Question:
Consider two operators \(\hat{A}\) and \(\hat{B}\) which are related as \(\hat{A} = \exp(i\theta \hat{B})\). If \(\theta\) is a non-zero real number, which of the following statements is/are true?
Consider two operators \(\hat{A}\) and \(\hat{B}\) which are related as \(\hat{A} = \exp(i\theta \hat{B})\). If \(\theta\) is a non-zero real number, which of the following statements is/are true?
Show Hint
Test each case using \((e^{\hat{X}})^\dagger = e^{\hat{X}^\dagger}\) and \(\text{Det}(e^{\hat{X}}) = e^{\text{Tr}(\hat{X})}\).
A Hermitian generator times \(i\) becomes anti-Hermitian, so its exponential is unitary; an anti-Hermitian generator times \(i\) becomes Hermitian, so its exponential is Hermitian.
A Hermitian generator times \(i\) becomes anti-Hermitian, so its exponential is unitary; an anti-Hermitian generator times \(i\) becomes Hermitian, so its exponential is Hermitian.
Updated On: Jul 28, 2026
- If \(\hat{B}\) is Hermitian, then \(\hat{A}\) is unitary.
- If \(\hat{B}\) is anti-Hermitian, then \(\hat{A}\) is unitary.
- If \(\hat{B}\) is Hermitian, then \(|\text{Det}(\hat{A})| = 1\).
- If \(\hat{B}\) is anti-Hermitian, then \(\hat{A}\) is Hermitian.
Show Solution
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The Correct Option is A, C, D
Solution and Explanation
Step 1: Understand what each term means.
A Hermitian (self-adjoint) operator satisfies \(\hat{B}^\dagger = \hat{B}\), so all its eigenvalues are real. An anti-Hermitian operator satisfies \(\hat{B}^\dagger = -\hat{B}\), so its eigenvalues are purely imaginary. A unitary operator satisfies \(\hat{A}^\dagger \hat{A} = \hat{I}\), which forces every eigenvalue of \(\hat{A}\) to have magnitude 1.
Step 2: Key formula.
For any operator \(\hat{X}\), the adjoint of its exponential follows the rule \((e^{\hat{X}})^\dagger = e^{\hat{X}^\dagger}\), and its determinant follows \(\text{Det}(e^{\hat{X}}) = e^{\text{Tr}(\hat{X})}\). We apply both rules to \(\hat{A} = e^{i\theta \hat{B}}\) with \(\theta\) real.
Step 3: Check statement (A).
If \(\hat{B}\) is Hermitian, \(\hat{B}^\dagger = \hat{B}\), so \((i\theta\hat{B})^\dagger = -i\theta\hat{B}^\dagger = -i\theta\hat{B}\). The generator \(i\theta\hat{B}\) is anti-Hermitian. Using Step 2, \(\hat{A}^\dagger = e^{(i\theta\hat{B})^\dagger} = e^{-i\theta\hat{B}}\), so \(\hat{A}^\dagger \hat{A} = e^{-i\theta\hat{B}} e^{i\theta\hat{B}} = \hat{I}\). So \(\hat{A}\) is unitary, and (A) is TRUE.
Step 4: Check statement (B).
If \(\hat{B}\) is anti-Hermitian, \(\hat{B}^\dagger = -\hat{B}\), so \((i\theta\hat{B})^\dagger = -i\theta\hat{B}^\dagger = -i\theta(-\hat{B}) = i\theta\hat{B}\). The generator \(i\theta\hat{B}\) is now Hermitian, not anti-Hermitian, so its exponential gives a Hermitian operator, not a unitary one. So (B) is FALSE.
Step 5: Check statement (C).
If \(\hat{B}\) is Hermitian, its eigenvalues are real, so \(\text{Tr}(\hat{B})\) is a real number. Then \(\text{Tr}(i\theta\hat{B}) = i\theta\,\text{Tr}(\hat{B})\) is purely imaginary. Using \(\text{Det}(\hat{A}) = e^{\text{Tr}(i\theta\hat{B})}\), this is \(e^{i(\text{real number})}\), a complex number of magnitude 1. So \(|\text{Det}(\hat{A})| = 1\), and (C) is TRUE.
