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?


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.

  • $\begingroup$ Is the momentum a fixed vector? How come it does not depend on position? $\endgroup$ – Lone Wolf Feb 26 '15 at 2:15
  • $\begingroup$ Momentum is not explicit function of coordinate, only velocity and mass. Partial derivative by coordinate will yield 0. Full derivative will most likely be non-zero. $\endgroup$ – aaaaa says reinstate Monica Feb 26 '15 at 2:56
  • $\begingroup$ does that make things clearer? $\endgroup$ – aaaaa says reinstate Monica Feb 26 '15 at 8:56
  • $\begingroup$ Momentum is an implicit function of position and time. Thank you kindly. $\endgroup$ – Lone Wolf Feb 26 '15 at 13:28

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