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.

  • $\begingroup$ basically each H operator is an observable $\endgroup$ – Wolphram jonny May 12 '19 at 13:04
  • 2
    $\begingroup$ but see physics.stackexchange.com/q/27038 $\endgroup$ – Wolphram jonny May 12 '19 at 13:09
  • $\begingroup$ @Wolphramjonny can you have a H operator without a complete set of orthogonal eigenstates? $\endgroup$ – user572780 May 12 '19 at 13:09
  • $\begingroup$ I dont remember the proof, but I am 99% sure the answer is no $\endgroup$ – Wolphram jonny May 12 '19 at 13:10
  • $\begingroup$ you should wait for someone with better knowledge to confirm $\endgroup$ – Wolphram jonny May 12 '19 at 13:12

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 $p=\hbar k$.

| cite | improve this answer | |
  • $\begingroup$ What do you mean observable is a number? All observables are operators. You can measure an observable and get a real number. $\endgroup$ – user572780 May 12 '19 at 12:50
  • $\begingroup$ 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 $p$. $\endgroup$ – SuperCiocia May 12 '19 at 12:53
  • $\begingroup$ 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". $\endgroup$ – user572780 May 12 '19 at 13:08
  • $\begingroup$ 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". $\endgroup$ – SuperCiocia May 12 '19 at 13:10
  • 1
    $\begingroup$ Then it depends on what you mean by “physical quantity”. I would say it’s the result of the measurement. $\endgroup$ – SuperCiocia May 12 '19 at 13:19

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.

| cite | improve this answer | |
  • $\begingroup$ All observables are Hermitian operators but are all Hermitian operators also observables? $\endgroup$ – user572780 May 12 '19 at 18:18
  • 2
    $\begingroup$ 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. $\endgroup$ – eyeballfrog May 12 '19 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.