Question:
If the perturbation \(V = \lambda x^3\) is added to the Hamiltonian of a one dimensional harmonic oscillator, the matrix element \(\langle m|V|0\rangle\) is/are non-zero for which of the following states? Here, the eigenstates of the harmonic oscillator are denoted by \(|n\rangle\).
If the perturbation \(V = \lambda x^3\) is added to the Hamiltonian of a one dimensional harmonic oscillator, the matrix element \(\langle m|V|0\rangle\) is/are non-zero for which of the following states? Here, the eigenstates of the harmonic oscillator are denoted by \(|n\rangle\).
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Hint:
Write \(x\) using ladder operators \(a, a^{\dagger}\) and check which states \((a+a^{\dagger})^3\) can reach starting from \(|0\rangle\); three \(\pm 1\) steps can only give a net change of \(\pm 1\) or \(\pm 3\).
Write \(x\) using ladder operators \(a, a^{\dagger}\) and check which states \((a+a^{\dagger})^3\) can reach starting from \(|0\rangle\); three \(\pm 1\) steps can only give a net change of \(\pm 1\) or \(\pm 3\).
Updated On: Jul 28, 2026
- \(|m=3\rangle\)
- \(|m=1\rangle\)
- \(|m=2\rangle\)
- \(|m=5\rangle\)
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The Correct Option is A, B
Solution and Explanation
Step 1: Understanding the Concept:
The perturbation is \(V = \lambda x^3\), and we want to know for which excited state \(|m\rangle\) the matrix element \(\langle m|V|0\rangle\) is non-zero. We write the position operator \(x\) in terms of the harmonic oscillator ladder operators \(a\) (lowering) and \(a^{\dagger}\) (raising), since these operators tell us exactly how many quanta a given term can add or remove.
Step 2: Key Formula or Approach:
The position operator is \(x = \sqrt{\dfrac{\hbar}{2m\omega}}(a + a^{\dagger})\), so
\[ x^3 = \left(\dfrac{\hbar}{2m\omega}\right)^{3/2} (a + a^{\dagger})^3 \]
To find \(\langle m|x^3|0\rangle\) we act with \((a+a^{\dagger})^3\) on the ground state \(|0\rangle\) and see which states \(|m\rangle\) appear, using \(a|n\rangle = \sqrt{n}\,|n-1\rangle\) and \(a^{\dagger}|n\rangle = \sqrt{n+1}\,|n+1\rangle\).
Step 3: Detailed Explanation:
Apply the operator one factor at a time. First, \((a+a^{\dagger})|0\rangle = |1\rangle\), since \(a|0\rangle = 0\) and \(a^{\dagger}|0\rangle = |1\rangle\).
Next, \((a+a^{\dagger})|1\rangle = a|1\rangle + a^{\dagger}|1\rangle = |0\rangle + \sqrt{2}\,|2\rangle\), so \((a+a^{\dagger})^2|0\rangle = |0\rangle + \sqrt{2}\,|2\rangle\).
Apply the operator a third time:
\[ (a+a^{\dagger})^3|0\rangle = (a+a^{\dagger})\big(|0\rangle + \sqrt{2}\,|2\rangle\big) = a|0\rangle + a^{\dagger}|0\rangle + \sqrt{2}\big(a|2\rangle + a^{\dagger}|2\rangle\big) \]
\(= 0 + |1\rangle + \sqrt{2}\big(\sqrt{2}\,|1\rangle + \sqrt{3}\,|3\rangle\big) = |1\rangle + 2|1\rangle + \sqrt{6}\,|3\rangle\)
\[ (a+a^{\dagger})^3|0\rangle = 3|1\rangle + \sqrt{6}\,|3\rangle \]
So \(x^3|0\rangle\) is built only from \(|1\rangle\) and \(|3\rangle\), each with a non-zero coefficient. This means \(\langle m|x^3|0\rangle\) is non-zero only when \(m=1\) or \(m=3\), and is exactly zero for every other value of \(m\), including \(m=2\) and \(m=5\).
