Lamb shift statistics in mesoscopic quantum ensembles

Abstract

Ensembles of two-level systems interacting with a single-mode electromagnetic field are increasingly being used for the development of quantum information processors and mesoscopic quantum devices. These systems are characterized by a set of energy-level hybridizations, split by Lamb shifts, that occur when the ensemble and field mode interact coherently with high cooperativity. For high dimensions simulating system dynamics while including the complete set of Lamb shifts is generally intractable. We present a compact description of Lamb shift statistics and behavior across all collective angular momentum and excitation subspaces of the ensemble without resorting to overly restrictive approximations of the state space. We also calculate the averaged effective Lamb shift weighted over the degeneracies of these subspaces to connect our results to experiments.

Appendices

Appendix: Explicit Calculations for $N=1,2,3$

In this appendix, we provide explicit calculations of the energy-level structure, state hybridization, and Lamb shifts for small ensembles of $N=1,2,$ and 3 spins.

1.1 N = 1

A single spin coupled to a single–mode electromagnetic field is described by the Jaynes–Cummings Hamiltonian [26]:

$$\begin{aligned} \hat{{\mathcal {H}}}_0 + \hat{{\mathcal {H}}}_{int} = \omega _{0} ({\hat{a}}^\dag {\hat{a}} + {\hat{\sigma }}_z) + g_0 \left( {\hat{a}}^\dag \sigma _{-} +{\hat{a}} \sigma _{+} \right) . \end{aligned}$$

(A.1)

Since $\hat{{\mathcal {H}}}_{int}$ couples spins with equal energy in the unperturbed spectrum, the Hilbert space decouples into blocks of constant total excitation, indexed by the good quantum number k:

$$\begin{aligned} \bigoplus _k |\psi _k\rangle \Big [ \langle \psi _k| \mathcal {{\hat{H}}}_0+\mathcal {{\hat{H}}}_{int} |\phi _k\rangle \Big ] \langle \phi _k|, \text { with }\mathcal {{\hat{H}}}_0\phi _k=E_k\phi _k \text { and } \mathcal {{\hat{H}}}_0\psi _k=E_k\psi _k.\nonumber \\ \end{aligned}$$

(A.2)

The ground state with no excitations ($k=0$), $|0\rangle |\downarrow \rangle$, is unique and is not hybridized by $H_{int}$. For the remaining states, hybridization occurs and we utilize conservation of excitation number by $H_{int}$ to calculate the eigenstructure. Consider the two states with finite excitation, $k > 0$, defined by $\lbrace \left| k\right\rangle \left| \downarrow \right\rangle , \left| k-1\right\rangle \left| \uparrow \right\rangle \rbrace$. The interaction Hamiltonian represented in this basis is given by the direct sum over all two–dimensional excitation spaces:

$$\begin{aligned} \mathcal {{\hat{H}}}_{int}\cong \bigoplus _k g_0\begin{bmatrix} 0 &{} \sqrt{k}\\ \sqrt{k} &{} 0 \end{bmatrix}. \end{aligned}$$

(A.3)

By direct diagonalization, the energy eigenvalues of $H_{int}$ in the manifold with k excitations are given by

$$\begin{aligned} E_{k,\pm }= k\omega _0\pm g_0\sqrt{k}, \end{aligned}$$

(A.4)

which correspond to hybridized energy eigenstates

$$\begin{aligned} |\psi _{k,\pm }\rangle =\frac{1}{\sqrt{2}}(|k\rangle |\downarrow \rangle \pm |k-1\rangle |\uparrow \rangle ). \end{aligned}$$

(A.5)

All Lamb shifts that occur for $N=1$ are summarized in the following table:

Subspace	Excitations	Lamb shifts	Numerical values
$j=1/2$	$k=1$	$g_0$	$g_0$
	$k=2$	$g_0 \sqrt{2}$	$1.41 g_0$
	$k=3$	$g_0 \sqrt{3}$	$1.73 g_0$
	k	$g_0 \sqrt{k}$

1.2 N = 2

Multiple spins coupled to a single-mode electromagnetic field are described by the Tavis–Cummings Hamiltonian [32]:

$$\begin{aligned} \hat{{\mathcal {H}}}_{TC} = \omega _0 ({\hat{a}}^\dagger {\hat{a}} + {\hat{J}}_z) + g_0({\hat{a}}^\dagger {\hat{J}}_- + {\hat{a}}{\hat{J}}_+), \end{aligned}$$

(A.6)

The case of two spins can be solved in two different ways: one where the full matrix is solved in the Zeeman basis of the spins, and one where the matrix is written in a direct sum representation of a spin-0 space (singlet) and spin-1 (triplet) space. We will solve this case in the Zeeman basis to contrast with the direct-sum representation used for the $N=3$ case. Inclusion of a second spin leads to a richer energy-level structure than for $N=1$:

Similar to the $N=1$ case, the ground state with no excitations ($k=0$), $|0\rangle |\downarrow \downarrow \rangle$, remains unperturbed by $H_{int}$:

$$\begin{aligned} |1,0:\alpha \rangle =|0\rangle |1,-1\rangle =|0\rangle |\downarrow \downarrow \downarrow \rangle \end{aligned}$$

(A.7)

The single-excitation manifold ($k=1$) undergoes hybridization, determined by the coupling matrix L(1, 1):

$$\begin{aligned} \begin{bmatrix} 0 &{} 1 &{} 1 \\ 1 &{} 0 &{} 0 \\ 1 &{} 0 &{} 0\\ \end{bmatrix} \end{aligned}$$

(A.8)

Through direct diagonalization, the hybridized cavity–dressed states and perturbed energies are

$$\begin{aligned} |1,1:\beta \rangle= & {} \frac{1}{\sqrt{2}}[|1\rangle |1;-1\rangle +|0\rangle |1;0\rangle ] =\frac{1}{2}[\sqrt{2}|1\rangle |\downarrow \downarrow \rangle +|0\rangle |\downarrow \uparrow \rangle +|0\rangle |\uparrow \downarrow \rangle ],\nonumber \\ E= & {} \omega _0+g_0\sqrt{2} \end{aligned}$$

(A.9)

$$\begin{aligned} |1,1:\alpha \rangle= & {} \frac{1}{\sqrt{2}}[|1\rangle |1;-1\rangle -|0\rangle |1;0\rangle ] =\frac{1}{2}[\sqrt{2}|1\rangle |\downarrow \downarrow \rangle -|0\rangle |\downarrow \uparrow \rangle -|0\rangle |\uparrow \downarrow \rangle ],\nonumber \\ E= & {} \omega _0-g_0\sqrt{2} \end{aligned}$$

(A.10)

$$\begin{aligned} |0,1:\alpha \rangle= & {} |0\rangle |0;0\rangle =\frac{1}{\sqrt{2}}[|0\rangle |\downarrow \uparrow \rangle -|0\rangle |\uparrow \downarrow \rangle ],\nonumber \\ E= & {} \omega _0+0, \end{aligned}$$

(A.11)

where for collective angular momentum j with secondary quantum number m, we have written $|j;m\rangle$. For our figure, we will call these dressed states $|1,1:\beta \rangle$, $|1,1:\alpha \rangle$, and $|0,1:\alpha \rangle$, respectively, where we have labeled our states with $|j,k:\text {index}\rangle$ with index being a Greek letter, alphabetically used in order of increasing energy.

