An operational measure for squeezing

Consider a bosonic system with n-modes, each of which is characterised by a pair of canonical variables $\{{Q}_{k},{P}_{k}\}$. Setting $R={({Q}_{1},{P}_{1},\ldots ,{Q}_{n},{P}_{n})}^{T}$ the canonical commutation relations (CCRs) take on the form $[{R}_{k},{R}_{l}]={\rm{i}}{\sigma }_{{kl}}$ with the standard symplectic form

$$\begin{eqnarray*}\sigma =\underset{i=1}{\overset{n}{\displaystyle \bigoplus }}\left(\begin{array}{cc}0 & 1\\ -1 & 0\end{array}\right).\end{eqnarray*}$$

Since it will sometimes be convenient, we also introduce another basis of the canonical variables: let $\tilde{R}={({Q}_{1},{Q}_{2},\ldots ,{Q}_{n},{P}_{1},{P}_{2},\ldots ,{P}_{n})}^{T}$, then the symplectic CCRs take on the form $[{\tilde{R}}_{k},{\tilde{R}}_{l}]={{\rm{i}}{J}}_{{kl}}$ with the symplectic form

$$\begin{eqnarray*}J=\left(\begin{array}{cc}0 & {{\mathbb{1}}}_{n}\\ -{{\mathbb{1}}}_{n} & 0\end{array}\right).\end{eqnarray*}$$

Clearly, J and σ differ only by a permutation, since R and $\tilde{R}$ differ only by a permutation. From functional analysis, it is well-known that the operators Q_k and P_k cannot be represented by bounded operators on a Hilbert space. In order to avoid complications associated to unbounded operators, it is usually easier to work with a representation of the CCR-relations on some Hilbert space ${ \mathcal H }$, instead. The standard representation is known as the Schrödinger representation and defines the Weyl system, a family of unitaries ${W}_{\xi }$ with $\xi \in {{\mathbb{R}}}^{2n}$ and

$$\begin{eqnarray*}&&{W}_{\xi }:= \exp ({\rm{i}}\xi \sigma R),\quad \xi \in {{\mathbb{R}}}^{2n}\end{eqnarray*}$$

fulfiling the Weyl relations ${W}_{\xi }{W}_{\eta }={\exp }^{-{\rm{i}}/2\xi \sigma \eta }{W}_{\xi +\eta }$ for all $\xi ,\eta$. Such a system is unique up to isomorphism under further assumptions of continuity and irreducibility as obtained by the Stone–von Neumann theorem. Given ${W}_{\xi }$ it is important to note that

$$\begin{eqnarray}&&{W}_{\xi }{R}_{k}{W}_{\xi }^{* }={R}_{k}+{\xi }_{k}{\mathbb{1}}\qquad \forall \xi \in {{\mathbb{R}}}^{2n}.\end{eqnarray} \tag{ 4 }$$

In this paper, we will not use many properties of the Weyl system, since instead, we can work with the much simpler moments of the state: given a quantum state $\rho \in {{ \mathcal S }}_{1}({L}^{2}({{\mathbb{R}}}^{2n}))$ (trace-class operators on L²), its first and second centred moments are given by

$$\begin{eqnarray}&&{d}_{k}:= \mathrm{tr}(\rho {R}_{k}),\end{eqnarray} \tag{ 5 }$$

$$\begin{eqnarray}&&{\gamma }_{{kl}}:= \mathrm{tr}(\rho {\{{R}_{k}-{d}_{k}{\mathbb{1}},{R}_{l}-{d}_{l}{\mathbb{1}}\}}_{+})\end{eqnarray} \tag{ 6 }$$

with $\{\cdot ,\,\cdot \}{}_{+}$ the regular anticommutator. We will write Γ instead of γ for the covariance matrix, if we work with $\tilde{R}$ instead of R. Again, a simple permutation relates the two.

An important question one can ask is when a matrix γ can occur as a covariance matrix of a quantum state. The answer is given by Heisenberg's principle, which here takes the form of a matrix inequality:

Proposition 2.1. Let $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$, then there exists a quantum state $\rho$ with covariance matrix $\gamma$ if and only if

$$\begin{eqnarray*}&&\gamma \geqslant {\rm{i}}\sigma ,\end{eqnarray*}$$

where $\geqslant$ denotes the standard partial order on matrices (i.e. $\gamma \geqslant {\rm{i}}\sigma$ if $\gamma -{\rm{i}}\sigma$ is positive semidefinite). Note that we leave out the usual factor of ${\hslash }/2$ to simplify notation.

Another question one might ask is when a covariance matrix belongs to a pure quantum state. This question cannot be answered without more information about the higher order terms If we however require the state to be uniquely determined by its first and second moments, i.e. if we consider the so called Gaussian states, we have an answer (see [ASI04]):

Proposition 2.2. Let $\rho$ be an $n$-mode Gaussian state (i.e. completely determined by its first and second moments), then $\rho$ is pure if and only if $\det ({\gamma }_{\rho })=1$.

A very important set of operations on a quantum system are those, that leave the CCRs invariant, i.e. linear transformations S such that $[{{SR}}_{k},{{SR}}_{l}]={\rm{i}}{\sigma }_{{kl}}$. Such transformations are called symplectic transformations.

Definition 2.3. Given a symplectic form σ on ${{\mathbb{R}}}^{2n\times 2n}$, the set of matrices $S\subset {{\mathbb{R}}}^{2n\times 2n}$ such that ${S}^{T}\sigma S=\sigma$ is called the linear symplectic group and is denoted by ${Sp}(2n,{\mathbb{R}},\sigma )$.

We will usually drop both σ and ${\mathbb{R}}$ in the description of the symplectic group since this will be clear from the context. The linear symplectic group is a Lie group and as such contains a lot of structure. For more information on the linear symplectic group and its connection to physics, we refer the reader to [Gos06, MS98] chapter 2. An overview for physicists is also found in [Arv+95a]. All of the following can be found in that paper:

Definition 2.4. Let $O(2n,{\mathbb{R}})$ be the real orthogonal group, Then we define the following three subsets of ${Sp}(2n)$:

$$\begin{eqnarray*}\begin{array}{rcl}K(n) & := & {Sp}(2n,{\mathbb{R}})\cap O(2n,{\mathbb{R}}),\\ Z(n) & := & \{{{\mathbb{1}}}_{2(j-1)}\oplus \mathrm{diag}({s}_{i},{s}_{i}^{-1})\oplus {{\mathbb{1}}}_{2(n-(j+1))}| s\geqslant 0,j=1,\ldots ,n\},\\ {\rm{\Pi }}(n) & := & \{S\in {Sp}(2n,{\mathbb{R}})| S\geqslant 0\}\mathrm{.}\end{array}\end{eqnarray*}$$

The first subset is the maximally compact subgroup of ${Sp}(2n)$, the second subset is the subset of single-mode-squeezers. It generates the multiplicative subgroup ${ \mathcal A }(2n)$, a maximally abelian subgroup of ${Sp}(2n)$. The third set is the set of positive definite symplectic matrices.

In addition, since ${Sp}(2n)$ is a Lie group, it possesses a Lie algebra. Let us collect a number of relevant facts about the Lie algebra and some subsets:

Proposition 2.5. The Lie algebra ${\mathfrak{sp}}(2n)$ of ${Sp}(2n)$ is given by

$$\begin{eqnarray*}&&{\mathfrak{sp}}(2n):= \{T\in {{\mathbb{R}}}^{2n\times 2n}| \sigma T+T\sigma =0\}\end{eqnarray*}$$

together with the commutator as Lie bracket. Certain other Lie algebras or subsets of Lie algebras are of relevance to us:

• (1)  
  ${\mathfrak{so}}(2n):= \{A\in {{\mathbb{R}}}^{2n\times 2n}| A+{A}^{T}=0\}$ the Lie algebra of ${SO}(2n)$.
• (2)  
  ${\mathfrak{k}}(n):= \{A\in {{\mathbb{R}}}^{2n\times 2n}| A=\left(\begin{array}{cc}a & b\\ -b & a\end{array}\right),a=-{a}^{T},b={b}^{T}\}$ the Lie algebra of $K(n)$.
• (3)  
  $\pi (n):= \{A\in {{\mathbb{R}}}^{2n\times 2n}| A=\left(\begin{array}{cc}a & b\\ b & -a\end{array}\right),a={a}^{T},b={b}^{T}\}$ the subspace of the Lie algebra ${\mathfrak{sp}}(2n)$ corresponding to ${\rm{\Pi }}(n)$.

Since the Lie algebra is a vector space, it is spanned by a set of vectors, the generators. A standard decomposition is given by taking the generators of ${\mathfrak{k}}(n)$, the so called passive transformations as one part and the generators of $\pi (n)$, the so called active transformations as the other part. That these two sets together determine the Lie algebra completely can be seen with the polar decomposition:

Proposition 2.6 For every symplectic matrix $S\in {Sp}(2n)$ there exists a unique $U\in K(n)$ and a unique $P\in {\rm{\Pi }}(n)$ such that $S={UP}$.

A basis for the Lie algebras ${\mathfrak{k}}(n)$ and $\pi (n)$ therefore characterises the complete Lie algebra ${\mathfrak{sp}}(2n)$. Elements of the Lie algebras are also called generators and a basis of generators therefore fixes the Lie algebra. Via the polar decomposition, this implies that they also generate the whole Lie group. We will need a set of generators ${g}_{{ij}}^{(p)}\in {\mathfrak{k}}(n)$ and ${g}_{{ij}}^{(a)}\in \pi (n)$ later on, which we will fix via the metaplectic representation:

Proposition 2.7 Let ${W}_{\xi }$ be the continuous irreducible Weyl system defined above and let $S\in {Sp}(2n)$. Then there exists an up to a phase unique unitary ${U}_{S}$ with

$$\begin{eqnarray*}&&\forall \xi \,:\quad {U}_{S}{W}_{\xi }{U}_{S}^{\dagger }={W}_{S\xi }.\end{eqnarray*}$$

Since we have the liberty of a phase, this is not really a representation of the symplectic group, but of its two-fold cover, the metaplectic group. We can also study the generators of this representation, which are given by $1/2{\{{R}_{k},{R}_{l}\}}_{+}$.

For the reader familiar with annihilation and creation operators, if we denote by ${a}_{i},{a}_{i}^{\dagger }$ the annihilation and creation operators of the n bosonic modes, the generators of the metaplectic representation are given by

$$\begin{eqnarray}&&{G}_{{ij}}^{p(1)}:= {\rm{i}}({a}_{j}^{\dagger }{a}_{{i}}-{a}_{i}^{\dagger }{a}_{j})\qquad {G}_{{ij}}^{p(2)}:= {a}_{j}^{\dagger }{a}_{j}+{a}_{j}^{\dagger }{a}_{{i}},\end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray}&&{G}_{{ij}}^{a(3)}:= {\rm{i}}({a}_{j}^{\dagger }{a}_{{i}}^{\dagger }-{a}_{{\rm{i}}}{a}_{j})\qquad {G}_{{ij}}^{a(4)}:= {a}_{{i}}^{\dagger }{a}_{j}^{\dagger }+{a}_{{i}}{a}_{j},\end{eqnarray} \tag{ 8 }$$

where the p stands for 'passive' and the a for 'active'. The passive generators are also frequently called linear transformations in the literature (see [Kok+07]). We can now define a set of generators of the symplectic group ${Sp}(2n)$ by using the set of metaplectic generators G_ij above and take corresponding generators g_ij in the Lie algebra ${\mathfrak{sp}}(2n)$ in a consistent way. As one would expect from the name, the passive metaplectic generators correspond to a set of passive generators of ${\mathfrak{k}}(n)$ and the set of active metaplectic generators corresponds to a set of active generators of $\pi (n)$. The details of the correspondence are irrelevant (they are explicitly spelled out in equation (6.6b) in [Arv+95a]), except for the fact that the set ${G}_{{ii}}^{a(3)}$, i = 1,...,n corresponds to the generators ${g}_{{ii}}^{a(3)}$ generating matrices in Z_n.

Given a Hamiltonian, the associated time evolution corresponds to a path on the Lie group: for a (sufficiently regular) path $\gamma \,:[0,1]\to {Sp}(2n)$ we can find a function $A(t)\in {\mathfrak{sp}}(2n)$ such that

$$\begin{eqnarray}&&{\gamma }^{\prime }(t)=A(t)\gamma (t).\end{eqnarray} \tag{ 9 }$$

Instead of directly studying Hamiltonians with time-dependent coefficients as in equation (1), it is equivalent to study functions $A\,\,:\,[0,1]\to {\mathfrak{sp}}(2n)$.

