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Quantum Chemistry Eigenfunction

  1. Feb 1, 2014 #1
    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.
     
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  3. Feb 1, 2014 #2
    [itex]\textbf{H}\psi = E\psi[/itex]

    This is the eigenvalue problem that you are solving. If psi is an eigenfunction of H, Hψ will be equal to a constant and this constant, E, is the eigenvalue. For a particle in a three-dimensional box, H = -hbar2∇ and E you already stated. So, solve Hψ and confirm that it is equal to Eψ.

    H is a linear operator, so if two functions are eigenfunctions, then any linear combination of them will also be. You can show this by applying H to the linear combination just as you did for the individual functions. However, ψ(1,1,5) and ψ(3,3,3) are constants, not functions, so I'm not really sure what is being asked here.
     
  4. Feb 2, 2014 #3

    Redbelly98

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    It is probably supposed to be ψ1,1,5(x,y,z) and ψ3,3,3(x,y,z). That is the only way I think the question could make any sense.
     
  5. Feb 2, 2014 #4
    The numbers correspond to the n values. (aka ψ1,1,5 = nx = 1, ny = 1, nz=5) What I am confused about is how to solve this equation if I am given the boundries in terms of L and not in terms of an actual unit of length, should I just assume that L=1?
     
    Last edited: Feb 2, 2014
  6. Feb 2, 2014 #5
    Length is arbitrary. Using L instead of an actual number just means that your solution will be more general. So instead of sin(n*pi*x), which you would get form a box of length 1, your function would be sin(n*pi*x/L), so that when x = L you have sin(n*pi) = 0 and the boundary conditions are still met.

    In the case of a three dimensional box, you have three lengths (a,b,c), one for each side, although in this case, a = b = c = L.
     
  7. Feb 2, 2014 #6
    But if sin(n*pi)=0, then wouldn't the whole psi function = 0? In that case it wouldn't be an eigenvalue, unless its a trivial solution.

    Do you think its only giving Psi to list the quantum numbers and I just need to solve for E and then list the result in terms of an eigenvalue?
     
  8. Feb 2, 2014 #7
    Well psi isn't going to be equal to sin(n*pi) across the whole box- remember, the actual function is sin(n*pi*x/L), so this will only be the cause when x = L (i.e. at the boundary).

    Psi is given so that you can apply the Hamiltonian to it. If it is an eigenfunction, that this will give you the same value as if you had multiplied Psi by E. Similarly, if Psi1 + Psi2 is an eigenfunction, applying the Hamiltonian to (Psi1 + Psi2) will give you the same value as E(Psi1 + Psi2).
     
  9. Feb 3, 2014 #8
    So I ended up getting the function: -50*sin(Pi*x)*Pi^6*sin(Pi*y)*sin(5*Pi*z)-1458*sin(3*Pi*x)*Pi^6*sin(3*Pi*y)*sin(3*Pi*z) after applying the Hamiltonian to the Psi functions, but the two Psi terms now have different coefficients. Does that mean that Psi1 + Psi 2 is not an Eigenfunction?
     
  10. Feb 3, 2014 #9
    It looks like you might be multiplying the derivatives instead of adding. Remember,∇2 = d/dx2 + d/dy2 + d/dz2, so taking the derivative twice with respect to each coordinate and adding should give you the same coefficients as E (although in E, the values should actually all be squared).

    Sorry to have to leave, but I really need to get some sleep. Do you have a book that might cover this? Levine goes over this exact problem (three dimensional box) in section 3.5.
     
  11. Feb 3, 2014 #10
    Its okay, I already handed it in. I messed up though. It turns out I just needed to find out that the two energy functions were degenerate and that means they are an eigenfunction of the hamiltonian.
     
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