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Being non-commuatable operators, momentum and position cannot have simultaneous eigenfunctions. But in "Theoretical Minimum: QM" by Lenny Susskind and Artsy Friedman, in explaining Heisenberg's Uncertainty principle: enter image description here

I don't understand why they assume that the eigenstate of momentum (Psi with subscript p) is going to have a localized poistion component. I would love some clarification on this topic. Thanks!

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    $\begingroup$ I don't understand what your objection is. Actually the given eigenstate of momentum $\psi_p(x)$ is as delocalized as it can be, since $|\psi_p(x)|^2$ has the same value for all $x$. $\endgroup$ Commented Jul 2, 2019 at 11:04
  • $\begingroup$ Wowza! I interpreted Lenny's words wrong. Thanks for pointing out that delocalization part. $\endgroup$ Commented Jul 2, 2019 at 11:38

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You confuse different things. An eigenstate of p or x is defined with the equation $A \psi = \lambda \psi$, where A is the momentum/position operator and $\lambda$ the corresponding eigenvalue. Psi here is an abstract mathematical element of the QM Hilbert space. What you have written down here as $\psi_p(x)$ is an eigenstate of the momentum operator in the x representation. It is a specific representation of the Hilbert space vector, but not an eigenstate of the x operator.

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  • $\begingroup$ Haha yh! What Lenny was trynna show was that eigenstate of momenta is perfectly delocalized in terms of position ( being 1/2pi for all x values). I thought momenta itself was delocalized. Thanks for pointing out that "ψp(x) is an eigenstate of the momentum operator in the x representation"! That was really helpful! $\endgroup$ Commented Jul 2, 2019 at 11:42
  • $\begingroup$ Ah okay, so it was just a missed formulation :) $\endgroup$ Commented Jul 3, 2019 at 15:04

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