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Creation operator

  1. Mar 9, 2010 #1
    Suppose a problem requires me to find the result of [(a+a#)^n ]ln> , where a is the annhilation operator , and a# is the creation operator , now n can be any number , suppose if it is 10 then will i write (a+a#) 10 times and then multiply each of these terms to see which are non zero , isn't there some general form of getting result of (a+a#)^n ?

    The normal method of multiplying will become too cumbersome if n is a large number .
     
  2. jcsd
  3. Mar 9, 2010 #2

    f95toli

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    Why not simply use Pascal's triangle?

    edit: just realized that this might not actually works since the operators do not commute...
     
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