- $\sqrt{2\pi}$
- $2\pi$
- $(2\pi)^2$
- $\dfrac{1}{\sqrt{2\pi}}$
- $\dfrac{1}{2\pi}$

Solution :

Eigenfunctions for a rigid dumbbell rotating about its centre have a $\phi$ dependence of the form $\psi(\phi)=Ae^{im\phi}$, where $m$ is a quantum number and $A$ is a constant. Which of the following values of $A$ will properly normalize the eigenfunction?

**Choice 4 is the answer.**

For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.

Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]

- $\sqrt{2\pi}$
- $2\pi$
- $(2\pi)^2$
- $\dfrac{1}{\sqrt{2\pi}}$
- $\dfrac{1}{2\pi}$

Solution :

For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.

Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]

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