commute with $p_m$, their action on the state $|p,n\rangle$ of a single particle with momentum $p_m$ and spin/species $n$ can only yield a linear combination:

$$B_A |p,n\rangle = \sum b_A |p,m\rangle$$

the fact that $B_A$ commutes with boosts implies that $b_A$ is independent of momentum, and the fact that it commutes with rotations implies that $b_A$ act as unit matrices on spin indices, so the $B_A$ are the generators of an ordinary internal symmetry, as was to be proved.

share|cite|improve this answer
You might want to consider adding a summary of the book's proof, rather than pointing OP & readers to the book; otherwise this is basically a link-only answer except without the link. – Kyle Kanos Oct 15 '15 at 12:37
the question was: can someone give a simple sketch of CM, i can't find any in the popular texts? so mmy answer was just that: yes, someone can, he's called steven, and he has written some popular stuff too ;). but ok, i'll try to develop the "simpler proof" sketch, where no S-matrix is involved, so it will give an answer with a point of view different from lawrence's sketch. – mmanu F Oct 15 '15 at 19:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.