Question

Show that$\mathit{\Theta }\mathbf{=}\mathit{A}\mathit{I}\mathit{n}\mathbf{\left[}\mathit{t}\mathit{a}\mathit{n}\mathbf{\right(}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{\left)}\mathbf{\right]}$satisfies the $\mathit{\theta }$equation (Equation 4.25), for l = m = 0. This is the unacceptable "second solution" -- whats wrong with it?

Short answer

$\Theta \left(\theta \right)=AIn\left[\tan \frac{\theta }{2}\right]$does satisfy the equation but this is the unacceptable second solution since $\Theta$blows up at $\theta =0$and at$\theta =\mathrm{\pi }$

Step-by-step solution

Define the Schrödinger equation

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

Calculation

We need to show that,

$\Theta \left(\theta \right)=AIn\left[\tan \frac{\theta }{2}\right]$

Satisfies the equation

$\sin \left(\theta \right)\frac{d}{d\theta }(\sin \left(\theta \right)\frac{d\Theta }{d\theta }+[I\left(I+1\right){\sin }^2\left(\theta \right)-m^2]\Theta =0$

So now take the derivative then we get,

$$\begin{gathered} \frac{d\Theta }{d\theta }=\frac{A}{\tan \left({\displaystyle \frac{\theta }{2}}\right)}\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right) \\ \frac{d\Theta }{d\theta }=\frac{A}{2}\frac{1}{\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)}=\frac{A}{\sin \theta } \end{gathered}$$

Therefore,

$\frac{d}{d\theta }\left(\sin \theta \frac{d\Theta }{d\theta }\right)=\frac{d}{d\theta }A=0$

Then, we get,

role="math" localid="1656064432888" $$\begin{gathered} \Theta \left(0\right)=AIn\left(0\right)=A\left(-\infty \right) \\ \Theta \left(\mathrm{\pi }\right)=AIn\left(\tan \frac{\mathrm{\pi }}{2}\right)=AIn\left(-\infty \right)=A\left(\infty \right) \end{gathered}$$