Probability flux: spatial variation of the phase equal to momentum? We can write any wave function as
$$\psi(\vec x, t) = \sqrt{\rho(\vec x,t)}\exp{\left[\frac{iS(\vec x,t)}{\hbar}\right]}$$
for $S$ real and $\rho >0$. Here we interpret $\rho$ as the probability density. With the definition of the probability flux as
$$\vec j(\vec x,t) \propto \psi^*\nabla\psi ,$$
Sakurai shows that for the wavefunction above
$$\vec j = \frac{\rho\nabla S}{m}.$$
The point here being that the probability flux depends on the spatial variation of the phase. Next he states the direction of $\vec j$ at some point $\vec x$ is always normal to the surface of a constant phase that goes through that point. He then gives the example of a plane wave:
$$\psi(\vec x,t) \propto \exp{\left(\frac{i\vec p\cdot\vec x}{\hbar}-\frac{iEt}{\hbar}\right)}$$
in which it is stated that 
$$\nabla S = \vec p.$$
Question: How can we show that the last equation is true? In the context of the first equation, I intrepret
$$S(\vec x,t) = \vec p\cdot\vec x-Et$$
and thus
$$\nabla S = \nabla(\vec p\cdot\vec x).$$
Surely we need not use a vector dot product identity. What am I missing?
 A: Consider 2D (XY) case for simplicity, $\vec x = x\vec e_x + y\vec e_y$. $\vec e_x$ is unit vector along x-axis.
By definition, gradient is the following operator upon scalar:
$\nabla=(\frac{\partial}{\partial x}\vec{e_x} + \frac{\partial}{\partial y}\vec{e_y})$
Apply this operator to $\vec p\cdot\vec x$:
$\nabla(\vec{p}\cdot \vec x)=(\frac{\partial}{\partial x}\vec{e_x} + \frac{\partial}{\partial y}\vec{e_y})(p_x x + p_y y)$
Then we consider first part: $\frac{\partial p_x x}{\partial x}\vec{e_x}=\frac{\partial p_x}{\partial x}x\vec{e_x}+\frac{\partial x}{\partial x}p_x\vec{e_x}=\frac{\partial x}{\partial x}p_x\vec{e_x}$, since $p_x$ is not explicitly function of $x$. Also, $\frac{\partial x}{\partial x}=1$.
Also, $p_y$ is not explicitly function of $x$, so partial derivative $\frac{\partial(p_y y)}{\partial x}$ is zero. Same story goes for $\frac{\partial }{\partial y}$.
Which yields: $\nabla(\vec{p}\cdot \vec x)=\frac{\partial x}{\partial x}p_x\vec{e_x}+\frac{\partial y}{\partial y}p_y\vec{e_y}=\vec p$
As I remember, such transitions are widely used across the field.
