GATE PH 2022 • Q45 MCQ • 2 Marks

Section 4: Quantum Mechanics

Harmonic oscillator

Subtopic: Algebraic Method with Ladder Operators

Question Statement:

For a one-dimensional harmonic oscillator, the creation operator $a^\dagger$ acting on the $n$-th state $|\psi_n\rangle$ gives $a^\dagger |\psi_n\rangle = \sqrt{n+1} |\psi_{n+1}\rangle$. The position operator $x$ in the basis of the first three states is proportional to

Options:

  • (A) $\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{3} \end{pmatrix}$
  • (B) $\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$
  • (C) $\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{pmatrix}$
  • (D) $\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 1 & 0 & \sqrt{3} \\ 0 & 0 & 0 \\ \sqrt{3} & 0 & 1 \end{pmatrix}$

Official Answer Key:

C

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.45)** **1. Quantum Harmonic Oscillator Operator Matrix Representation**: - Ladder operators $a, a^\dagger$ acting on number states $|n\rangle$: $a |n\rangle = \sqrt{n} |n-1\rangle, \quad a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle$ - Position operator $\hat{x}$ in terms of ladder operators: $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}} (a + a^\dagger)$ **2. Detailed Matrix Element Evaluation in Basis $\{|0\rangle, |1\rangle, |2\rangle\}$**: - Matrix elements $x_{mn} = \langle m | \hat{x} | n \rangle = \sqrt{\frac{\hbar}{2m\omega}} (\langle m | a | n \rangle + \langle m | a^\dagger | n \rangle)$: - $\langle 0 | \hat{x} | 0 \rangle = 0$ - $\langle 0 | \hat{x} | 1 \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle 0 | a | 1 \rangle = \sqrt{\frac{\hbar}{2m\omega}} (1)$ - $\langle 1 | \hat{x} | 0 \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle 1 | a^\dagger | 0 \rangle = \sqrt{\frac{\hbar}{2m\omega}} (1)$ - $\langle 1 | \hat{x} | 2 \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle 1 | a | 2 \rangle = \sqrt{\frac{\hbar}{2m\omega}} (\sqrt{2})$ - $\langle 2 | \hat{x} | 1 \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle 2 | a^\dagger | 1 \rangle = \sqrt{\frac{\hbar}{2m\omega}} (\sqrt{2})$ - All diagonal elements are zero: $x_{00} = x_{11} = x_{22} = 0$. - Constructing the $3 \times 3$ matrix: $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{pmatrix}$ **3. Official Verification**: The verified official IIT Master Answer is **Option (C)**.

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