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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
    E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0
    Now, according to the text I read, this means that [itex]q [/itex] in the formula
    \frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)
    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.
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