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I have a quantum mechanical system of two interacting electrons in one spatial dimension. The Hamiltonian of the system is of the form $H = h + \frac{1}{|x_1 - x_2|}$, where $h$ is a one-electron part irrelevant to the current question, and $x_1, x_2 \in [0, a],\, a > 0$ are spatial coordinates of the electrons in the system. I have decided to use configuration interaction method(Ritz method if you will) to solve the ground state energy of this system numerically. This involves expanding the exact two-electron wave function in a basis of Slater determinants $$\frac{1}{\sqrt{2}}\begin{vmatrix} \chi_i(\mathbf{x}_1) & \chi_j(\mathbf{x}_1) \\ \chi_i(\mathbf{x}_2) & \chi_j(\mathbf{x}_2) \end{vmatrix}.$$ $\chi_i$ is a spin-"orbital", which is simply a product of a one-electron wave function $\psi_i$ and either spin eigenfunction $\alpha$ or $\beta$ (corresponding to spin-up and spin-down, this notation is from Modern Quantum Chemistry by Szabo and Ostlund). $\alpha$ and $\beta$ are functions of a spin-variable $\omega$ and they are orthonormal wrt an inner product which is denoted by an integral, so e.g. $\int \alpha^*(\omega)\beta(\omega) \,d\omega = 0 $ (bit of an abuse of notation). In addition, $\mathbf{x}_i = (x_i, \omega_i)$. Below I use overbar to denote the spin-part of a spin-orbital so that $\chi_i = \psi_i\alpha$ and $\bar{\chi}_i = \psi_i\beta$.

The problem I have lies in the evaluation of the integrals needed to calculate the Hamiltonian matrix elements for the configuration interaction method. One type of these integrals is $$I(\bar{\chi}_i, \chi_j, \bar{\chi}_k, \chi_l) \equiv \int d\mathbf{x}_1d\mathbf{x}_2 \big[\bar{\chi}_i(\mathbf{x}_1)\chi_j(\mathbf{x}_2) - \chi_j(\mathbf{x}_1)\bar{\chi}_i(\mathbf{x}_2)\big] \frac{\bar{\chi}_k(\mathbf{x}_1)\chi_l(\mathbf{x}_2) - \chi_l(\mathbf{x}_1)\bar{\chi}_k(\mathbf{x}_2)}{|x_1-x_2|} \tag{1},$$ which involve Slater determinants formed by both a spin-orbital with spin-up and a spin-orbital with spin-down. Now, if I were to directly use the orthonormality of $\alpha$ and $\beta$, I would get $$I(\bar{\chi}_i, \chi_j, \bar{\chi}_k, \chi_l) = \int \frac{dx_1dx_2}{|x_1-x_2|} \big[\psi_i(x_1)\psi_j(x_2)\psi_k(x_1)\psi_l(x_2) + \psi_j(x_1)\psi_i(x_2)\psi_l(x_1)\psi_k(x_2)\big] \tag{2},$$ which very much looks like a diverging integral to me since $\frac{1}{|x_1-x_2|}$ has a singularity at $x_1 = x_2$, while the nominator of the integrand does not vanish at $x_1 = x_2$. Integrals such as $I(\chi_i, \chi_j, \chi_k, \chi_l)$ and $I(\bar{\chi}_i, \bar{\chi}_j, \bar{\chi}_k, \bar{\chi}_l)$ can be simplified to $$ \int dx_1dx_2 \big[\psi_i(x_1)\psi_j(x_2) - \psi_j(x_1)\psi_i(x_2)\big] \frac{\psi_k(x_1)\psi_l(x_2) - \psi_l(x_1)\psi_k(x_2)}{|x_1-x_2|}.$$ which is finite thanks to the antisymmetrized wavefunctions, which vanish at $x_1=x_2$, in the integrand.

