I have trouble getting my head around manipulating Dirac notation, it's still new to me and I'm not used to it. I'm following the rotating wave approximation derivation for Rabi oscillations and light matter interactions.

I have $|e\rangle$, $|g\rangle$ as excited and ground state of a two level atom, my interaction Hamiltonian is $\hat{H}_I$.

I have calculated $$|e\rangle \langle e| \hat{H}_I |g\rangle \langle g| + |g\rangle \langle g| \hat{H}_I |e\rangle \langle e| = \frac{1}{2} \hbar Xe^{i\omega t} |g\rangle \langle e| + \frac{1}{2}\hbar X^*e^{-i\omega t} |e\rangle \langle g|, $$

where $X=-eE_0d_{12}/\hbar$ and $d_{12}$ is the dipole matrix element. This should reduce down to

$$\hat{H}_I = \frac{1}{2} \hbar Xe^{i\omega t} |g\rangle \langle e| + \frac{1}{2}\hbar X^*e^{-i\omega t} |e\rangle \langle g| $$

given that $|e\rangle \langle e| +|g\rangle \langle g| =1 $ by completeness. I am sure this is just a simple step but I am unsure of what to do and what not to do manipulating Dirac notation, I don't know how to proceed. Spent a few hours getting this far, would be grateful if someone dug me out.


2 Answers 2


For a two-level system with ground state $|g\rangle$ and excited state $|e\rangle$, any operator $\hat O$ can be written

$$\hat O = \mathbb I\cdot \hat O \cdot \mathbb I = \big(|g\rangle\langle g| + |e\rangle\langle e|\big)\hat O \big(|g\rangle\langle g| + |e\rangle\langle e|\big)$$

This has four terms in general. However, if it so happens that your operator has vanishing diagonal elements (so $\langle g | \hat O |g\rangle = \langle e|\hat O|e\rangle = 0$) then you would have

$$\hat O = |g\rangle\langle g| \hat O |e \rangle\langle e| + |e\rangle\langle e| \hat O |g\rangle\langle g|$$

which is exactly what you wrote above.

The interaction Hamiltonian you describe is specifically proportional to the dipole operator $\hat d$, which has negative parity (i.e. $\mathcal P \hat d \mathcal P = -\mathcal d$). If the Hamiltonian of the system is invariant under parity inversion (so $\mathcal P \hat H_0 \mathcal P = \hat H_0 \implies [\mathcal P,\hat H_0]=0$), then you can choose your energy eigenbasis to consist of states with definite parity as well as definite energy.

However, this implies that the dipole operator has vanishing diagonal elements, because if $|\phi\rangle$ is an energy eigenstate with parity $\pm 1$, then

$$\langle \phi |\hat d|\phi \rangle = \langle \phi|\mathcal P^2 \hat d \mathcal P^2 |\phi\rangle = \big(\langle\phi|\mathcal P\big)\big(\mathcal P \hat d \mathcal P\big)\big(\mathcal P|\phi\rangle\big)$$ $$=\big(\pm \langle\phi| \big)\big(-\hat d\big)\big(\pm |\phi\rangle\big) = -\langle\phi|\hat d|\phi\rangle$$

which implies that $\langle \phi|\hat d|\phi\rangle = 0$.

  • $\begingroup$ Thanks a lot for explaining! Makes clear sense. I had a feeling using and understanding Dirac notation is just the tip of the theoretical quantum mechanics iceberg. Back to the books for me! $\endgroup$
    – Samalama
    May 23, 2019 at 8:39

In a two level system, $|e\rangle$ meant excited states, and $|g\rangle$ meant ground states.

I guess you are using undergraduate quantum, in that case, most of the time one only need to know that $\langle g|$ was complex conjugate of $|g\rangle$. Both $|e\rangle$ and $|g\rangle$ was orthonormal basis( unitary vector and $\delta_i^j$). Basically, you may think of basis as the position of a vector, and it's coefficient as the coordinates.


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