Timeline for How can I solve this quantum mechanical "paradox"?
Current License: CC BY-SA 3.0
7 events
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| Feb 2, 2016 at 21:04 | comment | added | higgsss | @AndreaDiBiagio Yes it does. So we have $\sigma_x \sigma_p\ge 0$ for a plane wave. | |
| Feb 2, 2016 at 20:21 | comment | added | Andrea | @higgsss from what you wrote, doesn't the expression in the absolute value read $1-a/|\sqrt{a}|^2 = 0$ for a plane wave? | |
| Feb 2, 2016 at 15:32 | comment | added | higgsss | @ACuriousMind You are right, but my purpose is to keep the discussion at the level of undergraduate QM. Considering domains of operators and distinguishing between Hermiticity and self-adjointness are somewhat too mathematically involved for many physicists including myself. | |
| Feb 2, 2016 at 15:01 | comment | added | ACuriousMind♦ | You don't need to modify the position operator, it is perfectly self-adjoint without the sawtooth function added to it. What you need to do is think about the domain of validity of the uncertainty principle, in particular the commutator. | |
| Feb 2, 2016 at 9:38 | comment | added | higgsss | This is a nonrigorous version of more mathematical answers other users might provide. | |
| Feb 2, 2016 at 9:38 | history | edited | higgsss | CC BY-SA 3.0 |
added 18 characters in body
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| Feb 2, 2016 at 9:32 | history | answered | higgsss | CC BY-SA 3.0 |