Dicke State and Holstein–Primakoff transformation

Collective spin operators§

Define§

$$\hat S_\pm = \sum_i{\hat\sigma_{(i)}^\pm} \qquad \hat S_e=\sum_i{\hat\sigma_e^{(i)}}\qquad \hat S_g=\sum_i{\hat\sigma_g^{(i)}} \qquad \hat S_z=\frac{1}{2}\sum_i{\hat\sigma_z^{(i)}}$$

Commutation§

With linearity of commutator, easily we see that this definition of collective spin makes sense, and also obeys the single atom spin commutation relation.

$$\left[\hat S_+, \hat S_-\right]=2\hat S_z \qquad \text{etc...}$$

Effective z-value component spin§

Just as single atom spin could have a z-component, here, easily we note that from

$$S_z=\frac{1}{2}\sum_i{\hat\sigma_z^{(i)}} = \sum_{i}\hat\sigma_e^{(i)}-\frac{N}{2}$$

z-component will be in $\{N/2, N/2 - 1, \dots,-N/2\}$, call this $m$

Dicke state Basis§

So, sufficiently by $S$ and $m$ (assuming all states act coherently), we define a basis

$$\ket{S,m}=\frac{1}{\sqrt{\binom{N}{n_e}}}\sum_{\text{perm}}\ket{\underbrace{eee\dots}_{n_e} \underbrace{ggg\dots}_{n_g}}$$

Where $n_e=S+m$, $n_g=S-m$

Raising and Lowering§

$$\hat S_-\ket{S,m}=\sqrt{S(S+1)-m(m-1)}\ket{S,m-1}\qquad \hat S_+\ket{S,m}=\sqrt{S(S+1)-m(m+1)}\ket{S,m+1}$$

For large enough $N$,

$$\sqrt{S^2-m^2+S+m}=\sqrt{(S+m)(S-m+1)}=\sqrt{n_e(n_g+1)}$$ $$\sqrt{S^2-m^2+S-m}=\sqrt{(S+m+1)(S-m)}=\sqrt{(n_e+1)n_g}\approx\sqrt{n_en_g}$$

Thus, $\braket{\hat S^+\hat S^-}=n_e(n_g+1)$. The point here is the coherence of the Dicke-state depends on both $n_e$ and $n_g$, differing from the expectation if all atoms go out of phase.

Holstein–Primakoff Transformation§

We aim to treat this ensemble as a single bosonic wave when interacting with a single photon, or at least approximately. Doing so provides practical use for the formalism developed above.

Thus, we take the ansatz. We want

$$\hat S_z = \hat c^\dagger \hat c - \frac{N}{2}\qquad \hat S_+ = \hat c^\dagger \sqrt{N-\hat c^\dagger \hat c} \qquad S_- = \sqrt{N-\hat c^\dagger \hat c}\,\hat c\qquad [\hat c,\hat c^\dagger]=1$$

Then, we will have

$$\left[\hat S_+, \hat S_-\right]=2\hat S_z \qquad \text{etc...}$$

And if $\hat c^\dagger \hat c \ll N$, we see that

$$\hat S_z = \hat c^\dagger\hat c-\frac{N}{2}\qquad \hat S_+ = \sqrt{N}\,\hat c^\dagger \qquad S_- = \sqrt{N}\,\hat c\qquad [\hat c,\hat c^\dagger]=1$$

Clearly now the ensemble of atoms interacting with a single mode will be a perfectly bosonic wave.

Treat this as a fundamental structural change. Having $c$ fundamentally changes the structure of the algebra while retaining the known physical meaning. With $c$, we are now able to use the single atom-photon interaction framework to treat the Dicke state ensemble.

Weak Excitation Regime§

The approximation $\hat c^\dagger \hat c \ll N$ defines the weak excitation regime. In this regime, the collective spin behaves strictly as a harmonic oscillator. This justifies the use of linear optical tools, such as standard beam splitter relations, to fully describe the light-matter interface.

Beam splitter§

With this, the light-matter interaction with the ensemble can be written as

$$\hat H_\text{int} = \hbar (g_\text{eff}\,\hat a\hat c^\dagger + g_\text{eff}^*\,\hat a^\dagger \hat c)$$

Where $g_\text{eff} = g\sqrt{N}$. This, is a beam splitter model.

Collective Rabi Oscillations and Quantum Memory§

Because the interaction takes the exact form of a beam splitter, the system undergoes collective Rabi oscillations between the photonic state $\ket{1}$ and the symmetric single-excitation Dicke state $\ket{S, m=-S+1}$ at a frequency enhanced by $\sqrt{N}$. This directly yields the famous vacuum Rabi splitting scaled by $\sqrt{N}$.

Furthermore, this beam splitter interaction provides the natural framework for quantum memory. A single photon in the cavity mode can be completely mapped into the collective atomic excitation through a $\pi$-pulse generated by this Hamiltonian.

Superradiance and Subradiance from Decay Rates§

Now that we have the beam splitter model and the exact expectation value $\braket{\hat S^+\hat S^-}=n_e(n_g+1)$, calculating the collective decay rate is straightforward.

Assuming the atoms are confined to a volume much smaller than the wavelength (the standard Dicke limit so all atoms see the exact same radiation field), the total emission intensity of the ensemble is proportional to $\braket{\hat S^+\hat S^-}$. If a single isolated atom decays at a rate $\Gamma$, the total decay rate $\Gamma_\text{tot}$ for a symmetric Dicke state $\ket{S,m}$ becomes:

$$\Gamma_\text{tot} = \Gamma \braket{\hat S^+\hat S^-} = \Gamma n_e(n_g+1)$$

To find the physical decay rate per excitation, we divide by the number of excited atoms $n_e$:

$$\Gamma_\text{per} = \frac{\Gamma n_e(n_g+1)}{n_e} = \Gamma (n_g+1)$$

Superradiance and the Superradiant Burst§

Let us look at the fully symmetric state with exactly a single excitation. Here, $n_e = 1$ and $n_g = N-1$. Plugging this into our formula gives $\Gamma_\text{per} = \Gamma N$, meaning the single excitation decays at a rate enhanced by a factor of $N$ compared to a single isolated atom.

Furthermore, consider the case where the ensemble is half excited such that $n_g = n_e = N/2$. The total intensity becomes $\Gamma_\text{tot} = \Gamma \frac{N}{2}(\frac{N}{2} + 1) \approx \frac{\Gamma N^2}{4}$. This demonstrates the characteristic $N^2$ intensity scaling of superradiance derived directly from the full coherent Dicke state expectation value.

As the system decays, $n_g$ grows while $n_e$ shrinks. The per-excitation decay rate $\Gamma(n_g+1)$ increases as the atoms drop to the ground state. This positive feedback causes the emission to accelerate over time, concentrating the emitted energy into a short, intense pulse known as a superradiant burst.

Subradiance§

For symmetric Dicke states $\ket{S,m}$ with excitations, the ground state population is $n_g \ge 1$, keeping the per-excitation decay rate proportional to $\Gamma (n_g+1)$. Subradiance corresponds to decay rates scaled below $\Gamma$ and arises from states with total spin $S' < N/2$. In these asymmetric states, the dipoles exhibit destructive interference. For example, with $N=2$, the anti-symmetric singlet state $\frac{1}{\sqrt{2}}(\ket{eg} - \ket{ge})$ has $S'=0$. This interference yields $\braket{\hat S^+\hat S^-} = 0$ and a decay rate of zero.