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?
 A: 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 $0$.

However: this convention is not universal, and you always need to check what (anti)commutation relations are being imposed into any such operators when they are introduced, whatever it is you're reading. The commutation relations fully specify the behaviour, though, so that if you have an object that satisfies the bosonic set of relations then you can call it a "ladder operator" or an "annihilation operator" according to personal taste, and no one will bat an eyelid at either term - they both have contexts where either can be preferable over the other.
