I would like to ask this following question. In the picture below we have a spherical and plane wave. For the plane wave we have the angle between k and the x axis, that is θ. I would like to know what this angle is for the spherical wave, 𝑘 =𝑘0(𝑐𝑜𝑠𝜃 𝚤𝑥̂ + 𝑠𝑖𝑛𝜃 𝚤𝑦̂).
1 Answer
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A plane wave:
$$ \psi(\mathbf{x}) = Ae^{i({\mathbf{ k \cdot x}}-\omega t)} $$
is an eigenstate of the wave-vector operator:
$$ {\mathbf{\hat k}} =-i\mathbf{\nabla} $$
per:
$$ {\mathbf{\hat k}}\psi(\mathbf{x}) =-i{\mathbf{\nabla}}Ae^{i({\mathbf{ k \cdot x}}-\omega t)} $$ $$ = -iAe^{i({\mathbf{ k \cdot x}}-\omega t)}i{\mathbf{\nabla}}({\mathbf{ k \cdot x}}-\omega t )$$
$$ = {\mathbf{k}}\psi(\mathbf{x}) $$
Hence, one can define the angle of $\mathbf k$ with respect to the $x$-axis.
Moreover, this eigenvalue reflects the translation symmetry of the plane wave: if the wave is translated in the direction $\mathbf k$ by $2\pi/||k||$, the wave is unchanged.
You can't to that with a spherical wave. It has some value for $k_r$, which points in all directions, while the natural angular eigenstates have rotational symmetry (the spherical harmonics). They do not have a well defined Cartesian wave vector with which to find an angle with respect to the $x$-axis.
