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May 7, 2021 at 5:54 answer added user41025 timeline score: 2
Apr 13, 2017 at 12:40 history edited CommunityBot
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Aug 24, 2014 at 9:49 answer added Sergio Lantzman timeline score: 0
Feb 14, 2014 at 11:45 vote accept qfzklm
Feb 14, 2014 at 11:38 vote accept qfzklm
Feb 14, 2014 at 11:45
Feb 14, 2014 at 11:38 vote accept qfzklm
Feb 14, 2014 at 11:38
Feb 14, 2014 at 11:37 comment added qfzklm @JánLalinský Thank you very much, and I have to apologize for my fault to everyone.
Feb 11, 2014 at 6:09 answer added Rococo timeline score: 6
Feb 10, 2014 at 20:20 answer added Urgje timeline score: 3
Feb 10, 2014 at 16:20 comment added Ján Lalinský qfzklm, you've got it wrong. The Hamiltonian function for particle in magnetic field is $H(\vec p, \vec r) = \frac{(\vec p - q\vec A(\vec r))^2}{2m}$, and lower-case $\vec p$ is called canonical momentum, while upper-case $\vec P = \vec p - q\vec A$ is kinetic momentum ($m\vec v$).
Feb 10, 2014 at 15:38 review Close votes
Feb 11, 2014 at 9:49
Feb 10, 2014 at 15:19 comment added Kyle Kanos possible duplicate of A question about canonical momentum and arbitrariness for potential in magnetism
Feb 10, 2014 at 15:01 answer added ZachMcDargh timeline score: 0
Feb 10, 2014 at 14:56 answer added jinawee timeline score: 0
Feb 10, 2014 at 14:23 comment added qfzklm @CarlWitthoft $H=\frac{P^2}{2m}$ and $[P,H]=0$, I think $P$ is conserved. Btw, I do not have this book and my textbook do not discuss this or other similar questions, could you tell me more details?
Feb 10, 2014 at 14:04 comment added Carl Witthoft But why do you think that? -- not that your linked question has a very clear answer. I'd dig into L&L or similar textbooks for details rather than just re-asking the question.
Feb 10, 2014 at 14:01 history edited Qmechanic CC BY-SA 3.0
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Feb 10, 2014 at 13:50 history asked qfzklm CC BY-SA 3.0