0
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ 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 $x$. $\endgroup$ – Thomas Fritsch Jul 2 at 11:04
  • $\begingroup$ Wowza! I interpreted Lenny's words wrong. Thanks for pointing out that delocalization part. $\endgroup$ – Eesh Starryn Jul 2 at 11:38
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Eesh Starryn Jul 2 at 11:42
  • $\begingroup$ Ah okay, so it was just a missed formulation :) $\endgroup$ – roran_physician Jul 3 at 15:04

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.