(also self-adjoint operator or Hermitian operator)
A coordinate free expression (and an infinite dimensional generalization) of self adjoint matrixs.
I think that in finite dimensional case, their matrix with respect to an orthonormal basis is a self adjoint matrix.
By the way, observe that given an operator $A$, the sum or product of $A$ with its conjugate transpose $A^{\dagger}$ is a Hermitian operator (easily checked by computation).
There is a version of spectral theorem for these operators.
One important property of such operators is that the eigenvalues of a self-adjoint operator are necessarily real. Indeed, if $k$ is any eigenvalue with corresponding (normalized) eigenvector $v$, we see
$$ k = k\langle v,v \rangle = \langle kv, v \rangle = \langle Sv, v \rangle = \langle v,Sv \rangle = \langle v, kv \rangle = \overline k \langle v, v \rangle = \overline k $$showing that $k$ is real.
If we have a basis of eigenvectors we can construct the associated resolution of identity.
They correspond to observables in Quantum Mechanics.
If we multiply them by the complex unit $i$ we obtain antiHermitian operators, which constitute the "Lie algebra" of the "group" of unitary operators. That is, given a self-adjoint operator $A$ in a Hilbert space then $e^{iA}$ is an unitary operator. In fact, its Hermitian adjoint is $e^{-iA^\dagger} = e^{-iA}$ because $A$ is Hermitian ($A = A^\dagger$). Now, we compute $e^{iA} e^{-iA}$ and $e^{-iA} e^{iA}$:
$$ e^{iA} e^{-iA} = e^{iA - iA} = e^0 = I $$ $$ e^{-iA} e^{iA} = e^{-iA + iA} = e^0 = I $$where we use the fact that $[A,-A]=0$.
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Author of the notes: Antonio J. Pan-Collantes
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