Quantum de Finetti Theorems

The bosonic quantum de Finetti theorem represents an infinitely extendible hierarchy of symmetric density matrices as a mixture of pure tensor powers. A finite-dimensional quantitative version controls the error for fixed marginals of a large finite system. This is a representation of correlations by a probability measure; it does not assert that every bosonic state is a single condensate.

Required background. Symmetric tensor products supplies the bosonic subspace. Partial trace supplies consistency of marginals. Mean-field limits gives the dynamical use of factorized hierarchies.

Consistent bosonic hierarchies are mixtures of powers

Let $\mathfrak h$ be a separable complex Hilbert space. Suppose that for every $k\geq1$, $\gamma^{(k)}$ is a positive trace-one operator supported on $\operatorname{Sym}^k\mathfrak h$, and

$$\operatorname{Tr}_{k+1}\gamma^{(k+1)}=\gamma^{(k)}.$$

The strong bosonic de Finetti theorem states that there is a unique probability measure $\mu$ on unit one-particle rays such that

$$\gamma^{(k)} =\int |u^{\otimes k}\rangle\langle u^{\otimes k}|\,d\mu(u) \qquad(k\geq1).$$

The integral is a trace-class Bochner integral. Equivalently, use the unique phase-invariant measure on the unit sphere. Phase must be quotiented or averaged because all displayed projectors are unchanged by $u\mapsto e^{i\theta}u$.

Consistency for every order is essential. A single finite entangled state need not be a mixture of powers. Also, support on the symmetric subspace is stronger than merely commuting with permutations. General exchangeable hierarchies use mixtures of powers of one-particle density matrices; the bosonic support restriction reduces them to pure one-particle projectors.

A finite-dimensional approximation has an explicit measure

Now let $\dim\mathfrak h=d<\infty$, and let $\Gamma_N$ be any trace-one state on $\operatorname{Sym}^N\mathfrak h$. Denote its marginals by $\gamma_N^{(k)}$. Let $du$ be normalized unitary-invariant measure on the unit sphere and define

$$D_N=\dim\operatorname{Sym}^N\mathbb C^d =\binom{N+d-1}{d-1},$$ $$d\mu_N(u)=D_N \langle u^{\otimes N},\Gamma_Nu^{\otimes N}\rangle\,du.$$

This is a probability measure. Indeed, unitary invariance and irreducibility of the symmetric-power representation give

$$D_N\int |u^{\otimes N}\rangle \langle u^{\otimes N}|\,du=I_{\operatorname{Sym}^N}.$$

Taking the trace against $\Gamma_N$ proves normalization.

The quantitative theorem gives, for $1\leq k\leq N$,

$$\left\|\gamma_N^{(k)} -\int|u^{\otimes k}\rangle \langle u^{\otimes k}|\,d\mu_N(u)\right\|_1 \leq\min\left\{2,\frac{4dk}{N}\right\}.$$

Here $\|\cdot\|_1=\operatorname{Tr}|\cdot|$, without the factor $1/2$ sometimes used for trace distance. This bound is the convenient consequence (2.7) of Lewin, Nam, and Rougerie’s Theorem 2.1. Their proof, based on finite symmetric-power identities and positivity, is imported; the exact measure and error dependence are part of the statement.

The bound is useful at fixed $d,k$ with $N$ large. It does not tend to zero if the dimension or observed particle count grows too rapidly with $N$.

Compactness turns the finite estimate into the infinite theorem

In finite dimension, apply the quantitative theorem to $\Gamma_N=\gamma^{(N)}$. Consistency ensures that its $k$-body marginal is exactly $\gamma^{(k)}$ whenever $N\geq k$.

The unit-ray space is compact. Extract a weakly convergent subsequence of the probability measures $\mu_N$, with limit $\mu$. For each fixed $k$, matrix elements of $|u^{\otimes k}\rangle\langle u^{\otimes k}|$ are continuous functions of the ray. The integral projectors therefore converge in finite-dimensional trace norm. The error bound tends to zero, proving the representation.

For uniqueness, the hierarchy determines every polynomial moment in the matrix entries of $|u\rangle\langle u|$. These functions form a self-adjoint algebra containing constants and separating rays. Stone–Weierstrass makes them dense in the continuous functions on the compact ray space, so two representing measures must agree.

The separable infinite-dimensional theorem requires additional localization and tightness arguments; compactness of the unit sphere is no longer available. The stated trace-one consistency hypotheses still give the sphere representation, as explained in the cited treatment.

A rigorous mean-field ground-energy consequence

In finite dimension take the Hamiltonian from mean-field limits, with one-particle $h$ and bounded exchange-symmetric pair operator $V$. For any bosonic $\Gamma_N$,

$$\frac{\operatorname{Tr}(H_N\Gamma_N)}N =\operatorname{Tr}(h\gamma_N^{(1)}) +\frac12\operatorname{Tr}(V\gamma_N^{(2)}).$$

Define the Hartree energy

$$\mathcal E_{\mathrm H}(u) =\langle u,hu\rangle+ \frac12\langle u\otimes u,V(u\otimes u)\rangle, \qquad \|u\|=1.$$

The same measure $\mu_N$ approximates both marginals. Trace duality and the finite theorem imply

$$\frac{\operatorname{Tr}(H_N\Gamma_N)}N \geq\inf_{\|u\|=1}\mathcal E_{\mathrm H}(u) -\frac{4d}{N}(\|h\|+\|V\|).$$

A pure tensor trial state gives the opposite variational upper bound with no error. Thus the ground energy per particle converges to the Hartree minimum, with the displayed finite-dimensional estimate. The proof does not require that the minimizer be unique or that all approximate ground states select one pure condensate.

For infinite-dimensional unbounded Hamiltonians, energy localization and control of the discarded modes are additional steps. One cannot insert $d=\infty$ into the finite bound.

Exercises

A cat state has mixed limiting marginals. Let $u,v$ be orthogonal unit vectors and $\Psi_N=(u^{\otimes N}+v^{\otimes N})/\sqrt2$. Find its fixed-$k$ marginal for $N>k$.

Solution

Every cross term contains the factor $\langle v,u\rangle^{N-k}=0$. Hence the marginal is one half of each pure tensor projector. The de Finetti measure is $(\delta_{[u]}+\delta_{[v]})/2$, not one point mass. Bosonic symmetry alone does not establish complete condensation into a specified orbital.

Loss of mass in infinite dimension. Let $(e_N)$ be orthonormal and take $\Gamma_N=|e_N^{\otimes N}\rangle\langle e_N^{\otimes N}|$. What do fixed marginals converge to against compact operators?

Solution

For each $k$, the vectors $e_N^{\otimes k}$ are orthonormal, so the expectations of compact operators tend to zero. The weak-star trace-class limit is the zero operator, despite every marginal having trace one. There is no trace-norm convergence and no trace-one consistent limit hierarchy to which the strong theorem applies. Additional compactness or energy confinement is needed to prevent escape into new modes.

References

• M. Lewin, P. T. Nam, and N. Rougerie, “Remarks on the quantum de Finetti theorem for bosonic systems,” arXiv:1310.2200, version 3, 2014, especially Theorem 2.1 and Section 1, author text.