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Hi, I'm reading the article "Galilean Conformal Algebra's and AdS/CFT", 0902.1385v2, and have a quick question about the Schrodinger symmetry group.
The symmetry group consists of the usual Galilean group G(d,1) with rotations, boosts and (space and time) translations. On top of that we have a dilatation symmetry of the form
[tex]
x \rightarrow \lambda x, \ \ \ \ \ \ \ \ t \rightarrow \lambda^2 t
[/tex]
This one is clear; just plug it in the free Schrodinger equation [itex]S=i\partial_t + \frac{1}{2m}\partial_i^2 [/itex]. Also, we have some sort of "time component special conformal symmetry",
[tex]
x \rightarrow x'= \frac{x}{1+\mu t}, \ \ \ \ \ \ \ \ t \rightarrow t'= \frac{t}{1+\mu t}
[/tex]
where mu parametrizes the transformation. So again I want to check whether this is really a symmetry. So I calculate
[tex]
\partial_{x} = \frac{\partial x'}{\partial x}\frac{\partial }{\partial x'} + \frac{\partial t'}{\partial x}\frac{\partial}{\partial t'} = \frac{1}{(1+\mu t)} \frac{\partial}{\partial x'}
[/tex]
and
[tex]
\partial_t = \frac{\partial x'}{\partial t}\frac{\partial}{\partial x'} + \frac{\partial t'}{\partial t}\frac{\partial}{\partial t'} = \frac{1}{(1+ \mu t)^2}[\frac{\partial}{\partial t'} - \mu x \frac{\partial}{\partial x'}]
[/tex]
Now the question: what is that [itex] -\mu x \partial_{x'}[/itex] term in the last equation doing there; it spoils the symmetry, right? Am I making a stupid calculational mistake or am I misunderstanding something here?
The symmetry group consists of the usual Galilean group G(d,1) with rotations, boosts and (space and time) translations. On top of that we have a dilatation symmetry of the form
[tex]
x \rightarrow \lambda x, \ \ \ \ \ \ \ \ t \rightarrow \lambda^2 t
[/tex]
This one is clear; just plug it in the free Schrodinger equation [itex]S=i\partial_t + \frac{1}{2m}\partial_i^2 [/itex]. Also, we have some sort of "time component special conformal symmetry",
[tex]
x \rightarrow x'= \frac{x}{1+\mu t}, \ \ \ \ \ \ \ \ t \rightarrow t'= \frac{t}{1+\mu t}
[/tex]
where mu parametrizes the transformation. So again I want to check whether this is really a symmetry. So I calculate
[tex]
\partial_{x} = \frac{\partial x'}{\partial x}\frac{\partial }{\partial x'} + \frac{\partial t'}{\partial x}\frac{\partial}{\partial t'} = \frac{1}{(1+\mu t)} \frac{\partial}{\partial x'}
[/tex]
and
[tex]
\partial_t = \frac{\partial x'}{\partial t}\frac{\partial}{\partial x'} + \frac{\partial t'}{\partial t}\frac{\partial}{\partial t'} = \frac{1}{(1+ \mu t)^2}[\frac{\partial}{\partial t'} - \mu x \frac{\partial}{\partial x'}]
[/tex]
Now the question: what is that [itex] -\mu x \partial_{x'}[/itex] term in the last equation doing there; it spoils the symmetry, right? Am I making a stupid calculational mistake or am I misunderstanding something here?