Physics Asked on March 3, 2021
In studying the Dirac equation, I often come across the unit momentum operator $hat{textbf{p}}$ which is defined as $$hat{textbf{p}} ={textbf{p} over |textbf{p}|},$$where $textbf{p}$ is supposedly the three-momentum vector operator and $|textbf{p}|$ is the magnitude of momentum of the particle. For example, the helicity operator is written as $$h=frac{textbf{S}cdottextbf{p}}{|textbf{p}|}.$$
But $|textbf{p}|$ can also be written in operator form as $$|textbf{p}|=sqrt{hat{p}_x^2+hat{p}_y^2+hat{p}_z^2}.$$
So are we trying to divide the vector operator $textbf{p}$ by another operator? That does not make sense as there is no such thing as operator division. What am I getting wrong?
The unit momentum operator is simply the operator with the same exact eigenstates as the usual momentum operator (i.e. the momentum eigenstates), but with the eigenvalues normalized to be unit vectors.
Correct answer by ReasonMeThis on March 3, 2021
I think you're confusing unit vectors and operators, as they're both denoted by the ^ symbol.
In your $hat p =dfrac{vec p}{|vec p|}$ equation, the $hat p$ is simply a unit vector pointing in the direction of your momentum $vec p$. It is not an operator.
On the other hand, the momentum operator is defined as $$hat p = i hbar dfrac partial {partial x}$$ in which case $hat p$ is indeed an operator, and not a unit vector.
$hat p$ can be used to denote the momentum unit vector and the momentum operator, but the former is definitely not the same as the latter.
Answered by user256872 on March 3, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP