Timeline for Why can I ignore $R(\rho)$?
Current License: CC BY-SA 4.0
7 events
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| Jan 19, 2023 at 18:54 | comment | added | ZeroTheHero | In other words, I could modify $C\to C_k$; I guess the very last step of finding $\vert C\vert^2$ must be incorrect although this does not affect the probabilities, i.e. the relative probabilities computed with different $k$'s remains the same so upon normalizing the probabilities so they sum to $1$ you still get an answer independent of $k$. | |
| Jan 19, 2023 at 18:50 | comment | added | ZeroTheHero | There's something very weird about $C$ being the same irrespective of $R(\rho)$. Consider the family $R_{km}(\rho,\phi)=r^k e^{-r^2/2} \sqrt{2/k!}e^{i m\phi}/\sqrt{2\pi}$. The $R_{km}(\rho,\phi)$ are suitably normalized but the overlap $\langle \psi\vert R_{km}\rangle$ depends on $k$, yet relative probabilities do not. In other words, for two different $R_{km}$, the overlap with your $\psi$ will be different but this does not affect the relative probabilities of finding the various $m$ states. | |
| Jan 18, 2023 at 2:43 | history | edited | ZeroTheHero | CC BY-SA 4.0 |
edited body
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| Jan 17, 2023 at 23:42 | comment | added | Clayton | I didn't think of letting that be just some number common to all states! This really helped my understanding. Thank you so much! | |
| Jan 17, 2023 at 23:40 | vote | accept | Clayton | ||
| Jan 17, 2023 at 13:22 | history | edited | ZeroTheHero | CC BY-SA 4.0 |
added 117 characters in body
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| Jan 17, 2023 at 12:46 | history | answered | ZeroTheHero | CC BY-SA 4.0 |