Classical Limit of coherent State in Jaynes Cummings Model

Im dealing with an exercise on the Jaynes Cummings model in a resonat single mode approximation. The interaction Hamiltonian in rotating wave approximation is

$$H_{int}=g\, \sigma_+\,a\,+g^*\,\sigma_-\,a^\dagger \tag{I}$$

where $a$ and $a^\dagger$ are the annihilation and creation operators for the bosonic state, which is assumed to be in a coherent state $|\alpha\rangle$. $\sigma_+$ and $\sigma_-$ are the raising and lowering operators for the atom in the cavity.

Now im required to do some calculations and for that I get the hint that $H_{int}$ can be replaced by the expectation value $$H_{int}\to H_c=\langle\alpha|H_{int}|\alpha\rangle=g\, \sigma_+\,\alpha\,+g^*\,\sigma_-\,\alpha^* \tag{II}$$

in the classical limit $|\alpha|>>1$.

I want to know why this approximation of the Hamiltonian is justified (Why we can take (II) instead of (I) for our Hamiltonian), in this limit.

My thoughts: Since $a|\alpha\rangle=\alpha|\alpha\rangle$, our exact interaction Hamiltonian differs from the approximation only by the fact that $a^\dagger$ has been replaced by $a^*$. This seems not too far fetched since this is just the complex conjugate of the eigenvalue that we get by applying the annihilation operator on $|\alpha\rangle$. Then I think we can somehow argue that since $|\alpha|>>1$, the creation operator will not change the coherent state much. But I cannot really cook up a solid argument and the mathematics.

EDIT

I was thinking that maybe in the classical limit, the standard deviation of $\alpha^\dagger$ is neglectable vs. the mean value. But if I havnt done a mistake, we have

$$SD_\alpha(a^\dagger)=\sqrt{\langle\alpha|a^\dagger a^\dagger|\alpha\rangle-\langle\alpha|a^\dagger|\alpha\rangle^2}=\sqrt{a^*a^*-(a^*)^2}=0, \tag{III}$$

independent of the value of $|\alpha|$. Which confuses me even more.

• What are you actually trying to calculate? It's unclear when it would be possible to replace a hamiltonian by a particular matrix element without knowing the context in which you're applying the hamiltonian. – d_b Oct 28 '19 at 0:29

Since $a\vert \alpha \rangle=\alpha\vert \alpha \rangle$ for the coherent state, it follows that \begin{align} \langle \alpha \vert a\vert\alpha\rangle=\alpha \tag{1} \end{align} by normalization and that (as you suspect) by taking the complex conjugate: $$\alpha^*=\langle \alpha \vert a\vert \alpha\rangle^*=\langle \alpha \vert a^\dagger\vert\alpha\rangle \tag{2}$$ which is true for any $\alpha$. The condition $\vert\alpha\vert\gg 1$ must enter elsewhere since (1) and (2) are exact, irrespective of $\vert \alpha\vert$.
• Ive clarified the question a bit. I know why the expectation value is equal to the expression (II). But im wondering why we can use (II) instead of (I). This must invole the classical limit, since $a^\dagger|α⟩$ is not $a^*|α⟩$ right? – curio Jul 22 '18 at 8:06
• @curiosity no but the point is $\langle \alpha\vert a^\dagger=\alpha^*\langle \alpha \vert$. The approximation is in replacing some operators by their averages, not in the way the averages are calculated. – ZeroTheHero Jul 22 '18 at 8:25
• I know that. What I want to know is why we are allowed to do this replacement. There has to be some way to see that the action of $a^\dagger|\alpha\rangle \to \alpha^*|\alpha\rangle$ in the limit $|\alpha| \to \infty$. – curio Jul 22 '18 at 10:52
• @curiosity : $a^\dagger\vert\alpha\rangle \ne \alpha^* \vert\alpha\rangle$ in any limit. The coherent state is not an eigenstate of $a^\dagger$, period. By taking the average value, you are not doing any "replacement" of the ket but you are instead doing a replacement of the bra $\langle\alpha\vert a^\dagger\to \alpha^*\langle\alpha\vert$. That's the key: you never have $a^\dagger$ acting on a ket, but you have it acting on a bra; To "generate" a bra you need to take the average value. – ZeroTheHero Jul 22 '18 at 12:52