0.6 Operators

A linear map between two vector spaces \mathcal{H} and \mathcal{K} is a function A\colon\mathcal{H}\to\mathcal{K} that respects linear combinations: A(c_1|v_1\rangle+c_2|v_2\rangle)=c_1 A|v_1\rangle+c_2 A|v_2\rangle for any vectors |v_1\rangle,|v_2\rangle and any complex numbers c_1,c_2. We will focus mostly on endomorphisms, that is, maps from \mathcal{H} to \mathcal{H}, and we will call them operators. The symbol \mathbf{1} is reserved for the identity operator that maps every element of \mathcal{H} to itself (i.e. \mathbf{1}|v\rangle=|v\rangle for all |v\rangle\in\mathcal{H}). The product AB of two operators A and B is the operator obtained by first applying B to some ket |v\rangle and then A to the ket which results from applying B: (AB)|v\rangle = A(B|v\rangle). The order does matter: in general, AB\neq BA. In the exceptional case in which AB=BA, one says that these two operators commute. The inverse of A, written as A^{-1}, is the operator that satisfies AA^{-1}=\mathbf{1}=A^{-1}A. For finite-dimensional spaces, one only needs to check one of these two conditions, since any one of the two implies the other, whereas, on an infinite-dimensional space, both must be checked. Finally, given a particular basis, an operator A is uniquely determined by the entries of its matrix, defined by A_{ij}=\langle i|A|j\rangle. The adjoint, or Hermitian conjugate, of A, denoted by A^\dagger, is defined by the relation \begin{gathered} \langle i|A^\dagger|j\rangle = \langle j|A|i\rangle^\star \\\text{for all $|i\rangle,|j\rangle\in\mathcal{H}$}. \end{gathered}

An operator A is said to be

  • normal if AA^\dagger = A^\dagger A,
  • unitary if AA^\dagger = A^\dagger A = \mathbf{1},
  • Hermitian (or self-adjoint) if A^\dagger = A.

Any physically admissible evolution of an isolated quantum system is represented by a unitary operator. Note that unitary operators preserve the inner product: given a unitary operator U and two kets |a\rangle and |b\rangle, and defining |a'\rangle=U|a\rangle and |b'\rangle=U|b\rangle, we have that \begin{gathered} \langle a'|=\langle a|U^\dagger \\\langle b'|=\langle b|U^\dagger \\\langle a'|b'\rangle=\langle a|U^\dagger U|b\rangle=\langle a|\mathbf{1}|b\rangle=\langle a|b\rangle. \end{gathered} Preserving the inner product implies preserving the norm induced by this product, i.e. unit state vectors are mapped to unit state vectors, i.e. unitary operations are the isometries of the Euclidean norm.