- #1
Gerardum
- 5
- 0
1. Consider a particle of mass m in a cubic (3-dimensional) box with V(x,y,z) = 0 for 0 < x < L, 0 < y < L, and 0 < z < L and V(x,y,z) = ∞elsewhere. Is 1/[itex]\sqrt{2}[/itex] * (ψ(1,1,5)+ψ(3,3,3)) an eigenfunction of the Hamiltonian for this system? If so, what is the eigenvalue? Explain your reasoning
2. Hamiltonian: E= [itex]\frac{h^2}{8m}[/itex]*([itex]\frac{n_{x}}{a}[/itex]+[itex]\frac{n_{y}}{b}[/itex]+[itex]\frac{n_{z}}{c}[/itex])
Psi: [itex]\sqrt{\frac{8}{abc}}[/itex]*sin([itex]\frac{n_{x}*pi*x}{a}[/itex])*sin([itex]\frac{n_{y}*pi*y}{b}[/itex])*sin([itex]\frac{n_{z}*pi*z}{c}[/itex])
3. So I tried solving for the two Psis added together and divided by the constant. But I don't see how they are in any way related to the Hamiltonian and how I could factor a constant out to obtain the Hamiltonian. I am right stumped.
2. Hamiltonian: E= [itex]\frac{h^2}{8m}[/itex]*([itex]\frac{n_{x}}{a}[/itex]+[itex]\frac{n_{y}}{b}[/itex]+[itex]\frac{n_{z}}{c}[/itex])
Psi: [itex]\sqrt{\frac{8}{abc}}[/itex]*sin([itex]\frac{n_{x}*pi*x}{a}[/itex])*sin([itex]\frac{n_{y}*pi*y}{b}[/itex])*sin([itex]\frac{n_{z}*pi*z}{c}[/itex])
3. So I tried solving for the two Psis added together and divided by the constant. But I don't see how they are in any way related to the Hamiltonian and how I could factor a constant out to obtain the Hamiltonian. I am right stumped.