There are a number of decompositions of the Lie group and its subgroup in addition to the polar decomposition. We will mostly be concerned with the so called Euler decomposition (sometimes called Bloch–Messiah decomposition) and Williamson's decomposition:

Proposition 2.8 Let $S\in {Sp}(2n)$, then there exist $K,{K}^{\prime }\in K(n)$ and $A\in { \mathcal A }(n)$ such that $S={{KAK}}^{\prime }$.

Proposition 2.9 Let $M\in {{\mathbb{R}}}^{2n\times 2n}$ be a positive definite matrix, then there exists a symplectic matrix $S\in {Sp}(2n,{\mathbb{R}})$ and a diagonal matrix $D\in {{\mathbb{R}}}^{n\times n}$ such that

$$\begin{eqnarray*}&&M={S}^{T}\tilde{D}S,\end{eqnarray*}$$

where $\tilde{D}=\mathrm{diag}(D,D)$ is diagonal. The entries of $D$ are also called symplectic eigenvalues.

In particular, for $M\in {\rm{\Pi }}(n)$, this implies that M has a symplectic square root. Since covariance matrices are always positive definite, this implies also that a Gaussian state is pure if and only if its covariance matrix is symplectic. Heisenberg's uncertainty principle has also a Williamson version:

Corollary 2.10. A positive definite matrix $M$ is a covariance matrix of a quantum state if and only if all symplectic eigenvalues are larger or equal to one.

The proof is simple and therefore omitted.

We have already noted that an important class of operations are those, which leave the CCR-relations invariant, namely the symplectic transformations. Given a quantum state ρ, the action of the symplectic group on the canonical variables R descends to a subgroup of unitary transformations on ρ via the metaplectic representation (see [Arv+95b]). Its action on the covariance matrix ${\gamma }_{\rho }$ of ρ is even easier: Given $S\in {Sp}(2n)$,

$$\begin{eqnarray}&&{\gamma }_{\rho }\mapsto {S}^{T}{\gamma }_{\rho }S.\end{eqnarray} \tag{ 10 }$$

In quantum optics, symplectic transformations can be implemented by the means of

• (1)  
  beam splitters and phase shifters, implementing operations in K(n) ([Rec+94])
• (2)  
  single-mode squeezers, implementing operations in Z(n).

Via the Euler decomposition, this implies that any symplectic transformation can be implemented (approximately) by a combination of those three elements.

Definition 2.11. An $n$-mode bosonic state $\rho$ is called squeezed, if its covariance matrix ${\gamma }_{\rho }$ possesses an eigenvalue $\lambda \lt 1$.

Especially in the early literature, squeezing is usually defined differently: a state ρ is squeezed if there exists a unitary transformation $K\in K(n)$ such that ${K}^{T}{\gamma }_{\rho }K$ has a diagonal entry smaller than one. This again comes from the physical definition of squeezed states being states where the Heisenberg uncertainty relations are satisfied with equality for at least one mode. These definitions however are well-known to be equivalent (see [SMD94]).

We will now define a first operational squeezing measure for symplectic transformations, which will later be used to define a measure for operational squeezing.

Definition 3.1. Define the function $F\ :{{\mathbb{R}}}^{2n\times 2n}\to {\mathbb{R}}$

$$\begin{eqnarray}&&F(A)=\displaystyle \sum _{i=1}^{n}\mathrm{log}({s}_{i}^{\downarrow }(A)),\end{eqnarray} \tag{ 11 }$$

where ${s}_{i}^{\downarrow }$ are the decreasingly ordered singular values of A.

Note that we sum only over half of the singular values. Restricting this function to symplectic matrices will yield an operational squeezing measure for symplectic transformations: recall that the symplectic group is generated by symplectic orthogonal matrices and single-mode squeezers. The orthogonal matrices are easy to implement and therefore will be considered a free resource. The squeezers have singular values s and ${s}^{-1}$ and they are experimentally hard to implement and should therefore be assigned a cost that depends on the squeezing parameter s. Using this, the amount of squeezing seems to be characterised by the largest singular values. Here, we quantify the amount of squeezing by a cost $\mathrm{log}(s)$, which can be seen as the interaction strength of the Hamiltonian needed to implement the squeezing.

Let us make this more precise: define the map

$$\begin{eqnarray*}\begin{array}{rcl}{\rm{\Delta }}\ :{Sp}(2n) & \to & \bigcup _{m\in {\mathbb{N}}}{Sp}{(2n)}^{\times m}\\ S & \mapsto & \bigcup _{m\in {\mathbb{N}}}\{({S}_{1},\ldots ,{S}_{m})| S={S}_{1}\cdots {S}_{m},{S}_{i}\in K(n)\cup Z(n)\}\mathrm{.}\end{array}\end{eqnarray*}$$

The image of Δ for a given symplectic matrix contains all possible ways to construct S as a product of matrices from K(n) or Z(n). We define:

Definition 3.2. Let $\overline{F}\ :{Sp}(2n)\to {\mathbb{R}}$ be a map defined via

$$\begin{eqnarray}&&\overline{F}(S):= \mathrm{log}\,\inf \left\{\displaystyle \prod _{i=1}^{m}{s}_{1}^{\downarrow }({S}_{i})\,| \,({S}_{1},\ldots ,{S}_{m})\in {\rm{\Delta }}(S)\right\}.\end{eqnarray} \tag{ 12 }$$

Proposition 3.3. If $S\in {Sp}(2n)$ then $F(S)=\overline{F}(S)$.

Proof. Let $S={{KAK}}^{\prime }$ be the Euler decomposition of S with $K,{K}^{\prime }\in K(n)$ and $A\in { \mathcal A }(n)$. Assume without loss of generality that $A=\mathrm{diag}({a}_{1},{a}_{1}^{-1},\ldots ,{a}_{n},{a}_{n}^{-1})$ and ${a}_{1}\geqslant {a}_{2}\geqslant \ldots \,\geqslant \,{a}_{n}\geqslant 1$ and define ${A}_{i}=\mathrm{diag}(1,\ldots ,1,{a}_{i},{a}_{i}^{-1},1,\ldots ,1)$. By construction $A={A}_{1}\cdots {A}_{n}$ and ${A}_{i}\in Z(n)$. Since $K,{K}^{\prime }\in K(n)$, $(K,{A}_{1},\ldots ,{A}_{n},{K}^{\prime })\in {\rm{\Delta }}(S)$. Using that ${s}_{i}^{\downarrow }(K)={s}_{i}^{\downarrow }({K}^{\prime })=1$ and the fact that the Euler decomposition is actually equivalent to the singular value decomposition of S, we obtain:

$$\begin{eqnarray*}&&\overline{F}(S)\leqslant \mathrm{log}\left({s}_{1}^{\downarrow }(K)\displaystyle \prod _{i=1}^{n}{s}_{1}^{\downarrow }({A}_{i}){s}_{1}^{\downarrow }({K}^{\prime })\right)=\mathrm{log}\displaystyle \prod _{i=1}^{n}{s}_{i}^{\downarrow }(S)=F(S).\end{eqnarray*}$$

Conversely, consider $({S}_{1},\ldots ,{S}_{m})\in {\rm{\Delta }}(S)$. Using that by definition for each ${S}_{j}\in K(n)\cup Z(n)$ we have ${\prod }_{i=1}^{n}{s}_{i}^{\downarrow }({S}_{j})={s}_{1}^{\downarrow }({S}_{j})$, we conclude:

$$\begin{eqnarray*}&&F(S)=\mathrm{log}\left(\displaystyle \prod _{i=1}^{n}{s}_{i}^{\downarrow }(S)\right)\mathop{\leqslant }\limits^{(\ast )}\mathrm{log}\left(\displaystyle \prod _{j=1}^{m}\displaystyle \prod _{i=1}^{n}{s}_{i}^{\downarrow }({S}_{j})\right)=\mathrm{log}\left(\displaystyle \prod _{j=1}^{m}{s}_{1}^{\downarrow }({S}_{j})\right),\end{eqnarray*}$$

where in $(\ast )$ we used a special case of a theorem by Gel'fand and Naimark ([Bha96], theorem III.4.5 and equation (III.19)). Taking the infimum on the right-hand side gives $F(S)\leqslant \overline{F}(S)$. □

Let us write the last observation in $(\ast )$ as a small lemma for later use:

Lemma 3.4. Let $S,{S}^{\prime }\in {Sp}(2n)$. Then $F({{SS}}^{\prime })\leqslant F(S)+F({S}^{\prime })$.

Up to now, we have only considered products of symplectic matrices, which would correspond to a chain of beam splitters, phase shifters and single-mode squeezers. The goal of this section is to prove that one cannot improve the results with arbitrary paths on ${Sp}(2n)$, corresponding to general Hamiltonians of the form of equation (1) as described in section 2.

Let ${{ \mathcal C }}^{r}(S)$ be the set of absolutely continuous paths $\alpha \ :[0,1]\to {Sp}(2n)$ with a derivative which is bounded almost everywhere such that $\alpha (0)={\mathbb{1}}$ and $\alpha (1)=S$. Such paths seem to capture most if not all physically relevant cases.

Recall the set of generators g of ${\mathfrak{sp}}(2n)$ defined in section 2 and order them in a single vector. Usin equation (9), any $\alpha \in {{ \mathcal C }}^{r}(S)$ corresponds to a $A\in {L}^{\infty }([0,1],{\mathfrak{sp}}(2n))$. Since the generators g form a basis, we can write $A(t)={c}_{\alpha }(t)\cdot g$ with a function ${c}_{\alpha }\in {L}^{\infty }([0,1],{\mathfrak{sp}}(2n))$. Both A or ${c}_{\alpha }$ together with the condition $\alpha (0)={\mathbb{1}}$ uniquely define α.

The goal of this section is to prove that this does not give us any better way to avoid squeezing:

Theorem 3.5. For any $S\in {Sp}(2n)$, we have

$$\begin{eqnarray}&&F(S)=\inf \left\{\left.{\int }_{0}^{1}\parallel {\vec{c}}_{\alpha }^{a}(t){\parallel }_{1}\,{\rm{d}}t\right|\alpha \in {{ \mathcal C }}^{r}(S),\dot{\alpha }(t)={({\vec{c}}_{\alpha }^{p}(t){g}^{p}(\alpha (t)),{\vec{c}}_{\alpha }^{a}(t){g}^{a}(\alpha (t)))}^{T}\right\},\end{eqnarray} \tag{ 13 }$$

where we introduced the notation $\vec{c}$ to clarify that ${g}^{p/a}$ are actually vectors containing a set of generators each, and the coefficients might differ for each of these generators.

The proof of this theorem is quite lengthy in details, thus we split it up into several lemmata. The general idea is easy to relate: we first show that paths corresponding to products of symplectic matrices of type Z(n) or K(n) produce the same outcome in (13) and (12). We then use an approximation argument: given any path, we can approximate it by a path of products of symplectic matrices to arbitrary precision.

To start, we prove the following lemma:

Lemma 3.6. Let $A\in {\mathfrak{sp}}(2n)$ and write $A=1/2(A+{A}^{T})+1/2(A-{A}^{T})=: {A}_{+}+{A}_{-}$. Then ${A}_{+}\in \pi (2n)$ and ${A}_{-}\in {\mathfrak{k}}(n)$ and we have $F(\exp (A))\leqslant F(\exp ({A}_{+}))$.

Proof. First note that F is continuous in S since the singular values are. Using the Trotter-formula, we obtain:

$$\begin{eqnarray*}\begin{array}{rcl}F(\exp (A)) & = & F(\mathop{\mathrm{lim}}\limits_{n\to \infty }{(\exp ({A}_{+}/n)\exp ({A}_{-}/n))}^{n})\\ & \mathop{\leqslant }\limits^{(* )} & \mathop{\mathrm{lim}}\limits_{n\to \infty }({nF}(\exp ({A}_{+}/n))+{nF}(\exp ({A}_{-}/n)))\\ & = & \mathop{\mathrm{lim}}\limits_{n\to \infty }{nF}(\exp ({A}_{+}/n))=F(\exp ({A}_{+})),\end{array}\end{eqnarray*}$$

where we used that $F(\exp ({A}_{-}))=0$ since ${A}_{-}\in {\mathfrak{k}}(n)$ and in $(\ast )$, we used a version of a theorem by Gel'fand and Naimark again (see [Bha96], equation (III.20)). □

Let us define yet another version of F which we call $\hat{F}$ in the following way:

$$\begin{eqnarray*}{C}^{N}(S) & := & \left\{({\vec{c}}_{1}^{a},{\vec{c}}_{1}^{p},\ldots ,{\vec{c}}_{N}^{a},{\vec{c}}_{N}^{p})\left|S=\displaystyle \prod _{j=1}^{N}\exp ({\vec{c}}_{j}^{a}{g}^{a}+{\vec{c}}_{j}^{p}{g}^{p}),{\vec{c}}_{j}\in {{\mathbb{R}}}^{4{n}^{2}}\right.\right\},\\ C(S) & := & \bigcup _{N\in {\mathbb{N}}}{C}^{N}(S),\\ \hat{F}(S) & := & \inf \left\{\left.\displaystyle \sum _{i}\parallel {\vec{c}}_{i}^{a}{\parallel }_{1}\right|\vec{c}\in C(S)\right\}.\end{eqnarray*}$$

This definition is of course reminiscent of the definition of $\overline{F}$ in equation (12):

Lemma 3.7. For $S\in {Sp}(2n)$, we have $\hat{F}(S)=\overline{F}(S)$.

