Operator exponential ($e^A$)

For a bounded linear operator, the series $e^A={\sum }_{n=0}^{\infty }{A}^n/n!$ converges in operator norm. For an unbounded self-adjoint operator $H$, the spectral theorem instead defines the bounded unitary operator $e^{-iHt/\hslash }$ by applying the scalar function $e^{-iEt/\hslash }$ to its projection-valued measure. This yields unitary time evolution without assuming that every vector lies in the domain of every power of $H$. In finite dimensions it is the matrix exponential.

For a bounded skew-adjoint operator on a complex Hilbert space, the operator exponential converges in norm because its term norms are bounded by $\Vert T{\Vert }^n/n!$. Continuity of the adjoint operator gives $(e^T{)}^{\ast }=e^{T^{\ast }}=e^{-T}$. Absolute convergence permits the Cauchy product of $e^T$ and $e^{-T}$: its degree-$n$ coefficient is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]$, zero for $n>0$ and the identity for $n=0$. Both multiplication orders therefore give the identity, proving unitarity. The boundedness hypothesis makes this elementary power-series proof valid without domain issues for unbounded generators.

For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.