# Helioid atoms | Quantum Mechanics

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Let $\varphi$ be a normalized, well-behaved function that satisfies the boundary conditions of the problem (called test function). The variational theorem says that: \begin{equation}\label{ecu1} \int\varphi^{\ast}\hat{H}\varphi dq\geq E_0 \end{equation} Where $E_0$ is the energy of the state fundamental.

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Let $f_1, f_2, ......., f_n$ be linearly independent functions. We build the test function $\varphi$ by linearly combining the $f_i$ functions. \begin{equation}\label{ec5} \varphi=c_1 f_1 + c_2 f_2 +........+c_n f_n =\sum_{j}c_i f_i \end{equation} Where $\varphi$ is the test variational function, $c_i$ are variational parameters that must minimize the variational integral. $c_i$ and $f_i$ are real. We establish the variational integral: \begin{equation}\label{ecu6} W=\frac{\int\varphi^{\ast}\hat{H}\varphi dq}{\int\varphi^{\ast}\varphi dq} \end{equation}

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The second approximate method that we are going to study to solve the Schrödinger equation, in two-electron systems, is the perturbation method. This method compares the system with no solution (perturbed system) with another, for which an analytical solution is available (unperturbed system). The difference between the two Hamiltonians is called a disturbance. For the error to be low, the disturbance must be as small as possible. That is, we must choose the unperturbed system that has a Hamiltonian as similar as possible to the perturbed system. Let be a time-independent Hamiltonian, $\hat{H}$, whose equation, $\hat{H}\phi_n=E_n\psi_n$, has no analytical solution (perturbed system).

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Helium has two electrons and a nucleus with a +2e charge. We place the origin of coordinates in the nucleus, giving coordinates ($x_1,y_1,z_1$), ($x_2,y_2,z_2$) to electrons 1 and 2. From nuclear charge we take +Ze to include helioid ions, $H ^-$, $Li+$, $Be^{2+}$...

We write the Hamiltonian of Helium.

\begin{equation} \hat{H}=-\frac{\hbar^2}{2m}\nabla_1^2-\frac{\hbar^2}{2m}\nabla_2^2-\frac{Ze^2 }{r_1}-\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \end{equation}

Due to the term $\frac{e^2}{r_{12} }$ the Schrödinger equation is not separable in any coordinate system and we must use approximate methods to solve it.

We apply the perturbation theory, choosing as the unperturbed system the atom with two electrons without interelectronic interaction, whose Hamiltonian is given by:

\begin{equation} \hat{H}^{0}=-\frac{\hbar^2} {2m}\nabla_1^2-\frac{\hbar^2}{2m}\nabla_2^2-\frac{Ze^2}{r_1}-\frac{Ze^2}{r_2} \end{equation}

Read more: Treatment of perturbations for the ground state of helium

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We apply the variational theorem, $W=\int\varphi^{\ast}\hat{H}\varphi dq\geq E_{fund.}$, using as proof function: \begin{equation} \varphi=\frac {1}{\sqrt{\pi}}\zeta^{3/2}e^{-\zeta r_1}\frac{1}{\sqrt{\pi}}\zeta^{3/2}e^ {-\zeta r_2}=\phi(1)\phi(2) \end{equation} We have obtained this test function from hydrogenoid 1s orbitals for both electrons, changing the atomic number Z by the variational parameter $\zeta$ (zeta).

The physical interpretation of $\zeta$ is simple, one electron tends to shield the other in front of the nucleus, each electron is subjected to an effective nuclear charge less than the total nuclear charge Z.

The Hamiltonian for the Helium atom, in atomic units, has the form: \begin{equation} \hat{H}=\frac{1}{2}\nabla_1^2-\frac{1}{2}\nabla_2 ^2-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}} \end{equation} Substituting into the variational integral:

\begin{equation} W= \left\langle\varphi\left|\hat{H}\right|\varphi\right\rangle=\left\langle\phi(1)\phi(2)\left|\frac{1}{2}\nabla_1^2-\frac{1}{2}\nabla_2^2-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}}\right|\phi(1)\phi(2)\right\rangle \end{equation}

Read more: Variational treatment for the ground state of helium

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The spin operators are established analogously to those for orbital angular momentum.