Proof. To prove $\hat{F}\leqslant \overline{F}$, consider the Euler decomposition $S={K}_{1}{A}_{1}\ldots {A}_{n}{K}_{2}$ with ${A}_{i}\in Z(n)$ and ${K}_{1},{K}_{2}\in {Sp}(2n)$. Since K(n) is compact, the exponential map is surjective and there exist ${\vec{c}}_{1}^{p}$ and ${\vec{c}}_{2}^{p}$ such that $\exp ({\vec{c}}_{1}^{p}{g}^{p})={K}_{1}$ and $\exp ({\vec{c}}_{2}^{p}{g}^{p})={K}_{2}$. Recall that we ordered the vector g^a in such a way that the generators g^a_i generate the matrices in Z(n) for i = 1,...,n, hence we know that there exist ${\vec{c}}_{i}^{a}={(0,\ldots ,0,({\vec{c}}_{i}^{a})}_{(i)},0,\ldots ,0)$ for i = 1,...,n such that

$$\begin{eqnarray*}&&S=\exp ({\vec{c}}_{1}^{p}{g}^{p})\displaystyle \prod _{i=1}^{n}\exp ({\vec{c}}_{i}^{a}{g}^{a})\exp ({\vec{c}}_{2}^{p}{g}^{p}).\end{eqnarray*}$$

This implies

$$\begin{eqnarray*}&&\hat{F}(S)\leqslant \displaystyle \sum _{i}\parallel {\vec{c}}_{i}^{a}{\parallel }_{1}=\displaystyle \sum _{i}F(\exp ({\vec{c}}_{i}^{a}{g}^{a}))=\displaystyle \sum _{i}F(\exp ({({\vec{c}}_{i}^{a})}_{(i)}{g}_{i}^{a}))=\displaystyle \sum _{i}\mathrm{log}{s}_{1}({A}_{i})=\overline{F}(S).\end{eqnarray*}$$

Here we used that ${({\vec{c}}_{i}^{a})}_{(i)}$ is also the largest singular value of $\exp {(({\vec{c}}_{i}^{a})}_{(i)}{g}_{i}^{a})\in Z(n)$, as $F{(\exp {(({\vec{c}}_{i}^{a})}_{(i)}{g}_{i}^{a}))=({\vec{c}}_{i}^{a})}_{(i)}$ by normalisation of g.

For the other direction $\hat{F}\geqslant \overline{F}$, let S be arbitrary. Let $c\in C(S)$ and consider each vector ${\vec{c}}_{i}$ separately. We drop the index i for readability, since we need to consider the entries of the vector ${\vec{c}}_{i}$. To make the distinction clear, we denote the jth entry of the vector $\vec{c}$ by ${\vec{c}}_{(j)}$. Recall that the active generators are exactly those generating the positive matrices. Then:

$$\begin{eqnarray*}\begin{array}{rcl}F(\exp (\vec{c}g)) & \mathop{\leqslant }\limits^{\mathrm{Lemma}\ 3.6} & F(\exp ({\vec{c}}^{a}{g}^{a}))=\mathop{\mathrm{lim}}\limits_{n\to \infty }F({(\displaystyle \prod _{i}\exp ({\vec{c}}_{(i)}^{a}{g}_{i}^{a}/n))}^{n})\\ & \leqslant & \displaystyle \sum _{i}F(\exp ({\vec{c}}_{(i)}^{a}{g}_{i}^{a}))=\displaystyle \sum _{i}| {\vec{c}}_{(i)}^{a}| =\parallel {\vec{c}}^{a}{\parallel }_{1},\end{array}\end{eqnarray*}$$

where we basically redid the calculations we used to prove lemma 3.6, using the continuity of F and the Trotter formula from matrix analysis. Until now, we have considered only one ${\vec{c}}_{i}$ of $c\in C(S)$. Now, if we define ${S}_{i}=\exp ({\vec{c}}_{i}g)$, then we have ${\prod }_{i}{S}_{i}=S$ and hence, using lemma 3.4, we find:

$$\begin{eqnarray*}&&\overline{F}(S)\mathop{\leqslant }\limits^{\mathrm{Lemma}\ 3.4}\displaystyle \sum _{i}F({S}_{i})\leqslant \displaystyle \sum _{i}F(\exp ({\vec{c}}_{i}g))\leqslant \displaystyle \sum _{i}\parallel {\vec{c}}_{i}^{a}{\parallel }_{1}\qquad \forall c\in C(S).\end{eqnarray*}$$

But this means $\overline{F}(S)\leqslant \hat{F}(S)$, as we claimed.□

We can now prove the first half of the theorem:

Lemma 3.8. For $S\in {Sp}(2n)$ we have

$$\begin{eqnarray*}&&F(S)\geqslant \inf \left\{{\int }_{0}^{1}\parallel {\vec{c}}_{\alpha }^{a}(t){\parallel }_{1}\,{\rm{d}}t| \alpha \in {{ \mathcal C }}^{r}(S),\dot{\alpha }(t)\right\}.\end{eqnarray*}$$

Proof. Let $S\in {Sp}(2n)$ and consider the Euler decomposition $S={K}_{1}{S}_{1}\cdots {S}_{n}{K}_{2}$. We can define a function $A\ :[0,1]\to {\mathfrak{sp}}(2n)$ via:

$$\begin{eqnarray}A(t):= \left\{\begin{array}{ll}(n+2)\cdot {\vec{c}}_{1}^{\,p}{g}^{p} & t\in [0,1/(n+2)),\\ (n+2)\cdot {({\vec{c}}_{i+1}^{a})}_{(i)}{g}_{i}^{a} & t\in [i/(2(n+2)),(i+1)/(n+2)),i=1,\ldots ,n\\ (n+2)\cdot {\vec{c}}_{n+2}^{\,p}{g}^{p} & t\in [(n+1)/(n+2),1],\end{array}\right.,\end{eqnarray} \tag{ 14 }$$

where $({\vec{c}}_{1}^{p},0,0,{\vec{c}}_{2}^{a},\ldots ,0,{\vec{c}}_{n+1}^{a},{\vec{c}}_{n+2}^{p},0)$ denotes the element in ${C}^{n+2}(S)$ for the Euler decomposition and vector indices are denoted by a subscript ${}_{(i)}$ as before. Let $U(s,t)$ be the propagator corresponding to A, then for $t\in [0,1/(n+2))$ according to proposition A.1, since A does not depend on t on this interval, it is given by $U(t,s)=\exp ((t-s)A)$. In particular, $U(1/(n+2),0)=\exp ({\vec{c}}_{n+2}^{p}{g}^{p})={K}_{2}$.

Iterating the procedure above, using $U(0,1)=U(0,1/(n+2))$ $\cdots U((n+1)/(n+2),1)$, we can see that by construction, $U(0,1)={K}_{1}{S}_{1}\cdots {S}_{n}{K}_{2}=S$. Hence A defines a continuous path on ${Sp}(2n)$ via $U(s,t)$. We can calculate:

$$\begin{eqnarray*}\begin{array}{rcl}{\displaystyle \int }_{0}^{1}\parallel {\vec{c}}^{a}(t){\parallel }_{1}\,{\rm{d}}t & = & \displaystyle \sum _{i=1}^{n}{\displaystyle \int }_{i/(n+2)}^{(i+1)/(n+2)}| (n+2)\cdot {({\vec{c}}_{i+1}^{a})}_{(i)}| \,{\rm{d}}t\\ & = & \displaystyle \sum _{i=1}^{n}| {({\vec{c}}_{i+1}^{a})}_{(i)}| \mathop{=}\limits^{\mathrm{Lemma}\,3.7}F(S),\end{array}\end{eqnarray*}$$

where we used that the integral over the interval $[0,1/(n+2))$ and $[(n+1)/(n+2),1]$ is empty due to the fact that all active components are zero. In the last step, we used that for the Euler decomposition, which takes the minimum in $\hat{F}$, this value is exactly ${\sum }_{i}| {({\vec{c}}_{i+1}^{a})}_{(i)}| ={\sum }_{i}\parallel {\vec{c}}_{i+1}^{a}{\parallel }_{1}$, since ${({\vec{c}}_{i+1}^{a})}_{(j)}=0$ for $j\ne i$. Taking the infimum on the left-hand side only decreases the value.□

For the other direction, we need some facts about ordinary differential equations that are collected in appendix A.

Lemma 3.9. For $S\in {Sp}(2n)$ we have

$$\begin{eqnarray}&&F(S)\leqslant \inf \left\{{\int }_{0}^{1}\parallel {\vec{c}}_{\alpha }^{a}(t){\parallel }_{1}\,{\rm{d}}t\,| \,\alpha \in {{ \mathcal C }}^{r}(S),\dot{\alpha }(t)\right\}.\end{eqnarray} \tag{ 15 }$$

Proof. Let $S\in {Sp}(2n)$ be arbitrary. Combining the proof of lemma 3.8 with proposition 3.3 and lemma 3.7 we have already proved:

$$\begin{eqnarray*}F(S) & = & \inf \left\{\left.{\displaystyle \int }_{0}^{1}\parallel {\vec{c}}_{\alpha }^{a}(t){\parallel }_{1}\,{\rm{d}}t\right|\alpha \in {{ \mathcal C }}^{r}(S),\right.\\ & & \dot{\alpha }(t)={({\vec{c}}_{\alpha }^{p}(t){g}^{p}(\alpha (t)),{\vec{c}}_{\alpha }^{a}(t){g}^{a}(\alpha (t)))}^{T},\vec{c}\,\mathrm{step}\,\mathrm{fct}.\}.\end{eqnarray*}$$

The only thing left to prove is that we can drop the step-function assumption. This will be done by a standard approximation argument: let $\tilde{F}(S)$ denote the right-hand side of equation (15). Let $\varepsilon \gt 0$ and consider an arbitrary $A\in {L}^{\infty }$ such that

$$\begin{eqnarray}&&\left|{\int }_{0}^{1}\parallel {\vec{c}}_{\alpha }^{a}(t){\parallel }_{1}\,{\rm{d}}t-\tilde{F}(S)\right|\lt \varepsilon \end{eqnarray} \tag{ 16 }$$

i.e. A corresponds to a path that is close to the infimum in the definition of $\tilde{F}$. We can now approximate ${c}_{\alpha }$ by step-functions ${c}_{\alpha ^{\prime} }$ (corresponding to a function ${A}^{\prime }$, see lemma A.2) such that

$$\begin{eqnarray}&&\left|{\int }_{0}^{1}\parallel {c}_{\alpha }(t)-{c}_{\alpha ^{\prime} }{\parallel }_{1}\,{\rm{d}}t\right|\lt \varepsilon .\end{eqnarray} \tag{ 17 }$$

Using the fact that the propagators ${U}_{A},{U}_{A^{\prime} }$ are differentiable almost everywhere (proposition A.1) and absolutely continuous when one entry is fixed, we can define a function $f(s):= {U}_{A}(0,s){U}_{{A}^{\prime }}(s,t)$, which is also differentiable almost everywhere. Furthermore, the fundamental theorem of calculus holds for f(s) (see [Rud87], theorems 6.10 and 7.8).

$$\begin{eqnarray*}&&\displaystyle \frac{{\rm{d}}}{{\rm{d}}s}f(s)=-{U}_{A}(0,s)A(s){U}_{{A}^{\prime }}(s,t)+{U}_{A}(0,s){A}^{\prime }(s){U}_{{A}^{\prime }}(s,t)\end{eqnarray*}$$

almost everywhere, which implies:

$$\begin{eqnarray*}{U}_{{A}^{\prime }}(0,t)-{U}_{A}(0,t) & = & f(t)-f(0)={\displaystyle \int }_{0}^{t}\displaystyle \frac{{\rm{d}}}{{\rm{d}}s}f(s){\rm{d}}s\\ & = & {\displaystyle \int }_{0}^{t}{U}_{A}(0,s)({A}^{\prime }(s)-A(s)){U}_{{A}^{\prime }}(s,t){\rm{d}}s.\end{eqnarray*}$$

Since U and g are bounded in $\parallel \cdot {\parallel }_{\infty }$, we obtain

$$\begin{eqnarray}&&\parallel {U}_{{A}^{\prime }}(0,t)-{U}_{A}(0,t){\parallel }_{1}\leqslant M{\int }_{0}^{t}\parallel {c}_{\alpha ^{\prime} }(s)-{c}_{\alpha }(s){\parallel }_{1}\,{\rm{d}}s\leqslant M\varepsilon .\end{eqnarray} \tag{ 18 }$$

M can explicitly be computed by the bounds given in proposition A.1.

