# Dirac equation for solving $\rm H^-$ anion

Let's try to solve 1 dimensional Dirac equation for a $$\rm H^-$$ anion. If you solve the time independent Dirac equation and allow motion of electrons only in the $$x$$ axis you get: $$-\chi\frac{d\Psi_{4}(x_{1},x_{2})}{dx_{1}}-\chi\frac{d\Psi_{3}(x_{1},x_{2})}{dx_{2}}+(V(x_{1},x_{2})+m_{e}c^{2}-E)(\Psi_{2}+\Psi_{1}) = 0.$$

By comparing the parts of the 2 sides of the equations we get that $$-\chi\frac{d\Psi_{4}(x_{1},x_{2})}{dx_{1}}=0,$$ $$-\chi\frac{d\Psi_{3}(x_{1},x_{2})}{dx_{2}}=0,$$ $$(V(x_{1},x_{2})+m_{e}c^{2}-E)\Psi_{2}=0,$$ $$(V(x_{1},x_{2})+m_{e}c^{2}-E)\Psi_{1}=0.$$ If we solve the 1st equation we will get $$\Psi_{4} = g(x_{2})$$ . Now if 1 electron is in the same position with the nucleus (1 proton) the other electron wont feel any electrostatic force so its solution of $$\Psi_{4}(0,x_{2})=c$$, where $$\int_{-\infty}^{\infty}c^{2}dx = 1$$ in order for the normalization of the wavefunction. So that means $$g(x_{2}) = c$$

The same is true when calculating $$\Psi_{3}$$.

However for $$\Psi_{1}$$ and $$\Psi_{2}$$ I get two solutions: one that $$\Psi_{1}=0$$ and one that $$(V(x_{1},x_{2})+m_{e}c^{2}-E)=0$$. Should I choose the second solution in order to find a relationship between $$x_{1}$$ and $$x_{2}$$ right?

Source for time independent Dirac equation:https://www.sciencedirect.com/topics/physics-and-astronomy/dirac-equation

• I may be out of my depth because I've never looked at solutions to the dirac equation. But I'm confused. You start by saying you want to solve the H- ion (there is no analytic solution). Then you make a crazy "approximation" "if 1 electron is in the same position as the nucleus" (doesn't that just defeat the purpose of finding a 2-particle dirac solution?) I don't understand the point of upgrading the precision by using dirac instead of schrodinger then taking a massive step down in precision by making an approximation doesnt make sense and eliminates any possibility of a meaningful solution. Apr 13, 2023 at 20:55
• I do this for 2 reasons:Electrons have half-integer spin and arent described precisely by the Schrodinger equation and the order of the differential in the Dirac equation is 1 while in Schrodinger it is 2 meaning when you include many particles the partial differential equation is 2 for Schrodinger and 1 for Dirac and 1st order partial differential equations are much easier to solve than 2nd order partial differential equations. Apr 13, 2023 at 20:59
• Also can you clarify where this version of the Dirac equation comes from? Where'd all the gamma matrices go that would ordinarily mix the four components of the Dirac spinor? Note that, as is ordinarily the case in the dirac equation, a first order differential equation involving matrices and vectors can secretly be a second order differential equation by doing something like $dx_1/dt=x_2$, $dx_2/dt=f(x_1)$. Again - maybe I just don't know enough about the Dirac equation - I'm just curious what's going on here. Apr 13, 2023 at 21:11
• If you try working from the non-relativistic model for such a system you'll begin to see the problems moving to a fully relativistic one. You might want to look up "Dirac Fock" for one technique to approach this problem. You might also try looking for approaches to the Helium ion as well. Apr 14, 2023 at 9:36
• @StephenG-HelpUkraine I think you meant the helium atom. But yeah. If you have a code that can solve the helium atom, you also have a code that can solve the hydrogen anion, and the lithium+ ion. It's just changing one number (the factor in front of the nuclear potential) Apr 14, 2023 at 23:20

Your problem runs straight into serious headaches. I am not sure why you are paying attention to the 1D case, but humanity had already thrown quite a lot of work into the 3+1D case.

Essentially, this is like dealing with the Helium ground state problem with Z = 1 rather than Z = 2.

If you want to solve this properly, you would have to look up things like the Bethe-Salpeter equation, two-body Dirac equations, and other similar headaches. The solution must obey plenty of crazy symmetry properties. Namely, when you swap the two electrons, you must get a minus sign in the wavefunction. You can easily get this if you use a Slater determinant, but Slater determinants, by definition, cannot capture the correlations. That is, the actually correct wavefunction of just the two-electrons, is a function of $$\frac{\vec r_1 + \vec r_2}2$$ and $$\vec r_2 - \vec r_1$$, not a Slater determinant of functions of $$\vec r_2$$ and $$\vec r_1$$, but of course that is so difficult to compute that nobody ever does that.

If you look up Bethe-Salpeter of two electrons, you will see that it is a 4x4=16 component equation, and has dependence on the relative time coördinate. There are work out there that set up constraints and so forth, and eliminate that unphysical time coördinate. i.e. there is an 8-component wavefunction solution for this.

And we have yet to include the nucleus, which is yet another Dirac particle.

Under no circumstance would you get that the wavefunction factorises like you got. That is, even when one of the electrons is right on top of the proton, you won't get a lack of binding. For one, you must at least get polarisation based binding.

• I am using the 1D model to get rid of some terms and your answer doesn't answer my question at all, it just says politely "You are wrong". Apr 14, 2023 at 17:01
• ... errm, because you are? I think you kinda should start with Schrödinger + spin half, solve approximately for this, and then do it with Pauli equation, before tacking the Dirac case. Then you will have a better appreciation of how and why we are telling you what we are telling you. And solve for general Z, because then you can compare results with Helium, which is very well-studied. Apr 14, 2023 at 17:09
• You start with an equation that seems like an incorrect formulation of the Dirac equation (should be four eqns). You restrict it to varying in only one dimension for a reason we don't understand. Then you make an "approximation" which is both a bad approximation and just an invalid thing to do with a wavefunction (the electrons are quantum mechanical - they can't be located "at the nucleus" that's the whole point). It also immediately yields a nonsensical result: the wavefunction being constant throughout space. We cant help you with the next step. The next step is fix the first step. Apr 14, 2023 at 23:14
• @appliedSciences The Dirac electron is a set of 4 interacting components. Even in the case of free particle plane waves, removing the "positron" parts of the wavefunction causes the equation to fail. You can try doing what you do, but experience of just everybody who have done this before you, are telling you that your resulting "answer" will be vastly wrong. There is no such thing as just adding equation of spin up to spin down either. That is, your approach is so wrong that even in 1D it is wrong. Apr 15, 2023 at 11:37
• @appliedSciences Quantum wavefunctions are not IVP. They are BVP. Apr 15, 2023 at 12:20