PlanetPhysics/Lie Algebras

Lie algebras in quantum theories

Continuous symmetries often have a special type of underlying continuous group, called a Lie group . Briefly, a Lie group $G$ is generally considered having a (smooth) $C^{\infty }$ manifold structure, and acts upon itself smoothly. Such a globally smooth structure is surprisingly simple in two ways: it always admits an Abelian fundamental group, and seemingly also related to this global property, it admits an associated, unique--as well as finite--Lie algebra that completely specifies locally the properties of the Lie group everywhere. There is a finite Lie algebra of quantum commutators and their unique (continuous) Lie groups. Thus, Lie algebras can greatly simplify quantum computations and the initial problem of defining the form and symmetry of the quantum Hamiltonian subject to boundary and initial conditions in the quantum system under consideration. However, unlike most regular abstract algebras, a Lie Algebra is not associative, and it is in fact a vector space. It is also perhaps this feature that makes the Lie algebras somewhat compatible, or consistent, with quantum logics that are also thought to have non-associative, non-distributive and non-commutative lattice structures.

General Lie algebra definition and Examples

A Lie algebra over a field $k$ is a vector space ${\mathfrak {g}}$ together with a bilinear map Failed to parse (syntax error): {\displaystyle [\ ,\] : \mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}} , called the Lie bracket and defined by the association $(x,y)\mapsto [x,y]$. The bracket is subject to the following two conditions:

1. $[x,x]=0$ for all $x\in {\mathfrak {g}}$.
2. The Jacobi identity: $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$ for all $x,y,z\in {\mathfrak {g}}$.

{\mathbf Examples:}

Any vector space can be made into a Lie algebra simply by setting $[x,y]=0$ for all vectors $x,y$. Such a Lie algebra is an Abelian Lie algebra.

If $G$ is a Lie group, then the tangent space at the identity forms a Lie algebra over the real numbers.

$\mathbb {R} ^{3}$ with the cross product operation is a non-Abelian three dimensional (3D) Lie algebra over $\mathbb {R}$.

Consider next the annihilation operator $a$ and the creation operator $a\dagger$ in quantum theory. Then, the Hamiltonian $H$ of a harmonic quantum oscillator, together with the operators $a$ and $a\dagger$ generate a 4--dimensional (4D) Lie algebra with commutators: Failed to parse (syntax error): {\displaystyle [H, a] = −a} , $[H,a\dagger ]=a\dagger ,$ and $[a,a\dagger ]=I$. This Lie algebra is solvable and generates after repeated application of $a\dagger$ all of the eigenvectors of the quantum harmonic oscillator.