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Two interacting electrons in 1D infinite well

  1. Aug 6, 2008 #1
    Hi all,
    I've been looking at the problem of two interacting electrons in a 1D infinite well and wanted to run my conclusions past other people to see if I'm on the right track. The potential is 0 from x=0 to x=1 and infinite outside of this. The 1 particle solutions to this potential are [tex]\psi(x)=2^{1/2}sin(m\pi x)[/tex] where m=1,2,3... For solutions to the two particle interacting Schrodinger equation, we will use linear combinations of spatially asymmetrical and symmetrical combinations of the one particle solutions. Schrodinger's equation is now:
    [tex]H=-\frac{1}{2}\frac{\partial ^2\psi(x_1,x_2)}{\partial x_1^2}-\frac{1}{2}\frac{\partial ^2\psi(x_1,x_2)}{\partial x_2^2}+\frac{\psi(x_1,x_2)}{|x_1-x_2|}=E\psi(x_1,x_2)[/tex]
    The added fun is the interaction between the electrons. The functions I will be taking linear combinations of are:
    [tex]\phi_{m,n\pm}(x_1,x_2)=N_{m,n}[sin(m \pi x_1)sin(n \pi x_2) \pm sin(n \pi x_2)sin(m \pi x_2)][/tex]
    where the + corresponds to a symmetric function and - corresponds to an asymmetric function. The normalizing parameter is:
    [tex]N_{m,n}=\left( \begin{array}{cc} 1 & m=n\\ \sqrt{2} & m \neq n\end{array}\right)[/tex].
    So our solutions will be:
    [tex]\psi_{\pm}(x_1,x_2)=\sum_{m,n}c_{m,n \pm} \phi_{m,n\pm}(x_1,x_2)[/tex].
    where the c's are the coefficients to be determined.

    Now we plug into our hamiltonian and we will get an eigenvalue equation:
    [tex]\sum_{o,p}H_{m,n,o,p}c_{o,p \pm}=E_{\pm}c_{m,n \pm}[/tex]

    So now that I've set up the problem, here are my thoughts that I've been looking into. The problem is with solving this is with the Coulomb (aka Hartree) term. It has a singularity whenever [tex]x_1=x_2[/tex] unless the wave function is 0 when [tex]x_1=x_2[/tex], but this is only true for the asymmetric functions. Does this mean that I shouldn't consider the symmetric functions? I know that by Pauli's exclusion principle that I need to have a symmetric spatial function paired with an asymmetric spin function and vice versa. There is the possibility that from a linear combination of the symmetric functions we can find a combination that will make the function zero along the x1=x2 line.

    I have also considered another way to make a spatially symmetric wave function that will satisfy the potential, but I think there is a problem with calculating the kinetic term. The function is:
    [tex]\phi_{m,n}(x_1,x_2)=sign(x_1-x_2)[sin(m\pi x_1)sin(n\pi x_2) - sin(n\pi x_1)sin(m\pi x_2)][/tex].
    Here [tex]sign(x_1-x_2)[/tex] is basically just the sign of the expression so it will have values of 1, -1 and 0.The problem that arises from this function is that the second partial derivative in either variable isn't really defined when [tex]x_1=x_2[/tex]. I think it would be the derivative of a delta function which isn't good.

    So my questions are:
    Is it okay to eliminate the (original) symmetric functions from my basis set for this problem?
    Therefore, is the ground state spatially asymmetric?
    Can I use my alternate symmetric functions to form a basis set?
    Any other thoughts on this problem?

    I've done a lot of work on this in Mathematica and have been able to calculate the energies and wave functions for the spatially asymmetric wave functions just fine.

    Thanks for any thoughts and for reading this. Also, please let me know if my explanation isn't making sense.
  2. jcsd
  3. Aug 8, 2008 #2
    Well, no one weighed in on this, but I think I've figured it out myself. In case anyone cared I thought I'd share what I found. I found another way to form symmetric waves functions for this case. I take the asymmetric wave functions, square them and then take the square root. This is essentially the same as taking the absolute value, but the derivative behaves better using the square root. There isn't any problem with the second derivatives anywhere. The problem is that these symmetric wave functions aren't orthogonal to each other. They are orthogonal to the asymmetric wave functions however.

    All I did then was use Gram-schmidt process to form an orthogonal symmetric basis and then did the same process I described above to find the linear combination of these wave functions that gives the lowest energy. So far it looks like the asymmetric wave functions give the ground state with just slightly lower energy than the symmetric wave function. I'll post again if I find a different result. I'm also planning on using density functional theory on this toy problem and comparing results.
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