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GATE 2026 PH – Question 51

Quantum Mechanics · linear vectors and operators in Hilbert space · 2 marks · Multiple select

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?

  1. If $\hat B$ is Hermitian, then $\hat A$ is unitary.
  2. If $\hat B$ is anti-Hermitian, then $\hat A$ is unitary.
  3. If $\hat B$ is Hermitian, then $|\operatorname{Det}(\hat A)|=1$.
  4. If $\hat B$ is anti-Hermitian, then $\hat A$ is Hermitian.

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Show answer and explanation

Correct answer: (A) If $\hat B$ is Hermitian, then $\hat A$ is unitary.; (C) If $\hat B$ is Hermitian, then $|\operatorname{Det}(\hat A)|=1$.; (D) If $\hat B$ is anti-Hermitian, then $\hat A$ is Hermitian.

Explanation

$\hat A=e^{i\theta\hat B}$ with real $\theta\neq0$.

**Hermitian $\hat B$ ($\hat B^\dagger=\hat B$):**
$$\hat A^\dagger=e^{-i\theta\hat B^\dagger}=e^{-i\theta\hat B}=\hat A^{-1},$$
so $\hat A$ is **unitary**. ✓ (A)

The determinant: if $\hat B$ has real eigenvalues $b_k$, then $\det\hat A=\prod_ke^{i\theta b_k}=e^{i\theta\sum b_k}$, whose modulus is 1. ✓ (C)

**Anti-Hermitian $\hat B$ ($\hat B^\dagger=-\hat B$):** then $i\theta\hat B$ is Hermitian: $(i\theta\hat B)^\dagger=-i\theta\hat B^\dagger=i\theta\hat B$. The exponential of a Hermitian operator is Hermitian (with positive eigenvalues), so
$$\hat A^\dagger=\hat A,$$
i.e. $\hat A$ is **Hermitian**. ✓ (D) It is not unitary in general (its eigenvalues are real and positive, not of unit modulus), so B is false.

Answer **A, C and D**.