Question

In section 7.5,${\mathbf{e}}^{{\mathbf{im}}_{\mathbf{l}}\mathbf{\phi }}$is presented a sour preferred solution to the azimuthal equation, but there is more general one that need not violate the smoothness condition, and that in fact covers not only complex exponentials but also suitable redelinitions of multiplicative constants, sine, and cosine,

$\mathit{\Phi }{\mathit{m}}_{\mathbf{1}}\left(\Phi \right)\mathbf{=}\mathit{A}{\mathbf{e}}^{\mathbf{+}{\mathbf{im}}_{\mathbf{l}}\mathbf{\phi }}\mathbf{+}{\mathbf{Be}}^{\mathbf{+}{\mathbf{im}}_{\mathbf{l}}\mathbf{\phi }}$

(a) Show that the complex square of this function is not, in general, independent of $\mathbf{\phi }$.

(b) What conditions must be met by A and/or B for the probability density to be rotationally symmetric – that is, independent of $\mathbf{\phi }$ ?

Short answer

(a) In above equation, in general, the complex square of the azimuthal equation is not independent of $\mathrm{\phi }$.

(b) When the value of ${\mathrm{m}}_1$takes both positive and negative integral values, they will be equivalent.

Step-by-step solution

Complex Square of azimuthal equation:

The equation ${\mathbf{\Phi m}}_{\mathbf{1}}\mathbf{\left(}\mathbf{\Phi }\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{{\mathbf{im}}_{\mathbf{l}}\mathbf{\phi }}$is the improved form of the solution of the equation governing azimuthal motion.

Consider the known equation as below.

${\mathrm{\Phi m}}_1\left(\mathrm{\Phi }\right)=A{\mathrm{e}}^{+{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}+{\mathrm{Be}}^{+{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}$

By taking the complex square of the azimuthal equation, we get,

localid="1659178898885" $$\begin{gathered} {\mathrm{\Phi }}^{*}\mathrm{m}\left(\mathrm{\Phi }\right){\mathrm{\Phi }}_{\mathrm{ml}}\left(\mathrm{\Phi }\right)=\left({\mathrm{A}}^{*{\mathrm{e}}^{-{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}}+{{\mathrm{B}}^{*\mathrm{e}}}^{+{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}\right)\left({\mathrm{Ae}}^{{}^{+{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}}+{\mathrm{Be}}^{-{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}\right) \\ {\mathrm{\Phi }}^{*}\mathrm{m}\left(\mathrm{\Phi }\right){\mathrm{\Phi }}_{\mathrm{ml}}\left(\mathrm{\Phi }\right)={\left|\mathrm{A}\right|}^2+{\left|\mathrm{B}\right|}^2+\mathrm{A}*{\mathrm{Be}}^{-2{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }}+\mathrm{B}*{\mathrm{Ae}}^{+2{\mathrm{im}}_{\mathrm{l}}\mathrm{\phi }} \\ \left(1\right) \end{gathered}$$

As it can be clearly shown in the above eq. (1) that in general, the complex square of the azimuthal equation is not independent of $\mathrm{\phi }$.

(b) Conclusion:

If we make A=0 and B=0 to make the equation independent of $\mathrm{\phi }$, the solution itself will be zero, hence, we cannot do that.

But when the value of ${\mathrm{m}}_{\mathrm{l}}$takes both positive and negative integral values, they will be equivalent.