What is the correct name for this operator I am calling the 'basis label operator' which returns the constant function of the eigenvalue for all vectors in a (momentum) eigenspace?

$$\hat{O} : \hat{O}(\psi) \rightarrow \frac{-i\hbar}{\psi}\frac{\partial\psi}{\partial x}$$

for those who would like me to be a little more formal / explicit in my formulation, I could write exactly the same definition as:

$$\hat{O} : \hat{O}(\psi) \rightarrow \left( \phi: \phi(a) \rightarrow \left. \frac{-i\hbar}{\psi(a)}\frac{\partial\psi}{\partial x} \right \rvert_{x=a}\right)$$

  • 3
    $\begingroup$ Assuming it’s extended to all vectors by linearity, this is the expectation value of momentum. $\endgroup$ – knzhou Jul 9 '18 at 9:50
  • 1
    $\begingroup$ Hi @knzhou Is there a theorem that this will give the expectation value of momentum for non-eigenfunctions? I wouldn't have expected that because I don't think it is a linear operator. I'm not sure what you meant by "extended to all vectors by linearity" but I think this operator is defined for all states in the Hilbert space without being linear. $\endgroup$ – user183966 Jul 9 '18 at 14:26
  • $\begingroup$ This is a very unusual object, in what context would this come up? I don't think that it has a name. $\endgroup$ – Noiralef Jul 9 '18 at 14:40
  • $\begingroup$ @user183966 Sorry, I shouldn't have said "extended by linearity" because it's not linear. $\endgroup$ – knzhou Jul 9 '18 at 14:42
  • 1
    $\begingroup$ What happens when $\psi(x)=0$? $\endgroup$ – J. Murray Jul 9 '18 at 16:56

Maybe it helps that $$ \hat O = \hat p \circ \log_x $$ where $\log_x$ is defined by its action in position space: $$ \log_x( |\psi\rangle ) = \int |x'\rangle\, \log(\langle x' | \psi \rangle)\, \mathrm dx' . $$

However. Be warned that, if something like this comes up in your calculations, and you are not 100% sure what you are doing, I think it is likely that what you are doing does not make sense at all. Non-linear operators come up only very rarely. Also note that I was careful in specifying that the action of the operator depends on the position space representation of $|\psi\rangle$. It is very unnatural to consider an operator whose action depends on a specific representation of the Hilbert space.

  • $\begingroup$ Thank you, this is extremely interesting. I am exploring auxiliary tools/formulations for quantum mechanics that do not rely on linear operators. As you know normal observable operators multiply each eigencomponent by it's 'labelled' value. I was considering operators that return the value instead of multiplying by the value (so as to use a 'traditional property' language rather than a linear operator language.) I know the expectation value also has that characteristic but I rarely see this expressed explicitly a an operator. $\endgroup$ – user183966 Jul 9 '18 at 15:16
  • 1
    $\begingroup$ @AccidentalFourierTransform Thanks, I updated the answer. $\endgroup$ – Noiralef Jul 9 '18 at 15:34
  • $\begingroup$ @AccidentalFourierTransform this made the answer clearer to me $\endgroup$ – user183966 Jul 9 '18 at 15:35

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.