opposite sign: If operators are related by anticommutators as well as commutators, we say they are part of a Lie superalgebra. Let's say we have a quantum system described by a Hamiltonian and a set of operators . We shall call this system supersymmetric if the following anticommutation relation is valid for all : If this is the case, then we call the system's supercharges. Example Let's look at the example of a one-dimensional nonrelativistic particle with a 2D (i.e., two states) internal degree of freedom called "spin" (it's not really spin because "real" spin is a property of 3D particles). Let be an operator which transforms a "spin up" particle into a "spin down" particle. Its adjoint then transforms a spin down particle into a spin up particle; the operators are normalized such that the anticommutator . And of course, . Let be the momentum of the particle and be its position with . Let (the "superpotential") be an arbitrary complex analytic function of and define the supersymmetric operators Note that and are self-adjoint. Let the Hamiltonian where W' is the derivative of W. Also note that {Q1,Q2}=0. This is nothing other than N = 2 supersymmetry. Note that acts like an electromagnetic vector potential. Let's also call the spin down state "bosonic" and the spin up state "fermionic". This is only in analogy to quantum field theory and should not be taken literally. Then, Q1 and Q2 maps "bosonic" states into "fermionic" states and vice versa. Let's reformulate this a bit: Define and of course, and An operator is "bosonic" if it maps "bosonic" states to "bosonic" states and "fermionic" states to "fermionic" states. An operator is "fermionic" if it maps "bosonic" states to "fermionic" states and vice versa. Any operator can be expressed uniquely as the sum of a bosonic operator and a fermionic operator. Define the supercommutator [,} as follows: Between two bosonic operators or a bosonic and a fermionic operator, it is none other than the commutator but between two fermionic operators, it is an anticommutator. Then, x and p are bosonic operators and b, , Q and are fermionic operators. Let's work in the Heisenberg picture where x, b and are functions of time. Then, This is nonlinear in general: i.e., x(t), b(t) and do not form a linear SUSY representation because isn't necessarily linear in x. To avoid this problem, define the self-adjoint