Up to now, we have taken a path α to S close to the infimum and approximated it by a path $\alpha ^{\prime}$. It is immediate by equations (16) and (17) that

$$\begin{eqnarray}&&\left|{\int }_{0}^{1}\parallel {c}_{\alpha ^{\prime} }(t){\parallel }_{1}\,{\rm{d}}t-\tilde{F}(S)\right|\lt 2\varepsilon .\end{eqnarray} \tag{ 19 }$$

Since ${c}_{\alpha ^{\prime} }\in {C}^{N}({S}^{\prime })$ for some $N\in {\mathbb{N}}$ and ${S}^{\prime }={U}_{{A}^{\prime }}(0,1)$, we would be done if ${S}^{\prime }=S$. To remedy this, we want to extend ${\alpha }^{\prime }$ to a path $\tilde{\alpha }$ such that it ends at S. This is where equation (18) enters: set $\tilde{S}:= {U}_{{A}^{\prime }}{(0,1)}^{-1}{U}_{A}(0,1)$, then

$$\begin{eqnarray}&&\parallel \tilde{S}-{\mathbb{1}}{\parallel }_{1}\leqslant \displaystyle \frac{M}{\parallel S{\parallel }_{1}}\varepsilon \end{eqnarray} \tag{ 20 }$$

hence $\tilde{S}\approx {\mathbb{1}}$ for ε small enough. Using the polar decomposition, we can write $\tilde{S}=\exp ({\vec{c}}_{N+1}^{p})\exp ({\vec{c}}_{N+2}^{a})$. A quick calculation yields

$$\begin{eqnarray}&&\parallel \mathrm{log}\tilde{S}{\parallel }_{1}\leqslant n\mathrm{log}\left(\varepsilon \displaystyle \frac{M}{\parallel S{\parallel }_{1}}\right)\leqslant n\varepsilon \displaystyle \frac{M}{\parallel S{\parallel }_{1}}=: C\varepsilon .\end{eqnarray} \tag{ 21 }$$

This lets us construct a new $\tilde{A}\,:[0,2]\to {\mathfrak{sp}}(2n)$:

$$\begin{eqnarray*}t\mapsto \left\{\begin{array}{ll}{A}^{\prime }(t) & t\in [0,1],\\ 2\cdot {\vec{c}}_{N+1}^{p}{g}^{p} & t\in (1,3/2)\\ 2\cdot {\vec{c}}_{N+2}^{a}{g}^{a} & t\in (3/2,2]\mathrm{.}\end{array}\right.,\end{eqnarray*}$$

By construction, for the corresponding propagator we have ${U}_{\tilde{A}}(0,2)=S$ and $\tilde{\alpha }$ is a feasible path for $\tilde{F}(S)$ (at least after reparameterisation) fulfiling:

$$\begin{eqnarray*}\begin{array}{rcl}\left|{\displaystyle \int }_{0}^{2}\parallel {c}_{\tilde{\alpha }}^{a}(t){\parallel }_{1}\,{\rm{d}}t-\tilde{F}(S)\right| & \leqslant & \left|{\displaystyle \int }_{0}^{1}\parallel {c}_{\tilde{\alpha }}^{a}(t){\parallel }_{1}\,{\rm{d}}t-\tilde{F}(S)\right|+\left|{\displaystyle \int }_{0}^{1}\parallel {c}_{\tilde{\alpha }}^{a}(t){\parallel }_{1}\,{\rm{d}}t\right|\\ & \mathop{\leqslant }\limits^{(19)+(21)} & (2+C)\varepsilon \mathrm{.}\end{array}\end{eqnarray*}$$

Since, ${c}_{\tilde{\alpha }}\in {C}^{N+2}(S)$, $\tilde{\alpha }$ is a valid path for $\hat{F}(S)$, which implies that for any $\epsilon \gt 0$, choosing $\varepsilon := \epsilon /(2+C)$, we have seen:

$$\begin{eqnarray}&&\hat{F}(S)\lt \tilde{F}(S)+\epsilon .\end{eqnarray} \tag{ 22 }$$

For $\epsilon \to 0$, $\hat{F}(S)\leqslant \tilde{F}(S)$, which implies the lemma via lemma 3.7.□

We will now give several reformulations of the squeezing measure and prove some of its properties. In particular, G is convex and one of the crucial steps towards proving convexity of G is given by a reformulation of G with the help of the Cayley transform. For the reader unfamiliar with the Cayley transform, a definition and basic properties are provided in appendix B.

Proposition 4.2. Let ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ and ${\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}$ symmetric. Then:

$$\begin{eqnarray}&&G({\rm{\Gamma }})=\inf \{F({{\rm{\Gamma }}}_{0}^{1/2})| {\rm{\Gamma }}\geqslant {{\rm{\Gamma }}}_{0}\geqslant {\rm{i}}{J}\}\end{eqnarray} \tag{ 24 }$$

$$\begin{eqnarray}&&=\,\inf \left\{\displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{n}\mathrm{log}\left(\displaystyle \frac{1+{s}_{i}(A+{\rm{i}}{B})}{1-{s}_{i}(A+{\rm{i}}{B})}\right)\,\,| \,{{ \mathcal C }}^{-1}({\rm{\Gamma }})\geqslant H,H\in { \mathcal H }\right\},\end{eqnarray} \tag{ 25 }$$

where ${ \mathcal H }$ is defined via:

$$\begin{eqnarray}{ \mathcal H }=\left\{H=\left(\begin{array}{cc}A & B\\ B & -A\end{array}\right)\in {{\mathbb{R}}}^{2m\times 2m}\,\text{}| \,{A}^{T}=A,{B}^{T}=B,\mathrm{spec}(H)\subset (-1,1)\right\}.\end{eqnarray} \tag{ 26 }$$

Proof. First note that the infimum in all three expressions is actually attained. We can see this most easily in the definition (23): the matrix inequalities ${\rm{\Gamma }}\geqslant {S}^{T}S(\geqslant {\rm{i}}{J})$ imply that the set of feasible S in the minimisation is compact, hence its minimum is attained. To see (23) = (24), first note that (24) ≤ (23) since any $S\in {Sp}(2n)$ also fulfils ${S}^{T}S\geqslant {\rm{i}}{J}$, hence ${\rm{\Gamma }}\geqslant {S}^{T}S\geqslant {\rm{i}}{J}$. For equality, note that for any ${\rm{\Gamma }}\geqslant {{\rm{\Gamma }}}_{0}\geqslant {\rm{i}}{J}$, using Williamson's theorem we can find $S\in {Sp}(2n)$ and a diagonal $\tilde{D}\geqslant {\mathbb{1}}$ (via corollary 2.10) such that ${{\rm{\Gamma }}}_{0}={S}^{T}{DS}\geqslant {S}^{T}S\geqslant {\rm{i}}{J}$. But since $F({{\rm{\Gamma }}}_{0}^{1/2})\geqslant F({({S}^{T}S)}^{1/2})=F(S)$ via the Weyl monotonicity principle, the infimum is achieved on symplectic matrices.

Finally, let us prove equality with (25). First observe that we can replace ${Sp}(2n)$ by ${ \mathcal H }$ using proposition B.1(4).

Using the fact that ${s}_{i}^{\downarrow }{(S)={\lambda }_{i}^{\downarrow }{({S}^{T}S)}^{1/2}={\lambda }_{i}^{\downarrow }({ \mathcal C }(H))}^{1/2}$ and the fact that H is diagonalised by the same unitary matrices as ${ \mathcal C }{(H)=({\mathbb{1}}+H)\cdot ({\mathbb{1}}-H)}^{-1}$ whence its eigenvalues are

$$\begin{eqnarray*}&&{\lambda }_{i}^{\downarrow }({ \mathcal C }(H))=\displaystyle \frac{1+{\lambda }_{i}^{\downarrow }(H)}{1-{\lambda }_{i}^{\downarrow }(H)},\end{eqnarray*}$$

we have:

$$\begin{eqnarray*}&&\inf \{F(S)| {\rm{\Gamma }}\geqslant {S}^{T}S,S\in {Sp}(2n)\}=\inf \left\{\mathrm{log}\displaystyle \prod _{i=1}^{n}{\left(\displaystyle \frac{1+{\lambda }_{i}^{\downarrow }(H)}{1-{\lambda }_{i}^{\downarrow }(H)}\right)}^{\tfrac{1}{2}}\,| {\rm{\Gamma }}\geqslant { \mathcal C }(H),H\in { \mathcal H }\right\}.\end{eqnarray*}$$

Next we claim ${\lambda }_{i}^{\downarrow }(H)={s}_{i}^{\downarrow }(A+{\rm{i}}{B})$ for i = 1,...,n. To see this note:

$$\begin{eqnarray}\displaystyle \frac{1}{2}\left(\begin{array}{cc}{\mathbb{1}} & {\rm{i}}{\mathbb{1}}\\ {\mathbb{1}} & -{\rm{i}}{\mathbb{1}}\end{array}\right)\cdot \left(\begin{array}{cc}A & B\\ B & -A\end{array}\right)\cdot \left(\begin{array}{cc}{\mathbb{1}} & {\mathbb{1}}\\ -{\rm{i}}{\mathbb{1}} & {\rm{i}}{\mathbb{1}}\end{array}\right)=\left(\begin{array}{cc}0 & A+{\rm{i}}{B}\\ A-{\rm{i}}{B} & 0\end{array}\right).\end{eqnarray} \tag{ 27 }$$

The singular values of the matrix on the right-hand side of equation (27) are the eigenvalues of $\mathrm{diag}{({(A+{\rm{i}}{B})}^{\dagger }{(A+{\rm{i}}{B}),(A+{\rm{i}}{B})(A+{\rm{i}}{B})}^{\dagger })}^{1/2}$, which are the singular values of A + iB with double multiplicity. From the structure of H, it is immediate that the eigenvalues of the right-hand side of equation (27) and thus of H come in pairs $\pm {s}_{i}(A+{\rm{i}}{B})$. Hence ${\lambda }_{i}^{\downarrow }(H)={s}_{i}^{\downarrow }(A+{\rm{i}}{B})$ for i = 1,...,n and we have:

$$\begin{eqnarray*}&&\begin{array}{l}\inf \{F(S)| {\rm{\Gamma }}\geqslant {S}^{T}S,S\in {Sp}(2n)\}\\ \quad =\inf \left\{\displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{n}\mathrm{log}\left(\displaystyle \frac{1+{s}_{i}(A+{\rm{i}}{B})}{1-{s}_{i}(A+{\rm{i}}{B})}\right)\,\,| \,{\rm{\Gamma }}\geqslant { \mathcal C }(H),H\in { \mathcal H }\right\}\mathrm{.}\end{array}\end{eqnarray*}$$

To see that that the right-hand side equals (25), we only need to use the fact that ${\rm{\Gamma }}\geqslant { \mathcal C }(H)\iff {{ \mathcal C }}^{-1}({\rm{\Gamma }})\geqslant H$ for all $H\in { \mathcal H }$ and ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ since the Cayley transform and its inverse are operator monotone.□

The reformulation (25) will allow us to prove:

Theorem 4.3.  $G$ is convex on the set of covariance matrices $\{{\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}| {\rm{\Gamma }}\geqslant {\rm{i}}{J}\}$.

The crucial part of the proof is the following lemma:

Lemma 4.4. Consider the map $f\,:{{\mathbb{R}}}^{n\times n}\times {{\mathbb{R}}}^{n\times n}\to {\mathbb{R}}$:

$$\begin{eqnarray}&&f(A,B)=\displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{n}\mathrm{log}\left(\displaystyle \frac{1+{s}_{i}(A+{\rm{i}}{B})}{1-{s}_{i}(A+{\rm{i}}{B})}\right).\end{eqnarray} \tag{ 28 }$$

If we restrict $f$ to symmetric matrices $A$ and $B$ such that ${s}_{i}(A+{\rm{i}}{B})\lt 1$ for all $i=1,\ldots ,n,f$ is jointly convex in $A,B$, i.e.

$$\begin{eqnarray*}&&f({tA}+(1-t){A}^{\prime },{tB}+(1-t){B}^{\prime })\leqslant {tf}(A,B)+(1-t)f({A}^{\prime },{B}^{\prime })\qquad \forall t\in [0,1].\end{eqnarray*}$$

Proof. Let $\tilde{A}:= {tA}+(1-t){A}^{\prime }$ and $\tilde{B}:= {tB}+(1-t){B}^{\prime }$. Note that $\tilde{A}$ and $\tilde{B}$ are also symmetric, and the largest singular value of $\tilde{A}+{\rm{i}}\tilde{B}$ fulfils ${s}_{1}^{\downarrow }(\tilde{A}+{\rm{i}}\tilde{B})\leqslant {{ts}}_{1}^{\downarrow }(A+{\rm{i}}{B})+(1-t){s}_{1}^{\downarrow }({A}^{\prime }+{{\rm{i}}{B}}^{\prime })$. Therefore, the singular values of any convex combination of A + iB and ${A}^{\prime }+{{\rm{i}}{B}}^{\prime }$ also lie in the interval $[0,1)$. This makes our restriction well-defined under convex combinations.

