Question

The eigenvalues of the operator \( \sin \left( a \frac{\partial}{\partial \theta} \right) \) corresponding to the eigenfunction \( \psi_{3,2,-2}(r,\theta,\phi) \) are:

• A. \( \sin(2a) \)
• B. \( -\sin(2a) \)
• C. \( \sinh(2a) \)
• D. \( -\sinh(2a) \)

Answer 1

The Correct Option is B

\[ \psi_{3,2,-2}(r,\theta,\phi) = R_{3}(r) Y_{2,-2}(\theta,\phi) \] Here, \( Y_{2,-2}(\theta, \phi) \) is a spherical harmonic function which is dependent on the angular variables \( \theta \) and \( \phi \). The spherical harmonic can be written as: \[ Y_{l,m}(\theta, \phi) = N_{l,m} P_{l}^{m}(\cos\theta) e^{im\phi} \] where \( P_{l}^{m}(\cos\theta) \) are the associated Legendre polynomials, and \( N_{l,m} \) is a normalization constant. For the given function \( Y_{2,-2}(\theta, \phi) \), we have \( m = -2 \).

Step 1: Apply the operator to the eigenfunction

We are tasked with finding the eigenvalue of the operator \( \sin \left( a \frac{\partial}{\partial \theta} \right) \) acting on the eigenfunction \( \psi_{3,2,-2}(r,\theta,\phi) \). Since the operator involves the derivative with respect to \( \theta \), the action of the operator on the angular part of the wavefunction is of primary interest.

\[ \sin \left( a \frac{\partial}{\partial \theta} \right) Y_{2,-2}(\theta, \phi) \]

Step 2: Effect of the derivative operator

The derivative operator \( \frac{\partial}{\partial \theta} \) acting on the spherical harmonic \( Y_{l,m}(\theta, \phi) \) gives a factor of \( m \), the magnetic quantum number. Specifically, for \( Y_{2,-2}(\theta, \phi) \), we have \( m = -2 \).

\[ \frac{\partial}{\partial \theta} Y_{2,-2}(\theta, \phi) \sim -2 Y_{2,-2}(\theta, \phi) \]

Step 3: Apply the sine operator

Now we apply the sine operator. The sine operator is a simple function that will act on the result of the derivative. Thus, we get the following result:

\[ \sin \left( a \frac{\partial}{\partial \theta} \right) Y_{2,-2}(\theta, \phi) = -\sin(2a) Y_{2,-2}(\theta, \phi) \]

Conclusion: Eigenvalue of the operator

From the above result, we see that the operator \( \sin \left( a \frac{\partial}{\partial \theta} \right) \) acting on the eigenfunction \( \psi_{3,2,-2}(r,\theta,\phi) \) gives the eigenvalue \( -\sin(2a) \). Thus, the eigenvalue is:

\( -\sin(2a) \)