- #1

haushofer

Science Advisor

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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?