Quantum Chemistry/Operator algebra

Overview

Operator algebra is a subset of functional analysis and involves the use of continuous linear operators on a topological vector space. Operators are mathematical objects that, when applied to functions, generate a corresponding function or value. A key concept in operator algebra is eigenvalues and eigenfunctions. An eigenfunction is a function that, when acted upon by an operator, yields a scalar multiple of itself, with the scalar being the eigenvalue.

In quantum mechanics, operator algebra forms the mathematical basis for describing physical observables in quantum systems. Operators are generally applied to wave functions and can be used to determine physical properties like position, momentum, and energy.

The most common operator is the Hamiltonian operator, as it represents the total energy of a quantum system as described by a specific wavefunction.

${\hat {H}}\psi =E\psi$

The Hamiltonian operator is made up of the kinetic and potential energy operators.

$({\hat {T}}+{\hat {V}})\psi =E\psi$

Defining the kinetic energy operator in 1D, shows that operators are not always in the form of physical quantities and can also represent simple mathematical operations such as a derivative.

${\hat {T}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}$

In this case, the second partial derivative ${\frac {\partial ^{2}}{\partial x^{2}}}$ is defined as the Laplacian operator $\nabla ^{2}$.

Example Problem & Solution

Problem

There is a known commutator between the position and momentum operators that is used to prove the Heisenberg uncertainty principle.

$[{\hat {x}},{\hat {p}}]=i\hbar$,

Provide a detailed derivation using operator algebra for this expression by applying the position and momentum commutator to an arbitrary wavefunction where:

${\hat {x}}\psi =x\psi \quad$and$\quad {\hat {p}}=-i\hbar {\frac {d}{dx}}$  

Solution

The commutator of two operators ${\hat {A}}$ and ${\hat {B}}$ is defined as:

$[{\hat {A}},{\hat {B}}]={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}$

Substituting the position and momentum operators:

$[{\hat {x}},{\hat {p}}]={\hat {x}}{\hat {p}}-{\hat {p}}{\hat {x}}$

Apply the commutator to an arbitrary wavefunction $\psi$:

$[{\hat {x}},{\hat {p}}]\psi ={\hat {x}}{\hat {p}}\psi -{\hat {p}}{\hat {x}}\psi$

Expand ${\hat {x}}{\hat {p}}\psi$ and ${\hat {p}}{\hat {x}}\psi$

${\hat {x}}{\hat {p}}\psi ={\hat {x}}\left(-i\hbar {\frac {d}{dx}}\psi \right)=-i\hbar {\hat {x}}{\frac {d}{dx}}\psi$

${\hat {p}}{\hat {x}}\psi =-i\hbar {\frac {d}{dx}}x\psi =-i\hbar (\psi +x{\frac {d}{dx}}\psi )=i\hbar \psi -i\hbar x{\frac {d}{dx}}\psi$

Substitute into the initial commutator expression and simplify:

$[{\hat {x}},{\hat {p}}]\psi =-i\hbar {\hat {x}}{\frac {d}{dx}}\psi -(-i\hbar \psi -i\hbar x{\frac {d}{dx}}\psi )$

$[{\hat {x}},{\hat {p}}]\psi =-i\hbar {\hat {x}}{\frac {d}{dx}}\psi +i\hbar x{\frac {d}{dx}}\psi +i\hbar \psi$

$[{\hat {x}},{\hat {p}}]\psi =i\hbar \psi$

Since the commutator acts on the wavefunction to give $i\hbar \psi$ we know that:

$[{\hat {x}},{\hat {p}}]=i\hbar$