For any j = 1,...,n, by Thompson's theorem (see [Tho76]), which states that for every complex $A,B$, there exist unitary matrices $U,V$ such that $| A+B| \leqslant U| A| {U}^{* }+V| B| {V}^{* }$, we have

$$\begin{eqnarray*}&&{s}_{j}(\tilde{A}+{\rm{i}}\tilde{B})={\lambda }_{j}(| \tilde{A}+{\rm{i}}\tilde{B}| )\leqslant {\lambda }_{j}(U| t(A+{\rm{i}}{B})| {U}^{* }+V| (1-t)({A}^{\prime }+{{\rm{i}}{B}}^{\prime })| {V}^{* }).\end{eqnarray*}$$

Using Lidskii's theorem ([Bha96], chapter III with explicit formulation in exercise III.4.3), we have

$$\begin{eqnarray}{s}_{j}(\tilde{A}+{\rm{i}}\tilde{B}) & \leqslant & {\lambda }_{j}(U| t(A+{\rm{i}}{B})| {U}^{* })+\displaystyle \sum _{\pi }{p}_{\pi }{\lambda }_{\pi (j)}(V| (1-t)({A}^{\prime }+{{\rm{i}}{B}}^{\prime })| {V}^{* })\\ & \mathop{=}\limits^{(* )} & {\lambda }_{j}(| t(A+{\rm{i}}{B})| )+\displaystyle \sum _{\pi }{p}_{\pi }{\lambda }_{\pi (j)}(| (1-t)({A}^{\prime }+{{\rm{i}}{B}}^{\prime })| )\\ & = & t{\lambda }_{j}(| A+{\rm{i}}{B}| )+(1-t)\displaystyle \sum _{\pi }{p}_{\pi }{\lambda }_{\pi (j)}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )\end{eqnarray} \tag{ 29 }$$

with ${p}_{\pi }\geqslant 0$ and ${\sum }_{\pi }{p}_{\pi }=1$. In $(\ast )$, we used that unitaries do not change the spectrum. Now each summand in equation (28) is the Cayley transform of a singular value. We can use the log-convexity of the Cayley-transform to prove the joint convexity of f:

$$\begin{eqnarray*}\begin{array}{rcl}f(\tilde{A},\tilde{B}) & = & \displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{s}_{i}(\tilde{A}+{\rm{i}}\tilde{B})]\\ & \leqslant & \displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }\left[t{\lambda }_{i}(| A+{\rm{i}}{B}| )+(1-t)\displaystyle \sum _{\pi }{p}_{\pi }{\lambda }_{\pi (i)}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )\right]\\ & \leqslant & \displaystyle \sum _{i=1}^{n}\left(t\mathrm{log}{ \mathcal C }[{\lambda }_{i}(| A+{\rm{i}}{B}| )]+(1-t)\displaystyle \sum _{\pi }{p}_{\pi }\mathrm{log}{ \mathcal C }[{\lambda }_{\pi (i)}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )]\right)\\ & = & \displaystyle \sum _{i=1}^{n}t\mathrm{log}{ \mathcal C }[{\lambda }_{i}(| A+{\rm{i}}{B}| )]+(1-t)\displaystyle \sum _{\pi }{p}_{\pi }\left(\displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{\lambda }_{\pi (i)}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )]\right)\\ & \leqslant & t\displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{\lambda }_{i}(| A+{\rm{i}}{B}| )]+(1-t)\displaystyle \sum _{\pi }{p}_{\pi }\cdot \mathop{\max }\limits_{\pi }\left(\displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{\lambda }_{\pi (i)}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )]\right)\\ & \mathop{=}\limits^{(* * )} & t\displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{\lambda }_{i}(| A+{\rm{i}}{B}| )]+(1-t)\displaystyle \sum _{i=1}^{n}\mathrm{log}{ \mathcal C }[{\lambda }_{i}(| {A}^{\prime }+{{\rm{i}}{B}}^{\prime }| )]\\ & = & {tf}(A,B)+(1-t)f({A}^{\prime },{B}^{\prime }),\end{array}\end{eqnarray*}$$

where in $(\ast \ast )$ we use that the sum of all eigenvalues is of course not dependent on the order of the eigenvalues.□

This lemma will later allow us to calculate G as a convex programme.

Proof of theorem 4.3. We can now finish the proof of the convexity of G.

First note that using the definition of f in lemma 4.4 we can reformulate (25) to

$$\begin{eqnarray}&&G({\rm{\Gamma }})=\inf \{f(A,B)| {{ \mathcal C }}^{-1}({\rm{\Gamma }})\geqslant H,H\in { \mathcal H }\}.\end{eqnarray} \tag{ 30 }$$

Let ${\rm{\Gamma }}\geqslant {\rm{i}}{J},{{\rm{\Gamma }}}^{\prime }\geqslant {\rm{i}}{J}$ be two covariance matrices and let $H,{H}^{\prime }\in { \mathcal H }$ be the matrices that attain the minimum of $G({\rm{\Gamma }}),G({{\rm{\Gamma }}}^{\prime })$ respectively. Then, in particular, ${tH}+(1-t){H}^{\prime }\in { \mathcal H }$. Furthermore, since ${{ \mathcal C }}^{-1}({\rm{\Gamma }})\geqslant H$ and ${{ \mathcal C }}^{-1}({{\rm{\Gamma }}}^{\prime })\geqslant {H}^{\prime }$ we have

$$\begin{eqnarray*}&&{{ \mathcal C }}^{-1}(t{\rm{\Gamma }}+(1-t){{\rm{\Gamma }}}^{\prime })\mathop{\geqslant }\limits^{(* )}t{{ \mathcal C }}^{-1}({\rm{\Gamma }})+(1-t){{ \mathcal C }}^{-1}({{\rm{\Gamma }}}^{\prime })\geqslant {tH}+(1-t){H}^{\prime },\end{eqnarray*}$$

where we used the operator concavity of ${{ \mathcal C }}^{-1}$ in $(\ast )$. This means that ${tH}+(1-t){H}^{\prime }$ is a feasible matrix for the minimisation in G, which implies using equation (30)

$$\begin{eqnarray*}&&G(t{\rm{\Gamma }}+(1-t){{\rm{\Gamma }}}^{\prime })\leqslant f({tA}+(1-t){A}^{\prime },{tB}+(1-t){B}^{\prime }).\end{eqnarray*}$$

The convexity now follows directly from lemma 4.4 and the fact that we chose H and ${H}^{\prime }$ to attain $G({\rm{\Gamma }})$ and $G({{\rm{\Gamma }}}^{\prime })$.□

From the convexity of G on the set of covariance matrices, it follows from general arguments in convex analysis that G is continuous on the interior of the set of covariance matrices (see [Roc97], theorem 10.1). What more can we say about the boundary?

Theorem 4.5.  $G$ is lower semicontinuous on the set of covariance matrices $\{{\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}| {\rm{\Gamma }}\geqslant {\rm{i}}{J}\}$ and continuous on its interior. Moreover, $G({\rm{\Gamma }}+\varepsilon {\mathbb{1}})\to G({\rm{\Gamma }})$ for $0\lt \varepsilon \to 0$ for any ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$.

The ultimate goal is to extend continuity from the interior to the exterior, which we do not know how to do at present. The proof will need a few notions from set-valued analysis that we review in appendix C.

Proof of theorem 4.5. As already observed, G is continuous on the interior. Let ${{\rm{\Gamma }}}_{0}\geqslant {\rm{i}}{J}$ be arbitrary and suppose

$$\begin{eqnarray*}&&{ \mathcal A }({\rm{\Gamma }}):= \{\hat{{\rm{\Gamma }}}| ({\rm{\Gamma }}-2\parallel {{\rm{\Gamma }}}_{0}\parallel {\mathbb{1}})\leqslant \hat{{\rm{\Gamma }}}\leqslant {\rm{\Gamma }}\}.\end{eqnarray*}$$

By definition, ${ \mathcal A }$ is compact and convex for any Γ. Moreover, it defines a set-valued function on the set of covariance matrices with non-empty values. Let $\varepsilon \gt 0$, then for all ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ with $\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt \varepsilon$, we have that for any $\hat{{\rm{\Gamma }}}\in { \mathcal A }({\rm{\Gamma }})$, $\tilde{{\rm{\Gamma }}}:= \hat{{\rm{\Gamma }}}+({\rm{\Gamma }}-{{\rm{\Gamma }}}_{0})\in { \mathcal A }({{\rm{\Gamma }}}_{0})$ and $\parallel \hat{{\rm{\Gamma }}}-\tilde{{\rm{\Gamma }}}\parallel \lt \varepsilon$. This is the condition in lemma C.2 hence the set-valued function defined by ${ \mathcal A }$ is upper semicontinuous at ${{\rm{\Gamma }}}_{0}$, which implies that ${ \mathcal A }({\rm{\Gamma }})\cap \{X| {\rm{i}}{J}\leqslant X\}$ is also upper semicontinuous by proposition C.3. If ε is small enough (e.g. $\varepsilon \lt 1$), this implies

$$\begin{eqnarray*}&&{ \mathcal A }({\rm{\Gamma }})\cap \{X| {\rm{i}}{J}\leqslant X\}=\{X| {\rm{i}}{J}\leqslant X\leqslant {\rm{\Gamma }}\}=: { \mathcal G }({\rm{\Gamma }}),\end{eqnarray*}$$

hence this set is upper semicontinuous at ${{\rm{\Gamma }}}_{0}$.

Since F is continuous on positive definite matrices, it is absolutely continuous if we restrict to a small neighbourhood of the covariance matrix ${{\rm{\Gamma }}}_{0}$. This means that for every $\varepsilon \gt 0$ there exists an $\epsilon \gt 0$ such that

$$\begin{eqnarray}&&F(\tilde{{\rm{\Gamma }}})-\varepsilon \lt F(\hat{{\rm{\Gamma }}})\lt F(\tilde{{\rm{\Gamma }}})+\varepsilon \end{eqnarray} \tag{ 31 }$$

for all $\parallel \tilde{{\rm{\Gamma }}}-\hat{{\rm{\Gamma }}}\parallel \lt \epsilon$ and all $\tilde{{\rm{\Gamma }}},\hat{{\rm{\Gamma }}}\in {\bigcup }_{\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt 1}{ \mathcal G }({\rm{\Gamma }})$.

Assuming without loss of generality that $\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt 1$, the set ${ \mathcal G }({\rm{\Gamma }})$ is exactly the set for the minimisation in the definition of G. The upper semicontinuity of ${ \mathcal G }({\rm{\Gamma }})$ implies by lemma C.2 that for every $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for all $\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt \delta$ we have: for all $\hat{{\rm{\Gamma }}}\in { \mathcal G }({\rm{\Gamma }})$ there exists a $\tilde{{\rm{\Gamma }}}\in { \mathcal G }({{\rm{\Gamma }}}_{0})$ such that $\parallel \hat{{\rm{\Gamma }}}-\tilde{{\rm{\Gamma }}}\parallel \lt \epsilon$. In particular, this is true for all minimisers $\hat{{\rm{\Gamma }}}$ with $G({\rm{\Gamma }})=F({\hat{{\rm{\Gamma }}}}^{1/2})$, where $\hat{{\rm{\Gamma }}}$ and $\tilde{{\rm{\Gamma }}}\in {\bigcup }_{\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt 1}{ \mathcal G }({\rm{\Gamma }})$. Using equation (31) we obtain: for every $\varepsilon \gt 0$ there exists a $\delta \gt 0$ such that for all $\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt \delta$, we have a pair $\hat{{\rm{\Gamma }}},\tilde{{\rm{\Gamma }}}$ with $\hat{{\rm{\Gamma }}}\in { \mathcal G }({\rm{\Gamma }})$ minimising $G({\rm{\Gamma }})$ and $\tilde{{\rm{\Gamma }}}\in { \mathcal G }({{\rm{\Gamma }}}_{0})$ such that

$$\begin{eqnarray*}&&F(\tilde{{\rm{\Gamma }}})-\varepsilon \lt F(\hat{{\rm{\Gamma }}})=G({\rm{\Gamma }}).\end{eqnarray*}$$

This implies that for all $\varepsilon \gt 0$ there exists a $\delta \gt 0$ such that

$$\begin{eqnarray*}&&G({{\rm{\Gamma }}}_{0})\leqslant G({\rm{\Gamma }})+\varepsilon \end{eqnarray*}$$

for all $\parallel {\rm{\Gamma }}-{{\rm{\Gamma }}}_{0}\parallel \lt \delta$.

