Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple roots Lie algebra

  1. May 2, 2014 #1
    If [itex]\alpha [/itex] and [itex]\beta [/itex] are simple roots, then [itex]\alpha-\beta [/itex] is not. This means that
    [tex]
    E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0
    [/tex]
    Now, according to the text I read, this means that [itex]q [/itex] in the formula
    [tex]
    \frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)
    [/tex]
    is zero, where [itex]\vec{\mu} [/itex] is a weight, and [itex]p[/itex] and [itex]q [/itex] are integers. I couldn't understand why [itex]q=0 [/itex], if someone could explain to me.
     
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. May 5, 2014 #3
    Eα is treated as a ladder operator, when it acts on a state of some weight say |μ> it adds to it and give something proportional to|μ+α>. E will do the opposite.

    In general, ##(E_α)^{P+1}|μ> =C E_α|μ+pα>=0## and ##(E_{-α})^{q+1}|μ> =C E_α|μ-qα>=0## for positive integers p and q.

    comparing to your equation,
    ##E_{-\vec{\alpha}}|μ\rangle = 0##, we have ##q=0##.
     
  5. May 6, 2014 #4
    So q=0 is the lowest state. Thanks...
     
  6. May 7, 2014 #5
    In this case only, not in general.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple roots Lie algebra
Loading...