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Solving 1D 2-particle/2D 1-particle Schrodinger equation

  1. Jan 27, 2010 #1
    Hi

    Most of my work in the past has been on 1D independent particle systems where I use matrix diagonalisation to solve the time-independent Schrodinger equation.

    How do I go about this for 2D? Or, more accurately, for a two-body wf in 1D? It strikes me as straightforward to construct a Hamiltonian matrix for 2D, but what will that give me when I diagonalise it? Am I going to have to bite the bullet and go to iterative procedures? Please don't make me use a Slater determinant basis!!! :frown:

    Thanks in advance.

    El Hombre
     
  2. jcsd
  3. Jan 28, 2010 #2
    First of all notice that now you have two position operator [tex]\hat{x},\hat{y}[/tex] and two momentum operators [tex]\hat{p_{x}},\hat{p_{y}}[/tex]
    So that operators from different coordinates commute, and a position operator doesn't commute with his conjugate momentum (as excpected).

    Usually, in simple systems, your coordinates will be uncoupled, that is, x coordinate is independent of y coordinate. This is the case where you Hamiltonian operator H can be seperated additively into two Hamiltonians, each corresponds to another dimension:
    [tex]\hat{H}=\hat{H_{x}}+\hat{H_{y}}[/tex]
    Due to commutativity properties, those two Hamiltonians commute. Therefore you know that they are both can be diagonalized simultanousley. In a functional form, your eigenstates will be [tex]|i,j>=\phi^{x}_{i}(x)\phi^{y}_{j}(y)[/tex], where i,j are indexing the basis (need not be discrete), and [tex]\phi_{x}[/tex] are eigenstates of [tex]\hat{H_{x}}[/tex] : [tex]\hat{H_{x}}\phi^{x}_{i}=E_{i}\phi^{x}_{i}[/tex]
    And the same holds for [tex]\phi_{y}[/tex].
    The total energy of each states (which are the eigenvalues of the total Hamiltonian operator) will be [tex]E_{i,j}=E_{i}+E_{j}[/tex] So the angular frequency of the time developement will be [tex]w_{i,j}=\frac{E_{i,j}}{\hbar}[/tex]

    And each state will be given by [tex]|\psi>=\sum_{i,j}|i,j>e^{-iw_{i,j}t[/tex]

    So you can solve simple problems with an harmonic potential in the y direction and a potential well in the x direction, without trouble.

    Notice that this introduces a new problem in quantum mechanics: degenerate states. If you have two pairs of [tex](E_{i},E_{j})[/tex] which sums up to the same total energy, then you got yourself two different eigenstates, with the same eigenvalue. This means that now when you measure the energy, the state is projected into a subspace of this energy's eigenstates, but this state is still some superposition of more basic eigenstates which are orthogonal to each other. This is also explored in the notion of "a complete set of commuting operators".

    A more complicated case will be when your coordinates are coupled, so the Hamiltonian is not seperable. Then, it really depends on the situation. You can either simply solve the PDE (unlikely), or you can introduce a change of variables which exerts symmetry in the problem (f.e. the analysis of a central potential is better done in spherical coordinates) in which the new variables are uncoupled or introduce simpler equations. Of course you must remember that when coordinates are changed, so does the momentum operators (their change is given by the chain rule if you look at them in position representation)
     
    Last edited: Jan 28, 2010
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