Angular momentum representation It is well know that, using position representation
$$\langle r\lvert L\rvert \psi\rangle     =r \times (-i\hbar\nabla\langle r|\psi\rangle   )=r \times (-i\hbar\nabla\psi(r)).$$
However, I read from some books that if $L$ is acting on some position ket directly, then
$$L|r\rangle ~=~ r \times (i\hbar\nabla|r\rangle).$$  
Can anyone explain the latter equation regarding the missing "-" sign?
 A: As Richard points out, you can derive the second equation by setting $\psi$ to be a position eigenstate in the first one. Doing that, you turn the general case
$$\langle \mathbf{r}\lvert \mathbf{L}\rvert \psi\rangle     =\mathbf{r} \times (-i\hbar\nabla\langle \mathbf{r}|\psi\rangle   )$$
into the relation
$$\langle \mathbf{r}\lvert \mathbf{L}\rvert \mathbf{r}'\rangle     
=\mathbf{r} \times (-i\hbar\nabla_\mathbf{r}\langle \mathbf{r}|\mathbf{r}'\rangle)
=\mathbf{r} \times \left(-i\hbar\nabla_\mathbf{r}\delta(\mathbf{r}-\mathbf{r}')\right).
$$
In here, you can change the $\mathbf{r}$'s into $\mathbf{r}'$s using the fact that both vectors are equal at the support of the delta function. Thus you can change $\mathbf{r}\times$ for  $\mathbf{r}'\times$, but the derivative is a bit trickier: since the argument of the delta fuction is $\mathbf{r}-\mathbf{r}'$, its derivatives w.r.t. $\mathbf{r}$ differ from its derivatives w.r.t. $\mathbf{r}'$ by a sign, and you must change $\nabla_\mathbf{r}$ for  $-\nabla_{\mathbf{r}'}$. With this, then,
$$\langle \mathbf{r}\lvert \mathbf{L}\rvert \mathbf{r}'\rangle     
=\mathbf{r} \times \left(-i\hbar\nabla_\mathbf{r}\delta(\mathbf{r}-\mathbf{r}')\right)
=\mathbf{r}' \times \left(+i\hbar\nabla_{\mathbf{r}'}\delta(\mathbf{r}-\mathbf{r}')\right)
=\mathbf{r}' \times \left(+i\hbar\nabla_{\mathbf{r}'}\langle \mathbf{r}|\mathbf{r}'\rangle\right).
$$
Once it is in this form, you simply have a global factor of $\langle\mathbf{r}|$, which you can simply "cancel out". (More formally, since the $|\mathbf{r}\rangle$ are a complete set, the projections on the $\langle \mathbf{r}|$ completely determine any vector. Or, if you prefer, simply multiply the equation by $|\mathbf{r}\rangle$ and integrate over all $\mathbf{r}$.)
Doing that, then, and dropping the primes, you get, finally
$$\mathbf{L}\rvert \mathbf{r}\rangle    
=\mathbf{r} \times \left(+i\hbar\nabla_{\mathbf{r}}|\mathbf{r}\rangle\right)
\tag1$$
as you wanted to get.

I must say, though that this relation is not particularly useful. What is useful, though, is its adjoint relation, which you can get from the original
$$
\langle \mathbf{r}\lvert \mathbf{L}\rvert \psi\rangle 
=\left(\mathbf{r} \times (-i\hbar\nabla)\langle \mathbf{r}|\right)|\psi\rangle   
$$
by simply "cancelling out" $|\psi\rangle$. (Or, more formally, by noting that the linear functionals on both sides coincide for all $|\psi\rangle$, and must therefore be equal as linear functionals.) This gives simply
$$
\langle \mathbf{r}\lvert \mathbf{L}
=\mathbf{r} \times (-i\hbar\nabla_\mathbf{r})\langle \mathbf{r}|, \tag 2$$
which is evidently the adjoint of (1). (What's remarkable is that the vector calculus remains valid.)
The reason I say that this is the form that's actually useful is that you very, very rarely deal with position ket $|\mathbf{r}\rangle$, as they are very much not physical states, but you do deal regularly with position bras $\langle \mathbf{r}|$, as they are an essential ingredient in well-written position representations. The form (2) then lets you find the position-representation wavefunction of the transformed vector $\mathbf{L}|\psi\rangle$ from the original wavefunction $\langle \mathbf{r}|\psi\rangle$.
This is analogous to the way to make precise the intuition that $\mathbf{p}$ equals the derivative $-i\hbar \nabla$, by considering its actions on bras instead of kets, to get
$$\langle \mathbf{r}|\mathbf{p}=-i\hbar\nabla_\mathbf{r}\langle \mathbf{r}|,$$
as I've said before in this answer. While this looks slightly unintuitive at first, it is actually more useful if you use it right.
A: Just take $|\psi\rangle=|r_0\rangle$ in the first equation and use $\langle r|r_0\rangle=\delta(r-r_0)$ so $\psi(r)=\delta(r-r_0)$.
