Question

Wavefunctions and energies for a particle confined in a cubic box are \( \psi_{n_x,n_y,n_z} \) and \( E_{n_x,n_y,n_z} \), respectively. The functions \( \phi_1, \phi_2, \phi_3 \), and \( \phi_4 \) are written as linear combinations of \( \psi_{n_x,n_y,n_z} \). Among these functions, the eigenfunction(s) of the Hamiltonian operator for this particle is/are \[ \phi_1 = \frac{1}{\sqrt{2}} \psi_{1,4,1} - \frac{1}{\sqrt{2}} \psi_{2,2,3} \] \[ \phi_2 = \frac{1}{\sqrt{2}} \psi_{1,5,1} + \frac{1}{\sqrt{2}} \psi_{3,3,3} \] \[ \phi_3 = \frac{1}{\sqrt{2}} \psi_{1,3,8} + \frac{1}{\sqrt{2}} \psi_{3,8,1} \] \[ \phi_4 = \frac{1}{2} \psi_{3,3,1} + \frac{\sqrt{3}}{2} \psi_{2,4,1} \]

• A. \( \phi_2 \)
• B. \( \phi_4 \)
• C. \( \phi_3 \)
• D. \( \phi_1 \)

Hint:

A linear combination of eigenfunctions of a Hamiltonian is also an eigenfunction of the Hamiltonian if and only if all the eigenfunctions in the linear combination have the same eigenvalue (energy, in this case). Check the indices \( (n_x, n_y, n_z) \) for each component of the linear combination and calculate the corresponding energy (proportional to \( n_x^2 + n_y^2 + n_z^2 \)). If the energies are the same, the linear combination is an eigenfunction.

Answer 1

The Correct Option is A, C

The energy of a particle in a cubic box of side L is given by:

E_nx,ny,nz = (h² / 8mL²) × (n_x² + n_y² + n_z²)

For a function to be an eigenfunction of the Hamiltonian operator, applying the Hamiltonian to it must yield the same function multiplied by a constant (the eigenvalue). If a wavefunction is a linear combination of eigenfunctions with different energies, it is not itself an eigenfunction. If it's a linear combination of eigenfunctions with the same energy (i.e., degenerate states), then it remains an eigenfunction of the Hamiltonian with that same energy.

Analysis of the φ_i functions:

• φ₁:
  E_1,4,1 ∝ 1 + 16 + 1 = 18
  E_2,2,3 ∝ 4 + 4 + 9 = 17
  → Not degenerate → φ₁ is not an eigenfunction
• φ₂:
  E_1,5,1 ∝ 1 + 25 + 1 = 27
  E_3,3,3 ∝ 9 + 9 + 9 = 27
  → Degenerate → φ₂ is an eigenfunction
• φ₃:
  E_1,3,8 ∝ 1 + 9 + 64 = 74
  E_3,8,1 ∝ 9 + 64 + 1 = 74
  → Degenerate → φ₃ is an eigenfunction
• φ₄:
  E_3,3,1 ∝ 9 + 9 + 1 = 19
  E_2,4,1 ∝ 4 + 16 + 1 = 21
  → Not degenerate → φ₄ is not an eigenfunction

Conclusion:
The functions φ₂ and φ₃ are eigenfunctions of the Hamiltonian. This corresponds to options (A) and (C).