Hermitian Operator

linear-algebra

Definition

Hermitian Operator

Let $T:\mathcal{H}\to \mathcal{H}$ be a linear operator on a complex inner product space $\mathcal{H}$. The operator is hermitian if it is equal to its adjoint, that is,

$$T=T^{†}$$

Equivalently:

$$⟨Tx,y⟩=⟨x,Ty⟩x,y\in \mathcal{H}$$

Spectral Representation

If $T:V\to V$ is self-adjoined, then $T$ has a spectral representation of form:

$$T=\sum _{i\ge 0}{\lambda }_i{P}_i$$

where ${\lambda }_i$ are the (real-valued) eigenvalues of of $T$ and each $P_i$ is an orthogonal projection.

Note that the convergence of this series is in the sense of the Hilbert-space norm, i.e., it holds that:

$$||Tx-\sum _{i=0}^k{\lambda }_i{P}_{i}x||\to 0\text{for }k\to \infty \forall x\in V$$

Operator Function

On the basis of the spectral representation, operator functions for self-adjoined operators can be defined as follows:
For a complex-valued function $f:\mathbb{C}\to \mathbb{C}$ and a self-adjoined operator $T$, define

$$f(T):=\sum _{i\ge 0}f({\lambda }_i)P_i$$

In particular, for $\lambda \ge 0$, $e^{i\lambda T}$ can be represented as series:

$$e^{i\lambda T}=\sum _{n\ge 0}\frac{(i\lambda T{)}^n}{n!}$$

It holds that $e^{i\lambda T}$ is unitary and

$$(e^{i\lambda T}{)}^{†}=e^{-i\lambda T}$$

where $(\cdot {)}^{†}$ denotes the adjoint operator.