Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 43 / 2

Use the usual normalization of the Super-Poincaré algebra, with ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ under the corresponding index convention:The supercharges are odd operators: they turn bosonic states into fermionic states and vice versa. If $\Pi =(-1{)}^F$ is fermion parity, this statement is $\Pi Q_{\alpha }{\Pi }^{-1}=-Q_{\alpha }$, soThe same relation holds for the conjugate supercharges.

For a finite-dimensional physical supermultiplet at fixed four-momentum with energy $E>0$, let $n_B,n_F$ count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation implySumming over the two spinor indices givesThis supertrace pairing at positive energy proves boson-fermion degeneracy in a supermultiplet for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement $E>0$ matters: a zero-energy supersymmetric vacuum can be a bosonic singlet without a paired fermionic vacuum.

If supersymmetry-breaking operators are explicitly added to the Lagrangian, the original supercharges generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical supermultiplets; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by $4E\Pi$ on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.

This is explicit versus spontaneous supersymmetry breaking. In spontaneous breaking the action still has conserved supercharges, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.

Sum the diagonal spinor entries of the supercharge anticommutator. Since $\mathrm{tr}{\sigma }^i=0$ and $\mathrm{tr}{\sigma }^0=2$, the Hamiltonian isFor a normalized vacuum, energy positivity in global supersymmetry follows fromIn the unbroken case, all supercharges annihilate the vacuum, giving $E_{\mathrm{vac}}=0$. Conversely, zero energy forces each nonnegative norm to vanish, so the vacuum is supersymmetric. The algebra fixes the additive zero of energy here. For an infinite homogeneous vacuum, use a finite-volume regulator and interpret the result as its vacuum energy density.

For spontaneous breaking of exact global supersymmetry, at least one supercharge does not annihilate the vacuum. Its norm in the preceding expression is positive, henceWith canonical kinetic terms, this is also seen in the nonnegative scalar potential, $V={\sum }_i|F_i{|}^2+\frac{1}{2}{\sum }_a{D}_a^2$: nonzero auxiliary expectation values signal breaking and positive energy density. The associated massless fermion is the Goldstino.

If “broken” instead means arbitrary explicit breaking by added operators, the exact Super-Poincaré algebra no longer fixes the full Hamiltonian, and the positive-norm argument does not constrain its vacuum energy. One can, for example, shift that explicitly broken Hamiltonian by a constant. The strict positivity conclusion therefore uses spontaneous breaking of an otherwise exact global theory, not a blanket assertion about all breaking terms. It also is not a statement about supergravity, whose scalar potential contains additional terms.

Take canonical charged chiral superfield kinetic terms, gauge coupling $e>0$, and no Fayet–Iliopoulos term, since none is specified. Elimination of the three complex auxiliary fields givesThe Abelian gauge auxiliary field obeys $D=-e(|{\phi }_{+}{|}^2-|{\phi }_{-}{|}^2)$ in this normalization. Thus the F-term scalar potential and the gauge D-term giveAll terms are nonnegative. The normalization of $e$ can be changed together with the vector-field normalization, but the relative charges and the zero-potential conditions cannot. A noncanonical Kähler potential would change the inverse-metric factors in the F-term potential; adding a Fayet–Iliopoulos term would shift $D$ and define a different model. Neither is silently introduced here.

For nonzero $\lambda$, zero potential requires both F-flatness and D-flatness:The product condition and equality of charged magnitudes together force ${\phi }_{+}={\phi }_{-}=0$. There is no condition on the neutral scalar. Consequently the global minima formOnly the neutral field can acquire a vacuum expectation value. It does not give the gauge vector a mass: its scalar kinetic term has no charged covariant derivative. Hence the gauged $U(1)$ remains unbroken in every global minimum for $\lambda \ne 0$. This is a neutral flat direction with oppositely charged chiral fields; a continuous vacuum family is not automatically gauge-symmetry breaking.

The exceptional uncoupled case $\lambda =0$ should be separated. Then only D-flatness remains, allowing $|{\phi }_{+}|=|{\phi }_{-}|=r$ and arbitrary ${\phi }_0$. For $r>0$, the charged expectations Higgs the $U(1)$; for $r=0$ it remains unbroken. The usual interacting answer assumes $\lambda \ne 0$.

Every global minimum found for nonzero $\lambda$ has $F_{+}=F_0=F_{-}=D=0$. By energy positivity in global supersymmetry, its zero energy means all supercharges annihilate it. ThereforeThe arbitrary neutral expectation $v$ is a supersymmetric modulus; the nonzero derivatives that would break supersymmetry vanish even though $v$ itself need not vanish. In the exceptional $\lambda =0$ case, every D-flatness minimum likewise has zero F- and D-auxiliaries, so supersymmetry remains unbroken even on the gauge-Higgsed branch. The supersymmetric Higgs mechanism permits internal gauge breaking without supersymmetry breaking. No extra constant or linear superpotential term is needed to find a zero-energy vacuum in this model.