Question

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 \( \hat{L}_z \) for the same state function is/are

• A. \( 10 \hbar^2 \)
• B. \( 16 \hbar^2 \)
• C. \( 25 \hbar^2 \)
• D. \( 0 \)

Hint:

The eigenvalues of \( \hat{L}^2 \) are \( l(l+1) \hbar^2 \), and the eigenvalues of \( \hat{L}_z \) are \( m_l \hbar \), where \( -l \le m_l \le l \). First, determine the value of \( l \) from the eigenvalue of \( \hat{L}^2 \). Then, find the possible values of \( m_l \) and the corresponding eigenvalues of \( \hat{L}_z \). If the options are in terms of \( \hat{L}_z^2 \), then consider \( m_l^2 \hbar^2 \).

Answer 1

The Correct Option is B, C, D

The operator ℒ² (denoted as ̂L²) is the square of the total angular momentum operator, and its eigenvalues are given by:

L²ψ = l(l + 1)ℐ²ψ

Here, l is the azimuthal quantum number: l = 0, 1, 2, ...

The z-component of angular momentum, ̂L_z, has eigenvalues:

L_zψ = m_lℐψ

Where m_l is the magnetic quantum number and ranges from −l to +l in integer steps.

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Given: Eigenvalue of ̂L² is 30ℐ², so:

l(l + 1) = 30

Solving:

l² + l − 30 = 0 → (l + 6)(l − 5) = 0 → l = 5 (since l must be ≥ 0)

For l = 5, possible values of m_l are:

−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5

Thus, eigenvalues of L_z are:

m_lℐ = −5ℐ, −4ℐ, ..., 0, ..., +5ℐ

If the question asks for possible values of L_z², then we look at:

(m_lℐ)² = m_l²ℐ² = 0, 1ℐ², 4ℐ², 9ℐ², 16ℐ², 25ℐ²

Comparison with options:

• (A) 10ℐ² → ❌ Not a perfect square
• (B) 16ℐ² → ✅ m_l = ±4
• (C) 25ℐ² → ✅ m_l = ±5
• (D) 0 → ✅ m_l = 0

Final Answer: The possible eigenvalues of L_z² are: (B), (C), and (D).