Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 42 (Introduction to Supersymmetry)

The canonical anticommutation relations realize the internal states as a fermionic Fock space ${\Lambda }^{\bullet }{\mathbb{C}}^3$. Choosing a Clifford vacuum $|0⟩$ annihilated by all ${\psi }_a$, the states are functions multiplying $|0⟩$, ${\psi }_a^{†}|0⟩$, ${\psi }_a^{†}{\psi }_b^{†}|0⟩$ with $a<b$, or ${\psi }_1^{†}{\psi }_2^{†}{\psi }_3^{†}|0⟩$. Thus the four fermion-number sectors in supersymmetric quantum mechanics arewith internal dimensions $1,3,3,1$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$. The sectors $n=0,2$ have even fermion parity, and $n=1,3$ have odd fermion parity.

Write $q={\sum }_a{\psi }_a^{†}{A}_a$, so that the supercharge is $Q=q+q^{†}$. The commutator $[A_a,A_b]$ is zero because the two mixed partial derivatives of $\chi$ agree. The canonical anticommutation relations then imply $q^2=(q^{†}{)}^2=0$ andThe Hamiltonian operator preserves each sector, while $q$ raises and $q^{†}$ lowers its degree by one. For a state in its operator domain, $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2\ge 0$, so the spectrum is nonnegative. If $Hv=Ev$ with $E>0$, then $Qv/\sqrt{E}$ is a state of the same energy and opposite fermion parity, and $Q(Qv/\sqrt{E})=\sqrt{E}v$. This is an odd involution pairing bosonic and fermionic states and proves equal even and odd multiplicities at every positive eigenvalue.

There is also a more precise adjacent-sector pairing. On the positive-energy eigenspace,The two summands are orthogonal since $q^2=(q^{†}{)}^2=0$. A state $u$ in the image of $q^{†}$ obeys $q^{†}u=0$, $q^{†}qu=Eu$, so $q/\sqrt{E}$ maps it to a normalized partner in the next sector, with inverse $q^{†}/\sqrt{E}$. Thus positive-energy doublets lie in ${\mathcal{H}}_0\oplus {\mathcal{H}}_1$, ${\mathcal{H}}_1\oplus {\mathcal{H}}_2$, or ${\mathcal{H}}_2\oplus {\mathcal{H}}_3$. In particular, every positive-energy state of ${\mathcal{H}}_0$ has a partner in ${\mathcal{H}}_1$, and every one of ${\mathcal{H}}_3$ has a partner in ${\mathcal{H}}_2$; the remaining middle-sector states pair with each other. There is no requirement that all four individual sector spectra coincide. Zero-energy states satisfy $qv=q^{†}v=Qv=0$ and can be unpaired. No confinement assumption was given: a continuous spectrum is possible, with the same even/odd pairing on its positive spectral subspaces.

For real $\chi$, the adjoint operator is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. In ${\mathcal{H}}_0$, every ${\psi }_a^{†}{\psi }_b$ vanishes; in the one-dimensional internal space of ${\mathcal{H}}_3$, it equals ${\delta }_{ab}$. The two scalar Hamiltonian operators are consequentlyThese are the extreme sectors of the gradient superpotential Hamiltonian in supersymmetric quantum mechanics.

For radial $\chi$ and a normalized state $v_0=f(r)|0⟩$ with $E>0$, $q^{†}{v}_0=0$ and $A_{a}f=(x_a/r)(f^{\prime }-{\chi }^{\prime }f)$. The radial superpartner in three-dimensional supersymmetric quantum mechanics therefore has the simple formIt belongs to ${\mathcal{H}}_1$; $H{v}_1=E{v}_1$ follows from $[H,Q]=0$, and its wavefunction normalization follows from $\Vert Q{v}_0{\Vert }^2=⟨v_0,H{v}_0⟩=E$. Also $Q{v}_1=\sqrt{E}{v}_0$. At the origin the formula is interpreted by regular extension; its degree-one components carry the radial unit vector rather than being three independent radial scalars.

