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Heisenberg algebra Isomorphic to Galilean algebra

  1. Dec 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Given for one-dimensional Galilean symmetry the generators ##K, P,## and ##H##, with the following commutation relations: $$[K, H] = iP$$ $$[H,P] = 0$$ $$[P,K] = 0$$
    2. Relevant equations
    Show that the Lie algebra for the generators ##K, P,## and ##H## is isomorphic to the Heisenberg algebra $$[X, P] = i \hbar I$$.

    3. The attempt at a solution
    The Heisenberg algebra is a non-trivial central extension of the Galilean algebra. I don't know how to prove how they are isomorphic.
     
    Last edited by a moderator: Dec 12, 2017
  2. jcsd
  3. Dec 12, 2017 #2

    fresh_42

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    What is a Galilean algebra? The multiplication rules you gave under section 1. is directly how the three dimensional Heisenberg algebra is defined. If you switch the names, then what is the Heisenberg algebra to you?
     
  4. Dec 12, 2017 #3
    I think by isomorphic, the commutation relations I have for my so-called "Galilean" algebra really are for the Heisenberg algebra. By Galilean algebra, I mean the Lie algebra for the Galilean group, which is the set of all Galilean transformations. I think Heisenberg algebra is the same as Bargmann Lie algebra, which as I understood it, is a non-trivial central extension of the Galilean algebra.
     
  5. Dec 12, 2017 #4

    fresh_42

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    Can you write down the multiplications within a given vector space basis of the Galilean algebra?

    One possible representation of the three dimensional Heisenberg algebra by matrices is the algebra of matrices of the form ##\begin{bmatrix}0&*&*\\0&0&*\\0&0&0\end{bmatrix}## with Lie multiplication ##[X,Y]=XY-YX##. It is nilpotent with a one-dimensional center. Can you do similar for the Galilean algebra, whatever this is? I don't know the term Galilean group either. Of course I can make some guesses what I think you mean, but it's probably better if you define it.
     
  6. Dec 12, 2017 #5
    The Galilean group, as what was discussed to us, is the set of Galilean transformations given in matrix form as:
    [​IMG]
    where R is for rotations, v is for Galilean boosts, a is spatial translations and s is for time translations. The commutation relations are the following.
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    I just took these equations from wikipedia. For the Bargmann algebra,
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    I think this one is the same as the Heisenberg algebra my professor is talking about with ##[C'_{i}, P'_{j}] = iM\delta_{ij}## equivalent to ##[X,P] = i\hbar I \delta_{ij}##. But I'm not really sure about this. A hint given to me is that I shouldn't really take the notations seriously and that ##P## in the Galilean algebra is the central generator. Does this mean that perhaps if I compare the two, maybe the isomorphism is ##K \to X, H \to P, P \to \hbar I##?

    Thank you very much for your interest in helping me!
     
  7. Dec 12, 2017 #6

    fresh_42

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    Yes, that's what Wikipedia says. But what dimension do we have here? Certainly a significantly higher number than three. So either we must consider generalized Heisenberg algebras (with increased dimension) or a low dimensional version of the Galilean group.

    We have as Heisenberg algebra ##\mathcal{H} = \left\{ \begin{bmatrix} 0&K&P\\0&0&H\\0&0&0\end{bmatrix} \right\}##. The same as we get by ##\{C_1,H,P_1\}## with ##C_1 \leftrightarrow K## and only one index, no ##L## and eventually a scalar correction for ##P=p\cdot P_1##. This means: no rotation and one-dimensional boost and translations, i.e. a different algebra than the large algebra you just described.
     
  8. Dec 12, 2017 #7
    Yes, I would only like to consider a smaller Galilean group. For 1D Galilean symmetry we have no rotations, right? So does that mean for a 1D physical space the two algebras are isomorphic (which now I think they are)? How about for 2D space with SO(2) rotational symmetry? I'm sorry I'm a little bit of a beginner here. Also thank you very much for your reply! I really appreciate it and I actually learned something new. ^_^
     
    Last edited by a moderator: Dec 12, 2017
  9. Dec 12, 2017 #8

    fresh_42

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    Yes.
    Yes. You don't even need a complicated basis transformation. The basis vectors only have different letters (in the versions above) and maybe one has to deal with a factor ##\pm i## or so for the central vector ##P##. I haven't checked.
    I admit my notation ##\begin{bmatrix}0&K&P\\0&0&H\\0&0&0\end{bmatrix}## has been a bit sloppy (lazy). It should have been:
    $$
    K=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\; , \; H=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\; , \;P=\begin{bmatrix}0&0&-i\\0&0&0\\0&0&0\end{bmatrix}
    $$
    I haven't checked the two-dimensional case. You could do it as an exercise.
     
  10. Dec 12, 2017 #9
    Wow! Thank you very much! I learned a lot of new things from your replies. Thank you again! ^_^
     
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