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Question 45
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) $\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 1 & 0 & \sqrt{3} \\ 0 & 0 & 0 \\ \sqrt{3} & 0 & 1 \end{pmatrix}$
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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