For a zero mode in the fermion-vacuum sector, the factorization of $H_0$ gives $0={\sum }_a\Vert A_{a}f{\Vert }^2$. Hence ${\partial }_{a}f=({\partial }_a\chi )f$, or ${\partial }_a(e^{-\chi }f)=0$. The requested wavefunction isA nonzero normalizable wavefunction exists only if this integral is finite. The positive sign in the exponential is fixed by the given convention $A_a={\partial }_a-{\partial }_a\chi$.

Odd elements of a Grassmann algebra anticommute, and in particular ${\psi }_i^2=0$. For variations with fixed endpoints, move each odd variation to the left before applying integration by parts. For example,The interaction varies as ${\delta }_{{\psi }_1}({\psi }_1{\psi }_2)=\delta {\psi }_1{\psi }_2$ and ${\delta }_{{\psi }_2}({\psi }_1{\psi }_2)=-\delta {\psi }_2{\psi }_1$. Thus variation of the action gives the Euler-Lagrange equationsAll products of $U$ and its derivatives commute with the odd variables. These signs are essential for this variant of Grassmann-valued classical supersymmetric mechanics.

The fermion bilinear is constant along a solution:For the energy, direct differentiation now givesSimilarly the given supercharge, denoted $Q_1$, satisfiesA second conserved supercharge issince ${\dot{Q}}_2=(\ddot{x}-U{U}^{\prime }){\psi }_2=U^{\prime \prime }{\psi }_1{\psi }_2{\psi }_2=0$. The two charges are independent functions on the unrestricted phase space: their coefficient matrix on $({\psi }_1,{\psi }_2)$ has determinant ${\dot{x}}^2-U^2$, and is invertible wherever its even body is nonzero. They can become dependent on a restricted family of trajectories.

Under $\dot{x}=U$ and ${\psi }_2=-{\psi }_1$, the bilinear ${\psi }_1{\psi }_2$ vanishes. Differentiating the first-order equation gives $\ddot{x}=U^{\prime }\dot{x}=U{U}^{\prime }$, which is the bosonic Euler-Lagrange equation. The first fermionic equation becomes ${\dot{\psi }}_1=-U^{\prime }{\psi }_1$; differentiating ${\psi }_2=-{\psi }_1$ gives ${\dot{\psi }}_2=U^{\prime }{\psi }_1$, so the second fermionic equation holds too. This establishes the first-order zero-energy solution in Grassmann supersymmetric mechanics without imposing an additional equation.

Let $x_0=x(0)$ and ${\psi }_{10}={\psi }_1(0)$. For $U=1+cx$ and $c\ne 0$, solve $d(x+1/c)/dt=c(x+1/c)$ and ${\dot{\psi }}_1=-c{\psi }_1$ to obtainFor $c=0$, the nonsingular limiting answer is $\boxed{x(t)=x_0+t,{\psi }_1(t)={\psi }_{10},{\psi }_2(t)=-{\psi }_{10}}$. These formulas also hold for even initial positions in the Grassmann algebra by ordinary polynomial evaluation. Finally, $U(x(t))=(1+c{x}_0)e^{ct}$, so the conserved quantities areThe individual odd supercharges generally do not vanish, although their sum vanishes on this restricted solution.

Use left Grassmann derivatives and define the supersymmetric covariant derivatives $D_{\pm }={\partial }_{{\theta }^{\pm }}-i{\theta }^{\pm }{\partial }_{\pm }$. Multiplication in the indicated order gives exactly the printed differential operator:Thus the equation is $i{D}_{-}{D}_{+}\Phi =W(\Phi )$ in two-dimensional N=(1,1) superspace.

Under a Lorentz boost, take $x^{\pm }\mapsto e^{\pm \omega }{x}^{\pm }$ and ${\theta }^{\pm }\mapsto e^{\pm \omega /2}{\theta }^{\pm }$. The derivatives ${\partial }_{\pm }$ have boost weights $\mp 1$, and $D_{\pm }$ have weights $\mp 1/2$. Consequently $D_{-}{D}_{+}$ has weight zero, and is a Lorentz scalar. Both $\Phi$ and the ordinary function $W(\Phi )$ are scalar superfields, so the equation is Lorentz invariant.

