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"The correspondence between fermions and bosons of identical mass that is postulated to have existed during the opening moments of the big bang and that relates gravity to the other forces of nature"


How and why does this correspondence relate gravity to the other forces of nature?

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Only with supersymmetry do the 3 forces found in the standard model of particle physics unify. A little more specifically, the running gauge coupling in the minimal supersymmetric standard model all converge at about 10^16 GeV or so. Thus it seems important if not essential in any unification scheme.


This does not include gravity. Now, if one localises supersymmetry one automatically gets a theory of gravity. This is because supersymmetry "contains" Poincare symmetry, the symmetries of flat space-time. It may be possible that a supergrvity theory is "large enough" to contain the symmetries of the standard model, for sure 11d supergravity is important from an M-theory/stringy angle.

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Yes, but it's the "correspondence between fermions and bosons of identical mass" that I don't understand.

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There is a morphism between bosonic and fermionic states of the theory. The SUSY algebra forces the paired states to have the same mass.


You can also discuss supersymmetry pseudoclassically.


I'd suggest you have a look in to supersymmetric quantum mechanics. You can get the main features of SUSY without getting in trouble with relativity or infinite dimensions.


If you give me a day or two, I'll post something more useful.

Merged post follows:

Consecutive posts merged

So, lets have a think about N=1 SUSY in 1-dimension. This is sufficient to get the important features.



A supersymmetric quantum mechanical system with N=1 SUSY is the triple [math]\{\mathcal{H}, Q, H \}[/math] where the Hilbert space decomposes as [math]\mathcal{H}= \mathcal{H}^{0}\oplus \mathcal{H}^{1}[/math] which we will call the bosonic (even) and fermionic (odd) sector respectively. The operator [math]Q[/math] is selfadjoint and odd: [math]Q = Q^{\dagger}[/math] and [math]Q : \mathcal{H}^{\pm} \rightarrow \mathcal{H}^{\mp}[/math]. The Hamiltonian [math]H[/math] is an even operator.


The SUSY algebra is


[math][Q,Q] = \frac{1}{2}Q^{2} = H[/math].


Here the brackets are the graded commutator, or in older language the anticommuator.


The first constraint from the above algebra is the the spectrum of the Hamiltonian is always non-negative. This is very important.


The second constraint imposed by the above algebra is


[math][H,Q] = 0[/math].


That is the supercharge [math]Q[/math] is a constant of motion.


Now to your question about the pairing of states. Let us assume we have a bosonic eigenstate of the Hamiltonian (there is no loss in generality here picking this to be bosonic rather than fermionic). That is


[math]H |\psi^{+}\rangle = E|\psi^{+}\rangle[/math].


So as the supercharge is odd we can always write its action on the above eigenstate as


[math]Q |\psi^{+}\rangle = 2E |\psi^{-}\rangle[/math].


I have explicitly assumed that [math]E \neq 0[/math], more about this in a moment.


From the eigenequation for the Hamiltonian this all makes sense if


[math]Q |\psi^{-}\rangle = |\psi^{+}\rangle[/math]. (apply [math]Q[/math] again in the above.)


So what about the energy? Remember than in a relativistic theory mass and energy and interchangeable. So in our non-relativistic toy model it is the energy we are worried about. So,


[math]H |\psi^{-}\rangle = \frac{1}{2} Q \left ( Q |\psi^{-}\rangle \right) = \frac{1}{2}Q |\psi^{+}\rangle = E |\psi^{-}\rangle[/math].


We see that the energy of the bosonic and fermionic states are identical. Again with the proviso of non-zero energy.


We no need to think about supersymmetry spontaneously breaking. That is the Lagrangian of the theory (which I have said nothing about) is invariant under the SUSY transformations but the vacuum/ground state is not annihilated by the action of the supercharges.



The above theory is said to posses a "good supersymmetry" if

[math]Q|\psi_{0}\rangle = 0[/math]


and a "bad supersymmetry" if

[math]Q|\psi_{0}\rangle > 0[/math].


As we are thinking about quantum mechanics in 1-d we can assume that the ground state is either bosonic or fermionic and is nondegenerate without any loss of generality.


Now it is quite easy to see that good supersymmetry requires that the ground state energy be identically zero. That is


[math]H|\psi_{0}\rangle = 0[/math].


Bad supersymmetry corresponds to


[math]H|\psi_{0}\rangle > 0[/math].


Thus we can think of the vacuum energy as the "order parameter".


So, our toy theory has a nice structure. If supersymmetry is good then there is a single (bosonic or fermionic) zero energy ground state and all the positive energy states are paired. If supersymmetry is bad then all the states have positive energy and are paired.


The situation is similar in supersymmetric field theories. But I hope the above is enough to give you a flavour of what is involved.

Edited by ajb
Consecutive posts merged.

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Tighter restrictions on supersymmetry (string theory) at the detector


It looks like it smells like a grave for string theory (and higher dimensional models). 

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I miss AJB...

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12 hours ago, Kuyukov Vitaly said:

It looks like it smells like a grave for string theory (and higher dimensional models). 

I don’t agree. The largest colliders we currently have operate at around ~14TeV, this is nowhere near what is needed to definitively rule out String Theory and most other models with compactified dimensions - it’s in fact off by many orders of magnitude.

But I do agree that the failure to detect any of the other Higgs particles within this energy range puts serious constraints on possible supersymmetry extensions to the Standard Model. In fact I don’t know of any SUSY models that contain Higgs bosons with more than around 10TeV, so if SUSY is really a thing, then we should have seen those signatures by now - but someone please correct me if I’m wrong on this. 

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