Step 4: Why the other options are wrong.
Option (C), \(|m=2\rangle\), needs a net change of 2 quanta from the ground state, but three factors of \(a\) or \(a^{\dagger}\), each changing the quantum number by exactly \(\pm 1\), can only add up to a net change of \(+3, +1, -1\) or \(-3\); a net change of 2 is impossible. Option (D), \(|m=5\rangle\), needs a net change of 5, even further out of reach of three \(\pm 1\) steps.
Final Answer:
The matrix element \(\langle m|V|0\rangle\) is non-zero only for \(m=1\) and \(m=3\). \[ \boxed{|m=3\rangle \text{ and } |m=1\rangle} \]
The perturbation is \(V = \lambda x^3\), and we want to know for which excited state \(|m\rangle\) the matrix element \(\langle m|V|0\rangle\) is non-zero. We write the position operator \(x\) in terms of the harmonic oscillator ladder operators \(a\) (lowering) and \(a^{\dagger}\) (raising), since these operators tell us exactly how many quanta a given term can add or remove.
Step 2: Key Formula or Approach:
The position operator is \(x = \sqrt{\dfrac{\hbar}{2m\omega}}(a + a^{\dagger})\), so
\[ x^3 = \left(\dfrac{\hbar}{2m\omega}\right)^{3/2} (a + a^{\dagger})^3 \]
To find \(\langle m|x^3|0\rangle\) we act with \((a+a^{\dagger})^3\) on the ground state \(|0\rangle\) and see which states \(|m\rangle\) appear, using \(a|n\rangle = \sqrt{n}\,|n-1\rangle\) and \(a^{\dagger}|n\rangle = \sqrt{n+1}\,|n+1\rangle\).
Step 3: Detailed Explanation:
Apply the operator one factor at a time. First, \((a+a^{\dagger})|0\rangle = |1\rangle\), since \(a|0\rangle = 0\) and \(a^{\dagger}|0\rangle = |1\rangle\).
Next, \((a+a^{\dagger})|1\rangle = a|1\rangle + a^{\dagger}|1\rangle = |0\rangle + \sqrt{2}\,|2\rangle\), so \((a+a^{\dagger})^2|0\rangle = |0\rangle + \sqrt{2}\,|2\rangle\).
Apply the operator a third time:
\[ (a+a^{\dagger})^3|0\rangle = (a+a^{\dagger})\big(|0\rangle + \sqrt{2}\,|2\rangle\big) = a|0\rangle + a^{\dagger}|0\rangle + \sqrt{2}\big(a|2\rangle + a^{\dagger}|2\rangle\big) \]
\(= 0 + |1\rangle + \sqrt{2}\big(\sqrt{2}\,|1\rangle + \sqrt{3}\,|3\rangle\big) = |1\rangle + 2|1\rangle + \sqrt{6}\,|3\rangle\)
\[ (a+a^{\dagger})^3|0\rangle = 3|1\rangle + \sqrt{6}\,|3\rangle \]
So \(x^3|0\rangle\) is built only from \(|1\rangle\) and \(|3\rangle\), each with a non-zero coefficient. This means \(\langle m|x^3|0\rangle\) is non-zero only when \(m=1\) or \(m=3\), and is exactly zero for every other value of \(m\), including \(m=2\) and \(m=5\).
Step 4: Why the other options are wrong.
Option (C), \(|m=2\rangle\), needs a net change of 2 quanta from the ground state, but three factors of \(a\) or \(a^{\dagger}\), each changing the quantum number by exactly \(\pm 1\), can only add up to a net change of \(+3, +1, -1\) or \(-3\); a net change of 2 is impossible. Option (D), \(|m=5\rangle\), needs a net change of 5, even further out of reach of three \(\pm 1\) steps.
Final Answer:
The matrix element \(\langle m|V|0\rangle\) is non-zero only for \(m=1\) and \(m=3\). \[ \boxed{|m=3\rangle \text{ and } |m=1\rangle} \]
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