Supersymmetric quantum mechanics (SUSY QM)

Supersymmetric quantum mechanics factors partner Hamiltonians through first-order operators. Its odd supercharges anticommute to the Hamiltonian; on a Riemannian manifold, the Hilbert space can be realized by square-integrable differential forms with supercharges given by a differential and its adjoint.

The candidate zero modes are $e^{-{\int }^{x}w}$ in the minus sector and $e^{+{\int }^{x}w}$ in the plus sector; this follows from the positive squared-norm expressions for their energy. Their membership in the Hilbert space determines whether supersymmetry is unbroken. For $w=x$ only the first is normalizable. For $w=x^2+a$, $a>0$, neither is normalizable, while both partner potentials grow quartically and have discrete spectra. Thus their strictly positive lowest energy is paired and supersymmetry is broken, with zero Witten index.

With a well-defined regulated supertrace, positive-energy states of supersymmetric quantum mechanics are paired by an odd invertible charge. Their opposite fermion parities cancel, leaving $I=n_B^0-n_F^0$, independent of $\beta$. A nonzero index proves the existence of a supersymmetric vacuum; zero does not prove its absence. Discrete-spectrum or suitable infrared conditions matter, because continuum thresholds and states escaping to infinity can invalidate naive trace manipulations or deformation invariance.

For a real smooth function $\chi$ on Euclidean space, put $A_a={\partial }_a-{\partial }_a\chi$ and $q={\sum }_a{\psi }_a^{†}{A}_a$. The adjoint operator for the usual $L^2$ inner product is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. Since $[A_a,A_b]=0$, the canonical anticommutation relations give $q^2=0$. Expanding $H=\{q,q^{†}\}$ and using $[A_a,A_b^{†}]=-2{\partial }_a{\partial }_b\chi$ givesIn the empty fermionic Fock space sector the last term vanishes; in the filled sector ${\psi }_a^{†}{\psi }_b={\delta }_{ab}$. Consequently the two scalar Hamiltonian operators are $H_0=-\Delta +|\nabla \chi {|}^2+\Delta \chi$ and $H_d=-\Delta +|\nabla \chi {|}^2-\Delta \chi$. On the compatible self-adjoint operator domain the Hamiltonian operator is nonnegative because $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2$.

For the gradient superpotential Hamiltonian in supersymmetric quantum mechanics, $H_0={\sum }_a{A}_a^{†}{A}_a$. A zero-energy normalizable wavefunction therefore satisfies $A_{a}f=0$ for every $a$, because its energy is ${\sum }_a\Vert A_{a}f{\Vert }^2$. Equivalently ${\partial }_a(e^{-\chi }f)=0$, so on a connected configuration space $f=C{e}^{\chi }$. It exists in the Hilbert space precisely when $e^{\chi }$ is square-integrable. In three dimensions with radial $\chi$, this requires $4\pi {\int }_0^{\infty }{r}^2{e}^{2\chi (r)}dr<\infty$. The sign depends on the convention $A=\partial -\partial \chi$; the filled sector instead has the possible zero mode $e^{-\chi }$.

For $d$ fermionic creation operators with the canonical anticommutation relations, the Hilbert space $L^2({\mathbb{R}}^d)\otimes {\Lambda }^{\bullet }{\mathbb{C}}^d$ splits into ${\mathcal{H}}_n=L^2({\mathbb{R}}^d)\otimes {\Lambda }^n{\mathbb{C}}^d$, $0\le n\le d$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$; its internal multiplicity is $\left(\genfrac{}{}{0ex}{}{d}{n}\right)$. An odd supercharge $Q=q+q^{†}$, with $q$ raising $n$ and $q^2=0$, gives $H=\{q,q^{†}\}$ and preserves these sectors.

On an eigenspace with $E>0$, $Q/\sqrt{E}$ is an odd involution pairing bosonic and fermionic states. More precisely,gives an orthogonal decomposition of each state into the images of $q$ and $q^{†}$: their inner product vanishes because $(q^{†}{)}^2=0$. If $u\in \mathrm{im}{q}^{†}$, then $q^{†}u=0$, $q^{†}qu=Eu$ and $\Vert qu{\Vert }^2=E\Vert u{\Vert }^2$. Thus $q/\sqrt{E}$ pairs such a state in degree $n$ with one in degree $n+1$, and $q^{†}/\sqrt{E}$ is the inverse. States with zero energy are killed by both operators and may remain unpaired. The pairing applies to positive spectral subspaces as well; discreteness is not implied.

