I am not sure if this is question is withing the scope of this page. I have an exam in quantum mechanics in a few weeks and I always confuse anhilation and creation operator.
$$ a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle $$ $$ a|n\rangle = \sqrt{n} |n-1\rangle $$
Does someone have a mnemonic for these relations?
For me this is most easily clinched by the fact that the annihilation operator annihilates the vacuum: $$\hat a|0\rangle=0.$$ If you're having trouble remembering which of $a$ and $a^\dagger$ is the annihilation/creator operator, this form may be useful; a-times-zero-is-zero sounds about as nice a mnemonic as you're likely to get, at least on the symbols side. (Of course, this requires you to beware of misremembering as $a^\dagger|0\rangle\stackrel{\text{!}}{=}0$. This form is more complicated - it's got more symbols -, so one way to remember it is to go for the simpler of the two.)
Further than that, if you know the general form of the relations, $$ \begin{align} a^\dagger |n\rangle &= \sqrt{n(+1?)} |n(\pm?)1\rangle, \\ a|n\rangle& = \sqrt{n(+1?)} |n(\pm?)1\rangle, \end{align} $$ but you're having trouble remembering their exact form, you can settle the ambiguity by noting that since $a|0\rangle$ must vanish, the coefficient must be $\sqrt{n}$; the coefficient of $a^\dagger$ must therefore be $\sqrt{n+1}$. The fact that $a$ 'kills' something also settles it as the annihilation operator, which means that it must take $|n\rangle$ to $|n-1\rangle$, and that leaves simply $|n+1\rangle$ for the final choice.
Of course, as with any mnemonics, it's a completely personal thing whether any particular mnemonic works for you or not. Also, as time goes on you will either use those relations often enough that you don't need a mnemonic, or you will stop using it and stop needing to remember.
I would say that I think $a$ for annihilate. Then since $a^\dagger$ has the dagger, it has to do the opposite thing i.e., create. That is how I would remember the difference between these two.
As far as the actual equation. $$ a|n\rangle = \sqrt{n} |n-1\rangle $$. $$ a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle $$
These two me always reminded me of differentiation and integration. The one with the annihilation operator is like differentiation (you bring down an $n$ and replace the original $n$ by $n-1$). The only difference is there is a square root.
Then the creation one looks like integration (you bring down an $n+1$ and then you increase the original $n$ to $n+1$). Here there are two differences: 1) the square root again, and 2) you multiply instead of divide.
If you have trouble remembering the differences from differentiation and integration, you can always check that $a^\dagger a = \hat{n}$. This equation will check out only if you remember about taking square roots and always multiplying.