For supersymmetry, $\{{\partial }_{\theta },\theta \}=1$ and anticommutation of distinct odd coordinates giveMoving either supercharge through the two odd derivatives introduces two minus signs, hence $[Q_t,D_{-}{D}_{+}]=0$. If $\delta \Phi ={ϵ}^{+}{Q}_{+}\Phi +{ϵ}^{-}{Q}_{-}\Phi$, with odd constant parameters, the chain rule for an even scalar superfield givesTherefore a solution is mapped to a solution, proving supersymmetry invariance.

To compute the supersymmetric Liouville equation, order the odd coordinates as ${\theta }^{-}{\theta }^{+}$ and anticommute the fermionic component fields with both coordinates. In particular, ${\partial }_{{\theta }^{+}}({\theta }^{-}{\theta }^{+})=-{\theta }^{-}$. Direct application of the operator givesLet $N=\Phi -\varphi$. Only the two products of its degree-one terms survive in $N^2$; for example, $(i{\theta }^{-}{\psi }_{+})(i{\theta }^{+}{\psi }_{-})={\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$. Hence $N^2=2{\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$ and $N^3=0$. The exponential function expands exactly asEquating the constant terms fixes the auxiliary field, $\boxed{F=e^{\varphi }}$. The linear and quadratic terms then give the requested component field equations of motion:The signs here follow from the printed operator, the printed component expansion, and left Grassmann derivatives; no convention change has been made in eliminating the auxiliary field.

A two-component Weyl spinor is a complex spinor transforming in one of the two inequivalent group representations $(1/2,0)$ and $(0,1/2)$ of the Lorentz spinor double cover $\mathrm{SL}(2,\mathbb{C})$. An undotted left-handed field ${\chi }_{\alpha }$ transforms by a matrix $M\in \mathrm{SL}(2,\mathbb{C})$; its complex conjugate ${\bar{\chi }}_{\dot{\alpha }}$ transforms by $M^{\ast }$. Dotted and undotted indices distinguish the two chirality representations. Raising a dotted index uses the invariant epsilon tensor and converts the complex conjugate transformation into the equivalent inverse-adjoint form. Rotations act by the same SU(2) generators on the two chiralities, while boosts have opposite generators, so the two complex representations cannot be identified by a complex-linear Lorentz invariant map.

The antisymmetric invariant tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}_{\dot{\alpha }\dot{\beta }}$ raise and lower spinor indices; $M^TϵM=ϵ$ proves invariance of the contracted bilinear $\chi \psi ={\chi }^{\alpha }{\psi }_{\alpha }$. The components of a fermionic Weyl spinor are odd Grassmann variables, so this bilinear is symmetric under exchange of the two spinors: the antisymmetric epsilon sign cancels the Grassmann sign. In particular, choosing ${ϵ}_{12}=1$ and ${ϵ}^{12}=-1$ gives $\chi \chi =2{\chi }_1{\chi }_2$, generally nonzero. This Weyl spinor bilinear exchange identity permits a Majorana mass term for a single suitable Weyl spinor. With mostly-minus metric signature and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, the free massless kinetic term is $i{\chi }^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi$, and its field equation of motion is $\boxed{{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi =0}$.

Use mostly-minus metric signature, $\eta =\mathrm{diag}(1,-1,-1,-1)$. The Weyl sigma matrices are ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, with the three Pauli matricesThe Pauli matrix multiplication law ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ immediately gives ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$: the $00$ case is $2I$, the mixed cases cancel, and the spatial cases give $-2{\delta }_{ij}I$. The spinor indices of ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ express the correspondence between a Lorentz four-vector and a Hermitian $2\times 2$ matrix. For $X=x^{0}I+x^i{\sigma }_i$, $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$; $X\mapsto MX{M}^{†}$ preserves this determinant for $M\in \mathrm{SL}(2,\mathbb{C})$, yielding the Lorentz spinor double cover.

