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?

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    $\begingroup$ I would not use lecture notes where the author is talking about "real" operators. $\endgroup$
    – Hyperon
    Feb 28 at 15:37
  • $\begingroup$ @Hyperon Why not? $\endgroup$ Feb 28 at 15:47
  • $\begingroup$ @AlbertusMagnus In a complex Hilbert space, such a "concept" (being basis dependent) does not make sense. $\endgroup$
    – Hyperon
    Feb 28 at 15:52
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    $\begingroup$ @Hyperon You're way too harsh, IMHO. Most QM classes speak of "functions in $L^2$", for example, too, which is wrong, but most physicists won't care or bother. Yes, I wouldn't use the terminology "real" either (for example because it leads to confusion), but at the end it is a matter of language. For example, Dirac uses this terminology. Yes, you can argue that his text is over $90$ years old, but at the end, as I said above, it does not really matter; what matters is the mathematical structure. $\endgroup$ Feb 28 at 16:22
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    $\begingroup$ @TobiasFünke The difference is, speaking of the elements of $L^2$ as "functions" (instead of equivalence classes of functions) is a well established facon de parler even in mathematics and I can hardly imagine a situation in QM where this habit could cause any harm. In contrast, using the terminology of a "real" operator (instead of hermitean or self adjoint, respectively) can be highly misleading already in a finite dimensional unitary vector space (e.g. Pauli matrix $\sigma_y$ vs. $i \sigma_y$). $\endgroup$
    – Hyperon
    Feb 28 at 22:04

2 Answers 2


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 $\sigma_y$.

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    $\begingroup$ Essentially, whether or not an operator is "real" or not is of little importance in Quantum Mechanics. $\endgroup$ Feb 29 at 0:13

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.

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    $\begingroup$ The phrase "real operator" makes no sense in an abstract Hilbert space... $\endgroup$ Feb 28 at 16:13
  • $\begingroup$ @TobiasFünke Yes, that was discussed in the above comments, perhaps you might take a look at the edited post and see how it holds up? Your acumen would be appreciated, thanks. $\endgroup$ Feb 28 at 16:17
  • $\begingroup$ @infinitezero Thanks! $\endgroup$ Feb 28 at 16:33
  • $\begingroup$ Thank you for your enlightening answer. Can you please explain a bit more how the momentum operator is real in the momentum basis? And is the property of being hermitian dependant on the basis as well or is it intrinsic to the operator? $\endgroup$
    – Andrea
    Feb 28 at 20:36
  • $\begingroup$ @Andrea The most important thing to take away from this stuff today is that operators in quantum mechanics should be Hermitian, which is the same as self-adjoint (whenever the underlying domain of the operator is a complex linear space). Personally, I would speak about Hermitian operators as opposed to real operators. I think that I have actually been mislead by a past professor and that I read that into what others were saying, however, it didn't do me any harm because I ultimately understood that it was the self-adjoint nature of the operator that was the heart of the issue. $\endgroup$ Feb 28 at 22:45

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