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##