For $k=2$, we have an interaction Hamiltonian given by:

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{2} &{} \sqrt{2} &{} 0 \\ \sqrt{2} &{} 0 &{} 0 &{} 1 \\ \sqrt{2} &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0\\ \end{bmatrix}. \end{aligned}$$

(A.12)

This matrix has eigenvectors (dressed states) and eigenvalues (energies with associated Lamb shift) of

$$\begin{aligned}&|1,2:\beta \rangle :=\frac{1}{\sqrt{3}}[-|2\rangle |1;-1\rangle +\sqrt{2}|0\rangle |1;1\rangle ]=\frac{1}{\sqrt{3}}[-|2\rangle |\downarrow \downarrow \rangle +\sqrt{2}|0\rangle |\uparrow \uparrow \rangle ] ,\nonumber \\ \end{aligned}$$

(A.13)

$$\begin{aligned}&E=2\omega _0+0 \end{aligned}$$

(A.14)

$$\begin{aligned}&|0,2:\alpha \rangle :=|1\rangle |0;0\rangle =\frac{1}{\sqrt{2}}[|1\rangle |\downarrow \uparrow \rangle -|1\rangle |\uparrow \downarrow \rangle ] , \end{aligned}$$

(A.15)

$$\begin{aligned}&E=2\omega _0+0 \end{aligned}$$

(A.16)

$$\begin{aligned}&|1,2:\gamma \rangle :=\frac{1}{\sqrt{6}}[\sqrt{2}|2\rangle |1;-1\rangle +\sqrt{3}|1\rangle |1;0\rangle +|0\rangle |1;1\rangle ] \end{aligned}$$

(A.17)

$$\begin{aligned}&\quad =\frac{1}{2\sqrt{3}}[2|2\rangle |\downarrow \downarrow \rangle +\sqrt{3}|1\rangle |\downarrow \uparrow \rangle +\sqrt{3}|1\rangle |\uparrow \downarrow \rangle +\sqrt{2}|0\rangle |\uparrow \uparrow \rangle ], \end{aligned}$$

(A.18)

$$\begin{aligned}&E=2\omega _0+g_0\sqrt{6} \end{aligned}$$

(A.19)

$$\begin{aligned}&|1,2:\alpha \rangle :=\frac{1}{\sqrt{6}}[\sqrt{2}|2\rangle |1;-1\rangle -\sqrt{3}|1\rangle |1;0\rangle +1|0\rangle |1;1\rangle ] \end{aligned}$$

(A.20)

$$\begin{aligned}&\quad =\frac{1}{2\sqrt{3}}[2|2\rangle |\downarrow \downarrow \rangle -\sqrt{3}|1\rangle |\downarrow \uparrow \rangle -\sqrt{3}|1\rangle |\uparrow \downarrow \rangle +\sqrt{2}|0\rangle |\uparrow \uparrow \rangle ] \end{aligned}$$

(A.21)

$$\begin{aligned}&E=2\omega _0-g_0\sqrt{6} \end{aligned}$$

(A.22)

For higher excitation manifolds, $k\ge 2$, there are always exactly four states that must be diagonalized, determined by the coupling matrix $H_{int}/g_0$ with exactly k excitations:

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{k} &{} \sqrt{k} &{} 0 \\ \sqrt{k} &{} 0 &{} 0 &{} \sqrt{k-1} \\ \sqrt{k} &{} 0 &{} 0 &{} \sqrt{k-1} \\ 0 &{} \sqrt{k-1} &{} \sqrt{k-1} &{} 0\\ \end{bmatrix} \end{aligned}$$

(A.23)

Again, through direct diagonalization, the hybridized cavity-dressed states and perturbed energies are:

$$\begin{aligned}&\frac{1}{\sqrt{2k-1}}[-\sqrt{k-1}|k\rangle |1;-1\rangle +\sqrt{k}|k-2\rangle |1;1\rangle ] \end{aligned}$$

(A.24)

$$\begin{aligned}&\quad =\frac{1}{\sqrt{2k-1}}[-\sqrt{k-1}|k\rangle |\downarrow \downarrow \rangle +\sqrt{k}|k-2\rangle |\uparrow \uparrow \rangle ], \end{aligned}$$

(A.25)

$$\begin{aligned}&\quad E=k\omega _0+0 \end{aligned}$$

(A.26)

$$\begin{aligned}&\quad |k-1\rangle |0;0\rangle \end{aligned}$$

(A.27)

$$\begin{aligned}&\quad =\frac{1}{\sqrt{2}}[|k-1\rangle |\downarrow \uparrow \rangle -|k-1\rangle |\uparrow \downarrow \rangle ], \end{aligned}$$

(A.28)

$$\begin{aligned}&\quad E=k\omega _0+0 \end{aligned}$$

(A.29)

$$\begin{aligned}&\quad \frac{1}{\sqrt{2}\sqrt{2k-1}}[\sqrt{k}|k\rangle |1;-1\rangle +\sqrt{2k-1}|k-1\rangle |1;0\rangle +\sqrt{k-1}|k-2\rangle |1;1\rangle ]\nonumber \\ \end{aligned}$$

(A.30)

$$\begin{aligned}&\quad =\frac{1}{2\sqrt{2k-1}}[\sqrt{2}\sqrt{k}|k\rangle |\downarrow \downarrow \rangle +\sqrt{2k-1}|k-1\rangle |\downarrow \uparrow \rangle +\sqrt{2k-1}|k-1\rangle |\uparrow \downarrow \rangle \nonumber \\&\qquad +\sqrt{2}\sqrt{k-1}|k-2\rangle |\uparrow \uparrow \rangle ], \end{aligned}$$

(A.31)

$$\begin{aligned}&\quad E=k\omega _0+g_0\sqrt{2}\sqrt{2k-1} \end{aligned}$$

(A.32)

$$\begin{aligned}&\quad \frac{1}{\sqrt{2}\sqrt{2k-1}}[\sqrt{k}|k\rangle |1;-1\rangle -\sqrt{2k-1}|k-1\rangle |1;0\rangle +\sqrt{k-1}|k-2\rangle |1;1\rangle ]\qquad \qquad \end{aligned}$$

(A.33)

$$\begin{aligned}&\quad =\frac{1}{2\sqrt{2k-1}}[\sqrt{2}\sqrt{k}|k\rangle |\downarrow \downarrow \rangle -\sqrt{2k-1}|k-1\rangle |\downarrow \uparrow \rangle -\sqrt{2k-1}|k-1\rangle |\uparrow \downarrow \rangle \nonumber \\&\qquad +\sqrt{2}\sqrt{k-1}|k-2\rangle |\uparrow \uparrow \rangle ] \end{aligned}$$

(A.34)

$$\begin{aligned}&\quad E=k\omega _0-g_0\sqrt{2}\sqrt{2k-1} \end{aligned}$$

(A.35)

Notice that within each eigenvector only terms involving a single collective spin value j is used for each term. If the direct sum representation is used instead of the Zeeman representation, such that the basis states are constant total angular momentum states (spin-1 and spin-0), the complexity of the problem is somewhat reduced. In this case, the only difference is replacing $|k\rangle |\uparrow \downarrow \rangle$ and $|k\rangle |\downarrow \uparrow \rangle$ with $|k\rangle (|\uparrow \downarrow \rangle +|\downarrow \uparrow \rangle )$ and $|k\rangle (|\uparrow \downarrow \rangle -|\downarrow \uparrow \rangle )$. In this representation, the singlet state $|\uparrow \downarrow \rangle -|\downarrow \uparrow \rangle$ is annihilated by $J_{\pm }$ and forms its own space. All Lamb shifts that occur for $N=2$ are summarized in the following table:

Subspace	Excitations	Lamb shifts	Numerical values
$j=1$	$k=1$	$g_0 \sqrt{2}$	$1.41 g_0$
	$k=2$	$g_0 \sqrt{6}$	$2.45 g_0$
	$k=3$	$g_0 \sqrt{10}$	$3.16 g_0$
	k	$g_0 \sqrt{2}\sqrt{2k-1}$
$j=0$	k	0	0

1.3 N = 3

The utility of the direct sum representation becomes even more apparent when $N=3$ and subspace degeneracy first arises (see Fig. 8). As for $N=1$ and 2, the ground state with no excitations ($k=0$), $|0\rangle |\downarrow \downarrow \downarrow \rangle$ remains unperturbed by $H_{int}$. When the number of excitations are such that $k \le 2$, the number of hybridized states are sub-maximal. For clarity, we divide the analysis into three cases that depend on the total number of excitations present in the joint spin–cavity system: $k=1$, $k=2$, and $k\ge 3$.

