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22:01

@Shashaank by the way, it does look much better now

Shashaank

@AccidentalFourierTransform Ok. I did not know that. I thought reposting will be be best. Nevertheless I will keep that in Mind from now onwards.

AccidentalFourierTransform

@Shashaank yeah, thanks :-)

it took me a lot of time to learn the rules of physics.SE

Shashaank

@AccidentalFourierTransform Yeah that's pretty much the problem with me too. I was requesting an answer from someone but he mistook it as a demand. I am a beginner probably that's why taking such a lot of time. Nevertheless Thank you . Hoping someone sees it and answers it !

AccidentalFourierTransform

22:38

anyone around?

ACuriousMind

::crickets::

AccidentalFourierTransform

hi ::blushes::

@AccidentalFourierTransform In the professor notes we keep it, that why I asked the question in the first place, cause I didn't get the difference between the case of $\vec{p}$ and $\vec{\pi}$. We keep that term and then we rewrite it in terms of the magnetic field. — Run like hell 7 mins ago

see here please ^

is $(\vec\sigma\cdot\vec \pi)^2=\vec \pi^2$ correct or not?

ACuriousMind

@AccidentalFourierTransform It depends on what the $\pi,p$ are. The second term vanishes for the canonical momentum but not for the kinematic momentum.

AccidentalFourierTransform

but why?

why doens't it vanish for both?

ACuriousMind

@AccidentalFourierTransform Because the kinematic momentum components no longer commute

It's $\sim \partial + A$, and $[\partial + A,\partial + A] f \neq 0$.

AccidentalFourierTransform

22:48

arrgh now I gotta go and tell OP that I forgot basic calculus and that he/she should forget everything I said

@ACuriousMind thanks

AccidentalFourierTransform

23:13