In four-dimensional N=1 supersymmetry, the odd supercharges $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ transform as conjugate Weyl spinors. Their Super-Poincaré algebra isThe ordinary translation and Lorentz generators satisfy the Poincare algebra, and the Lorentz generators act on the supercharges by the corresponding group representations on Weyl spinors. Tracing the mixed anticommutator gives $P_0=\frac{1}{4}{\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}$, proving energy positivity in global supersymmetry. For a massive rest-frame state, $P_{\mu }=(M,0,0,0)$, the rescaled charges $Q_{\alpha }/\sqrt{2M}$ obey the canonical anticommutation relations. Acting with their adjoints builds a supermultiplet containing states of alternating fermion parity.

For two independent left-handed Weyl spinors $\chi$ and $\eta$, a Dirac mass term isThe four-component Dirac field ${\Psi }_D=({\chi }_{\alpha },{\bar{\eta }}^{\dot{\alpha }}{)}^T$ has independent particle and antiparticle degrees of freedom; after a phase redefinition making $m$ real, the same term is $-m{\bar{\Psi }}_D{\Psi }_D$. The two left-handed spinors can carry opposite charges under a conserved U(1) symmetry, $\chi \mapsto e^{iq\omega }\chi$, $\eta \mapsto e^{-iq\omega }\eta$. A gauge-invariant Dirac mass term contracts conjugate gauge group representations.

For one left-handed Weyl spinor, a Majorana mass term instead has the formIts self-contraction need not vanish, by the Weyl spinor bilinear exchange identity. The factor $1/2$ avoids double counting identical fields in variation. The associated four-component Majorana fermion ${\Psi }_M=({\chi }_{\alpha },{\bar{\chi }}^{\dot{\alpha }}{)}^T$ satisfies ${\Psi }_M^c={\Psi }_M$, so its particle is its own antiparticle. A Majorana mass term breaks continuous fermion-number phase symmetry for $\chi$, and is allowed only if its gauge indices admit an invariant symmetric contraction, as for a neutral field or a real gauge group representation. An isolated field of nonzero U(1) charge cannot have it. More generally, the mass matrix of several left-handed Weyl spinors is symmetric, since ${\chi }_i{\chi }_j={\chi }_j{\chi }_i$.

The barred notation denotes the antichiral supersymmetric covariant derivative. Write barred spinor indices as dotted indices, and use the stated plus-sign convention with left Grassmann derivatives:Their supercovariant derivative algebra with left derivatives gives $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$ and $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=2i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The constraint $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ defines a chiral superfield, also called an L-superfield.

To solve the constraint, put $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$. For a left Grassmann derivative, ${\partial }_{{\bar{\theta }}^{\dot{\alpha }}}(\theta {\sigma }^{\mu }\bar{\theta })=-{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }$. Therefore ${\bar{D}}_{\dot{\alpha }}{y}^{\mu }=0$ and ${\bar{D}}_{\dot{\alpha }}{\theta }^{\beta }=0$, so $\bar{D}$ becomes ordinary differentiation with respect to $\bar{\theta }$ at fixed $y$. Its general solution is the chiral-superfield component expansionHere $A$ is a complex scalar, $\chi$ a Weyl spinor, and $F$ a complex auxiliary field. Before imposing field equations of motion this is four real bosonic and four real fermionic components. Chirality means independence of $\bar{\theta }$ at fixed $y$, rather than at fixed $x$. The differential supercharges $Q_{\alpha }={\partial }_{{\theta }^{\alpha }}-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}{\partial }_{\mu }$ and ${\bar{Q}}_{\dot{\alpha }}={\partial }_{{\bar{\theta }}^{\dot{\alpha }}}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$ anticommute with both covariant derivatives, so a supersymmetry transformation preserves the chiral constraint.

Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.