1.3.1 Case 1: $k=1$

In the single–excitation manifold, there are four basis states, with hybridization determined by the coupling matrix $H_{int}$ for $k=1$:

$$\begin{aligned} g_0\begin{bmatrix} 0 &{} 1 &{} 1 &{} 1\\ 1 &{} 0 &{} 0 &{} 0\\ 1 &{} 0 &{} 0 &{} 0\\ 1 &{} 0 &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

(A.36)

Solving this matrix through direct diagonalization gives the hybridized cavity-dressed states and perturbed energies of

$$\begin{aligned} |3/2,1:\beta \rangle:= & {} \frac{1}{\sqrt{2}}[|1\rangle |3/2;-3/2\rangle +|0\rangle |3/2;-1/2\rangle ] \end{aligned}$$

(A.37)

$$\begin{aligned}&=\frac{1}{\sqrt{6}}[\sqrt{3}|1\rangle |\downarrow \downarrow \downarrow \rangle +|0\rangle (|\downarrow \downarrow \uparrow \rangle +|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )],\nonumber \\ E= & {} \omega _0+g_0\sqrt{3} \end{aligned}$$

(A.38)

$$\begin{aligned} |3/2,1:\alpha \rangle:= & {} \frac{1}{\sqrt{2}}[-|1\rangle |3/2;-3/2\rangle +|0\rangle |3/2;-1/2\rangle ] \end{aligned}$$

(A.39)

$$\begin{aligned}= & {} \frac{1}{\sqrt{6}}[-\sqrt{3}|1\rangle |\downarrow \downarrow \downarrow \rangle +|0\rangle (|\downarrow \downarrow \uparrow \rangle +|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )],\nonumber \\ E= & {} \omega _0-g_0\sqrt{3} \end{aligned}$$

(A.40)

$$\begin{aligned}&|1/2,1:\alpha \rangle _1:=|0\rangle |1/2;-1/2\rangle _1 \end{aligned}$$

(A.41)

$$\begin{aligned}= & {} \frac{1}{\sqrt{2}}|0\rangle (|\uparrow \downarrow \downarrow \rangle -|\downarrow \downarrow \uparrow \rangle ),\nonumber \\ E= & {} \omega _0 \end{aligned}$$

(A.42)

$$\begin{aligned}&|1/2,1:\alpha \rangle _2:=|0\rangle |1/2;-1/2\rangle _2 \end{aligned}$$

(A.43)

$$\begin{aligned}= & {} \frac{1}{\sqrt{2}}|0\rangle (|\downarrow \uparrow \downarrow \rangle -|\downarrow \downarrow \uparrow \rangle ), E=\omega _0, \end{aligned}$$

(A.44)

where the subscript on $|1/2;-1/2\rangle$ indicates the degeneracy number of that subspace.

Fig. 8

Illustration of the resulting hybridization of energy levels in the Tavis–Cummings model for $N=3$, explicitly on resonance such that $\omega _0 = \omega _s = \omega _c$. Vertical single arrow lines (red) indicate transitions mediated by ${\hat{J}}_+$, meaning that the eigenstates represented by the horizontal bars have a non-zero ${\hat{J}}_+$ matrix element. Transitions are all–to–all between neighboring excitation subspaces of the same angular momentum, with some transitions between the $k=2$ and $k=3$ subspaces omitted for clarity. Note that there are no allowed transitions via collective spin or photon operators between distinct angular momentum subspaces, regardless of the value of j. Separation between excitation spaces is a constant $\omega _0$, denoted by bidirectional arrows (blue) between the pre-hybridized angular momentum states. Lamb shift splittings are denoted by bidirectional arrows (green) to the right of the hybridized states. In the $j=1/2$ subspaces, these splittings are given by $E_{1/2,1} = g_0$ and $E_{1/2,2} = g_0 \sqrt{2}$. In the $j=3/2$ subspace, the Lamb shifts are given by: $E_{3/2,1} = g_0 \sqrt{3} \approx 1.73 g_0$, $E_{3/2,2} = g_0 \sqrt{10} \approx 3.16 g_0$, $E_{3/2,3,1} = g_0\sqrt{10 - \sqrt{73}} \approx 1.21 g_0$, and $E_{3/2,3,2} = g_0\sqrt{10 + \sqrt{73}} \approx 4.31 g_0$

1.3.2 Case 2: $k=2$

When a second excitation is added to the joint spin–cavity system, the number of basis states increases to seven, with hybridization determined by the coupling matrix $H_{int}/g_0$ for $k=2$:

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{2} &{} \sqrt{2} &{} 0 &{} \sqrt{2} &{} 0 &{} 0 \\ \sqrt{2} &{} 0 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 \\ \sqrt{2} &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 \\ \sqrt{2} &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ 0 &{} 1 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 \\ \end{bmatrix} \end{aligned}$$

(A.45)

This matrix may be represented as a direct sum, $H_{int}/g_0=L(3/2,2)\oplus L(1/2,2)_1\oplus L(1/2,2)_2$, to yield

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{6} &{} 0 \\ \sqrt{6} &{} 0 &{} 2 \\ 0 &{} 2 &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix} \end{aligned}$$

(A.46)

By direct diagonalization, the first entry in the direct sum gives hybridized cavity-dressed states and perturbed energies of

$$\begin{aligned}&|3/2,2:\alpha \rangle :=\frac{1}{2}[\sqrt{\frac{1}{2}}|2\rangle |3/2;-3/2\rangle -\sqrt{\frac{5}{2}}|1\rangle |3/2;-1/2\rangle +|0\rangle |3/2;1/2\rangle ]= \end{aligned}$$

(A.47)

$$\begin{aligned}&\frac{1}{2\sqrt{3}}[\sqrt{\frac{3}{2}}|2\rangle |\downarrow \downarrow \downarrow \rangle -\sqrt{\frac{5}{2}}|1\rangle (|\downarrow \downarrow \uparrow \rangle +|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )+|0\rangle (|\downarrow \uparrow \uparrow \rangle +|\uparrow \downarrow \uparrow \rangle +|\uparrow \uparrow \downarrow \rangle )],\nonumber \\&E=2\omega _0-g_0\sqrt{10} \end{aligned}$$

(A.48)

$$\begin{aligned}&|3/2,2:\gamma \rangle :=\frac{1}{2}[\sqrt{\frac{1}{2}}|2\rangle |3/2;-3/2\rangle +\sqrt{\frac{5}{2}}|1\rangle |3/2;-1/2\rangle +|0\rangle |3/2;1/2\rangle ]= \end{aligned}$$

(A.49)

$$\begin{aligned}&\frac{1}{2\sqrt{3}}[\sqrt{\frac{3}{2}}|2\rangle |\downarrow \downarrow \downarrow \rangle +\sqrt{\frac{5}{2}}|1\rangle (|\downarrow \downarrow \uparrow \rangle +|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )+|0\rangle (|\downarrow \uparrow \uparrow \rangle +|\uparrow \downarrow \uparrow \rangle +|\uparrow \uparrow \downarrow \rangle )],\nonumber \\&E=2\omega _0+g_0\sqrt{10} \end{aligned}$$

(A.50)

$$\begin{aligned}&|3/2,2:\beta \rangle :=\frac{1}{\sqrt{11}}[-\sqrt{2}|2\rangle |3/2;-3/2\rangle +3|0\rangle |3/2;1/2\rangle ]= \end{aligned}$$

(A.51)

$$\begin{aligned}&\sqrt{\frac{3}{11}}[-\sqrt{\frac{2}{3}}|2\rangle |\downarrow \downarrow \downarrow \rangle +|0\rangle (|\downarrow \uparrow \uparrow \rangle +|\uparrow \downarrow \uparrow \rangle +|\uparrow \uparrow \downarrow \rangle )], E=2\omega _0 \end{aligned}$$

(A.52)

Similarly, the second and third entries of the direct sum give

$$\begin{aligned}&\frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _1\pm |1\rangle |1/2;-1/2\rangle _1] \end{aligned}$$

(A.53)

$$\begin{aligned}&\quad =\frac{1}{2}[|1\rangle [|\downarrow \uparrow \downarrow \rangle -|\uparrow \downarrow \downarrow \rangle ]\pm |0\rangle [|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle ]],E=2\omega _0 \pm g_0 \end{aligned}$$

(A.54)

$$\begin{aligned}&\quad \frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _2\pm |1\rangle |1/2;-1/2\rangle _2] \end{aligned}$$

(A.55)

$$\begin{aligned}&\quad =\frac{1}{2\sqrt{3}}[|1\rangle [2|\downarrow \downarrow \uparrow \rangle -|\uparrow \downarrow \downarrow \rangle -|\downarrow \uparrow \downarrow \rangle ]\pm |0\rangle [|\uparrow \downarrow \uparrow \rangle +|\downarrow \uparrow \uparrow \rangle -2|\uparrow \uparrow \downarrow \rangle ]],\nonumber \\&\quad E=2\omega _0 \pm g_0 \end{aligned}$$

