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.


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.

  • $\begingroup$ 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. $\endgroup$ – Roger Vadim 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. $\endgroup$ – Roger Vadim 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

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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