A Weird Contradiction about angular momentum operator in quantum mechanics I am starting with the standard definition of an angular momentum operator in quantum mechanics given as $$\mathbf{L} = k(\mathbf{r}\times\mathbf{p}) = k(\mathbf{r}\times\nabla),$$ 
where $k=-\mathbf{i}\hbar$ which is supposed to act on a complex scalar $\psi$ that comes as a solution of the Schroedinger eigenvalue problem. Hence, application of the said operator has the form 
$$\mathbf{L}\psi=k(\mathbf{r}\times\nabla\psi).$$
One can now recall that in the above the first part $\mathbf{r}$ is an integrable field and so one can write the previous in the form
$$\mathbf{L}\psi = k(\nabla\phi\times\nabla\psi), \quad \text{where}\; \phi = \frac{|\mathbf{r}|^2}{2}.$$
Whoever is familiar with old style vector analysis in Hydrodynamics or with current MHD, will immediately recognise in the above expression the so called "Clebsh Variables" ($\phi,\psi$) parametrization, also called "stream-flux" potentials. It is then trivial to derive an equivalent form of the above in the associated "Monge Representation" given as
$$\mathbf{L}\psi = \nabla\times(k\phi\nabla\psi).$$
But then, the above is a purely solenoidal field. Hence, any further application of the $\mathbf{L}$ operator as $\mathbf{L}\cdot\mathbf{L} = \mathbf{L}^2$ would necessarilly give zero! But this is in straight contradiction with the dictums of Quantum Mechanics where the square of the Angular Momentum is always associated with the eigenvalues $l(l+1)$. Where is the mistake in the above?
 A: The last formula $\mathbf{L}\psi=\nabla\times(k\phi\nabla\psi)$ is correct. What is incorrect is the statement which follows: "Hence, any further application of the $\mathbf{L}$ operator as $\mathbf{L}\cdot\mathbf{L} = \mathbf{L}^2$ would necessarilly give zero". 
A: If you don't like the physics way of computing $\mathbf{L}^2$ using indices, using commutator, or under spherical coordinates, here is the traditional multivariate calculus way.
First by the formula you gave:
$$
\mathbf{L}\psi = k\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\
\partial_x & \partial_y & \partial_z \\  \phi \partial_x\psi & \phi \partial_y\psi & \phi \partial_z\psi \end{vmatrix}
\\
=k\left(\begin{vmatrix}\partial_y & \partial_z \\ \phi \partial_y\psi & \phi \partial_z\psi \end{vmatrix}, 
-\begin{vmatrix} 
\partial_x &   \partial_z \\  \phi \partial_x\psi &    \phi \partial_z\psi \end{vmatrix},
\begin{vmatrix} 
\partial_x & \partial_y  \\  \phi \partial_x\psi & \phi \partial_y\psi \end{vmatrix}\right)
\\
=k\begin{pmatrix}\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi
\\
\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi
\\
\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi\end{pmatrix}
:= \begin{pmatrix}\mathbf{L}_x\psi
\\
\mathbf{L}_y\psi
\\
\mathbf{L}_z\psi\end{pmatrix}.
$$
Then 
$$
\begin{aligned}
\mathbf{L}\cdot \mathbf{L}\psi &= \mathbf{L}_x \mathbf{L}_x \psi+  \mathbf{L}_y\mathbf{L}_y \psi +  \mathbf{L}_z \mathbf{L}_z \psi
\\
&= k\mathbf{L}_x(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi)
\\
&\quad  + k\mathbf{L}_y(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi)  
\\
&\quad \;+k\mathbf{L}_z (\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi)
\\
&=k^2\Big[ \partial_y\phi \partial_z(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi) - \partial_z\phi \partial_y(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi)
\\
&\; +\partial_z\phi \partial_x(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi) - \partial_x\phi \partial_z(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi)
\\
&\; + \partial_x\phi \partial_y(\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi) - \partial_y\phi \partial_x(\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi)\Big].
\end{aligned}$$
And I immediately regretted doing this...anyway making use of $\phi = (x^2+y^2+z^2)/2$ and simplifying this you will have (spent two whole pieces of scrap paper):
$$
\begin{aligned}
\mathbf{L}\cdot \mathbf{L}\psi &= k^2\Big( (y^2+z^2)\partial_{xx}\psi + (z^2+x^2)\partial_{yy}\psi + (x^2+y^2)\partial_{zz}\psi 
\\
&\quad \;-2xy\partial_{xy}\psi - 2yz\partial_{yz}\psi - 2zx\partial_{zx} \psi -2x\partial_x\psi - 2y\partial_y\psi -2z\partial_z\psi\Big).
\end{aligned} 
$$
Using some algebra trick yields the traditional multivariate calculus formula:
$$
\mathbf{L}\cdot \mathbf{L}\psi = k^2\Big(\Delta \psi |\mathbf{r}|^2 - (\mathbf{r}\cdot \nabla)^2 \psi- \mathbf{r}\cdot \nabla \psi \Big),\tag{1}
$$
where 
$$
(\mathbf{r}\cdot \nabla)^2 = (x\partial_x+y\partial_y+z\partial_z)^2.
$$
So (1) is our ultimate formula for angular momentum squared. In quantum mechanics, $\mathbf{L}^2$ has $\hbar^2 l(l+1)$ as its eigenvalue, so let's check if the result above agrees with the deduction in index notation, when $\psi = \chi_{0,0} = x$:
$$
\mathbf{L}^2 \chi_{0,0} = i^2\hbar^2(\Delta x |\mathbf{r}|^2  - (\mathbf{r}\cdot \nabla)^2 x - \mathbf{r}\cdot \nabla x ) = 2\hbar^2 x= \hbar^2 l(l+1)\chi_{0,0}
$$
where $l=1$ so we are happy. Let's check one more $\psi =\chi_{1,-1} = (x-iy)/\sqrt{2}$:
$$
\mathbf{L}^2 \chi_{1,-1} = i^2\hbar^2\Big(0 |\mathbf{r}|^2  - (\mathbf{r}\cdot \nabla) (x-iy)/\sqrt{2} - \mathbf{r}\cdot  (1/\sqrt{2},-i/\sqrt{2},0) \Big) = 2\hbar^2 \chi_{1,-1},
$$
happy again.
A: Thank you all for your answers. The reason for posting the above was to show the dangers behind some definitions as the one mentioned in Wikipedia without any other comment
http://en.wikipedia.org/wiki/Angular_momentum_operator#Orbital_angular_momentum_operator
And of course, as Raskolnikov correctly pointed out, the operators do not commute here the same as in the classical cross product, hence the sign is not at all innocent as it introduces a serious "operator ordering" problem. But if we treat it the same way as in the case of the so called "Convective Derivative" with precedence from the left, it is quite a different case.