Let $\chi =\chi (r)$, $r=|x|$, and let $f(r)|0⟩$ be a normalized eigenstate of $H_0$ with $E>0$, where $|0⟩$ is the Clifford vacuum. Acting with the supercharge differentiates only its wavefunction:The normalized partner is $E^{-1/2}(f^{\prime }-{\chi }^{\prime }f){\psi }_r^{†}|0⟩$. Indeed $[H,Q]=0$, and $\Vert Qf|0⟩{\Vert }^2=E\Vert f|0⟩{\Vert }^2$; acting with $Q$ again returns $\sqrt{E}$ times the original state. The expression at $r=0$ is interpreted by the regular extension of the original state. Although its coefficient depends only on $r$, its degree-one components contain the radial unit vector.

The fermionic Fock space of $n$ modes is the exterior algebra ${\Lambda }^{\bullet }{\mathbb{C}}^n$. Creation acts by exterior multiplication and annihilation by contraction.

An orthogonal projection $P$ splits a one-particle Hilbert space into empty and filled modes. Take completed exterior Fock spaces of empty particles and of holes in the filled modes, with the graded tensor product enforcing the canonical anticommutation relations between both factors. The vacuum is killed by particle annihilation in $PH$ and by particle creation in $(I-P)H$. Only finite particle-hole excitations are needed for a dense basis. Normal ordering subtracts the filled-vacuum expectation from bilinear observables.

In charge $q$, let ${\Omega }_q$ fill all integer modes below $q$. Its rotation energy is $q(q-1)/2$. The negative Heisenberg current algebra of a fermion on the circle modes create mutually commuting oscillators $b_n^{\ast }=J_{-n}/\sqrt{n}$ with $[b_n,b_m^{\ast }]={\delta }_{nm}$. Their normalized occupation monomials applied to ${\Omega }_q$ are orthonormal. The Jacobi triple product identifies the fermionic charge-energy generating function with ${\sum }_q{u}^q{t}^{q(q-1)/2}{\prod }_{n>0}(1-t^n{)}^{-1}$, proving completeness in each finite-dimensional energy space. Thus the rotation Hamiltonian is $Q(Q-1)/2+{\sum }_{n>0}n{b}_n^{\ast }{b}_n$.

For integer modes filled at indices below zero, set $E_{rs}=c_r^{†}{c}_s-{\delta }_{rs}{1}_{r<0}$ and $J_n={\sum }_r{E}_{r,r+n}$ on the finite-energy domain. The canonical anticommutation relations give a central term ${\delta }_{st}{\delta }_{ur}(1_{r<0}-1_{s<0})$ in $[E_{rs},E_{tu}]$. Summing across the filled-empty boundary gives $[J_n,J_m]=n{\delta }_{n+m,0}I$, with $J_n^{\ast }=J_{-n}$. The current's level is the coefficient of this central term.

A Fourier mode of the normal-ordered fermion density is a fermionic current. On the finite-energy domain its defining sum has finitely many nonzero actions on each occupation vector. The modes satisfy $J_n^{\ast }=J_{-n}$ and $[J_n,J_m]=n{\delta }_{n+m,0}I$. The zero mode is charge; positive modes annihilate charge ground states and negative modes create the oscillators in the fermion-boson correspondence on the circle.

For the irreducible representation of the canonical anticommutation relations determined by $P$, a unitary one-particle map $u$ is unitarily implementable precisely when $P-uP{u}^{\ast }$ is Hilbert-Schmidt, equivalently when $[P,u]$ is Hilbert-Schmidt. The implementer is unique up to a scalar phase. The equivalent representation criterion for two polarizations is that their projections differ by a Hilbert-Schmidt operator. This yields a continuous projective unitary representation of the restricted unitary group.

A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

A Clifford vacuum is a state annihilated by all the chosen fermionic annihilation operators. In a massless supermultiplet it labels the starting helicity, and applying fermionic raising and lowering operators constructs the remaining states.

Fermionic creation and annihilation operators satisfy $\{a_i,a_j^{†}\}={\delta }_{ij}$ and vanishing equal-type anticommutators. These relations make each mode either empty or singly occupied.

For constant internal matrices acting on Dirac fields, the equal-time canonical anticommutation relations givewith the corresponding spatial delta functions for fields. The quartic terms cancel after fermion reordering. Integrating proves the displayed algebra for well-defined global Noether charges, with normal ordering and a regulator that preserves the nonanomalous internal symmetry. Orthogonal chiral projectors make the left/right mixed commutators zero. This reduces current-charge closure to a finite matrix calculation.

