# Ladder operators vs creation/annihilation operators

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?

• They are the same thing. The two names exist because in quantum field theory they represent adding or removing particles from the state of the system, but this interpretation does not make sense in all context. Nov 14, 2018 at 14:11
• 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. 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 $$0$$.