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My poorly written lecture notes say that any Hermitian operator does have a complete set of orthogonal eigenstates with real corresponding eigenvalues and is therefore an observable.

In the article Observables, it is said that in order for a Hermitian operator to be observable its eigenvectors must form a complete set.

user572780
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According to the postulates of quantum mechanics, each observable $p$ quantity is associated with an operator $\hat{p}$ that acts on the wavefunction $\psi$.

The relationship is given by the eigenvalue equation: $$ \hat{p}\psi = p\psi. $$

$\hat{p}$ is an operator, which means nothing on its own. $p$ is the eigenvalues, the observable which is a number.

For instance, if $p$ is the momentum:

  • $\hat{p} = \frac{\hbar}{i}\nabla $, i.e. a functional operator so quite useless on its own;

  • Acting of a plane wave $\psi = e^{ikx}$, $\hat{p}\psi = \hbar k\,\psi $. I.e. the observable momentum is $p=\hbar k$.

SuperCiocia
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  • What do you mean observable is a number? All observables are operators. You can measure an observable and get a real number. – user572780 May 12 '19 at 12:50
  • All observables are represented by operators. The actual observable is something that you measure, so it has to be a number, i.e. not the function $\hat{p}$ but its eigenvalue $p$. – SuperCiocia May 12 '19 at 12:53
  • The Observable Wikipedia article says "an observable is a physical quantity that can be measured." It then goes on to say "In quantum physics, it is an operator". – user572780 May 12 '19 at 13:08
  • Semantic differences. When you measure the physical quantity, you get a number. IN order to get a number from the wavefunction $\psi$, you need to act on it with an operator. You could say "observable=operator" then, I would prefer "observable $\leftrightarrow$ operator". – SuperCiocia May 12 '19 at 13:10
  • I have been taught that the physical quantity is not a number but your measurement of the physical quantity is a number. So an observable is an operator that you can measure using physical equipment. In your example: $$ \hat{p}\psi = p\psi. $$

    $\hat{p}$ is the observable.

     – user572780 May 12 '19 at 13:18
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    Then it depends on what you mean by “physical quantity”. I would say it’s the result of the measurement. – SuperCiocia May 12 '19 at 13:19
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An operator need not be hermitian. For instance, the harmonic oscillator creation operator $\hat a^\dagger$ is not hermitian, and neither is the angular momentum lowering operator $\hat L_-$. Yet both are perfectly legitimate (linear) operators, i.e. they act linearly on a state and produce a different state.

Setting aside subtle points about domains of operators and self-adjointness, observables must be hermitian (in the sense that their matrix representations are hermitian matrices) because eigenvalues of hermitian matrices are real, which is good since in a lab we measure real (rather than complex) quantities. Moreover, hermitian matrices have a complete set of eigenvectors that spans the entire space.

Note that it is important to realize that this doesn’t imply that non-hermitian operators cannot have eigenvalues or eigenvectors, just that there’s no guarantee the eigenvalues are real and the eigenvectors for a complete set. The usual example of this is the harmonic oscillator coherent state $\vert \alpha\rangle$ (where $\alpha$ is any complex number) which is an eigenvector of the annihilation operator $\hat a$, with complex eigenvalue $\alpha$. The eigenvalue need NOT be real since $\alpha$ can be complex, and the coherent states form an overcomplete set of vectors for the Hilbert space of the harmonic oscillator.

ZeroTheHero
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  • All observables are Hermitian operators but are all Hermitian operators also observables? – user572780 May 12 '19 at 18:18
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    It might be worth noting that $\hat{a}$ has an overcomplete basis of eigenfunctions because it is not only non-hermitian, but also non-normal ($[\hat{a},\hat{a}^\dagger]\ne 0$). A normal operator always has a nice orthonormal basis of eigenfunctions even if it is non-hermitian. – eyeballfrog May 12 '19 at 18:37
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First, a mathematical subtlety. In finite dimensional Hilbert spaces spaces, every self-adjoint operator can be represented by a Hermitian matrix, i.e. one that is equal to its conjugate transpose, and every Hermitian matrix corresponds to a self-adjoint operator. However, in infinite dimensional spaces, as it is often the case for the Hilbert spaces describing quantum systems, a symmetric operator $A$—which in finite dimensions is equivalent to a Hermitian matrix—is only self-adjoint if its domain is the same as the one of its adjoint $A^\dagger$, namely $\mathrm{dom}\ A = \mathrm{dom}\ A^\dagger$. A self-adjoint operator, though, is always symmetric. This being said, every observable corresponds to a self-adjoint operator.

Nevertheless, it is generally false to suppose the converse: not every self-adjoint operator is an observable, and a typical example of such is the density operator $\hat{\rho}$. More on the subject can be read here: not every self-adjoint operator is an observable.

Albert
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