Taking the limit inferior on both sides implies that G is lower semicontinuous at ${{\rm{\Gamma }}}_{0}$. Upper semicontinuity would follow for instance if ${ \mathcal G }({{\rm{\Gamma }}}_{0})$ is also lower semicontinuous.

Finally, let us prove that $G({{\rm{\Gamma }}}_{0}+\varepsilon {\mathbb{1}})\to 0$ for $\varepsilon \to 0$. To see this, consider the closed sets

$$\begin{eqnarray*}&&{C}_{n}:= \bigcup _{0\leqslant \xi \leqslant 1/n}{ \mathcal G }({{\rm{\Gamma }}}_{0}+\xi {\mathbb{1}})\end{eqnarray*}$$

for any $n\in {\mathbb{N}}$. It is easy to see that ${C}_{n+1}\subseteq {C}_{n}$ and that ${\bigcap }_{n\in \infty }{C}_{n}={ \mathcal G }({{\rm{\Gamma }}}_{0})$. Moreover, C₁ is compact. Now let ${{\rm{\Gamma }}}_{n}$ be the sequence of minimisers for $G({{\rm{\Gamma }}}_{0}+1/n{\mathbb{1}})$, then ${{\rm{\Gamma }}}_{n}\in {C}_{n}$ for all $n\in {\mathbb{N}}$. By compactness, a subsequence will converge to

$$\begin{eqnarray*}&&{\rm{\Gamma }}\in \bigcap _{n\in \infty }{C}_{n}={ \mathcal G }({{\rm{\Gamma }}}_{0}).\end{eqnarray*}$$

Therefore, $G({{\rm{\Gamma }}}_{0})\leqslant {\mathrm{lim}}_{\varepsilon \to 0}G({{\rm{\Gamma }}}_{0}+\varepsilon {\mathbb{1}})$, but since ${{\rm{\Gamma }}}_{0}\leqslant {{\rm{\Gamma }}}_{0}+\varepsilon {\mathbb{1}}$ for all $\varepsilon \gt 0$ we also have $G({\rm{\Gamma }})\geqslant {\mathrm{lim}}_{\varepsilon \to 0}G({{\rm{\Gamma }}}_{0}+\varepsilon {\mathbb{1}})$.□

Now we consider additivity properties of G. We switch our basis again and use γ and σ.

Proposition 4.6. For any covariance matrices ${\gamma }_{A}\in {{\mathbb{R}}}^{2{n}_{1}\times 2{n}_{1}}$ and ${\gamma }_{B}\in {{\mathbb{R}}}^{2{n}_{2}\times 2{n}_{2}}$, we have

$$\begin{eqnarray*}&&\displaystyle \frac{1}{2}(G({\gamma }_{A})+G({\gamma }_{B}))\leqslant G({\gamma }_{A}\oplus {\gamma }_{B})\leqslant G({\gamma }_{A})+G({\gamma }_{B}).\end{eqnarray*}$$

In particular, $G$ is subadditive.

Proof. For subadditivity, let ${S}^{T}S\leqslant {\gamma }_{A}$ and ${S}^{\prime T}{S}^{\prime }\leqslant {\gamma }_{B}$ obtain the minimum in $G({\gamma }_{A})$ and $G({\gamma }_{B})$ respectively. Then $S\oplus {S}^{\prime }$ is symplectic and ${(S\oplus {S}^{\prime })}^{T}(S\oplus {S}^{\prime })\leqslant {\gamma }_{A}\oplus {\gamma }_{B}$ hence, $G({\gamma }_{A}\oplus {\gamma }_{B})\leqslant G(A)+G(B)$.

To prove the lower bound, we need the following equation that we will only prove later on (see equation (46)):

$$\begin{eqnarray}&&a\geqslant 1\text{}:\qquad G({\gamma }_{A})\leqslant G({\gamma }_{A}\oplus a{{\mathbb{1}}}_{{n}_{2}}).\end{eqnarray} \tag{ 32 }$$

Assuming this inequality, let $a\geqslant 1$ be such that $a{{\mathbb{1}}}_{{n}_{2}}\geqslant {\gamma }_{B}$, then

$$\begin{eqnarray*}&&G({\gamma }_{A}\oplus a{{\mathbb{1}}}_{{n}_{2}})\leqslant G({\gamma }_{A}\oplus {\gamma }_{B})\end{eqnarray*}$$

hence $G({\gamma }_{A})\leqslant G({\gamma }_{A}\oplus {\gamma }_{B})$ and since we can do the same reasoning for ${\gamma }_{B}$, we have $G({\gamma }_{A})+G({\gamma }_{B})\leqslant 2G({\gamma }_{A}\oplus {\gamma }_{B})$.□

We do not know whether G is also superadditive, which would make it additive. At present, we can only prove:

Corollary 4.7. Let ${\gamma }_{A}\in {{\mathbb{R}}}^{2{n}_{1}\times 2{n}_{1}}$ and ${\gamma }_{B}\in {Sp}(2{n}_{2})$, be two covariance matrices (i.e. ${\gamma }_{B}$ is a covariance matrix of a pure state). Then $G$ is additive.

Proof. Subadditivity has already been proven in the lemma. For superadditivity, we use the second reformulation of the squeezing measure in equation (24): note that there is only one matrix ${\gamma }_{B}\geqslant \gamma \geqslant {\rm{i}}\sigma$, namely ${\gamma }_{B}$ itself. Now write

$$\begin{eqnarray*}{\gamma }_{A}\oplus {\gamma }_{B}\geqslant \left(\begin{array}{cc}\tilde{A} & C\\ {C}^{T} & \tilde{B}\end{array}\right)\geqslant {\rm{i}}\sigma \end{eqnarray*}$$

for $\tilde{A}\in {{\mathbb{R}}}^{2{n}_{1}\times 2{n}_{1}}$ and $\tilde{B}\in {{\mathbb{R}}}^{2{n}_{2}\times 2{n}_{2}}$. Then in particular ${\gamma }_{B}-\tilde{B}\geqslant 0$, but also $\tilde{B}\geqslant {\rm{i}}\sigma$, hence ${\gamma }_{B}\geqslant \tilde{B}\geqslant {\rm{i}}\sigma$ and therefore $\tilde{B}={\gamma }_{B}$. But then

$$\begin{eqnarray*}{\gamma }_{A}\oplus {\gamma }_{B}-\left(\begin{array}{cc}\tilde{A} & C\\ {C}^{T} & \tilde{B}\end{array}\right)=\left(\begin{array}{cc}{\gamma }_{A}-\tilde{A} & C\\ {C}^{T} & 0\end{array}\right)\end{eqnarray*}$$

hence also C = 0 and the matrix that takes the minimum in $G({\gamma }_{A}\oplus {\gamma }_{B})$ must be block-diagonal. Then ${\gamma }_{A}\oplus {\gamma }_{B}\geqslant \tilde{A}\oplus {\gamma }_{B}\geqslant 0$ and $\tilde{A}$ is in the feasible set of $G({\gamma }_{A})$. □

Corollary 4.8. For any covariance matrices ${\gamma }_{A}\in {{\mathbb{R}}}^{2{n}_{1}\times 2{n}_{1}}$ and ${\gamma }_{B}\in {{\mathbb{R}}}^{2{n}_{2}\times 2{n}_{2}}$,

$$\begin{eqnarray*}G({\gamma }_{A})+G({\gamma }_{B})\leqslant 2G\left(\left(\begin{array}{cc}{\gamma }_{A} & C\\ {C}^{T} & {\gamma }_{B}\end{array}\right)\right).\end{eqnarray*}$$

If $G$ is superadditive, then this inequality holds without the factor of two.

Proof. 

$$\begin{eqnarray*}G({\gamma }_{A})+G({\gamma }_{B}) & \leqslant & 2G\left(\left(\begin{array}{cc}{\gamma }_{A} & 0\\ 0 & {\gamma }_{B}\end{array}\right)\right)=2G\left(\displaystyle \frac{1}{2}\left(\begin{array}{cc}{\gamma }_{A} & C\\ {C}^{T} & {\gamma }_{B}\end{array}\right)+\displaystyle \frac{1}{2}\left(\begin{array}{cc}{\gamma }_{A} & -C\\ -{C}^{T} & {\gamma }_{B}\end{array}\right)\right)\\ & \mathop{\leqslant }\limits^{(* )} & G\left(\left(\begin{array}{cc}{\gamma }_{A} & C\\ {C}^{T} & {\gamma }_{B}\end{array}\right)\right)+G\left(\left(\begin{array}{cc}{\gamma }_{A} & -C\\ -{C}^{T} & {\gamma }_{B}\end{array}\right)\right)\,\mathop{=}\limits^{(* * )}\,2G\left(\left(\begin{array}{cc}{\gamma }_{A} & C\\ {C}^{T} & {\gamma }_{B}\end{array}\right)\right).\end{eqnarray*}$$

Here we used proposition 4.6 and then convexity of G in $(\ast )$. Finally, in $(\ast \ast )$ we used that for every

$$\begin{eqnarray}\left(\begin{array}{cc}{\gamma }_{A} & C\\ {C}^{T} & {\gamma }_{B}\end{array}\right)\geqslant \left(\begin{array}{cc}{S}_{A} & \tilde{C}\\ {\tilde{C}}^{T} & {S}_{B}\end{array}\right){\left(\begin{array}{cc}{S}_{A} & \tilde{C}\\ {\tilde{C}}^{T} & {S}_{B}\end{array}\right)}^{T}\in {Sp}(2({n}_{1}+{n}_{2}))\end{eqnarray} \tag{ 33 }$$

we also have:

$$\begin{eqnarray}\left(\begin{array}{cc}{\gamma }_{A} & -C\\ -{C}^{T} & {\gamma }_{B}\end{array}\right)\geqslant \left(\begin{array}{cc}{S}_{A} & -\tilde{C}\\ -{\tilde{C}}^{T} & {S}_{B}\end{array}\right){\left(\begin{array}{cc}{S}_{A} & -\tilde{C}\\ -{\tilde{C}}^{T} & {S}_{B}\end{array}\right)}^{T}\in {Sp}(2({n}_{1}+{n}_{2}))\end{eqnarray} \tag{ 34 }$$

and vice versa. Since the two matrices on the right-hand side of equations (33) and (34) have equal spectrum, the two squeezing measures of the matrices on the left-hand side need to be equal. □

Let us give a few simple bounds on G.

Proposition 4.9 Let ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ be a valid covariance matrix and ${\lambda }^{\downarrow }({\rm{\Gamma }})$ be the vector of eigenvalues in decreasing order. Then:

$$\begin{eqnarray}&&-\displaystyle \frac{1}{2}\displaystyle \sum _{{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})\lt 1}\mathrm{log}({\lambda }_{i}^{\downarrow }({\rm{\Gamma }}))\leqslant G({\rm{\Gamma }})\leqslant \displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{n}\mathrm{log}{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})=F({{\rm{\Gamma }}}^{1/2}).\end{eqnarray} \tag{ 35 }$$

Proof. According to the Euler decomposition, a symplectic positive definite matrix has positive eigenvalues that come in pairs $s,{s}^{-1}$ and we can find $O\in {SO}(2n)$ such that for any ${S}^{T}S\leqslant {\rm{\Gamma }}$

$$\begin{eqnarray*}&&{O}^{T}{\rm{\Gamma }}O\geqslant \mathrm{diag}({s}_{1},\ldots ,{s}_{n},{s}_{1}^{-1},\ldots ,{s}_{n}^{-1}).\end{eqnarray*}$$

But then, ${\lambda }_{k}^{\downarrow }({\rm{\Gamma }})\geqslant {\lambda }_{k}^{\downarrow }(\mathrm{diag}({s}_{1},\ldots ,{s}_{n},{s}_{1}^{-1},\ldots ,{s}_{n}^{-1}))$ via the Weyl inequalities ${\lambda }_{i}^{\downarrow }(A)\geqslant {\lambda }_{i}^{\downarrow }(B)$ for all i and $A-B\geqslant 0$ (see [Bha96], theorem III.2.3). This implies:

$$\begin{eqnarray*}&&G({\rm{\Gamma }})\leqslant \displaystyle \sum _{i=1}^{n}\mathrm{log}(\max \{{s}_{i},{s}_{i}^{-1}\})\leqslant \displaystyle \sum _{i=1}^{n}\mathrm{log}{\lambda }_{i}^{\downarrow }{({\rm{\Gamma }})}^{1/2}.\end{eqnarray*}$$

