How do you calculate the commutator $$[b_i^n, (b_j^\dagger)^m]$$ where $b_i$ are annihilators and $b_j^\dagger$ are creators in second quantization. All annihalators should be swapped to the right side.
Asked
Active
Viewed 190 times
0
-
1Possible duplicate: https://physics.stackexchange.com/q/45053/2451 – Qmechanic Jan 12 '18 at 20:26
-
3Possible duplicate of The cleverest way to calculate $\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]$ with $\left[\hat{a},\hat{a}^{\dagger}\right]=1$ – Punk_Physicist Jan 12 '18 at 21:51
1 Answers
4
Simply use the definition of the commutator $$[A B, C] = A B C - C A B = ABC - ACB + ACB - CAB \\= A (BC - CB) + (AC - CA)B = A[B, C] + [A,C]B$$ which generalizes to $$ [A^n, B] = \sum_{m=0}^{n-1} A^m[A,B]A^{n-m-1}$$ by repeated application. You should be able to do the rest yourself from here.
noah
- 10,324
-
1There's a problem here, since your formula with $n=1$ implies that $[A,B]=[A,B]A+A[A,B]$. I think it's that the upper bound should be $n-1$. – Chris Jan 12 '18 at 21:33