3
$\begingroup$

Let $\vec{p} = p_x \hat{x} + p_y \hat{y} + p_z \hat{z}$, and also use the notation $|\vec{p}| = p$, where $p^2 = p_x^2 + p_y^2 + p_z^2$.

What is the difference between the operator $\hat{p}^2 = \hat{\vec{p}} \cdot \hat{\vec{p}}$, and the operator $\widehat{p^2}$? And which one is the operator that correctly represents the magnitude of the momentum, squared?

$\endgroup$
2
  • $\begingroup$ How do you define $\widehat{p^2}$ ? $\endgroup$ Sep 20 '20 at 19:28
  • $\begingroup$ @AjayShanmugaSakthivasan If the operator for the square of the $i^\mathrm{th}$ component of momentum is $\widehat{p_i^2}$, then I suppose we would have $\widehat{p^2} = \widehat{p_x^2} + \widehat{p_y^2} + \widehat{p_z^2}$ by canonical quantisation of the equation $p^2 = p_x^2 + p_y^2 + p_z^2$. $\endgroup$
    – 13509
    Sep 20 '20 at 19:42
4
$\begingroup$

What is the difference between the operator $\hat{p}^2$, and the operator $\widehat{p^2}$?

For the sake of clarity let me call your operator $\widehat{P^2}$ as $\hat{M}$ (and let us first work with one dimension). Then by your definition (as stated in the comments), $$\hat{M}\vert\psi\rangle = p^2 \vert\psi\rangle,$$ where $\vert\psi\rangle$ is a momentum squared eigenstate. Now, $$p^2\vert\psi\rangle = p\cdot p \vert\psi\rangle = p\cdot \hat{P}\vert\psi\rangle = \hat{P}(p\vert\psi\rangle) = \hat{P}^2 \vert\psi\rangle.$$ Which means that $\hat{M} = \widehat{P^2} = \hat{P}^2$. Note that the operators $\widehat{P^2}$ and $\hat{P}^2$ share eigenstates (Why? It is because they commute. This can be arrived at by considering the Poisson bracket of classical $P^2$ and $P$).

But typically $P$ (from classical mechanics) is always quantized to give $\hat{P}$ (of quantum mechanics).

And which one is the operator that correctly represents the magnitude of the momentum, squared?

The above discussion should have made this clear. It really doesn't matter which one you use, as long as you define your $\widehat{P_i^2}$ to obey the eigenvalue equation that returns the ($i^{th}$ compononent of the) momentum squared value for any momentum eigenstate.

$\endgroup$
4
  • $\begingroup$ $``p\cdot p \vert\psi\rangle = p\cdot \hat{P}\vert\psi\rangle''$ -- This already assumes that $\vert \psi\rangle$ is an eigenstate of $ \hat{P}$. $\endgroup$
    – Dvij D.C.
    Sep 20 '20 at 20:27
  • $\begingroup$ Yes it does. Like I have mentioned later, "Note that the operators $\widehat{P^2}$ and $\hat{P}^2$ share eigenstates". This is because they commute. Again this can be arrived at by considering the Poisson bracket of classical $P^2$ and $P$. $\endgroup$ Sep 20 '20 at 20:38
  • $\begingroup$ Sure, it was a bit unclear what you were assuming and what you were claiming to have proven. As long as the logical flow is what you mention in the comment, I agree. $\endgroup$
    – Dvij D.C.
    Sep 20 '20 at 21:16
  • $\begingroup$ Thank you for your comment. I have added it to the answer to make it more clear. $\endgroup$ Sep 20 '20 at 21:19

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.