Question:
The energy \(E\) and degeneracy \(d\) of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency \(\omega\) are
The energy \(E\) and degeneracy \(d\) of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency \(\omega\) are
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Second excited state of a 3D isotropic oscillator is \(n=2\): use \(E_n=(n+3/2)\hbar\omega\) and degeneracy \(d(n)=(n+1)(n+2)/2\).
Updated On: Jul 28, 2026
- \(E = \dfrac{7}{2}\hbar\omega,\ d = 6\)
- \(E = \dfrac{7}{2}\hbar\omega,\ d = 3\)
- \(E = \dfrac{5}{2}\hbar\omega,\ d = 3\)
- \(E = \dfrac{5}{2}\hbar\omega,\ d = 6\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
A three-dimensional isotropic quantum harmonic oscillator is just three independent 1D oscillators of the same frequency \(\omega\), one along each axis. The energy of a single 1D oscillator with quantum number \(n_x\) is \((n_x+\tfrac{1}{2})\hbar\omega\), so the total energy only depends on the sum \(n = n_x+n_y+n_z\).
Step 2: Key Formula or Approach:
The energy of the level labeled by \(n\) is \(E_n = (n+\tfrac{3}{2})\hbar\omega\).
Its degeneracy is the number of ways three non-negative integers \(n_x,n_y,n_z\) can add up to \(n\), which is \(d(n) = \dfrac{(n+1)(n+2)}{2}\).
Step 3: Detailed Explanation:
The ground state is \(n=0\), the first excited state is \(n=1\), so the second excited state is \(n=2\).
Energy: \[ E_2 = \left(2+\frac{3}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega \]
Degeneracy: \[ d(2) = \frac{(2+1)(2+2)}{2} = \frac{3\times4}{2} = 6 \]
Checking this by listing every \((n_x,n_y,n_z)\) with \(n_x+n_y+n_z=2\): \((2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1)\), that is 6 distinct states, matching the formula.
Step 4: Why the other options are wrong.
Option (B) has the right energy but the wrong degeneracy, 3 is the degeneracy of the first excited state (\(n=1\)), not the second.
Options (C) and (D) both use \(E=\tfrac{5}{2}\hbar\omega\), which is the energy of the first excited state, not the second.
Final Answer:
The second excited state has energy \(\tfrac{7}{2}\hbar\omega\) and degeneracy 6. \[ \boxed{E = \frac{7}{2}\hbar\omega,\ d = 6} \]
A three-dimensional isotropic quantum harmonic oscillator is just three independent 1D oscillators of the same frequency \(\omega\), one along each axis. The energy of a single 1D oscillator with quantum number \(n_x\) is \((n_x+\tfrac{1}{2})\hbar\omega\), so the total energy only depends on the sum \(n = n_x+n_y+n_z\).
Step 2: Key Formula or Approach:
The energy of the level labeled by \(n\) is \(E_n = (n+\tfrac{3}{2})\hbar\omega\).
Its degeneracy is the number of ways three non-negative integers \(n_x,n_y,n_z\) can add up to \(n\), which is \(d(n) = \dfrac{(n+1)(n+2)}{2}\).
Step 3: Detailed Explanation:
The ground state is \(n=0\), the first excited state is \(n=1\), so the second excited state is \(n=2\).
Energy: \[ E_2 = \left(2+\frac{3}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega \]
Degeneracy: \[ d(2) = \frac{(2+1)(2+2)}{2} = \frac{3\times4}{2} = 6 \]
Checking this by listing every \((n_x,n_y,n_z)\) with \(n_x+n_y+n_z=2\): \((2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1)\), that is 6 distinct states, matching the formula.
Step 4: Why the other options are wrong.
Option (B) has the right energy but the wrong degeneracy, 3 is the degeneracy of the first excited state (\(n=1\)), not the second.
Options (C) and (D) both use \(E=\tfrac{5}{2}\hbar\omega\), which is the energy of the first excited state, not the second.
Final Answer:
The second excited state has energy \(\tfrac{7}{2}\hbar\omega\) and degeneracy 6. \[ \boxed{E = \frac{7}{2}\hbar\omega,\ d = 6} \]
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