I have been reading about raising and lowering operators, but I am a bit unsure of one thing. If you have a string of raising and lowering operators (or any kind of operator), do you apply them in order of right to left or left to right?
An example would be $$A^{\dagger}A^{\dagger}A^{\dagger}A$$ In this example, do you first apply the lowering operator then the three raising operators or do you do the three raising operators then the lowering operator?
In the usual situation, where the quantum state is on the right (a ket), like $A^{\dagger}A^{\dagger}A^{\dagger}A|\psi\rangle$, you do indeed apply the (annihilation) operator on the far right first, and then successively to the left. If you have trouble remembering this, think of it as only the operator immediately adjacent to the state vector can act on it. Once $A$ has operated on $|\psi\rangle$, you then have an new state vector $|\psi'\rangle=A|\psi\rangle$, upon which the remaining operators can act, like $A^{\dagger}A^{\dagger}A^{\dagger}|\psi'\rangle$.
However, it is also possible to have the state vector on the left (as a bra), like $\langle\psi|A^{\dagger}A^{\dagger}A^{\dagger}A$, in which case the operator are applied from left to right (so that the $A$ operator is last). Once again, the adjacent operator acts on the state vector. This is also required by the fact that $\left[\langle\psi|M\right]^{\dagger}=M^{\dagger}|\psi\rangle$, and if $M$ is a product of operators, (for example, $M=ABC$), the Hermitian conjugation reverses the order of the product, $M^{\dagger}=(ABC)^{\dagger}=C^{\dagger}B^{\dagger}A^{\dagger}$.
With any operators acting on a ket $| \psi \rangle$, you apply the one farthest to the right first. So in your case $A$.
