Question:
Let \( |m\rangle \) and \( |n\rangle \) denote the energy eigenstates of a one-dimensional simple harmonic oscillator. The position and momentum operators are \( \hat{X} \) and \( \hat{P} \), respectively. The matrix element \( \langle m | \hat{P}\hat{X} | n \rangle \) is non-zero when:
Let \( |m\rangle \) and \( |n\rangle \) denote the energy eigenstates of a one-dimensional simple harmonic oscillator. The position and momentum operators are \( \hat{X} \) and \( \hat{P} \), respectively. The matrix element \( \langle m | \hat{P}\hat{X} | n \rangle \) is non-zero when:
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For position and momentum operators, the matrix elements in the harmonic oscillator energy eigenbasis involve transitions between the same state or those differing by 2 quanta.
Updated On: Apr 8, 2025
- \( m = n \pm 2 \) only
- \( m = n \) or \( m = n \pm 2 \)
- \( m = n \pm 3 \) only
- \( m = n \pm 1 \) only
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The Correct Option is B
Solution and Explanation
Step 1: In the quantum harmonic oscillator, the position and momentum operators \( \hat{X} \) and \( \hat{P} \) can be expressed in terms of the raising and lowering operators. The matrix elements of these operators in the energy eigenbasis are non-zero only for certain transitions.
Step 2: The matrix element \( \langle m | \hat{P} \hat{X} | n \rangle \) is non-zero when \( m = n \) or when \( m = n \pm 2 \). This condition arises from the properties of the creation and annihilation operators.
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