# Basis representation of hamiltonian

I do not understand this representation of a Hamiltonian involving the basis projection operator and Identity matrix. \begin{align} \hat{H}_0&= \sum_{n_1,l_1,j_1,m_{j1} }E_{n_1 l_1 j_1}\left|n_1 l_1 j_1 m_{j1}\right> \left \left Can someone please explain what is going on? I have quoted the Hamiltonian of two non-interacting atoms.

• This has been downvoted (but not by me) because of the screenshotting. That's not well liked here because it's not SE-friendly.
– Gert
Nov 8, 2021 at 18:28
• Please use Mathjax to display math on the site. It is the site standard and using images or text or math is very strongly discouraged. Nov 8, 2021 at 18:49
• Sorry, i am new here, the screenshots will not happen again Nov 9, 2021 at 10:04

In this way, the atom is described by a state space $$\mathcal{E_{1,2}}$$ where the indeces refer to atom 1 and 2. Now, the full system, comprising of the two atoms, is obtained by performing the tensor product of the two states $$$$\mathcal{E}=\mathcal{E}_1 \otimes \mathcal{E}_2,$$$$ since the atoms are non-interacting. Let $$H_1$$ be the Hamiltonian of atom 1 and $$H_2$$ of atom 2. $$H_1$$ acts on $$\mathcal{E}_1$$ but not on $$\mathcal{E}_2$$, and $$H_2$$ acts on $$\mathcal{E}_2$$ but not on $$\mathcal{E}_1$$. Now we extend the action of $$H_1$$ to all $$\mathcal{E}$$ in such a way that nothing happens on $$\mathcal{E}_2$$, which can be done as taking $$H_1$$ to be $$\mathbb{1}$$ in $$\mathcal{E}_2$$. The same goes for $$H_2$$. As so the Hamiltonian in the full state space $$\mathcal{E}$$ is $$$$H=H_1 \otimes \mathbb{1}+ \mathbb{1}\otimes H_2.$$$$
Each Hamiltonian can be expanded in a basis of its eigenvectors, just as you can always do for any operator given a basis of its eigenvectors that span all space. In the case above we have considered that the system is fully described by the complete set $$\{H, L, J, J_z\}$$ for each atom. Thus $$$$H_i=\sum_{n_i, l_i, j_i, m_{ji}} E_{n_i l_i j_i m_{ji}}|n_i l_i j_i m_{ji}\rangle \langle n_i l_i j_i m_{ji}|.$$$$ In the equations you submitted we have further assumed that the energy is degenerate in $$m_j$$, which is usual in many systems. Now, recalling how we wrote the full Hamiltonian: $$$$H=\sum_{n_1, l_1, j_1, m_{j1}} E_{n_1 l_1 j_1 m_{j1}}|n_1 l_1 j_1 m_{j1}\rangle \langle n_1 l_1 j_1 m_{j1}| \otimes \mathbb{1}+ \mathbb{1}\otimes \sum_{n_2, l_2, j_2, m_{j2}} E_{n_2 l_2 j_2 m_{j2}}|n_2 l_2 j_2 m_{j2}\rangle \langle n_2 l_2 j_2 m_{j2}|.$$$$