(A.56)

where we define for the figure:

$$\begin{aligned} |1/2,2:\alpha \rangle _1:= & {} \frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _1- |1\rangle |1/2;-1/2\rangle _1], |1/2,2:\beta \rangle _1\nonumber \\:= & {} \frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _1+ |1\rangle |1/2;-1/2\rangle _1] \end{aligned}$$

(A.57)

$$\begin{aligned} |1/2,2:\alpha \rangle _2:= & {} \frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _2- |1\rangle |1/2;-1/2\rangle _2], |1/2,2:\beta \rangle _2\nonumber \\:= & {} \frac{1}{\sqrt{2}}[|0\rangle |1/2;1/2\rangle _2+ |1\rangle |1/2;-1/2\rangle _2] \end{aligned}$$

(A.58)

1.3.3 Case 3: $k=3$

When $k=3$, the number of spin states becomes maximal and all $2^3=8$ spin states participate in hybridization, determined by the coupling matrix $H_{int}/g_0$ at $k=3$:

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{3} &{} \sqrt{3} &{} 0 &{} \sqrt{3} &{} 0 &{} 0 &{} 0\\ \sqrt{3} &{} 0 &{} 0 &{} \sqrt{2} &{} 0 &{} \sqrt{2} &{} 0 &{} 0\\ \sqrt{3} &{} 0 &{} 0 &{} \sqrt{2} &{} 0 &{} 0 &{} \sqrt{2} &{} 0\\ 0 &{} \sqrt{2} &{} \sqrt{2} &{} 0 &{} 0 &{} 0 &{} 0 &{} 1\\ \sqrt{3} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{2} &{} \sqrt{2} &{} 0\\ 0 &{} \sqrt{2} &{} 0 &{} 0 &{} \sqrt{2} &{} 0 &{} 0 &{} 1\\ 0 &{} 0 &{} \sqrt{2} &{} 0 &{} \sqrt{2} &{} 0 &{} 0 &{} 1\\ 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 1 &{} 1 &{} 0 \end{bmatrix}, \end{aligned}$$

(A.59)

where the ordered basis states for this matrix representation are given by the set $\{|3\rangle |\downarrow \downarrow \downarrow \rangle ,|2\rangle |\downarrow \downarrow \uparrow \rangle ,\ldots |0\rangle |\uparrow \uparrow \uparrow \rangle \}$. This matrix admits a direct sum representation with coupling matrices $L(3/2,3)\oplus L(1/2,3)\oplus L(1/2,3)$, given by:

$$\begin{aligned} \begin{bmatrix} 0 &{} 3 &{} 0 &{} 0\\ 3 &{} 0 &{} 2\sqrt{2} &{} 0\\ 0 &{} 2\sqrt{2} &{} 0 &{} \sqrt{3}\\ 0 &{} 0 &{} \sqrt{3} &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} \sqrt{2}\\ \sqrt{2} &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} \sqrt{2}\\ \sqrt{2} &{} 0 \end{bmatrix}. \end{aligned}$$

(A.60)

The first matrix is written in the Dicke (fully symmetric) basis (normalized versions of $|k-m\rangle {\hat{J}}_+^m|\downarrow \downarrow \downarrow \rangle$ with $m\in \{0,1,2,3\}$), while the second and third are written in terms of the composite spin–1/2 bases, given by

$$\begin{aligned} \frac{1}{\sqrt{2}}|2\rangle (|\downarrow \uparrow \downarrow \rangle -|\uparrow \downarrow \downarrow \rangle )&,&\quad \frac{1}{\sqrt{2}}|1\rangle (|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle ) \end{aligned}$$

(A.61)

$$\begin{aligned} \frac{1}{\sqrt{6}}|2\rangle (2|\downarrow \downarrow \uparrow \rangle -|\uparrow \downarrow \downarrow \rangle -|\downarrow \uparrow \downarrow \rangle )&,&\quad \frac{1}{\sqrt{6}} |1\rangle (|\uparrow \downarrow \uparrow \rangle +|\downarrow \uparrow \uparrow \rangle -2|\uparrow \uparrow \downarrow \rangle ).\nonumber \\ \end{aligned}$$

(A.62)

The matrix representations for the degenerate spin-1/2 subspaces are identical, and thus indistinguishable under a collective operation or measurement. There is freedom in the choices of bases for the degenerate subspaces; the states provided are the standard basis states for these subspaces, as computed via a Clebsch–Gordon table [57].

We diagonalize each block individually, starting with the matrix representing the $j=3/2$ subspace, and find the resulting (non-normalized) Lamb-shifted dressed states are given by the following superpositions:

$$\begin{aligned} |3/2, 3; \pm _1, \pm _2 \rangle:= & {} \pm _1\sqrt{10{\mp }_2 \sqrt{73}}(1\pm _2 \sqrt{73})|3\rangle |3/2;-3/2\rangle \nonumber \\&+ 3(7{\mp }_2 \sqrt{73})|2\rangle |3/2;-1/2\rangle \nonumber \\&{\mp }_1 6\sqrt{10{\mp }_2 \sqrt{73}}\sqrt{2}|1\rangle |3/2;1/2\rangle +6\sqrt{6}|0\rangle |3/2;3/2\rangle \qquad \qquad \end{aligned}$$

(A.63)

$$\begin{aligned}= & {} \pm _1\sqrt{10{\mp }_2 \sqrt{73}}(1\pm _2 \sqrt{73})|3\rangle |\downarrow \downarrow \downarrow \rangle + (7{\mp }_2 \sqrt{73})3|2\rangle \nonumber \\&\frac{1}{\sqrt{3}}(|\downarrow \downarrow \uparrow \rangle +|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )\nonumber \\&{\mp }_1 6\sqrt{10{\mp }_2 \sqrt{73}}\sqrt{2}|1\rangle \frac{1}{\sqrt{3}}(|\downarrow \uparrow \uparrow \rangle +|\uparrow \downarrow \uparrow \rangle +|\uparrow \uparrow \downarrow \rangle )\nonumber \\&+6\sqrt{6}|0\rangle |\uparrow \uparrow \uparrow \rangle , \end{aligned}$$

(A.64)

each with associated energy

$$\begin{aligned} E_{3;\pm _1\pm _2}=3\omega _0 {\mp }_1 g_0 \sqrt{10{\mp }_2\sqrt{73}}. \end{aligned}$$

(A.65)

We have introduced a shorthand notation via a subscript on the ± sign, such that $\pm _1,\pm _2$ are a pair of sign choices (and ${\mp }_1$ indicates that the opposite sign as $\pm _1$ is used, and likewise for ${\mp }_2$) which allows for a more compact expression for all four dressed states. The four perturbed energy values are not equally spaced, though they still come in oppositely signed pairs of equal magnitude. In terms of the Greek letter notation, we have the following equivalences:

$$\begin{aligned} |3/2,3:\alpha \rangle :=|3/2,3; +,-\rangle&,\quad&|3/2,3:\beta \rangle :=|3/2,3; +,+\rangle , \end{aligned}$$

(A.66)

$$\begin{aligned} |3/2,3:\gamma \rangle :=|3/2,3; -,+\rangle&,\quad&|3/2,3:\delta \rangle :=|3/2,3;-,-\rangle . \end{aligned}$$

(A.67)

For the remaining two matrices with $j=\frac{1}{2}$ in the direct sum decomposition, these systems are algebraically equivalent to the single spin model. This equivalence allows us to immediately write down the diagonalized states and perturbed energies:

$$\begin{aligned} |1/2, 3; \pm \rangle _1&:= \frac{1}{2} [|2\rangle [|\downarrow \uparrow \downarrow \rangle -|\uparrow \downarrow \downarrow \rangle ] \pm |1\rangle [|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle ]],\\ E_{3;\pm }&=3\omega _0 \pm g_0\sqrt{2}\\ |1/2,\pm \rangle _2&:= \frac{1}{2\sqrt{3}} [|2\rangle [2|\downarrow \downarrow \uparrow \rangle -|\uparrow \downarrow \downarrow \rangle -|\downarrow \uparrow \downarrow \rangle ]\pm |1\rangle [|\uparrow \downarrow \uparrow \rangle +|\downarrow \uparrow \uparrow \rangle \\&\quad -2|\uparrow \uparrow \downarrow \rangle ]],\\ E_{3;\pm }&= 3\omega _0 \pm g_0\sqrt{2}. \end{aligned}$$

The subscript on the kets in the above equations indicate the arbitrarily chosen degeneracy label of that subspace.

