# Can momenta eigenstate written in term of $x$ be an eigenfunction of position?

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:

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!

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

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.