Linear Operator

Action of linear operators on each basis vector gives the action on any arbitrary vector.


A linear operator, \(L\), on \(\mathcal{V}\) is a linear transformation \(L:\mathcal{V}\to \mathcal{V}\).

In other words, a linear operator is a linear transformation that maps a vector space \(\mathcal{V}\) onto itself.

Special Operators

Assume \(A\), \(B\) are linear. Let \(|a\rangle\in\mathcal{V}\)

  • \(A=0\) means \(A|a\rangle = 0|a\rangle=|0\rangle = `0'\)
  • \(A=B\) implies \(A|a\rangle = B|a\rangle\)
  • \(C=A+B\) means \(C|a\rangle = A|a\rangle + B|a\rangle\)
  • \(D=AB\) means \(D|a\rangle = AB|a\rangle = A(B|a\rangle)\)
  • \(A^n|a\rangle = A\cdot A\cdots A|a\rangle\)
  • Identity: \(I=\mathbb{I}=E\) means \(I|a\rangle=|a\rangle\)

Knowing \(A^n\) allows us to define (almost) arbitrary functions of an operator from a power series, ’Functions’ Taylor series.

Functions of Operators


\(e^A = \sum_{n=0}^\infty \frac{A^n}{n!}\), \(A^0=I\).

\(\sin A, \cos A, \sqrt{A}, \ln A, \det A\)

Must show convergence of series.

Example in QM: \(U(t) = \exp\left(-\frac{i\mathcal{H}t}{\hbar}\right)\)

Representation of Operators

Suppose we have a basis \({|i\rangle}\). \(A|j\rangle\in\mathcal{V}\), so for every \(j\) we can expand \(A|j\rangle\) in our basis: \(A|j\rangle=\sum_{i=1}^nA^i_j|i\rangle\). \({A_j^i}\) is a set of \(n^2\) numbers completely characterize \(A\) but in a basis-dependent way.

So if, \(|\alpha\rangle = A|a\rangle = \sum_{i=0}^n\alpha^i|i\rangle\), \(|a\rangle=\sum_{i=0}^n a^i|i\rangle\).

\(\alpha^i = \sum_{j=0}^nA_j^ia^j = A_j^ia^j\) (in Einstein notation). In the matrix representation, \(i\) is the row and \(j\) is the column in \(A_j^i\). (I.e. the top index is the row and the bottom is the column)

Classes of Operators

Reitz Representation Theorem

Outer Products

Projection Operators

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:26