The equal-time canonical anticommutation relations implyApply $-i{\alpha }_{bc}^k{\partial }_{z^k}+m{\beta }_{bc}$, set $\mathbf{z}=\mathbf{y}$, and integrate over $\mathbf{y}$. The mass term cancels, leaving $i\int d^{3}y{\psi }^{†}{\alpha }^k\psi {\partial }_{y^k}{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. Integration by parts proves the title formula as a distribution identity. The Heisenberg equation of motion then yields ${\partial }_{t}n=-\nabla \cdot \mathbf{j}$ for the physical Dirac current $j^k=\bar{\psi }{\gamma }^k\psi$. The factor $-i$ is required because the commutator of Hermitian operators is anti-Hermitian.

For normalized triplets $(E^{+},n,e{)}_L$ and $(E^{+},N,e{)}_R$, let the raising matrix have ones in entries $(1,2)$ and $(2,3)$. Its commutator with its adjoint is $\mathrm{diag}(1,0,-1)$. Neutral mixing in $n=c_{\alpha }\nu +s_{\alpha }N$ leaves this result unchanged because $c_{\alpha }^2+s_{\alpha }^2=1$. With current normalization using $1\mp {\gamma }^5=2P_{L,R}$, fermionic bilinear charge algebra gives the displayed proportionality to the leptonic electric charge $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$. Other current normalizations change the numerical prefactor, not closure. Equal-time algebra does not itself imply conservation of a fermionic weak charge in a symmetry-broken theory.

For mode normalization $1/\sqrt{2E_{\mathbf{p}}}$, particle and antiparticle operators satisfy $\{a_{\mathbf{p}}^s,a_{\mathbf{q}}^{r†}\}=\{b_{\mathbf{p}}^s,b_{\mathbf{q}}^{r†}\}=(2\pi {)}^3{\delta }^{sr}{\delta }^3(\mathbf{p}-\mathbf{q})$. The Dirac spinor completeness relations imply ${\sum }_s[u_s(\mathbf{p})u_s^{†}(\mathbf{p})+v_s(-\mathbf{p})v_s^{†}(-\mathbf{p})]=2E_{\mathbf{p}}I$. Inserting this into the mode expansion of a Dirac field gives the displayed local anticommutator. The reversed momentum in the antiparticle completeness term is essential.

The fermion number operator counts occupied fermionic modes. Its parity $(-1{)}^F$ is positive on even exterior degree and negative on odd exterior degree.

A supertrace is the trace weighted by fermion parity. Supersymmetry makes paired positive-energy states cancel in a supertrace.

An equivariant supertrace inserts a symmetry $f$ as well as fermion parity, $\mathrm{Tr}((-1{)}^{F}fA)$. In a path integral, $f$ produces boundary conditions twisted by its action.

A fermionic path integral integrates Grassmann-valued histories. Its Gaussian integral is a determinant, in contrast with the inverse determinant produced by a bosonic Gaussian integral.

The Berezin integral is differentiation with respect to a Grassmann variable: $\int d\theta 1=0$ and $\int d\theta \theta =1$. A finite-dimensional fermionic Gaussian integral gives a determinant or Pfaffian.

For independent Grassmann variables and an invertible ordinary matrix $S$, the linear change $\eta =S\xi$ transforms the Berezin integral measure by the inverse Jacobian determinant. Indeed, the highest-degree monomial transforms by $\det S$, and coefficient extraction must undo that factor. For paired variables $\eta =S\xi$ and $\bar{\eta }=\bar{\xi }{S}^{-1}$ the two factors cancel. The order of odd variables and differentials fixes the overall sign and must be kept consistent.

The short-time limit rescales a Euclidean time circle toward zero size. Supersymmetry cancels paired nonzero-mode determinants, leaving an integral over constant or symmetry-fixed zero modes.

For a real function $h$, the twisted de Rham differential is $d_h=d+dh\wedge =e^{-h}d{e}^h$. Its adjoint and $d_h$ define a supersymmetric Hamiltonian, and its square-integrable cohomology describes zero-energy states.

Partner Hamiltonians reverse the order of two first-order factors and share corresponding positive-energy states.

A Pöschl-Teller potential is a solvable one-dimensional potential built from ${\mathrm{sech}}^{2}x$. The attractive memberis reflectionless when $n$ is a positive integer.

LetThenThus the one-soliton Schrodinger operator is paired with the free Schrodinger operator. Its normalizable state annihilated by $A$ is proportional to $\mathrm{sech}(\chi x)$.