$\Phi_1 = \frac{1}{\sqrt{2}}\psi_{1,4,1} - \frac{1}{\sqrt{2}}\psi_{2,2,3}$
$\Phi_2 = \frac{1}{\sqrt{2}}\psi_{1,5,1} + \frac{1}{\sqrt{2}}\psi_{3,3,3}$
$\Phi_3 = \frac{1}{\sqrt{2}}\psi_{1,3,8} + \frac{1}{\sqrt{2}}\psi_{3,8,1}$
$\Phi_4 = \frac{1}{2}\psi_{3,3,1} + \frac{\sqrt{3}}{2}\psi_{2,4,1}$
Particle in Cubic Box: Hamiltonian Eigenfunctions
For a particle confined in a three-dimensional cubic box, the energy eigenfunctions $\psi_{n_x,n_y,n_z}$ are characterized by quantum numbers ($n_x, n_y, n_z$). The corresponding energy eigenvalues are calculated using the formula:
$E_{n_x,n_y,n_z} = \frac{\pi^2 \hbar^2}{2ma^2}(n_x^2 + n_y^2 + n_z^2)$
A key principle is that a linear combination of wavefunctions forms an eigenfunction of the Hamiltonian operator only if all the individual wavefunctions within the combination possess the same energy. This means the sum of the squares of the quantum numbers ($n_x^2 + n_y^2 + n_z^2$) must be identical for all terms in the linear combination.
Analyzing Linear Combinations for Eigenfunctions
Function $\Phi_1$ Analysis
$\Phi_1 = \frac{1}{\sqrt{2}}\psi_{1,4,1} - \frac{1}{\sqrt{2}}\psi_{2,2,3}$
- Calculate energy contribution for $\psi_{1,4,1}$: $n_x^2 + n_y^2 + n_z^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$
- Calculate energy contribution for $\psi_{2,2,3}$: $n_x^2 + n_y^2 + n_z^2 = 2^2 + 2^2 + 3^2 = 4 + 4 + 9 = 17$
The energy contributions (18 and 17) differ. Therefore, $\Phi_1$ is not an eigenfunction of the Hamiltonian.
Function $\Phi_2$ Analysis
$\Phi_2 = \frac{1}{\sqrt{2}}\psi_{1,5,1} + \frac{1}{\sqrt{2}}\psi_{3,3,3}$
- Calculate energy contribution for $\psi_{1,5,1}$: $n_x^2 + n_y^2 + n_z^2 = 1^2 + 5^2 + 1^2 = 1 + 25 + 1 = 27$
- Calculate energy contribution for $\psi_{3,3,3}$: $n_x^2 + n_y^2 + n_z^2 = 3^2 + 3^2 + 3^2 = 9 + 9 + 9 = 27$
The energy contributions (27 and 27) are the same. Therefore, $\Phi_2$ is an eigenfunction of the Hamiltonian.
Function $\Phi_3$ Analysis
$\Phi_3 = \frac{1}{\sqrt{2}}\psi_{1,3,8} + \frac{1}{\sqrt{2}}\psi_{3,8,1}$
- Calculate energy contribution for $\psi_{1,3,8}$: $n_x^2 + n_y^2 + n_z^2 = 1^2 + 3^2 + 8^2 = 1 + 9 + 64 = 74$
- Calculate energy contribution for $\psi_{3,8,1}$: $n_x^2 + n_y^2 + n_z^2 = 3^2 + 8^2 + 1^2 = 9 + 64 + 1 = 74$
The energy contributions (74 and 74) are the same. Therefore, $\Phi_3$ is an eigenfunction of the Hamiltonian.
Function $\Phi_4$ Analysis
$\Phi_4 = \frac{1}{2}\psi_{3,3,1} + \frac{\sqrt{3}}{2}\psi_{2,4,1}$
- Calculate energy contribution for $\psi_{3,3,1}$: $n_x^2 + n_y^2 + n_z^2 = 3^2 + 3^2 + 1^2 = 9 + 9 + 1 = 19$
- Calculate energy contribution for $\psi_{2,4,1}$: $n_x^2 + n_y^2 + n_z^2 = 2^2 + 4^2 + 1^2 = 4 + 16 + 1 = 21$
The energy contributions (19 and 21) differ. Therefore, $\Phi_4$ is not an eigenfunction of the Hamiltonian.
Conclusion on Hamiltonian Eigenfunctions
The analysis shows that $\Phi_2$ and $\Phi_3$ are eigenfunctions of the Hamiltonian operator because they are constructed from wavefunctions with identical energy contributions.
Important Questions from Particle in a Box
Consider two non-interacting particles confined to a one-dimensional box with infinite potential barriers. Their wavefunctions are $\psi_1$ and $\psi_2$ and energies are $E_1$ and $E_2$, respectively. The INCORRECT statement(s) about this system is/are
The wave function of a particle in a cubic box (of side L) is given by
$\psi(x, y, z) = \sqrt{32/L^3} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L}$.
The ratio of the energy of the state corresponding to the above wave function to the ground state energy is ________.
(rounded off to the nearest integer)- The wavelength associated with a particle in one-dimensional box of length $L$ is ($n$ refers to the quantum number)
- Assume 1,3,5-hexatriene to be a linear molecule and model the $\pi$ electrons as particles in a one-dimensional box of length 0.70 nm. The wavelength (in nm) corresponding to the transition from the ground-state to the first excited-state is ________
The $\pi$ electrons in benzene can be modelled as particles in a ring that follow Pauli's exclusion principle. Given that the radius of benzene is 1.4 Å, the longest wavelength of light that is absorbed during an electronic transition in benzene is ____________ nm. (Up to one decimal place. Use $m_e =9.1\times10^{-31} \text{ kg}$, $h=6.6\times10^{-34} \text{ Js}$, $c=3.0\times10^8 \text{ m s}^{-1}$)