1.3.4 Case 4: $k \ge 3$

While the above considered $k=3$, we can solve this case as well as all higher cases at once. In the general case of $k \ge 3$, the number of manifold states become maximal (see Fig. 8) and all $2^3=8$ spin states participate in hybridization, determined by the coupling matrix L(3/2, k):

$$\begin{aligned} L(k) = \begin{bmatrix} 0 &{} \sqrt{k} &{} \sqrt{k} &{} 0 &{} \sqrt{k} &{} 0 &{} 0 &{} 0\\ \sqrt{k} &{} 0 &{} 0 &{} \sqrt{k-1} &{} 0 &{} \sqrt{k-1} &{} 0 &{} 0\\ \sqrt{k} &{} 0 &{} 0 &{} \sqrt{k-1} &{} 0 &{} 0 &{} \sqrt{k-1} &{} 0\\ 0 &{} \sqrt{k-1} &{} \sqrt{k-1} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{k-2}\\ \sqrt{k} &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{k-1} &{} \sqrt{k-1} &{} 0\\ 0 &{} \sqrt{k-1} &{} 0 &{} 0 &{} \sqrt{k-1} &{} 0 &{} 0 &{} \sqrt{k-2}\\ 0 &{} 0 &{} \sqrt{k-1} &{} 0 &{} \sqrt{k-1} &{} 0 &{} 0 &{} \sqrt{k-2}\\ 0 &{} 0 &{} 0 &{} \sqrt{k-2} &{} 0 &{} \sqrt{k-2} &{} \sqrt{k-2} &{} 0 \end{bmatrix},\qquad \quad \end{aligned}$$

(A.68)

where the ordered basis states for this matrix representation are given by the set $\{|k\rangle |\downarrow \downarrow \downarrow \rangle ,|k-1\rangle |\downarrow \downarrow \uparrow \rangle ,\ldots |k-3\rangle |\uparrow \uparrow \uparrow \rangle \}$. As before, this matrix admits a direct sum representation ($\frac{3}{2} \oplus \frac{1}{2}\oplus \frac{1}{2}$) with coupling matrices $L(3/2,k)\oplus L(1/2,k)\oplus L(1/2,k)$, given by:

$$\begin{aligned} \begin{bmatrix} 0 &{} \sqrt{3}\sqrt{k} &{} 0 &{} 0\\ \sqrt{3}\sqrt{k} &{} 0 &{} 2\sqrt{k-1} &{} 0\\ 0 &{} 2\sqrt{k-1} &{} 0 &{} \sqrt{3}\sqrt{k-2}\\ 0 &{} 0 &{} \sqrt{3}\sqrt{k-2} &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} \sqrt{k-1}\\ \sqrt{k-1} &{} 0 \end{bmatrix}\oplus \begin{bmatrix} 0 &{} \sqrt{k-1}\\ \sqrt{k-1} &{} 0 \end{bmatrix}.\nonumber \\ \end{aligned}$$

(A.69)

The first matrix is written in the Dicke (fully symmetric) basis (normalized versions of $|k-m\rangle {\hat{J}}_+^m|\downarrow \downarrow \downarrow \rangle$ with $m\in \{0,1,2,3\}$), while the second and third are written in terms of the composite spin–1/2 bases, given by

$$\begin{aligned} \frac{1}{\sqrt{2}}|k-1\rangle (|\downarrow \uparrow \downarrow \rangle -|\uparrow \downarrow \downarrow \rangle )&,&\quad \frac{1}{\sqrt{2}}|k-2\rangle (|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle ) \end{aligned}$$

(A.70)

$$\begin{aligned} \frac{1}{\sqrt{6}}|k-1\rangle (2|\downarrow \downarrow \uparrow \rangle -|\uparrow \downarrow \downarrow \rangle -|\downarrow \uparrow \downarrow \rangle )&,&\quad \frac{1}{\sqrt{6}} |k-2\rangle (|\uparrow \downarrow \uparrow \rangle +|\downarrow \uparrow \uparrow \rangle -2|\uparrow \uparrow \downarrow \rangle ).\nonumber \\ \end{aligned}$$

(A.71)

The matrix representations for the degenerate spin-1/2 subspaces are identical, and thus indistinguishable under a collective operation or measurement. There is freedom in the choices of bases for the degenerate subspaces; the states provided are the standard basis states for these subspaces, as computed via a Clebsch–Gordon table [57].

We diagonalize each block individually, starting with the matrix representing the $j=3/2$ subspace, and find the resulting (non-normalized) Lamb-shifted dressed states are given by the following superpositions:

$$\begin{aligned} |3/2, k; \pm _1, \pm _2 \rangle:= & {} \pm _1\sqrt{5k-5{\mp }_2 \sqrt{25-32k+16k^2}}(2k-5\pm _2\nonumber \\&\sqrt{25-32k+16k^2})|k\rangle |3/2;-3/2\rangle \nonumber \\&+ (1+2k{\mp }_2 \sqrt{25-32k+16k^2})\sqrt{3}\sqrt{k}|k-1\rangle |3/2;-1/2\rangle \nonumber \\&{\mp }_1 2\sqrt{5k-5{\mp }_2 \sqrt{25-32k+16k^2}}\sqrt{3} \sqrt{k-1}\sqrt{k}|k-2\rangle |3/2;1/2\rangle \nonumber \\&+6\sqrt{k-2}\sqrt{k-1}\sqrt{k}|k-3\rangle |3/2;3/2\rangle \end{aligned}$$

(A.72)

$$\begin{aligned}= & {} \pm _1\sqrt{5k-5{\mp }_2 \sqrt{25-32k+16k^2}}(2k-5\pm _2 \nonumber \\&\sqrt{25-32k+16k^2}) |k\rangle |\downarrow \downarrow \downarrow \rangle \nonumber \\&+ (1+2k{\mp }_2 \sqrt{25-32k+16k^2})\sqrt{3}\sqrt{k}|k-1\rangle \frac{1}{\sqrt{3}}(|\downarrow \downarrow \uparrow \rangle \nonumber \\&+|\downarrow \uparrow \downarrow \rangle +|\uparrow \downarrow \downarrow \rangle )\nonumber \\&{\mp }_1 2\sqrt{5k-5{\mp }_2 \sqrt{25-32k+16k^2}}\sqrt{3} \sqrt{k-1}\sqrt{k}|k-2\rangle \frac{1}{\sqrt{3}}(|\downarrow \uparrow \uparrow \rangle \nonumber \\&+|\uparrow \downarrow \uparrow \rangle +|\uparrow \uparrow \downarrow \rangle )\nonumber \\&+6\sqrt{k-2}\sqrt{k-1}\sqrt{k}|k-3\rangle |\uparrow \uparrow \uparrow \rangle , \end{aligned}$$

(A.73)

each with associated energy

$$\begin{aligned} E_{k;\pm _1\pm _2}=k\omega _0 {\mp }_1 g_0 \sqrt{5(k-1){\mp }_2\sqrt{16k^2-32k+25}}. \end{aligned}$$

(A.74)

We have introduced a shorthand notation via a subscript on the ± sign, such that $\pm _1,\pm _2$ are a pair of sign choices (and ${\mp }_1$ indicates that the opposite sign as $\pm _1$ is used, and likewise for ${\mp }_2$) which allows for a more compact expression for all four dressed states. The four perturbed energy values are not equally spaced, though they still come in oppositely signed pairs of equal magnitude. In terms of the Greek letter notation, we have the following equivalences:

$$\begin{aligned} |3/2,k:\alpha \rangle :=|3/2,k; +,-\rangle&,\quad&|3/2,k:\beta \rangle :=|3/2,k; +,+\rangle , \end{aligned}$$

(A.75)

$$\begin{aligned} |3/2,k:\gamma \rangle :=|3/2,k; -,+\rangle&,\quad&|3/2,k:\delta \rangle :=|3/2,k;-,-\rangle . \end{aligned}$$

