Difference between real operators and Hermitian operators in quantum mechanics

Question

I'm reading some lecture notes on quantum mechanics, while describing the rigid rotor in bra-ket notation, the author mentions the parity operator $\hat{P}$ acting on kets as $\hat{P} \left \lvert m \right \rangle = \left \lvert -m \right \rangle$, then the author says that the operator is real by showing that $\hat{P}=\hat{P}^+$. From what I learnt that is also a property (or even the definition) of Hermitian operators. So are real and Hermitian operators synonyms and is the "real" naming referred to the realness of the operator's eigenvalues?

Answer 1

In order to avoid unnecessary mathematical complications, let us consider a finite-dimensional complex Hilbert space $\mathcal H$, i.e. a complex vector space with dimension $n\lt \infty$ and a (sesquilinear) scalar product $\langle . | .\rangle : \mathcal H \times \mathcal H \to \mathbb{C}$. A linear operator $A: \mathcal H \to \mathcal H$ is called hermitean (or self-adjoint) if the relation $\langle f | Ag\rangle =\langle Af |g\rangle $ holds for all vectors $f,g \in \mathcal H$. As an immediate consequence, the eigenvalues of $A$ are real and there exists an orthonormal basis of eigenvectors of $A$ (spectral theorem for hermitean operators). The converse is not true: an operator with a purely real spectrum is not necessarily hermitean. As an example, take $\mathcal H = \mathbb C^2$ with the standard scalar product $\langle x | y\rangle = \sum\limits_{i=1}^2x_i^*y_i$ and the non-hermitean linear operator $B: \mathbb C^2 \to \mathbb C^2$ given by $$B= \begin{pmatrix} 1 & 1 \\0 & 1\end{pmatrix}, $$ having the real eigenvalue $1$.

The nonstandard terminology of "real" operator instead of hermitean (or self-adjoint) operator is highly confusing and therefore strongly discouraged. Take for instance the Pauli matrix $$\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix},$$ being a hermitean operator with real eigenvalues $\pm 1$ but purely imaginary off-diagonal matrix elements $\pm i$. Taking advantage of the spectral theorem, one can find a suitable unitary transformation $$U=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}, \qquad U^\dagger U = U U^\dagger = \mathbb{1}_2,$$ where the rows are (normalized) eigenvectors of $\sigma_y$, such that $$U^\dagger \sigma_y U = \begin{pmatrix} 1 & 0 \\0& -1 \end{pmatrix}.$$ In other words, the original operator has now become a real diagonal matrix with respect to the orthonormal basis of eigenvectors of $\sigma_y$.

Answer 2

A real linear operator is one whose matrix elements are real. For example, given an orthonormal basis $\{|\psi_k\rangle\}$,and an operator $\hat O$, suitably defined on a Hilbert space $\mathcal H$; when we calculate the expectation value of the operator: $\langle \psi_i|\hat O|\psi_j\rangle, \;\forall |\psi_i\rangle,|\psi_j\rangle\in \{|\psi_k\rangle\}$; we always get real numbers. Because we have assumed a particular basis, the realty of an operator is not guaranteed to be a property that holds for any arbitrary basis used to span the Hilbert space. Every Hermitian operator is also a real operator, however, not all real operators are Hermitian. Although the realty of an operator is a basis dependent concept, Hermiticity is intrinsic to the operator and thus holds in any basis, i.e. $\langle\hat O\psi |\psi\rangle=\langle\psi |\hat O\psi\rangle, \;\forall |\psi\rangle\in \mathcal H$. So Hermiticity is the crucial requirement needed to do quantum mechanics, it does no good for an operator to be real in this sense, without having Hermiticity.