91影视

Azimuthal quantum number specifies shape and angular momentum of the orbital.

Consider the given data as below.

$\mathrm{桅}\left(\mathrm{蠁}\right)={\mathrm{Ae}}^{\sqrt{-\mathrm{顿蠁}}}+{\mathrm{Be}}^{{}^{\sqrt{-\mathrm{顿蠁}}}}$ 鈥�.. (1)

Where, is the Azimuthal Angle, are the Arbitrary constants, and is the Azimuthal function.

$D=-\frac{1}{桅}\frac{{鈭�}^2桅}{鈭�{蠁}^2}$

Let, eq. (1) is continuous when $蠁=0$ and $蠁=2蟺$.

As you know that, for a function to be continuous at$蠁=0$ and$蠁=2蟺$, the value at$蠁=0$ andsh$蠁=2蟺$ould be equal.

Hence, for that to hold,

$$\begin{gathered} 桅\left(0\right)=桅\left(2蟺\right) \\ A{e}^{\sqrt{-D}脳0}+B{e}^{{-}^{\sqrt{-D}脳0}}=A{e}^{{}^{\sqrt{-D}脳2蟺}}+B{e}^{-\sqrt{-D}脳2蟺} \\ A+B=A{e}^{{}^{\sqrt{-D}2蟺}}+B{e}^{-\sqrt{-D}2蟺} \end{gathered}$$ 鈥�.. (2)

Also, if its derivative is continuous

$$\begin{gathered} \sqrt{-D}\left(A-B\right)=\sqrt{-D}\left(A{e}^{\sqrt{-D}2蟺}+B{e}^{-\sqrt{-D}2蟺}\right) \\ A-B=A{e}^{\sqrt{-D}2蟺}+B{e}^{-\sqrt{-D}2蟺} \end{gathered}$$ 鈥�.. (3)

Now, by adding equation (2) and (3), you get,

$A=A{e}^{\sqrt{-D}2蟺}$ 鈥�.. (4)

Also, by subtracting equation (3) from (2), you get,

$B=B{e}^{-\sqrt{-D}2蟺}$ 鈥�.. (5)

For the equation (4) to hold,

Either $A=0$ or $D=0$

And for eq. (5) to hold

Either $B=0$ or $D=0$

You can鈥檛 have both$A=0$ and $B=0$, wave function will not be possible if it holds.

And if $D=0$, the given equation$桅\left(蠁\right)=A{e}^{\sqrt{-D}蠁}+B{e}^{{-}^{\sqrt{-D}蠁}}$will be a constant.

Hence, the given function cannot meet itself smoothly when it finishes a round trip about the z-axis.

If $D=0$, the equation $桅\left(蠁\right)=A{e}^{\sqrt{-D}蠁}+B{e}^{{-}^{\sqrt{-D}蠁}}$ will be a constant and will be a linear function, it will only repeat itself after $2蟺$ if the slope is zero.