(A.76)

For the remaining two matrices with $j=\frac{1}{2}$ in the direct sum decomposition, these systems are algebraically equivalent to the single spin model. This equivalence allows us to immediately write down the diagonalized states and perturbed energies:

$$\begin{aligned} |1/2, k; \pm \rangle _1&:= \frac{1}{2}[|k-1\rangle [|\downarrow \uparrow \downarrow \rangle -|\uparrow \downarrow \downarrow \rangle ]\pm |k-2\rangle [|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle ]],\\ E_{k;\pm }&=k\omega _0 \pm g_0\sqrt{k-1}\\ |1/2,\pm \rangle _2&:= \frac{1}{2\sqrt{3}}[|k-1\rangle [2|\downarrow \downarrow \uparrow \rangle -|\uparrow \downarrow \downarrow \rangle -|\downarrow \uparrow \downarrow \rangle ]\pm |k-2\rangle [|\uparrow \downarrow \uparrow \rangle +|\downarrow \uparrow \uparrow \rangle \\&\quad -2|\uparrow \uparrow \downarrow \rangle ]],\\ E_{k;\pm }&= k\omega _0 \pm g_0\sqrt{k-1}. \end{aligned}$$

The subscript on the kets in the above equations indicates the arbitrarily chosen degeneracy label of that subspace. All Lamb shifts that occur for $N=3$ are summarized in the following table:

Subspace	Excitations	Lamb shifts	Numerical values
$j=3/2$	$k=1$	$g_0 \sqrt{3}$	$1.73 g_0$
	$k=2$	$g_0 \sqrt{10}$	$3.16 g_0$
	$k=3$	$g_0 \sqrt{10\pm \sqrt{73}}$	$1.21 g_0, 4.21 g_0$
	k	$g_0 \sqrt{5(k-1)\pm \sqrt{16k^2 -32k +25}}$
$j=1/2$	$k=1$	0	0
	$k=2$	$g_0$	$g_0$
	$k=3$	$g_0\sqrt{2}$	$1.41 g_0$
	k	$g_0 \sqrt{k-1}$
$j=1/2$	$k=1$	0	0
	$k=2$	$g_0$	$g_0$
	$k=3$	$g_0\sqrt{2}$	$1.41 g_0$
	k	$g_0 \sqrt{k-1}$

1.4 $N>3$

The following figure illustrates the general decomposition structure into the various angular momentum subspaces (Fig. 9):

Fig. 9

Schematic representation of the energy eigenstates of the Tavis-Cummings Hamiltonian with excitations $0 \le k \le 3$ along the vertical, and labelled horizontally by the number of degeneracies of each angular momentum subspace

Appendix: Mathematical Proofs

Proof of equation (30)

To derive the value of $j^*$, we begin by defining the degeneracy function in a convenient form,

$$\begin{aligned} f(j)=\frac{2j+1}{N/2+j+1} {N \atopwithdelims ()N/2+j}, \end{aligned}$$

(A.1)

where j takes integer or half-integer values $0 \le j \le N/2$, depending on the parity of N. To prepare for differentiation, the binomial coefficient can be extended to a continuous function

$$\begin{aligned} {N \atopwithdelims ()K} = \frac{\Gamma (N+1)}{\Gamma (K+1)\Gamma (N-K+1)}, \end{aligned}$$

(A.2)

such that (A.1) can be written as a continuous function in j

$$\begin{aligned} f(j) = \frac{2j+1}{N/2+j+1} \frac{\Gamma (N+1)}{\Gamma (N/2 + j+1)\Gamma (N/2 - j + 1)}. \end{aligned}$$

(A.3)

We may now differentiate and look for critical values:

$$\begin{aligned} \frac{d}{dj}f(j)=4{N\atopwithdelims ()N/2+j}\frac{\frac{1}{2}(2j+1)(N+2j+2)(H_{N/2-j}-H_{N/2+j})+N+1}{(N+2j+2)^2}.\nonumber \\ \end{aligned}$$

(A.4)

In equation (A.4), $H_x$ is the Harmonic series truncated at term x. The degeneracy is then maximal when

$$\begin{aligned} \frac{1}{2}(2j+1)(N+2j+2)(H_{N/2-j}-H_{N/2+j})+N+1=0. \end{aligned}$$

(A.5)

We can re-cast this result by utilizing the expression for $H_x=\log x+\gamma +O(x^{-1})$, where $\gamma$ is the Euler–Mascheroni constant ($\gamma \approx 0.577$). This allows us to take the difference of the Harmonic numbers as $\log$’s, which cancels the additive term. After simplifying, we are left with

$$\begin{aligned} \frac{1}{2}(2j+1)(N+2j+2)\log (\frac{N/2-j}{N/2+j})+N+1=0. \end{aligned}$$

(A.6)

This is a very tight approximation. If we examine the series expansions of this, we see that taking $j\approx \frac{\sqrt{N}}{2}$ will remove the leading error. We may repeat this procedure, noting that the errors are a Laurent series in $\sqrt{N}$, so we adjust by decreasing powers of $\sqrt{N}$ corrections. Using this, we take as a guess that $j=\frac{\sqrt{N}-1}{2}+\frac{1}{6\sqrt{N}}$, which yields:

$$\begin{aligned} (3N+1)(3(N^{3/2}+N+\sqrt{N})+1)\frac{\log (\frac{-6N+6\sqrt{N}-2}{3\sqrt{N}(N+\sqrt{N}-1)+1}+1)}{18N}+N+1.\qquad \end{aligned}$$

(A.7)

When expanded as a series in the limit of large N, we have:

$$\begin{aligned} \frac{1}{\sqrt{N}}+O(\frac{1}{N}), \end{aligned}$$

(A.8)

and thus our guess is equal to the true root in the limit of $N\longrightarrow \infty$. Thus,

$$\begin{aligned} j^*= \frac{\sqrt{N}-1}{2}+\frac{1}{6\sqrt{N}} + O(N^{-1}) \end{aligned}$$

(A.9)

is the collective spin space with the largest degeneracy, up to error $O(N^{-1})$. $\square$

Proof of equation (31)

We begin by considering the ratio:

$$\begin{aligned} \frac{d_{j^*}}{d_{j^*+1}}= & {} \frac{N! (2j^*+1)}{(N/2 - j^*)!(N/2+j^*+1)!}\cdot \frac{(N/2 - j^*-1)!(N/2+j^*+2)!}{N! (2j^*+3)}\nonumber \\ \end{aligned}$$

(A.10)

$$\begin{aligned}= & {} \frac{2j^*+1}{2j^*+3}\cdot \frac{N/2+j^*+2}{N/2-j^*} \end{aligned}$$

(A.11)

$$\begin{aligned}= & {} (1-\frac{2}{2j^*+3})\cdot \frac{1+\frac{2j^*}{N}+\frac{4}{N}}{1-\frac{2j^*}{N}} \end{aligned}$$

(A.12)

$$\begin{aligned}= & {} (1-\frac{2}{2j^*+3})(1+\frac{2j^*}{N}+\frac{4}{N})(1+\frac{2j^*}{N}+(\frac{2j^*}{N})^2+O(((j^*)/N)^3))\nonumber \\ \end{aligned}$$

(A.13)

where the last factor is a geometric series expansion with a ratio of $\frac{2j^*}{N}$. Grou** by powers, using $j^*=O(\sqrt{N})$, we have:

$$\begin{aligned} 1+[-\frac{2}{2j^*+3}+\frac{2j^*}{N}+\frac{2j^*}{N}]+[\frac{4}{N}-2\frac{4j^*}{(2j^*+3)N}+2(\frac{2j^*}{N})^2]+O(N^{-3/2})\nonumber \\ \end{aligned}$$

(A.14)

Now we utilize $j^*=\frac{\sqrt{N}}{2}-\frac{1}{2}+\frac{1}{6\sqrt{N}}$ to evaluate the above:

$$\begin{aligned}= & {} 1+0+[-\frac{2}{N}+\frac{4}{N}-\frac{4\sqrt{N}}{\sqrt{N}N}+2(\frac{1}{\sqrt{N}})^2]+O(N^{-3/2}) \end{aligned}$$

(A.15)

So the ratio of the degeneracies is $1+O(N^{-3/2})$. $\square$

Proof of Strong Support for $0\le j\le O(\sqrt{N})$

Recall the computation of the relative population between the maximal angular momentum subspace and its neighbor and consider the case when the leading term will contribute to the ratio of neighboring values of j. We found that when $\frac{2j}{N}\ll 1$,

$$\begin{aligned} \frac{d_j}{d_{j+1}}=1+[\frac{4j}{N}-\frac{2}{2j+3}]+O(N^{-1}), \end{aligned}$$

(A.16)

which cancelled when $j=j^*$ since $j^*=\frac{\sqrt{N}}{2}+O(1)$.