For the lower bound, given an optimal matrix S with eigenvalues s_i, we have

$$\begin{eqnarray*}&&G({\rm{\Gamma }})=\displaystyle \sum _{i}\max \{{s}_{i},{s}_{i}^{-1}\}\mathrm{.}\end{eqnarray*}$$

If ${S}^{T}S={O}^{T}\mathrm{diag}({s}_{1}^{2},\ldots ,{s}_{n}^{2},{s}_{1}^{-2},\ldots ,{s}_{n}^{-2})O$ with $O\in {SO}(2n)$ is the diagonalisation of ${S}^{T}S$, we can write:

$$\begin{eqnarray*}&&{O}^{-T}{{\rm{\Gamma }}}^{-1}{O}^{-1}\leqslant \mathrm{diag}({s}_{1}^{2},\ldots ,{s}_{n}^{2},{s}_{1}^{-2},\ldots ,{s}_{n}^{-2})\end{eqnarray*}$$

and again by Weyl's inequalites, we can find for all $k\leqslant n$:

$$\begin{eqnarray}&&-\displaystyle \frac{1}{2}\displaystyle \sum _{i=2n-k+1}^{2n}\mathrm{log}({\lambda }_{i}^{\downarrow }({\rm{\Gamma }}))\leqslant \displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{k}\mathrm{log}{\lambda }_{i}^{\downarrow }(\mathrm{diag}({s}_{1}^{2},\ldots ,{s}_{n}^{2},{s}_{1}^{-2},\ldots ,{s}_{n}^{-2}))\leqslant G({\rm{\Gamma }}).\end{eqnarray} \tag{ 36 }$$

Now, $-\tfrac{1}{2}{\sum }_{i=2n-k+1}^{2n}{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})$ can be upper bounded by restricting to eigenvalues ${\lambda }_{i}^{\downarrow }({\rm{\Gamma }})\lt 1$. This implies

$$\begin{eqnarray*}&&-\displaystyle \frac{1}{2}\displaystyle \sum _{{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})\lt 1}\mathrm{log}({\lambda }_{i}^{\downarrow }({\rm{\Gamma }}))\leqslant G({\rm{\Gamma }})\end{eqnarray*}$$

using that the number of eigenvalues ${\lambda }_{i}({\rm{\Gamma }})\lt 1$ can at most be n (hence $k\leqslant n$ in the inequality of equation (36)), since ${\rm{\Gamma }}\geqslant {S}^{T}S$ and S^TS has at least n eigenvalues bigger than one. □

Numerics suggest that the lower bound is often very good for low dimensions. In fact, it can sometimes be achieved:

Proposition 4.10. Let ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ be a covariance matrix, then $G$ achieves the lower bound in equation (35) if there exists an orthonormal eigenvector basis ${\{{v}_{i}\}}_{i=1}^{2n}$ of Γ with ${v}_{i}^{T}{{Jv}}_{j}={\delta }_{i,n+j}$. Conversely, if G achieves the lower bound, then ${v}_{i}^{T}{{Jv}}_{j}=0$ for all normalised eigenvectors ${v}_{i},{v}_{j}$ of ${\rm{\Gamma }}$ with ${\lambda }_{i},{\lambda }_{j}\lt 1$.

Proof. Suppose that the lower bound in equation (35) is achieved. Via Weyl's inequalities (see [Bha96] theorem III.2.3), for all ${S}^{T}S\leqslant {\rm{\Gamma }}$ in the definition of G we have ${\lambda }_{i}^{\downarrow }({S}^{T}S)\leqslant {\lambda }_{i}^{\downarrow }({\rm{\Gamma }})$. For the particular S achieving G, this implies that for all ${\lambda }_{i}^{\downarrow }({\rm{\Gamma }})\lt 1$ we have ${\lambda }_{i}^{\downarrow }({S}^{T}S)={\lambda }_{i}^{\downarrow }({\rm{\Gamma }})$. But then ${\rm{\Gamma }}\geqslant {S}^{T}S$ implies that S^TS and Γ share all eigenvectors to the smallest eigenvalue. Iteratively, every eigenvector of Γ with ${\lambda }_{i}({\rm{\Gamma }})\lt 1$ must be an eigenvector of S^TS with the same eigenvalue.

Since the matrix diagonalising S^TS also diagonalises ${{ \mathcal C }}^{-1}({S}^{T}S)$, the eigenvectors of the two matrices are the same. Now, since ${{ \mathcal C }}^{-1}({S}^{T}S)\in { \mathcal H }$ by reformulation (25), for any eigenvector v_i of any eigenvalue ${{ \mathcal C }}^{-1}({\lambda }_{i})\lt 0$, Jv_i is also an eigenvector of ${{ \mathcal C }}^{-1}({S}^{T}S)$ to the eigenvalue $-{{ \mathcal C }}^{-1}({\lambda }_{i})$, implying ${v}_{i}^{T}{{Jv}}_{j}=0$ for all $i,j$. By definition, this means that $\{{v}_{i},{{Jv}}_{j}\}$ forms a symplectic basis. Above, we already saw that the eigenvectors of Γ for ${\lambda }_{i}({\rm{\Gamma }})\lt 1$ are also eigenvalues of S^TS, hence ${v}_{i}^{T}{{Jv}}_{j}=0$ for all i such that ${\lambda }_{i}({\rm{\Gamma }})\lt 1$.

Conversely, suppose we have an orthonormal basis ${\{{v}_{i}\}}_{i=1}^{2n}$ such that ${v}_{i}^{T}{{Jv}}_{j}={\delta }_{i,j+n}$ (modulo $2n$ if necessary) for all eigenvectors of Γ, i.e. Γ is diagonalisable by a symplectic orthonormal matrix $\tilde{O}\in U(n)$. Then

$$\begin{eqnarray*}&&\tilde{O}{\rm{\Gamma }}{\tilde{O}}^{T}=\mathrm{diag}({\lambda }_{1},\ldots ,{\lambda }_{2n}).\end{eqnarray*}$$

Since ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ we have ${\lambda }_{i}{\lambda }_{2i}\geqslant 1$. Assume that ${\lambda }_{i}\geqslant {\lambda }_{n+i}$ for all i = 1,...,n and the ${\lambda }_{n+i}$ are ordered in decreasing order. Then ${\lambda }_{n+r}\lt 1\leqslant {\lambda }_{n+r-1}$ for some $r\leqslant n$ and

$$\begin{eqnarray*}&&{S}^{T}S={\tilde{O}}^{T}\mathrm{diag}(1,\ldots ,1,{\lambda }_{r}^{-1},\ldots ,{\lambda }_{n}^{-1},1,\ldots ,1,{\lambda }_{n+r},\ldots ,{\lambda }_{2n})\tilde{O}\end{eqnarray*}$$

fulfils ${S}^{T}S\leqslant {\rm{\Gamma }}$ and obviously achieves the lower bound in equation (35).□

In contrast to this, the upper bound can be arbitrarily bad. For instance, consider the thermal state ${\rm{\Gamma }}=(2N+1)\cdot {\mathbb{1}}$ for increasing N. It can easily be seen that $G({\rm{\Gamma }})=0$, since ${\rm{\Gamma }}\geqslant {\mathbb{1}}\in {\rm{\Pi }}(n)$ and $F({\mathbb{1}})=0$, hence $G({\rm{\Gamma }})\leqslant 0$. However, the upper bound in equation (35) is $n/2\mathrm{log}(2N+1)\to \infty$ for $N\to \infty$, therefore arbitrarily bad.

We can achieve better upper bounds by using Williamson's normal form:

Proposition 4.11 (Williamson bounds).  Let ${\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}$ be such that ${\rm{\Gamma }}\geqslant {\rm{i}}J$ and consider its Williamson normal form ${\rm{\Gamma }}={S}^{T}{DS}$. Then:

$$\begin{eqnarray}&&F(S)-\mathrm{log}(\sqrt{\det ({\rm{\Gamma }})})\leqslant G({\rm{\Gamma }})\leqslant F(S).\end{eqnarray} \tag{ 37 }$$

Proof. Since $D\geqslant {\mathbb{1}}$ via ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$, the upper bound follows directly from the definition. Also, $F(S)\leqslant F({{\rm{\Gamma }}}^{1/2})$, which makes this bound trivially better than the spectral upper bound in equation (35).

The lower bound follows from:

$$\begin{eqnarray*}G({\rm{\Gamma }}) & \mathop{\geqslant }\limits^{(36)} & \displaystyle \frac{1}{2}\mathrm{log}\left(\displaystyle \prod _{i=n+1}^{2n}{\lambda }_{i}^{\downarrow }{({\rm{\Gamma }})}^{-1}\right)=\displaystyle \frac{1}{2}\mathrm{log}\displaystyle \frac{{\displaystyle \prod }_{i=1}^{n}{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})}{{\displaystyle \prod }_{i=1}^{2n}{\lambda }_{i}^{\downarrow }({\rm{\Gamma }})}\\ & = & F({{\rm{\Gamma }}}^{1/2})-\mathrm{log}(\det {({\rm{\Gamma }})}^{1/2})\geqslant F({({S}^{T}S)}^{1/2})-\mathrm{log}(\sqrt{\det ({\rm{\Gamma }})})\\ & = & F(S)-\mathrm{log}(\sqrt{\det ({\rm{\Gamma }})})\end{eqnarray*}$$

using Weyl's inequalities once again, implying that since ${S}^{T}S\leqslant {\rm{\Gamma }}$, we also have $F{(S)}^{2}=F({S}^{T}S)\leqslant F({\rm{\Gamma }})$.□

The upper bound here can also be arbitrarily bad. One just has to consider ${\rm{\Gamma }}:= {S}^{T}(N\cdot {\mathbb{1}})S$ with ${S}^{2}=\mathrm{diag}(N-1,\ldots ,N-1,{(N-1)}^{-1},\ldots ,{(N-1)}^{-1})\in {Sp}(2n)$. Then ${\rm{\Gamma }}\geqslant {\mathbb{1}}$, i.e. $G({\rm{\Gamma }})=0$, but $F(S)\to \infty$ for $N\to \infty$.

Proposition 4.12. Let ${\rm{\Gamma }}\geqslant {\rm{i}}{J}$ be a covariance matrix. Then

$$\begin{eqnarray}&&G({\rm{\Gamma }})\geqslant \displaystyle \frac{1}{4}\inf \{\parallel {\gamma }_{0}{\parallel }_{1}| \mathrm{log}{\rm{\Gamma }}\geqslant {\gamma }_{0},{\gamma }_{0}\in \pi (n)\},\end{eqnarray} \tag{ 38 }$$

where $\pi (n)$ was defined in proposition 2.5 as the Lie algebra of the positive semidefinite symplectic matrices. This infimum can be computed efficiently as a semidefinite programme.

Proof. Recall that the logarithm is operator monotone on positive definite matrices. Using this, we have:

$$\begin{eqnarray*}\begin{array}{rcl}G({\rm{\Gamma }}) & = & \mathrm{log}\,\inf \left\{\displaystyle \prod _{i=1}^{n}{\lambda }_{i}^{\downarrow }{({S}^{T}S)}^{1/2}| {\rm{\Gamma }}\geqslant {S}^{T}S\right\}\\ & \geqslant & \inf \left\{\displaystyle \sum _{i=1}^{n}\mathrm{log}{\lambda }_{i}^{\downarrow }{(\exp ({\gamma }_{0}))}^{1/2}| \mathrm{log}{\rm{\Gamma }}\geqslant {\gamma }_{0},{\gamma }_{0}\in \pi (n)\right\}\\ & = & \inf \left\{\displaystyle \frac{1}{2}\displaystyle \sum _{i=1}^{n}{\lambda }_{i}^{\downarrow }({\gamma }_{0})| \mathrm{log}{\rm{\Gamma }}\geqslant {\gamma }_{0},{\gamma }_{0}\in \pi (n)\right\}\\ & = & \inf \left\{\displaystyle \frac{1}{4}\displaystyle \sum _{i=1}^{2n}{s}_{i}^{\downarrow }({\gamma }_{0})| \mathrm{log}{\rm{\Gamma }}\geqslant {\gamma }_{0},{\gamma }_{0}\in \pi (n)\right\}\mathrm{.}\end{array}\end{eqnarray*}$$

The last step is valid, because the eigenvalues of matrices ${\gamma }_{0}\in \pi (n)$ come in pairs $\pm {\lambda }_{i}$. Since the sum of all the singular values is just the trace-norm, we are done.

It remains to see that this can be computed by a semidefinite programme. First note that since the matrices $H\in \pi (n)$ are those symmetric matrices with ${HJ}+{JH}=0$, the constraints are already linear semidefinite matrix inequalities. The trace norm is an SDP by standard reasoning [RFP10, VB96]:

$$\begin{eqnarray*}\parallel {\gamma }_{0}{\parallel }_{1}=\min \left\{\left.\displaystyle \frac{1}{2}\mathrm{tr}(A+B)\right|\left(\begin{array}{cc}A & {\gamma }_{0}\\ {\gamma }_{0} & B\end{array}\right)\geqslant 0\right\}\end{eqnarray*}$$

which is clearly a semidefinite programme.□

Numerics for small dimensions suggest that this bound is mostly smaller than the spectral lower bounds.

