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Oct 20, 2017 at 16:09 comment added Luke Pritchett I have edited my answer to make it slightly clearer the distinction between the abstract generator of translations and the differential operator $-i\partial_x$. In any case, the statement that $[x,p]=i\hbar$ is exactly equivalent to the statement that $p$ generates translations.
Oct 20, 2017 at 16:07 history edited Luke Pritchett CC BY-SA 3.0
Added detail on meaning of generator of translations.
Oct 19, 2017 at 22:52 comment added John Doe @ZeroTheHero I think you may have misread what I wrote - we have defined $p$ in this way. I just didn't want to use the fact that $p=-i\hbar \frac\partial{\partial x}$.
Oct 19, 2017 at 20:28 comment added ZeroTheHero The OP wishes to prove without using the fact that $p$ generates translations...
Oct 19, 2017 at 17:59 history answered Luke Pritchett CC BY-SA 3.0