Step 6: Check statement (D).
From Step 4, when \(\hat{B}\) is anti-Hermitian, \(i\theta\hat{B}\) is Hermitian. The exponential of a Hermitian operator stays Hermitian, because \((e^{\hat{H}})^\dagger = e^{\hat{H}^\dagger} = e^{\hat{H}}\) whenever \(\hat{H}^\dagger = \hat{H}\). Taking \(\hat{H} = i\theta\hat{B}\), we get \(\hat{A}^\dagger = \hat{A}\). So \(\hat{A}\) is Hermitian, and (D) is TRUE.
Final Answer:
Statements (A), (C) and (D) follow correctly from the given relation; only (B) is wrong. \[ \boxed{\text{A, C, D}} \]
A Hermitian (self-adjoint) operator satisfies \(\hat{B}^\dagger = \hat{B}\), so all its eigenvalues are real. An anti-Hermitian operator satisfies \(\hat{B}^\dagger = -\hat{B}\), so its eigenvalues are purely imaginary. A unitary operator satisfies \(\hat{A}^\dagger \hat{A} = \hat{I}\), which forces every eigenvalue of \(\hat{A}\) to have magnitude 1.
Step 2: Key formula.
For any operator \(\hat{X}\), the adjoint of its exponential follows the rule \((e^{\hat{X}})^\dagger = e^{\hat{X}^\dagger}\), and its determinant follows \(\text{Det}(e^{\hat{X}}) = e^{\text{Tr}(\hat{X})}\). We apply both rules to \(\hat{A} = e^{i\theta \hat{B}}\) with \(\theta\) real.
Step 3: Check statement (A).
If \(\hat{B}\) is Hermitian, \(\hat{B}^\dagger = \hat{B}\), so \((i\theta\hat{B})^\dagger = -i\theta\hat{B}^\dagger = -i\theta\hat{B}\). The generator \(i\theta\hat{B}\) is anti-Hermitian. Using Step 2, \(\hat{A}^\dagger = e^{(i\theta\hat{B})^\dagger} = e^{-i\theta\hat{B}}\), so \(\hat{A}^\dagger \hat{A} = e^{-i\theta\hat{B}} e^{i\theta\hat{B}} = \hat{I}\). So \(\hat{A}\) is unitary, and (A) is TRUE.
Step 4: Check statement (B).
If \(\hat{B}\) is anti-Hermitian, \(\hat{B}^\dagger = -\hat{B}\), so \((i\theta\hat{B})^\dagger = -i\theta\hat{B}^\dagger = -i\theta(-\hat{B}) = i\theta\hat{B}\). The generator \(i\theta\hat{B}\) is now Hermitian, not anti-Hermitian, so its exponential gives a Hermitian operator, not a unitary one. So (B) is FALSE.
Step 5: Check statement (C).
If \(\hat{B}\) is Hermitian, its eigenvalues are real, so \(\text{Tr}(\hat{B})\) is a real number. Then \(\text{Tr}(i\theta\hat{B}) = i\theta\,\text{Tr}(\hat{B})\) is purely imaginary. Using \(\text{Det}(\hat{A}) = e^{\text{Tr}(i\theta\hat{B})}\), this is \(e^{i(\text{real number})}\), a complex number of magnitude 1. So \(|\text{Det}(\hat{A})| = 1\), and (C) is TRUE.
Step 6: Check statement (D).
From Step 4, when \(\hat{B}\) is anti-Hermitian, \(i\theta\hat{B}\) is Hermitian. The exponential of a Hermitian operator stays Hermitian, because \((e^{\hat{H}})^\dagger = e^{\hat{H}^\dagger} = e^{\hat{H}}\) whenever \(\hat{H}^\dagger = \hat{H}\). Taking \(\hat{H} = i\theta\hat{B}\), we get \(\hat{A}^\dagger = \hat{A}\). So \(\hat{A}\) is Hermitian, and (D) is TRUE.
Final Answer:
Statements (A), (C) and (D) follow correctly from the given relation; only (B) is wrong. \[ \boxed{\text{A, C, D}} \]
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