Suppose we still have $\frac{2j}{N}\ll 1$, but now we consider a subspace nearby the maximal angular momentum space, such that $j=j^*+\Omega (\sqrt{N})$. The ratio for this value of j is then $1+\Omega (N^{-1/2})$. This ratio will remain valid for increasing values of j, so long as $\frac{2j}{N}\ll 1$.

To find the ratio of the degeneracies of the next nearest neighbors, we apply this procedure twice, finding

$$\begin{aligned} \frac{d_j}{d_{j+2}}=(1+\Omega (1/\sqrt{N}))^2. \end{aligned}$$

(A.17)

To continue this argument to further subspace degeneracies, listed within the set $\{d_j,d_{j+1},\ldots , d_{j+O(\sqrt{N})}\}$, we can write a geometric series taken at the infinity limit since the other term will contribute a much smaller portion to the sum. Thus, this series limits to

$$\begin{aligned} \frac{1}{1-\frac{1}{1+\Omega (1/\sqrt{N})}}=O(\sqrt{N}). \end{aligned}$$

(A.18)

We have shown that while the total number of allowed values for j is O(N), the fractional contribution contained in this region is only $O(N^{-1/2})$. Thus, we have that of the $2^N$ possible angular momentum states, most of them are contained within the lowest, smallest values of j, $O(\sqrt{N})$ angular momentum subspaces. $\square$

Proof of equation (45)

We will make use of the following summation formulae,

$$\begin{aligned} \sum _{i=1}^{n} i&= \frac{n(n+1)}{2}\\ \sum _{i=1}^{n} i^2&= \frac{n(n+1)(2n+1)}{6}\\ \sum _{i=1}^{n} i^3&= \frac{n^2(n+1)^2}{4}. \end{aligned}$$

Recall that the trace of the square of the coupling matrix can be written exactly as

$$\begin{aligned} {\text {tr}}L(j,k)^2 = 2 \sum _{\alpha =1}^{n} l_\alpha (j,k)^2 = 2 \sum _{\alpha =1}^{n} \big (2\alpha j-\alpha (\alpha -1)\big )\big (k'-\alpha +1\big ). \end{aligned}$$

(A.19)

Now, grou** the summand by orders of $\alpha$ we have

$$\begin{aligned} l_\alpha ^2(j,k) = \alpha ^3 - \alpha ^2(2j+1+k'+1) +\alpha (2j+1)(k'+1). \end{aligned}$$

(A.20)

And so

$$\begin{aligned} {\text {tr}}L(j,k)^2&= \frac{1}{2}\left| {\mathcal {B}}_{j,k} \right| ^4 - \frac{2}{3}\left| {\mathcal {B}}_{j,k} \right| ^3(2j+k'+7/2) + \left| {\mathcal {B}}_{j,k} \right| ^2(4j+2k'+2jk'+7/2)\nonumber \\&\quad - \frac{2}{3}\left| {\mathcal {B}}_{j,k} \right| (3jk' + 4j + 2k + 5/2). \end{aligned}$$

(A.21)

The variance is then, by definition,

$$\begin{aligned} {\text {Var}}(\Lambda (j,k))&= \frac{1}{2}\left| {\mathcal {B}}_{j,k} \right| ^3 -\frac{1}{3}\left| {\mathcal {B}}_{j,k} \right| ^2(2k'+4j+7)\nonumber \\&\quad + \left| {\mathcal {B}}_{j,k} \right| (2jk'+2k'+4j+7/2) - \frac{1}{3}(6jk' + 8j + 4k' + 5). \end{aligned}$$

(A.22)

$\square$

Proof of equation (47)

We begin by transforming the entry values into a continuous function of $\alpha$ so that we may differentiate it. We will use Perron–Frobenius since we have a non-negative matrix and so can bound the maximal eigenvalue by the maximal row sum. To this end, we focus on maximizing a single $l_\alpha (j,k)$ entry and double it since the true maximum will occur within one entry of the optimal continuous value choice for $\alpha$ and there are two entries in that row.

Differentiating this we have:

$$\begin{aligned} \frac{d}{d\alpha }\left[ \sqrt{\alpha }\sqrt{2j+1-\alpha }\sqrt{k'-\alpha +1}\right] =\frac{3\alpha ^2-4\alpha +2j(-2\alpha +k'+1)-2\alpha k'+k'+1}{2\sqrt{\alpha (\alpha -2j-1)(\alpha -k'-1)}}\nonumber \\ \end{aligned}$$

(A.23)

and so this is optimized when:

$$\begin{aligned} \alpha= & {} \frac{1}{3}\left( 2j+k'+2\pm \sqrt{4j^2-2jk'+2j+(k')^2+k'+1}\right) \end{aligned}$$

(A.24)

$$\begin{aligned}= & {} \frac{1}{3}\left( 2j+k'+2\pm \sqrt{(2j+\frac{1}{2})^2+(k'+\frac{1}{2})^2-2jk'+\frac{1}{2}}\right) \end{aligned}$$

(A.25)

In the above, we must exclude the positive sign choice since this results in $\alpha \ge \left| {\mathcal {B}}_{j,k} \right|$, which is beyond the domain for $\alpha$. With the negative sign choice we note that $\alpha$ is linear in j and $k'$ to first order, so we remove the 1 shifts in our objective function. With this, we have:

$$\begin{aligned} \sqrt{j^2-(\alpha -j)^2}\sqrt{k'-\alpha }= & {} \sqrt{2j\alpha -\alpha ^2}\sqrt{k'-\alpha } \end{aligned}$$

(A.26)

$$\begin{aligned}= & {} \sqrt{\alpha }\sqrt{2j-\alpha }\sqrt{k'-\alpha } \end{aligned}$$

(A.27)

Solving for the roots again using this simplified expression provides:

$$\begin{aligned} \alpha= & {} \frac{1}{3}(2j+k'-\sqrt{4j^2-2jk'+(k')^2})+O(\sqrt{j}+\sqrt{k}) \end{aligned}$$

(A.28)

$$\begin{aligned}= & {} \frac{1}{3}(2j+k'-\sqrt{(2j+k')^2-6jk'}) \end{aligned}$$

(A.29)

$$\begin{aligned}= & {} \frac{1}{3}(2j+k'-(2j+k')\sqrt{1-\frac{6jk'}{(2j+k')^2}}) \end{aligned}$$

(A.30)

$$\begin{aligned}= & {} \frac{1}{3}(2j+k')(1-\sqrt{1-\frac{6jk'}{(2j+k')^2}}) \end{aligned}$$

(A.31)

Observe that $\max _{j,k'} \frac{6jk'}{(2j+k')^2}=\frac{3}{4}$ where $k'=2j$, and $\min _{j,k'}\frac{6jk'}{(2j+k')^2}=0$ when one is constant and the other approaches infinity. This means that:

$$\begin{aligned} 0< \alpha \le \frac{1}{6}(2j+k') \end{aligned}$$

(A.32)

Putting this into our expression for the largest eigenvalue, being sure to include the factor of two due to there being a second entry, provides:

$$\begin{aligned} \max \Lambda (j,k)< & {} 2 \sqrt{\frac{1}{6}(2j+k')}\sqrt{2j}\sqrt{k'} \end{aligned}$$

(A.33)

$$\begin{aligned}= & {} \frac{2}{\sqrt{3}}\sqrt{(2j+k')jk'}+O(j^{3/4}+(k')^{3/4}) \end{aligned}$$

(A.34)