$\hat{l}_x$ | $\hat{l}_y$ | $\hat{l}_z$ | $\hat{l}^2$ |

$\hat{s}_x$ | $\hat{s}_y$ | $\hat{s}_z$ | $\hat{s}^2$ |

Where, $\hat{s}^2$, is the square of the spin angular momentum and $\hat{s}_z$, the z component of the spin angular momentum.

Spin angular momentum has the same commutation relations as orbital angular momentum:

\begin{equation} [\hat{l}_x,\hat{l}_y]=i\hbar \hat{l}_z\; \;\rightarrow [\hat{s}_x,\hat{s}_y]=i\hbar \hat{s}_z \end{equation}

\begin{equation} [\hat{l}_y,\hat{ l}_z]=i\hbar \hat{l}_x\;\;\rightarrow [\hat{s}_y,\hat{s}_z]=i\hbar \hat{s}_x \end{equation}

\begin{equation} [\hat{l}_z,\hat{l}_x]=i\hbar \hat{l}_y\;\;\rightarrow [\hat{s}_z,\hat{s}_x ]=i\hbar \hat{s}_y \end{equation}

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The wave functions that specify the state of the electron depend not only on the x,y,z coordinates but also on the spin of the electron. The complete wave function is constructed by multiplying the spatial (orbital) part by the corresponding spin function. When introducing the spin, the doubling of the hydrogenoid quantum states is produced, without there being a change in energy. Thus, the first hydrogenoid energy level (ground state) will be formed by two degenerate quantum states, whose wave functions are: \begin{eqnarray} \psi_{1s}(x,y,z)\cdot\alpha(m_s) \\ \psi_{1s}(x,y,z)\cdot\beta(m_s) \end{eqnarray} The degeneracy of the energy levels of the hydrogen atom goes from $n^2$ to $2n^2$.

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In quantum mechanics the particles are indistinguishable, the uncertainty principle prevents knowing the path that each particle follows. Therefore, the wave function of a system of identical particles will not distinguish one particle from another. Let them be a set of particles, whose coordinates are $q_1,q_2,.....q_n$. The wave function will depend on the coordinates of all the particles that make up the system $\psi(q_1,q_2,.....q_n)$. This wave function can present two different behaviors when faced with the exchange of any two particles:

$\psi(q_2,q_1,.....q_n)=\psi(q_1,q_2,.....q_n)$, the function is said to be trade-symmetric. $\psi(q_2,q_1,.....q_n)=-\psi(q_1,q_2,.....q_n)$, the function is said to be antisymmetric with respect to swapping.

**Pauli's principle** , the wave function of a system of electrons (fermions) must be antisymmetric with respect to the exchange of any two electrons and must not distinguish between electrons.

In the case of bosons (particles with integer spin), the wave function is symmetric.

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We are going to consider the helium atom from the point of view of electron spin and the Pauli principle. We know that the ground state has a wave function 1s(1)1s(2), where 1s represents the hydrogenoid orbital and the number in brackets refers to the electron. To take spin into account, we must multiply this spatial function by a proper function of spin. Let's look at the possible eigenfunctions of spin:

\begin{equation} \alpha(1)\alpha(2);\;\;\beta(1)\beta(2);\;\;\alpha(1)\beta (2);\;\;\beta(1)\alpha(2) \end{equation}

The first two functions are symmetric with respect to the exchange of electrons and do not distinguish between them, therefore being valid to construct the wave function of helium. The third and fourth functions are not valid because they violate the principle of indistinguishability. Furthermore, they are neither symmetric nor antisymmetric towards the exchange of electrons. To solve this problem, we build linear combinations of the form:

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The construction of antisymmetric wave functions, using the Pauli antisymmetry principle, is very complex for atoms with more than two electrons. In these cases we use the Slater determinants, which generate antisymmetric functions and do not distinguish between electrons. To construct a Slater determinant, the spin-orbitals occupied by the electrons are placed in the columns, while each row corresponds to one electron.

As an example, we will start by constructing the ground state of lithium by using Slater determinants.

Step 1. Write the electron configuration, Li:$1s^22s^1$.

Step 2. Write the spin-orbitals, $1s\alpha$; $1s\beta$; $2s\alpha$; $2s\beta$

Step 3. Write the determinant by placing the spin-orbitals in the columns. The first row is for the electron (1), the second for the electron (2) and the third for the (3).