I am trying to figure out the difference between the ladder operators (for harmonic oscillator) $$a^\dagger$$, $$a$$ and the creating/annihilation operators $$c^\dagger$$, $$c$$. Are they the same? I have read that the number operator is given by $$N=c^\dagger c$$, but also by $$N=a^\dagger a$$, so I assume that they must be the same and therefor they commute? Is this right or?
• To such an extent that $a$ and $a^+$ are practically always called "creation/anihilation operators", in order to reserve the name "ladder operators" for the angular momentum ones $J_+$ and $J_-$. I recomment to do this, so that you not mix ideas. Commented Nov 14, 2018 at 15:08
It is a reasonably wide convention to reserve the symbols $$a$$ and $$a^\dagger$$ for bosonic annihilation and creation operators which satisfy the canonical commutation relations $$[a,a^\dagger] = 1, \qquad [a,a] = [a^\dagger,a^\dagger] = 0,$$ while reserving the symbols $$c$$ and $$c^\dagger$$ for fermionic annihilation and creation operators which satisfy canonical anticommutation relations of the form $$\{c,c^\dagger\} = 1, \qquad \{c,c\} = \{c^\dagger,c^\dagger\} = 0.$$ Under that convention, the fermionic operators still give a number operator as $$N=c^\dagger c$$, but their behaviour is very different since now the setting of anticommutators to zero implies that $$c^2 = (c^\dagger)^2 = 0$$, which restricts occupation numbers to only $$1$$ or $$0$$.