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The Hamiltonian operator for a particle with mass m on a sphere with radius $r_0$ can be written as: $\hat{H}=-\frac{\hbar^2}{2mr_0^2}\hat{\Lambda}^2$

where

$\hat{\Lambda}^2=\frac{1}{sin^2\theta}\frac{\partial ^2}{\partial \varphi^2}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}sin\theta\frac{\partial}{\partial\theta}$

The solutions for $\psi_{lm_l}(\theta\varphi)$ are $Y_{lm_l}(\theta\varphi)$.

Confirm that $Y_{11}(\theta\varphi)=-\frac{1}{2}\sqrt{\frac{3}{2\pi}}sin\theta e^{i\varphi}$ is a solution and find the energy $E_{11}$.

Any help?

ChemGuy
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1 Answers1

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What this means is you need to show that

$$ \hat{H}\psi_{11}(\theta,\phi) = E_{11}\psi_{11}(\theta,\phi) $$

So it boils down to calculate the effect of $\hat{\Lambda}$ on $\psi_{11}$,

\begin{eqnarray} \hat{\Lambda}^2\psi_{11}(\theta,\phi) &=& \frac{1}{\sin^2\theta}\frac{\partial^2 \psi_{11}}{\partial\phi^2} + \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi_{11}}{\partial\theta}\right) \\ &=& (\cdots) \\ &=& \left(\frac{3}{2\pi}\right)\sin\theta e^{i\phi} = -2\psi_{11}(\theta,\phi) \end{eqnarray}

That is

$$ \hat{H}\psi_{11} = -\frac{\hbar^2}{2mr_0^2}\hat{\Lambda}^2\psi_{11} = \frac{2\hbar^2}{2mr_0^2}\psi_{11} = E_{11}\psi_{11} $$

that is, $\psi_{11}$ is an eigenstate of $\hat{H}$ with eigenvalue

$$ E_{11} = \frac{\hbar^2}{mr_0^2} $$

caverac
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