# 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
Commented 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. Commented Nov 8, 2021 at 18:49
• Sorry, i am new here, the screenshots will not happen again Commented Nov 9, 2021 at 10:04

## 1 Answer

Each atom is described by an Hamiltonian, which has its own eigenvalues and eigenstates.

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$$, which is usual in many systems. Now, recalling how we wrote the full Hamiltonian: $$$$H=\sum_{n_1, l_1, j_1, m_{j1}} E_{n_1 l_1 j_1 m_{j1}}|n_1 l_1 j_1 m_{j1}\rangle \langle n_1 l_1 j_1 m_{j1}| \otimes \mathbb{1}+ \mathbb{1}\otimes \sum_{n_2, l_2, j_2, m_{j2}} E_{n_2 l_2 j_2 m_{j2}}|n_2 l_2 j_2 m_{j2}\rangle \langle n_2 l_2 j_2 m_{j2}|.$$$$

• Okay! That clears up my doubt, Thank you so much! And also sorry about the screen shot, it won't happen again! Commented Nov 9, 2021 at 10:03