Question:
The non-relativistic Hamiltonian for a single electron atom is
\[ H_0 = \frac{p^2}{2m} - V(r) \]
where \( V(r) \) is the Coulomb potential and \( m \) is the mass of the electron.Considering the spin-orbit interaction term:
\[ H' = \frac{1}{2m^2c^2} \frac{1}{r} \frac{dV}{dr} \overrightarrow{L} \cdot \overrightarrow{S} \]
added to \( H_0 \), which of the following statements is/are true?
The non-relativistic Hamiltonian for a single electron atom is
\[ H_0 = \frac{p^2}{2m} - V(r) \]
where \( V(r) \) is the Coulomb potential and \( m \) is the mass of the electron.Considering the spin-orbit interaction term:
\[ H' = \frac{1}{2m^2c^2} \frac{1}{r} \frac{dV}{dr} \overrightarrow{L} \cdot \overrightarrow{S} \]
added to \( H_0 \), which of the following statements is/are true?
\[ H_0 = \frac{p^2}{2m} - V(r) \]
where \( V(r) \) is the Coulomb potential and \( m \) is the mass of the electron.Considering the spin-orbit interaction term:
\[ H' = \frac{1}{2m^2c^2} \frac{1}{r} \frac{dV}{dr} \overrightarrow{L} \cdot \overrightarrow{S} \]
added to \( H_0 \), which of the following statements is/are true?
Updated On: Jul 12, 2024
- \( H' \) commutes with \( L^2 \).
- \( H' \) commutes with \( L_z \) and \( S_z \).
- For a given value of principal quantum number \( n \) and orbital angular momentum quantum number \( l \), there are \( 2(2l + 1) \) degenerate eigenstates of \( H_0 \).
- \( H_0 \), \( L^2 \), \( S^2 \), \( L_z \), and \( S_z \) have a set of simultaneous eigenstates.
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The Correct Option is A, C, D
Solution and Explanation
The correct Answers are (A):\( H' \) commutes with \( L^2 \).,(C):For a given value of principal quantum number \( n \) and orbital angular momentum quantum number \( l \), there are \( 2(2l + 1) \) degenerate eigenstates of \( H_0 \). ,(D): \( H_0 \), \( L^2 \), \( S^2 \), \( L_z \), and \( S_z \) have a set of simultaneous eigenstates,
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