This expression is mostly relevant when $2j\approx k'$. We now move to the cases of $2j\ll k'$ and $k'\ll 2j$. Returning to our prior expression this is:

$$\begin{aligned}&\sqrt{\frac{1}{3}(2j+k')(1-\sqrt{1-\frac{6jk'}{(2j+k')^2}})}\sqrt{2j-\frac{1}{3}(2j+k')(1-\sqrt{1-\frac{6jk'}{(2j+k')^2}})} \nonumber \\ \end{aligned}$$

(A.35)

$$\begin{aligned}&\times \sqrt{k'-\frac{1}{3}(2j+k')(1-\sqrt{1-\frac{6jk'}{(2j+k')^2}})} \end{aligned}$$

(A.36)

$$\begin{aligned}= & {} \sqrt{\frac{2}{27}(8j^3(\sqrt{1-\frac{6jk}{(2j+k)^2}}-1)+6j^2k+k^3(\sqrt{1-\frac{6jk}{(2j+k)^2}}-1)+3jk^2)}\nonumber \\ \end{aligned}$$

(A.37)

Taking a series expansion of this and doubling for there being two entries we have:

$$\begin{aligned} \max \Lambda (j,k)\le & {} {\left\{ \begin{array}{ll} 2\sqrt{j^2k'-j^3+\frac{j^4}{4k'}} &{} 2j\ll k'\\ 2\sqrt{\frac{j(k')^2}{2}-\frac{1}{8}(k')^3+\frac{1}{128}\frac{(k')^4}{j}} &{} k'\ll 2j \end{array}\right. } \end{aligned}$$

(A.38)

$$\begin{aligned}\approx & {} {\left\{ \begin{array}{ll} 2[j\sqrt{k'}-\frac{1}{2}\frac{j^2}{\sqrt{k}}+\frac{1}{8}\frac{j^4}{k^{5/2}}+O(j^5/(k')^{7/2})] &{} 2j\ll k'\\ 2[\frac{1}{\sqrt{2}}k'\sqrt{j}-\frac{1}{8\sqrt{2}}\frac{(k')^2}{\sqrt{j}}+\frac{1}{512}\frac{(k')^4}{j^{5/2}}+O((k')^5/j^{7/2})] &{} k'\ll 2j \end{array}\right. }\nonumber \\ \end{aligned}$$

(A.39)

$\square$

Proof of equation (55)

This relation can be seen as the weighted average of averages and can thus be derived as follows:

$$\begin{aligned} \left<\Lambda (k)^t\right>&= \frac{1}{D_k}\sum _{j} d_j \left| {\mathcal {B}}_{j,k} \right| \left<\Lambda (j,k)^t\right>\\&=\frac{1}{D_k}\sum _{j} d_j \left| {\mathcal {B}}_{j,k} \right| \frac{{\text {tr}}(L(j,k)^t}{\left| {\mathcal {B}}_{j,k} \right| }\\&=\frac{1}{D_k}\sum _{j} d_j {\text {tr}}(L(j,k)^t). \end{aligned}$$

$\square$

Lemma 1

The degeneracies in our system satisfy:

$$\begin{aligned} \frac{d_j}{2^N}=O(\frac{1}{N}) \end{aligned}$$

(A.40)

for all allowed values of j.

Proof

Since $d_j < d_{j^*}$ for each j, we particularize to $j=j^*$. Then, taking only the leading term of $j^* = \sqrt{N}/2$, we have

$$\begin{aligned} d_{j^*} = \frac{\sqrt{N}/2 + 1}{N/2 + \sqrt{N}/2 + 1}\left( {\begin{array}{c}N\\ N/2 + \sqrt{N}/2 + 1\end{array}}\right) . \end{aligned}$$

(A.41)

Focusing on the first factor,

$$\begin{aligned} \frac{2}{\frac{N+1}{\sqrt{N} + 1} + 1 } = O(1/\sqrt{N}). \end{aligned}$$

(A.42)

Since $k = N/2 + \sqrt{N}/2 + 1$, we have that $\left| N/2 - k \right| = o(n^{2/3})$, we can utilize the following asymptotic equivalence relation [58]:

$$\begin{aligned} \left( {\begin{array}{c}N\\ k\end{array}}\right) \sim \frac{2^N}{\sqrt{N\pi /2}}e^{-(N-2k)^2 / (2N)}. \end{aligned}$$

(A.43)

Then, using the fact that $N-2k = \sqrt{N} - 2$, we find that

$$\begin{aligned} e^{-(N-2k)^2 / (2N)}&= e^{-(\sqrt{N}-2)^2/(2N)}\nonumber \\&= e^{-1/2 + 2/\sqrt{N} - 2/N}\nonumber \\&= \frac{1}{\sqrt{e}} \big (1 + O(1/\sqrt{N})\big ). \end{aligned}$$

(A.44)

Putting together the leading term with the asymptotic equivalence relation, we find

$$\begin{aligned} \left( {\begin{array}{c}N\\ N/2 + \sqrt{N}/2 + 1\end{array}}\right)&\sim \frac{2^N}{\sqrt{N\pi e/2}} \big (1 + O(1/\sqrt{N})\big )\nonumber \\&= O(2^N/\sqrt{N}). \end{aligned}$$

(A.45)

This finally implies

$$\begin{aligned} d_{j^*} = O(2^N / N), \end{aligned}$$

(A.46)

and thus it hold that for all allowed j,

$$\begin{aligned} d_{j} 2^{-N} = O(1/N). \end{aligned}$$

(A.47)

$\square$

Proof of equation (57)

In order to derive our result, we particularize to $k>N$, since this fixes $D_k = 2^N$ and $k > N/2 + j$ is true for each value of j. Then,

$$\begin{aligned} {\text {Var}}(\Lambda (k)) = \frac{1}{2^N}\sum _{j} d_j {\text {tr}}(L(j,k)^2). \end{aligned}$$

(A.48)

Recall the trace of the square of a coupling matrix is given by,

$$\begin{aligned} {\text {tr}}(L(j,k)^2)&= \frac{1}{2}\left| {\mathcal {B}}_{j,k} \right| ^4 - \frac{1}{3}\left| {\mathcal {B}}_{j,k} \right| ^3(2k'+4j+7)+ \left| {\mathcal {B}}_{j,k} \right| ^2(2jk'+2k'+4j+7/2)\\&\quad - \frac{1}{3}\left| {\mathcal {B}}_{j,k} \right| (6jk' + 8j + 4k' + 5) \end{aligned}$$

Using the fact now that $\left| {\mathcal {B}}_{j,k} \right| = 2j+1$ and $k' = k - N/2 + j$, we find that

$$\begin{aligned} {\text {tr}}(L(j,k)^2)&= k\big (\frac{8}{3}j^3 + 4j^2 + \frac{4}{3}j\big )- N\big (\frac{4}{3}j^3 + 2j^2 + \frac{2}{3}j\big )+ \big (\frac{4}{3}j^3 + 2j^2 + \frac{2}{3}j\big ). \end{aligned}$$

We focus on only the terms of order k, thus the dominant part of the expression we wish to analyze is given by

$$\begin{aligned} \frac{4}{3}k(2j^3 + 3j^2 + j). \end{aligned}$$

(A.49)

It remains to determine the order k contribution to the entire variance, upon averaging over the degeneracies,

$$\begin{aligned} {\text {Var}}(\Lambda (k)) = \frac{4}{3} k \sum _{j} \frac{d_j}{2^N} (2j^3 + 3j^2 + j) + \cdots , \end{aligned}$$

(A.50)

where the terms of order $k^0$ will be dropped moving forward.

In order to make the sum over j tractable, we make use of Lemma 1,

$$\begin{aligned} \frac{d_j}{2^N} = O\big (\frac{1}{N}\big ). \end{aligned}$$

(A.51)

Given that the strong support of the weighting function is from 0 to $O(\sqrt{N})$, we have,

$$\begin{aligned} \sum _{j} \frac{4 k d_j}{2^N} \frac{2j^3 + 3j^2 - 2j}{3}&= \frac{4k}{3} \sum _{j} O\big (\frac{1}{N}\big ) (2j^3 + 3j^2 - 2j)\nonumber \\&\approx O\big (\frac{k}{N}\big ) \sum _{j=0}^{O(\sqrt{N})}(2j^3 + 3j^2 - 2j)\nonumber \\&= O\big (\frac{k}{N}\big ) O(N^2)\nonumber \\&= O(Nk). \end{aligned}$$

(A.52)

$\square$