Equation (5) in beautiful paper (link) on Jaynes-Cummings Hamiltonian gives energy levels

$$E_n = n \hbar \omega \pm 2 g \sqrt{n}, \text{ where } n=0, 1, 2, \ldots$$

But eigenstates in Equation (4) are in basis of excited and photon and ground and photon:

$$|e,n\rangle \text{ and } |g, n-1\rangle.$$

So, what is meaning of the state with $n=0$, since eigenstate is $|g, n-1\rangle = |g, -1\rangle$ photon?


The ground state is pretty easy - the ground state of the atom, with no photons. This is pretty trivial to verify directly.

To be a bit more explicit, if you take the paper's hamiltonian, $$ \mathcal H_\mathrm{sys} = \hbar\omega \sigma^+\sigma^- + (\hbar \omega-\Delta)a^\dagger a + g(\sigma^-a^\dagger+\sigma^+a), $$ and apply it to the ground state $|g,0\rangle$, then it is easy to check explicitly that $$ \mathcal H_\mathrm{sys}|g,0\rangle=0=0|g,0\rangle, $$ since $\sigma^-|g,0\rangle = a|g,0\rangle = 0$. In particular, in no way does $a|g,0\rangle$ invoke states with negative numbers of photons.

The paper makes it clear, when it says

with the exception of $n = 0$ which is unique,

that the ground state is not a part of any doublet of the form $|g,n\rangle \pm|e,n-1\rangle$, and that instead it sits on an invariant subspace of dimension $1$ (as opposed to the layering of the rest of the Hilbert space into the invariant subspaces $\mathrm{span}\mathopen{}\left(\{|g,n\rangle ,|e,n-1\rangle\}\right)$, of dimension $2$, for excitation number $n\geq1$).

The paper is maybe a bit terse when it explains it, which is par for the course because the material is standard and only needs to be laid out to fix the notation; as is, the explanation is sufficient by the normal standards of the literature. If you want an in-depth explanation of the diagonalization of the Jaynes-Cummings model, then the primary literature is not the place to look ─ instead, you should look for a suitable quantum optics textbook such as Gerry & Knight or Scully & Zubairy.

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    $\begingroup$ No, it doesn't. Act on it with the hamiltonian, directly, and keep in mind that the state with zero photons vanishes under the action of the annihilation operator. At no point should you need to invoke negative numbers of photons; if you do, you're doing things wrong. $\endgroup$ – Emilio Pisanty Sep 14 '17 at 7:51
  • $\begingroup$ Thanks! Paper gives eigenstates as Eq. (4). Paper gives eigenvalues as Eq. (5). Paper says it possible n=0. Sub of n=0 into Eq. (4) gives |g,−1⟩. Why paper allows - 1 photon? $\endgroup$ – Nigel1 Sep 14 '17 at 7:57
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    $\begingroup$ You're misreading the paper; it makes no such claim. $\endgroup$ – Emilio Pisanty Sep 14 '17 at 8:37
  • $\begingroup$ Thanks! Where does it say one may not take n=0 with presented solution of eigenstate Eq. (4)? $\endgroup$ – Nigel1 Sep 14 '17 at 12:15
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    $\begingroup$ See edited answer. $\endgroup$ – Emilio Pisanty Sep 14 '17 at 12:22

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