In a Fundamental of Quantum Optics and Quantum Information book which I am reading, it states without explanation that, in a two-level atomic configuration in a cavity system, the

  • Atomic Hamiltonian is given by, $$H^{\mathcal{A}}=\hbar\omega_{g}\lvert g\rangle\langle g\rvert+\hbar\omega_{e}\lvert e\rangle\langle e\rvert$$
  • Hamiltonian for cavity is, $$H^{\mathcal{F}}=\hbar\omega\hat{a}^{\dagger}\hat{a}$$

where, $\omega_{g}$ and $\omega_{e}$ is frequencies associated with atomic level $\lvert g\rangle$ and $\lvert e\rangle$ respectively, $\omega$ is the frequency of the cavity mode with near resonant with $\omega_{eg}=\omega_{e}-\omega_{g}$, and $\hat{a}^{\dagger}$ and $\hat{a}$ are creation and annihilation operators.

Could you tell me how to derive the relations of both atomic Hamiltonian and cavity field Hamiltonian? Two-Level Atomic State Configuration

P.S. I apology for the image. I can't find a way to zoom it out.

  • 1
    $\begingroup$ To typeset Dirac notation use, for example, \lvert a \rangle, which produces $\lvert a \rangle$. Derivation of the atomic Hamiltonian is trivial, it is just how one writes in Dirac notation a general $2\times 2$ matrix whose eigenvectors are $\lvert e\rangle$, $\lvert g\rangle$ with eigenvalues $\hbar \omega_{e,g}$. I'll write up a derivation of the field Hamiltonian later unless someone wants to jump in first. $\endgroup$ Jul 24, 2015 at 7:01
  • $\begingroup$ @MarkMitchison Thank for the tips. I thought latex style code can be used here as well. $\endgroup$
    – TBBT
    Jul 24, 2015 at 7:05
  • $\begingroup$ You can use LaTeX style code pretty much everywhere. As far as I know \ket{} is not a standard LaTeX macro and you would need to define it yourself. $\endgroup$ Jul 24, 2015 at 7:06
  • $\begingroup$ By relation, do you mean the Hamiltonians you've stated, or a possible interaction (which would "relate" the two systems)? $\endgroup$
    – Daniel
    Jul 24, 2015 at 7:50
  • $\begingroup$ @Daniel No not the interaction Hamiltonian $\mathcal{V}^{\mathcal{AF}}$ that I want to derive. Just the atomic and cavity field Hamiltonians relation that I stated in my question. $\endgroup$
    – TBBT
    Jul 24, 2015 at 7:57

1 Answer 1


As said in the commentaries, the first one comes from the Dirac formalism. Simply put, it deals with quantum states as vectors $\lvert a \rangle$ whose components contain the projections of the system into different eigenstates. For a system, the set of state eigenvectors $\lvert a_i \rangle$ must be linearly independent $\langle a_i \lvert a_j \rangle=\delta_{ij}$ which is the inner vector product, and is equivalent to $\int \psi_i \psi_j dx^3$. And then there is the transition or projection operator, which results from outer vector product $\lvert a \rangle\langle b \lvert$ and yields a matrix characterizing the transition between states. Finally if you have an operator $\hat A$, represented by a matrix, you can find its components like $\langle a \lvert \hat A \lvert b \rangle$, where is easy to see that when expressed in terms of its eigenvectors, all non-diagonal values are zero and the diagonal contains the eigen values ($\langle a_i \lvert \hat A \lvert a_j \rangle$ = 0, $\langle a_i \lvert \hat A \lvert a_i \rangle= a_i$)

Now the case of the atom is similar to the harmonic oscillator, and basically that of any system dealing with confined particles: discrete levels and discrete energy steps to change from one to the other. So this allows to write the Hamiltonian for this kind of systems like counting total energy: if level k is occupied, then add $E_k$ to the system energy. This is what the atomic Hamiltonian represents: if you have the system in a state $\lvert \psi \rangle = c_g \lvert g \rangle + c_e \lvert e \rangle$ you will get $\langle \psi \lvert H^A\lvert \psi \rangle = (c_g^* \langle g \lvert + c_e^* \langle e \lvert) (\hbar \omega_g \lvert g \rangle \langle g \lvert+ \hbar \omega_e \lvert e \rangle \langle e \lvert)(c_g \lvert g \rangle + c_e \lvert e \rangle)$ which you will see to get the mean values for the energy of the system in this state with $p_e=\sqrt{c_e c_e^*}$ and $p_g=\sqrt{c_g c_g^*}$ are the probabilities for each state, or the projections of $\lvert \psi \rangle$ into the eigenstates $\lvert e \rangle$ and $\lvert g \rangle$.

The second one is similar, but here there can be many photons in one state. So the total energy inside the cavity is a counting of how many photons are inside, and the creation and annihilation operators can be seen here for the harmonic oscillator (probably relevant here) and other examples.

  • $\begingroup$ This doesn't explain why only one light mode is considered (for which there is no explanation prior to interaction and rotating wave approximation, which should be explained imho) $\endgroup$
    – Daniel
    Jul 24, 2015 at 20:26
  • $\begingroup$ Since we are considering a two-level atom we can expect only one energy for the photons. And I don't think you can deduce that result since it is a consequence of the postulates of Quantum Mechanics. In other words, is one of the experimental facts taken without explanation, it does not fall into the area of explained phenomena within QM. $\endgroup$
    – rmhleo
    Jul 24, 2015 at 23:18

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