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Quick question on the Schrodinger symmetry group

  1. Sep 7, 2009 #1

    haushofer

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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?
     
  2. jcsd
  3. Sep 7, 2009 #2

    haushofer

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    Also, their notion of group contraction is not clear to me. At

    https://www.physicsforums.com/showthread.php?t=41723

    I found an excellent explanation of how this group contracting stuff works for SO(3,1), but somehow the authors take the non-relativistic scaling

    [tex]
    t \rightarrow \epsilon^r t, \ \ \ \ \ \ \ x_i \rightarrow x_i \epsilon^{r+1}
    [/tex]

    If I plug this in the relativistic generators, for instance [itex]J_{\mu\nu} = - (x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})[/itex] I obtain factors which go like [itex]\epsilon^{-1}[/itex] which in the non-relativistic limit blows up. Also, from the link above I understood that the rotation group SO(d) should be left untouched by this group contracting because it's a subgroup of SO(d,1), or in this case ISO(d,1), without any time parameter involved. Can anyone shed some light on this?
     
  4. Sep 7, 2009 #3

    haushofer

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    So let me elaborate on the group contraction of SO(3,1). We have rotation generators J and boost generators K, and

    [tex]
    [J_i, J_j] = i \epsilon_{ijk}J_k, \ \ \ \ [K_i,K_j] = -i\epsilon_{ijk}J_k, \ \ \ \ \ [J_i,K_j] = i \epsilon_{ijk}K_k
    [/tex]

    The idea is now to take the Galilean limit. The subgroup SO(3) is preserved by this limit, so the J operators are not scaled. The K generators however are scaled by a factor epsilon; the Galilean limit consists then in plugging this information into the algebra and take the limit epsilon --> 0.

    So I get then

    [tex]
    [J_i, J_j] = i \epsilon_{ijk}J_k
    [/tex]

    , which is preserved, and

    [tex]
    [K_i,K_j] \rightarrow \epsilon^{2}[K_i,K_j], -i\epsilon_{ijk}J_k \rightarrow \epsilon -i\epsilon_{ijk}J_k
    [/tex]

    so

    [tex]
    \epsilon^2 [K_i,K_j] = -i\epsilon \ \epsilon_{ijk}J_k \rightarrow 0
    [/tex]

    and

    [tex]
    [J_i,K_j] \rightarrow \epsilon [J_i,K_j], \ \ \ \ \epsilon_{ijk}K_k \rightarrow \epsilon \epsilon_{ijk}K_k
    [/tex]

    so this one is also left untouched (the infinitesimal epsilon of K of the left hand side cancels the infinitesimal epsilon of K on the right hand side) :

    [tex]
    [J_i,K_j] = i \epsilon_{ijk}K_k
    [/tex]

    Can I now conclude that the algebra of the Galilean symmetry (which leaves the origin untouched) is

    [tex]
    [J_i, J_j] = i \epsilon_{ijk}J_k, \ \ \ \ \ [K_i,K_j] = 0, \ \ \ \ \ [J_i,K_j] = i \epsilon_{ijk}K_k
    [/tex]

    ?
     
  5. Sep 7, 2009 #4

    haushofer

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    So the quick question has become a little deceiving now, I see. Well, any help will be appreciated, I'm not very strong on Lie groups and Lie algebras, but working hard to improve this.
     
  6. Sep 7, 2009 #5
    In this post I'll refer to your first question.
    An operator O represents a symmetry if it commutes with the Schrodinger operator
    S=-ih+p^2/2m acting on the wavefunction:
    [tex][S,O]\Psi(x)[/tex]. In general, though, the wavefunction that's a solution of S alone must be modified by a complex factor so that
    [tex][S,O]\Psi'(x)=0[/tex] where [tex]\Psi'=e^{i...}\Psi[/tex]. From this condition you can figure out what that factor has to be to ensure invariance.
     
