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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:

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

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    $\begingroup$ Assuming it’s extended to all vectors by linearity, this is the expectation value of momentum. $\endgroup$
    – knzhou
    Jul 9, 2018 at 9:50
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    $\begingroup$ 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, 2018 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, 2018 at 14:40
  • $\begingroup$ @user183966 Sorry, I shouldn't have said "extended by linearity" because it's not linear. $\endgroup$
    – knzhou
    Jul 9, 2018 at 14:42
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    $\begingroup$ What happens when $\psi(x)=0$? $\endgroup$
    – J. Murray
    Jul 9, 2018 at 16:56

1 Answer 1

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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.

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  • $\begingroup$ 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, 2018 at 15:16
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    $\begingroup$ @AccidentalFourierTransform Thanks, I updated the answer. $\endgroup$
    – Noiralef
    Jul 9, 2018 at 15:34
  • $\begingroup$ @AccidentalFourierTransform this made the answer clearer to me $\endgroup$
    – user183966
    Jul 9, 2018 at 15:35

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