Creation operator

  • Thread starter mkbh_10
  • Start date
  • #1
mkbh_10
222
0
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 .
 

Answers and Replies

  • #2
f95toli
Science Advisor
Gold Member
3,328
829
Why not simply use Pascal's triangle?

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

Suggested for: Creation operator

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
567
  • Last Post
Replies
4
Views
2K
Replies
4
Views
679
Replies
2
Views
935
Replies
6
Views
2K
Replies
4
Views
3K
Replies
2
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
12
Views
2K
Top