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Question

The spherical harmonic function, $Y_{l,m}(\theta,\phi)$, with appropriate values of $l$ and $m$, is an eigen function of $\hat{L}_x^2 + \hat{L}_y^2$ operator. The corresponding eigenvalue is

The correct answer is
$(l(l+1)-m^2)\hbar^2$

Understanding Spherical Harmonics and Angular Momentum Operators

Spherical harmonic functions, denoted as $Y_{l,m}(\theta,\phi)$, are fundamental solutions to the angular part of the Schrödinger equation in spherical coordinates. They are simultaneous eigenfunctions of the square of the total angular momentum operator ($\hat{L}^2$) and the z-component of the angular momentum operator ($\hat{L}_z$).

  • Eigenvalue of $\hat{L}^2$ for $Y_{l,m}$ is $\hbar^2 l(l+1)$.
  • Eigenvalue of $\hat{L}_z$ for $Y_{l,m}$ is $\hbar m$.

Deriving the Eigenvalue for $\boldsymbol{\hat{L}_x^2 + \hat{L}_y^2}$

The total angular momentum squared operator is related to its components by:

$ \hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 $

We need the eigenvalue for the operator $\hat{L}_x^2 + \hat{L}_y^2$. Rearranging the equation above, we get:

$ \hat{L}_x^2 + \hat{L}_y^2 = \hat{L}^2 - \hat{L}_z^2 $

Now, we apply this operator to the spherical harmonic eigenfunction $Y_{l,m}(\theta,\phi)$:

$ (\hat{L}_x^2 + \hat{L}_y^2) Y_{l,m}(\theta,\phi) = (\hat{L}^2 - \hat{L}_z^2) Y_{l,m}(\theta,\phi) $

Using the known eigenvalues:

$ = \hat{L}^2 Y_{l,m}(\theta,\phi) - \hat{L}_z^2 Y_{l,m}(\theta,\phi) $

$ = (\hbar^2 l(l+1)) Y_{l,m}(\theta,\phi) - (\hbar m)^2 Y_{l,m}(\theta,\phi) $

$ = \left( \hbar^2 l(l+1) - \hbar^2 m^2 \right) Y_{l,m}(\theta,\phi) $

$ = (l(l+1) - m^2)\hbar^2 Y_{l,m}(\theta,\phi) $

Final Eigenvalue Result

The eigenvalue corresponding to the operator $\hat{L}_x^2 + \hat{L}_y^2$ acting on the spherical harmonic $Y_{l,m}(\theta,\phi)$ is $(l(l+1) - m^2)\hbar^2$. This matches option A.

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Important Questions from Angular Momentum

  1. The orbital angular momentum (L) of an electron is $\sqrt{12}\hbar$. The minimum angle between L and its z-component (in degrees) is ______ (rounded off to one decimal place).
  2. If a particle's state function is an eigenfunction of the operator $\hat{L}^2$ with eigenvalue $30\hbar^2$, then the possible eigenvalue(s) of the operator $\hat{L}_z^2$ for the same state function is/are
  3. The angle of orientation (in degrees) of the angular momentum vector with respect to z-axis for $l = 2$ and $m_l = +2$ state of H-atom is ________ (up to two decimal places)
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