# Creation operator

1. Mar 9, 2010

### mkbh_10

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. Mar 9, 2010

### f95toli

Why not simply use Pascal's triangle?

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