Assuming steps between (1) and (2) aren't justified, I have tried other approaches for calculating $(1)$ and found that $$ \int dx_1dx_2 \big[\psi_i(x_1)\psi_j(x_2) - \psi_j(x_1)\psi_i(x_2)\big] \frac{\psi_k(x_1)\psi_l(x_2) - \psi_l(x_1)\psi_k(x_2)}{|x_1-x_2|} = I(\bar{\chi}_i, \chi_j, \bar{\chi}_k, \chi_l) + I(\bar{\chi}_i, \chi_j, {\chi}_k, \bar{\chi}_l) $$ but this doesn't seem to lead anywhere.

Is there a way to express $I(\bar{\chi}_i, \chi_j, \bar{\chi}_k, \chi_l)$ as a convergent integral involving only spatial wave functions $\psi_i$ or is this model/formalism unfounded applied to this system?

Edit: To give a bit more background, I am trying to get results similar to the ones obtained in this paper. One of the methods they use in that paper is actually Ritz method, but with a bit different basis set and and without spin.

I am also aware of the close relationship to 1D hydrogen atom which was pointed out by Vadim in their answer. I believe I should be able to get finite results in this case(see the earlier paper). And after thinking about it more, I think my best bet is to use a different set of basis functions.

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Firstly, it seems that you haven't correctly used the orthonormality of the spin functions, as your integral lacks the exchange term. This is probably the result of inconsistent notation, as $\alpha,\beta$ mean spin states of two electrons ($x_1$ and $x_2$) but not necessarily different spin states.

Secondly, in one dimension the Coulomb interaction may still diverge - this is certainly the case for the attractive interaction - see the classical paper of Loudon, much cited in connection to excitons in carbon nanotubes.

Finally, perhaps the Coulombn interaction in your problem is not at all $1/|x_1-x_2|$? E.g., in quantum wires this is certainly not the case - although the electrons may be confined to 1D, the Coulomb interaction remains three-dimensional; moreover, it may be screened.

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  • $\begingroup$ Thank you for the answer! As far as I understand the notation the "spin-variable" $\omega_i$ is used to indicate the electron whose state we are talking about. So, for example $\alpha(\omega_1)$ would refer to spin up and $\beta(\omega_1)$ to spin down state of electron 1. Can you clarify what you mean by the exchange term? I also added a bit more background to my question above. $\endgroup$
    – Confusee
    Mar 9 at 21:30
  • $\begingroup$ @Confusee do I understand correctly that you cobsider only the situation where one electron is spin-up and the other is down? In this case your calculation seems correct, as the electrons are not constrained by the Pauli principle. $\endgroup$ Mar 10 at 7:05
  • $\begingroup$ Yes and the reason I am currently only concerned with the Slater determinants that assign one electron spin up and other spin down is the fact that they are the only type of basis functions I am considering, which cause the matrix elements to diverge. That being said, I could exclude such determinants from the finite subset of basis functions used in practice to calculate the matrix elements. $\endgroup$
    – Confusee
    Mar 10 at 14:35
  • $\begingroup$ @Confusee These are legitimate wave functions for two electrons. However, as I mentioned in my answer, the integral does diverge in 1D (but not in 2d and 3D). How you deal with it depends on the problem - I can't say more, since I don't know what you are working on. $\endgroup$ Mar 10 at 14:38
  • $\begingroup$ I see, although clearly some additional restrictions must be set to the basis functions to get reasonable results. I'll accept your answer since it got me thinking :) and I have decided to just modify my basis set. Afterall, I only need to ensure that the set of functions spans a suitable space. E.g. in the paper I linked they expand the spatial part separately and use symmetric functions for the Ritz method, (see p.462-463, the ground state indeed seems to have symmetric spatial part so that's a reasonable choice). $\endgroup$
    – Confusee
    Mar 11 at 1:18
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An integral with a $1/|x|$ need not diverge. For example
$$ \int_{|x|<R} \frac 1 {|x|}d^3x = 4\pi \int_0^R \left(\frac 1 r\right) r^2 dr = 2\pi R^2 $$ does not diverge.

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