I do not understand this representation of a Hamiltonian involving the basis projection operator and Identity matrix. $$\begin{align} \hat{H}_0&= \sum_{n_1,l_1,j_1,m_{j1} }E_{n_1 l_1 j_1}\left|n_1 l_1 j_1 m_{j1}\right> \left<n_1 l_1 j_1 m_{j1}\right| \otimes\mathbb{I} \\ &+ \mathbb{I}\otimes \sum_{n_2,l_2,j_2,m_{j2}}E_{n_2 l_2 j_2} \left|n_2 l_2 j_2 m_{j2}\right> \left<n_2 l_2 j_2 m_{j2}\right| \end{align}$$ Can someone please explain what is going on? I have quoted the Hamiltonian of two non-interacting atoms.

  • $\begingroup$ This has been downvoted (but not by me) because of the screenshotting. That's not well liked here because it's not SE-friendly. $\endgroup$
    – Gert
    Commented Nov 8, 2021 at 18:28
  • $\begingroup$ 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. $\endgroup$ Commented Nov 8, 2021 at 18:49
  • $\begingroup$ Sorry, i am new here, the screenshots will not happen again $\endgroup$ Commented Nov 9, 2021 at 10:04

1 Answer 1


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 \begin{equation} \mathcal{E}=\mathcal{E}_1 \otimes \mathcal{E}_2, \end{equation} 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 \begin{equation} H=H_1 \otimes \mathbb{1}+ \mathbb{1}\otimes H_2. \end{equation}

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 \begin{equation} 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}|. \end{equation} 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: \begin{equation} 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}|. \end{equation}

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

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