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Again, as usual Schwinger leaves me startled as he writes, the Hermitian displacement operator in 2D is $$ G = p_1\delta x_1 +p_2 \delta x_2 $$ Now, we know clearly that this group is an Abelian group, therefore $ [p_1,p_2] = 0$. But the author suggests this is equivalent to saying that the commutator generates any unitary transformation without physical consequences, and goes on to write, $$ \hbar ^{-1}[p_1,p_2] = \mathbb{1} \tag{1}$$

Now, effect of an infinitesimal change in an operator($X$) is given by $\delta X = -i[X,G] $ where G is the generator of an unitary transformation ($U$)

Hence I obtain, $$ \delta p_1 = -i[p_1,G] = \delta x_2 $$ $$ \delta p_2 = -i[p_1,G] = -\delta x_1 $$

Which then means, $$ x_1 = -p_2 \equiv q \:\:\:\:and\:\:\:\: x_2 = p_1 = p $$ Now, we see the association to the single particle's phase space state $(q,p)$.

Further translations in this (phase-space) two-dimensional space are described by the three parameter group with the generator,

$$ G = p\delta q -q\delta p + \delta \phi \mathbb{1} $$

My questions are :

a) How is the choice of commutator in equation $(1)$?

b) What exactly is the author trying to do here, or what is he motivating from this exercise? Why does he connect the translation group in normal 2D space to phase space of a single particle?

c) What is this operator $\delta \phi \mathbb{1} $ in the last generator?

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I'd reckon that you will more likely get an answer if you clarify your notation (the notation used in your message is not conventional afaik). First are you discussing $U(1)$ or $SO(2)$ (my guess would be $U(1)$)? What do $p_i$ and $\delta x_i$ stand for? Overall, what is the context? – Hunter Feb 27 '14 at 15:16
Sorry, I cannot find any sense in what you (user35952) wrote. The final group seems to be the so called Heisenberg group which has nothing to do with the initial translation group in two-dimensional space... – Valter Moretti Feb 27 '14 at 15:24
@Hunter : This is the translation group in two dimensions. $p_i$ stands for generator of translations. I am not sure what precisely the context that Schwinger is trying to address, but he trying develop the Lie-algebra of this group. Secondly it also seems that he is trying to connect it to translations in phase-space. – user35952 Feb 27 '14 at 15:48
@V.Moretti : Thanks, That is a useful observation !! What about the first question ? How to justify it ? Also how is Heisenberg group connected translations in phase-space ? – user35952 Feb 27 '14 at 15:51
Oh, I'm sorry, I thought it was the rotation group. – Hunter Feb 27 '14 at 15:52

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