# 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?

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.

Just take $|\psi\rangle=|r_0\rangle$ in the first equation and use $\langle r|r_0\rangle=\delta(r-r_0)$ so $\psi(r)=\delta(r-r_0)$.