Multiplication of ladder-operators

[Philip Land]

Hi!

When calculating $(\hat{a} \hat{a}^{\dagger})^2$ i get $\hat{a} \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger}$ which is perfectly fine.

But how do I end up with the ultimate simplified expression $$\hat{ a}^{\dagger} \hat{a} \hat{a}^{\dagger} \hat{a} + \hat{a}^{\dagger} \hat_{a} + 2\hat{a}^{\dagger} \hat_{a} + 2 = N^2 + 3 N +1 $$

Are there any definitions, rules or framework I can use to carry out these calculations to make my life easier or do I simply need to write out the definitions of $\hat{a}^{\dagger}$ and $\hat{a}$ and tediously recognize each term?

 

[stevendaryl]

Well, $(\hat{a} \hat{a}^\dagger)^2 = \hat{a} \hat{a}^\dagger \hat{a} \hat{a}^\dagger$, not $\hat{a} \hat{a} \hat{a}^\dagger \hat{a}^\dagger$. If you want to write it in terms of $\hat{N}$, use $\hat{a}^\dagger \hat{a} = \hat{N}$ and $\hat{a} \hat{a}^\dagger = \hat{N} + 1$. So you have:

$(\hat{a} \hat{a}^\dagger)^2 = (\hat{N} + 1)^2 = \hat{N}^2 + 2 \hat{N} + 1$