Timeline for Why is $\hat{p} \circ \hat{p}$ the operator corresponding to $p^2$?
Current License: CC BY-SA 3.0
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| Sep 29, 2014 at 6:36 | history | edited | BeastRaban | CC BY-SA 3.0 |
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| Sep 29, 2014 at 6:05 | comment | added | BeastRaban | I added a bit to the answer, basically it's by looking at the momentum as the generator of translations. | |
| Sep 29, 2014 at 6:04 | history | edited | BeastRaban | CC BY-SA 3.0 |
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| Sep 29, 2014 at 5:52 | comment | added | David Zhang | More precisely, why $\langle p^2 \rangle$ is given by $$ \int_{-\infty}^\infty \Psi^* (-ih\,\partial_x)^2 \Psi\,\mathrm{d} x.$$ | |
| Sep 29, 2014 at 5:50 | comment | added | BeastRaban | So you're basically asking why $\frac{\partial^2}{\partial x^2}\propto p^2$? | |
| Sep 29, 2014 at 5:42 | comment | added | David Zhang | I don't think you've really addressed the question. What I'm trying to ask concerns the correspondence of quantum-mechanical operators to classical quantities. Given an operator $\hat{p}$ which corresponds to the classical momentum (in the sense I defined in the question), why does $\hat{p} \circ \hat{p}$ correspond to its square? I understand that the eigenvalues of $\hat{p}^2$ are the eigenvalues of $\hat{p}$ squared, but I don't see how this answers the question. | |
| Sep 29, 2014 at 5:32 | history | answered | BeastRaban | CC BY-SA 3.0 |