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Jun 12, 2021 at 15:23 answer added J.G. timeline score: 0
Jun 12, 2021 at 14:05 comment added Vladimir Kalitvianski A particular form of operators $a$ and $a^{\dagger}$ depends also on the scalar product definition.
Jun 12, 2021 at 13:53 history edited Qmechanic
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Jun 12, 2021 at 11:20 answer added Himanshu timeline score: 1
Jun 12, 2021 at 11:19 comment added FrodCube $d/dx$ is not Hermitean, but anti-Hermitean. The second piece of the last line takes a minus sign when you do the conjugate.
Jun 12, 2021 at 11:13 history asked Y2H CC BY-SA 4.0