# Helium atom: excited states

We do consider the first excited state, that is ${\displaystyle 1s2s}$. Since we have to dispone two electrons on two different orbital levels we can get four global wf:

• ${\displaystyle \psi ={\frac {1}{\sqrt {2}}}\left(\psi _{1s}(1)\psi _{2s}(2)-\psi _{1s}(2)\psi _{2s}(1)\right)\alpha (1)\alpha (2)}$
• ${\displaystyle \psi ={\frac {1}{\sqrt {2}}}\left(\psi _{1s}(1)\psi _{2s}(2)-\psi _{1s}(2)\psi _{2s}(1)\right)\beta (1)\beta (2)}$
• ${\displaystyle \psi _{-}={\frac {1}{\sqrt {2}}}\left(\psi _{1s}(1)\psi _{2s}(2)-\psi _{1s}(2)\psi _{2s}(1)\right){\underset {triplet\;state}{\underbrace {\left[{\frac {\alpha (1)\beta (2)+\alpha (2)\beta (1)}{\sqrt {2}}}\right]} }}}$
• ${\displaystyle \psi _{+}={\frac {1}{\sqrt {2}}}\left(\psi _{1s}(1)\psi _{2s}(2)+\psi _{1s}(2)\psi _{2s}(1)\right){\underset {singlet\;state}{\underbrace {\left[{\frac {\alpha (1)\beta (2)-\alpha (2)\beta (1)}{\sqrt {2}}}\right]} }}}$

For the mean value of the energy we have:

${\displaystyle \left\langle {\hat {H_{tot}}}\right\rangle =\left\langle \psi _{\pm }\left|{\frac {e^{2}}{4\pi \epsilon _{0}\mid {\overrightarrow {x}}_{1}-{\overrightarrow {x}}_{2}\mid }}\right|\psi _{\pm }\right\rangle =J\pm K}$

Where ${\displaystyle J}$ and ${\displaystyle K}$ are called direct and switch integrals, and take on the values:

${\displaystyle J={\frac {e^{2}}{4\pi \epsilon _{0}}}\int d^{3}{\overrightarrow {x}}_{1}d^{3}{\overrightarrow {x}}_{2}{\frac {\psi _{1s}^{\star }({\overrightarrow {x_{2}}})\psi _{1s}({\overrightarrow {x_{2}}})\psi _{2s}^{\star }({\overrightarrow {x_{1}}})\psi _{2s}({\overrightarrow {x_{1}}})}{\mid {\overrightarrow {x}}_{1}-{\overrightarrow {x}}_{2}\mid }}>0}$

${\displaystyle K=J({\overrightarrow {x}}_{i}\mapsto {\overrightarrow {x}}_{j})>0,\;i,j=1,2.\,i\neq j}$

Excited states allow us to introduce new energy levels, called orbitals. For the first excited state we have the configuration ${\displaystyle 1s2p}$; ${\displaystyle 2p}$ means that ${\displaystyle L=1}$, and so ${\displaystyle M_{L}=\pm 1,0}$. We thus have three possible wf for this state: ${\displaystyle \psi _{1,1},\,\psi _{1,0},\,\psi _{1,-1}}$. Actually only ${\displaystyle \psi _{1,0}}$ is a real wf. Since we want to have real functions in order to describe orbitals[1], we take linear combinations of them:

${\displaystyle {\begin{cases}p_{z}\mapsto \psi _{1,0}\\p_{x}\mapsto {\frac {1}{\sqrt {2}}}\left(\psi _{1,1}-\psi _{1,-1}\right)\\p_{y}\mapsto {\frac {1}{\sqrt {2}}}\left(\psi _{1,1}+\psi _{1,-1}\right)\\\end{cases}}}$

1. To be precise, here we should have taken only the radial part of the w.f. Since orbitals do describe the probability of occupation of a particular angular space region. Anyway, if you're given a state ${\displaystyle \psi }$ in terms of orbitals ${\displaystyle p_{i}}$ you can rewrite your state in terms of a w.f. using the combinations given below.