  7. Sep 8, 2009 #6

    haushofer

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    I understand that the group Sch(d,1) is given by all the elements O for which [S,O]=0, but ofcourse one can also explicitly check the invariance if you write down the coordinate changes under the specific transformation, and in this case I don't understand why I get something like the last line of my first post :)
     
  8. Sep 8, 2009 #7
    Well, my statement wasn't that [S,O]=0, since this is clearly not true for various operators that are in the Schrodinger group. The "symmetry" in this case requires that the wavefunction also be changed by a complex factor...hence [S,O]Psi'=0, where the prime indicates the new wavefunction. You have to figure out what this factor is in order to satisfy the vanishing of the above.

    In practice it's convenient to do this by inserting coordinate transformations into the operator S, but it's the same idea. You have the initial equation L*Psi=0 where L is the Schrodinger operator. Then the transformed equation you want to satisfy is L'*Psi'(x')=0. Let's define the complex factor such that [tex]\Psi'(x')=\Phi(x) \Psi(x)[/tex]; that is, the wavefunction in terms of transformed coordinates differs from that of the untransformed coordinates by a function [tex]\Phi(x)[/tex]. Write L'*Psi'=0 in terms of unprimed quantities and you'll find an equation: [tex]L'(x'\rightarrow x)\Phi(x)\Psi(x)[/tex]; then you can find [tex]\Phi(x)[/tex] so that this expression vanishes.

    As an example, consider the operator B_x in the article you're looking at, which is supposed to act as [tex]x=x'+Vt'[/tex] with the other coordinates unchanged (this is the inverted form of x'=x-Vt). Writing [tex]S=-iH+\frac{1}{2m}P^{2}_{x}[/tex], the commutator gives [S,B_{x}]=P_{x}(i-1) (2nd term is due to central extension of algebra). So Galilean shifts don't even satisfy the simple symmetry rule; rather the generalized symmetry rule in which the wavefunction must also be modified. In terms of our transformed coordinates you can write the transformed expression S'Psi'=0 and see that it is not invariant in the simple sense: you'll find that S'(x')=S(x)+F(x), where F(x) is some extra operator (the presence of F is another way of saying the commutator btwn B and S doesn't vanish). Then the condition for generalized symmetry is [tex][S(x)+F(x)]\Phi(x)\Psi(x)[/tex] where [tex]\Psi(x)=exp[-i(Et-px)][/tex]. then determine the form of Phi so that the expression vanishes (you will need to use the old condition [tex]S(x)\Psi(x)=0[/tex]). Anyway, in this case, we can write down Phi quickly because we know x'=x-Vt, v'=v-V: The wavefunction in terms of primes is [tex]exp[-i(E't'-p'x')][/tex] with p'=mv' and similarly for E'; then expand everything in unprimed quantities to get [tex]\Phi(x)[/tex].
     
    Last edited: Sep 8, 2009
  9. Sep 9, 2009 #8

    haushofer

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    Ok, I take a look at this, thanks a lot!
     
  10. Sep 9, 2009 #9
    By the way, the original papers on the dilatation/special conformal parts of the group were in 1972 (see references in the article you are reading), so you'd need access to a science library if you want to look at them. Otherwise, it seems hard to find an introductory discussion about these generalized symmetries.
     
  11. Sep 10, 2009 #10

    haushofer

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    You mean the "Scale and conformal transformations in Galilean-Covariant field theory" by Hagen in Phys.Rev. D? I already have that one, but some things of that article also weren't very clear to me.
     
  12. Sep 10, 2009 #11
    That's one of them, yeah. Sorry it didn't help.
     
  13. Sep 16, 2009 #12

    haushofer

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    Ok, so I basically found out how all these things work. My next question is: can I also play the same game with gravity?

    I read some interesting things about taking non-relativistic limits for GR in eg Misner, Thorne, Wheeler, in which space-time becomes a fibre-bundle in which the basemanifold is the universal time t, and the fibers are just 3-dimensional Cartesian spaces. Ofcourse, transporting vectors in such a fiber is not exciting, because they are flat. But transporting vectors via different fibers gives us some interesting geometry.

    However, I have the feeling that this can be done easier via these so called group contractions. In GR, we have as symmetry group the group of general coordinate transformations. The special-relativistic limit of these is the group of Lorentz transformations globally defined, and the non-relativistic limit of these LT's are again our Galilean transformations.

    So my question is: can I perform this group contraction on gravity (and eventually supergravity) in order to get non-relativistic limits?
     
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