We first specify the preparation procedure. Since we want to quantify squeezing, it seems natural that we allow to freely draw states from the vacuum or a thermal bath to start with. Furthermore, we can perform an arbitrary number of the following operations for free:

• (1)  
  Add ancillary states also from a thermal bath or the vacuum.
• (2)  
  Add Gaussian noise.
• (3)  
  Implement any gate from linear optics.
• (4)  
  Perform Weyl-translations of the state.
• (5)  
  Perform selective or non-selective Gaussian measurements such as homodyne or heterodyne detection.
• (6)  
  Forget part of the state.
• (7)  
  Create convex combinations of ensembles.In addition, the following operation comes with an associated cost:
• (8)  
  Implement single-mode squeezers at a cost of $\mathrm{log}(s)$, where s is the squeezing parameter.

All these operations are standard operations in quantum optics and they should capture all important Gaussian operations except for squeezing.

It is well-known that all of these operations are captured by the following set of operations on the covariance matrix (for a justification, see appendix D):

• We can always draw N-mode states with $\gamma \in {{\mathbb{R}}}^{2N\times 2N}$ for any dimension N from the vacuum $\gamma ={\mathbb{1}}$ or a bath fulfiling $\gamma \geqslant {\mathbb{1}}$.
• We can always add ancillary modes from the vacuum ${\gamma }_{\mathrm{anc}}={\mathbb{1}}$ or a bath $\gamma \geqslant {\mathbb{1}}$ and consider $\gamma \oplus {\gamma }_{\mathrm{anc}}$.
• We can freely add noise with ${\gamma }_{\mathrm{noise}}\geqslant 0$ to our state, which is simply added to the covariance matrix of a state.
• We can perform any beam splitter or phase shifter and in general any operation $S\in K(n)$, which translates to a map $\gamma \mapsto {S}^{T}\gamma S$ on covariance matrices of states.
• We can perform any single-mode squeezer $S=\mathrm{diag}(1,\ldots ,1,s,{s}^{-1},1\ldots ,1)$ for some $s\in {{\mathbb{R}}}_{+}$.
• We can perform any Weyl-translation leaving the covariance matrix invariant.
• Given two states with covariance matrices ${\gamma }_{1}$ and ${\gamma }_{2}$, we can always take their convex combination $p{\gamma }_{1}+(1-p){\gamma }_{2}$ for any $p\in [0,1]$.
• At any point, we can perform a selective measurement of the system corresponding to a projection into a finitely or infinitely squeezed state. Given a state with covariance matrix $\gamma =\left(\begin{array}{cc}A & B\\ {B}^{T} & C\end{array}\right)$, this maps
  $$\begin{eqnarray*}&&\gamma \mapsto A-C{(B-{\gamma }_{G})}^{{\rm{MP}}}{C}^{T},\end{eqnarray*}$$
  where ^MP denotes the Moore–Penrose pseudoinverse.

Only operation (O4) comes at a cost of $\mathrm{log}(s)$, all other operations are free.

We are now ready to state our main theorem, which states that the minimal squeezing cost for any possible preparation procedure consisting of operations (1)–(8). is given by G.

Theorem 5.1. Let $\rho$ be a quantum state with covariance matrix $\gamma$. Consider arbitrary sequences

$$\begin{eqnarray*}&&{\vec{\gamma }}_{N}:= {\gamma }_{0}\to {\gamma }_{1}\to \cdots \,\to \,{\gamma }_{N},\end{eqnarray*}$$

where ${\gamma }_{0}$ fulfils (O0) and every arrow corresponds to an arbitrary operation (O1)–(O5) or (O7). Using (O6), we can merge two sequences ${\vec{\gamma }}_{{N}_{1}}$ and ${\vec{\gamma }}_{{N}_{2}}$ to one resulting tree with ${\gamma }_{{N}_{1}+{N}_{2}+1}=\lambda {\gamma }_{{N}_{1}}+(1-\lambda ){\gamma }_{{N}_{2}}$ for some $\lambda \in (0,1)$. Iteratively, we can construct trees of any depth and width using operations (O1)–(O7).

Let ${{\mathfrak{O}}}_{N}(\gamma )$ be the set of such trees with $N$ operations ending with γ (i.e. ${\gamma }_{N}=\gamma$). Let ${\mathfrak{O}}(\gamma )={\bigcup }_{N=1}^{\infty }{{\mathfrak{O}}}_{N}(\gamma )$.

Furthermore, for any tree $\hat{\gamma }\in {{\mathfrak{O}}}_{N}(\gamma )$, let $\vec{s}={\{{s}_{i}\}}_{i=1}^{M}$ be the sequence of the largest singular values of any single-mode squeezer (O4) implemented along the sequence (in particular, $M\leqslant N$). Then

$$\begin{eqnarray}&&G(\gamma )=\inf \left\{\displaystyle \sum _{i}\mathrm{log}{s}_{i}| {s}_{i}\in \vec{s},\hat{\gamma }\in {\mathfrak{O}}(\gamma )\right\}.\end{eqnarray} \tag{ 39 }$$

Since we consider many different operations, the proof is rather lengthy, where the main difficulties will be in showing that measurements do not squeeze. In order to increase readability, the proof will be split into several lemmata.

Lemma 5.2. Let $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$ be a covariance matrix, ${\gamma }_{0}\geqslant {\mathbb{1}}$, let $N\in {\mathbb{N}}$ and

$$\begin{eqnarray}&&{\gamma }_{0}\to {\gamma }_{1}\to \cdots \,\to \,{\gamma }_{N}=\gamma \end{eqnarray} \tag{ 40 }$$

be any sequence of actions (O1)–(O5) or (O7). If we denote the cost (sum of the logarithm of the largest singular values of any symplectic matrix involved) of this sequence by c, then one can replace this sequence by:

$$\begin{eqnarray}{\gamma }_{0} & \mathop{\to }\limits^{(O1)} & {\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}\mathop{\to }\limits^{(O2)}{\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}}\mathop{\to }\limits^{(O3),(O4)}{S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S\\ & \mathop{\to }\limits^{(O7)} & { \mathcal M }({S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S)\end{eqnarray} \tag{ 41 }$$

with ${\gamma }_{\mathrm{anc}}\geqslant {\mathbb{1}}$, ${\gamma }_{\mathrm{noise}}\geqslant 0$, $S\in {Sp}(2n)$ and ${ \mathcal M }$ a partial Gaussian measurement of type specified in (O7). For this sequence, $c\geqslant F(S)$.

Proof. We prove the proposition by proving that given any chain ${\gamma }_{0}\to {\gamma }_{1}\to \cdots \,\to \,{\gamma }_{N}=\gamma$ as in (40), we can interchange all operations and obtain a chain as in equation (41). For readability, we will not always specify the size of the matrices and we will assume that $\gamma \geqslant {\rm{i}}\sigma$, ${\gamma }_{\mathrm{anc}}\geqslant {\mathbb{1}}$, ${\gamma }_{\mathrm{noise}}\geqslant 0$, and S a symplectic matrix, whenever the symbols arise:

• (1)  
  We can combine any sequence ${\gamma }_{i}\to {\gamma }_{i+1}\to \cdots \,\to \,{\gamma }_{i+m}$ for some $m\in {\mathbb{N}}$ where each of the arrows corresponds to a symplectic transformation S_j, j = 1,...,m as in (O3) or (O4), into a single symplectic matrix $S\in {Sp}(2n)$ such that ${\gamma }_{i+m}={S}^{T}{\gamma }_{i}S$. Furthermore lemma 3.4 implies $F(S)\leqslant {\sum }_{i}{s}_{1}^{\downarrow }({S}_{i})$, hence this recombination of steps only lowers the amount of squeezing.
• (2)  
  Any sequence $\gamma \to {S}^{T}\gamma S\to {S}^{T}\gamma S+{\gamma }_{\mathrm{noise}}$ can be converted into a sequence $\gamma \to {S}^{T}(\gamma +{\tilde{\gamma }}_{\mathrm{noise}})S$ with the same S and hence the same costs by setting ${\tilde{\gamma }}_{\mathrm{noise}}:= {S}^{-T}{\gamma }_{\mathrm{noise}}{S}^{-1}\geqslant 0$.
• (3)  
  Any sequence $\gamma \to {S}^{T}\gamma S\to {S}^{T}\gamma S\oplus {\gamma }_{\mathrm{anc}}$ can be converted into a sequence $\gamma \to \gamma \oplus {\gamma }_{\mathrm{anc}}\to {\tilde{S}}^{T}(\gamma \oplus {\gamma }_{\mathrm{anc}})\tilde{S}$ by setting $\tilde{S}=S\oplus {\mathbb{1}}$ with ${\mathbb{1}}$ of the same dimension as ${\gamma }_{\mathrm{anc}}$. Since we only add the identity, we have $F(\tilde{S})={\sum }_{i}\mathrm{log}{s}_{i}^{\downarrow }(\tilde{S})=F(S)$ and the costs do not increase.
• (4)  
  Any sequence $\gamma \to \gamma +{\gamma }_{\mathrm{noise}}\to (\gamma +{\gamma }_{\mathrm{noise}})\oplus {\gamma }_{\mathrm{anc}}$ can be converted into a sequence $\gamma \to \gamma \oplus {\gamma }_{\mathrm{anc}}\to \gamma \oplus {\gamma }_{\mathrm{anc}}+{\tilde{\gamma }}_{\mathrm{noise}}$ by setting ${\tilde{\gamma }}_{\mathrm{noise}}={\gamma }_{\mathrm{noise}}\oplus 0\geqslant 0$, which is again a valid noise matrix. As no operation of type (O4) is involved, the squeezing costs do not change.

In a next step we consider measurements. We will only consider homodyne detection, since the proof is exactly the same for arbitrary Gaussian measurements of type (O7). Given a covariance matrix γ, we assume a decomposition

$$\begin{eqnarray*}\gamma =\left(\begin{array}{cc}A & C\\ {C}^{T} & B\end{array}\right);\qquad { \mathcal M }(\gamma )=A-C{(\pi B\pi )}^{{\rm{MP}}}{C}^{T}\end{eqnarray*}$$

as in the definition of (O7) with $\pi =\mathrm{diag}(1,0)$.

• (5)  
  Any sequence $\gamma \to { \mathcal M }(\gamma )\to {S}^{T}{ \mathcal M }(\gamma )S$ can be converted into a sequence $\gamma \to {\tilde{S}}^{T}\gamma \tilde{S}\to { \mathcal M }({\tilde{S}}^{T}\gamma \tilde{S})$ by setting $\tilde{S}=S\oplus {{\mathbb{1}}}_{2}$. To see this, write ${S}^{T}{ \mathcal M }{(\gamma )S={S}^{T}{AS}-{S}^{T}C(\pi B\pi )}^{{\rm{MP}}}{C}^{T}S$ and

  $$\begin{eqnarray*}{ \mathcal M }\left({\left(\begin{array}{cc}S & 0\\ 0 & {\mathbb{1}}\end{array}\right)}^{T}\left(\begin{array}{cc}A & C\\ {C}^{T} & B\end{array}\right)\left(\begin{array}{cc}S & 0\\ 0 & {\mathbb{1}}\end{array}\right)\right) & = & { \mathcal M }\left(\left(\begin{array}{cc}{S}^{T}{AS} & {S}^{T}C\\ {C}^{T}S & B\end{array}\right)\right)\\ & = & {S}^{T}{AS}-{S}^{T}C{(\pi B\pi )}^{{\rm{MP}}}{C}^{T}S\end{eqnarray*}$$

  hence the final covariance matrices are the same. By the same reasoning as in (3), the costs are equivalent.
• (6)  
  Any sequence $\gamma \to { \mathcal M }(\gamma )\to { \mathcal M }(\gamma )+{\gamma }_{\mathrm{noise}}$ can be converted into a sequence $\gamma \to \gamma +{\tilde{\gamma }}_{\mathrm{noise}}\to { \mathcal M }(\gamma +{\tilde{\gamma }}_{\mathrm{noise}})$ by setting ${\tilde{\gamma }}_{\mathrm{noise}}={\gamma }_{\mathrm{noise}}\oplus 0$, with 0 on the last mode being measured. Since no symplectic matrices are involved, the costs are equivalent.
• (7)  
  Any sequence $\gamma \to { \mathcal M }(\gamma )\to { \mathcal M }(\gamma )\oplus {\gamma }_{\mathrm{anc}}$ can be changed into a sequence $\gamma \to \gamma \oplus {\gamma }_{\mathrm{anc}}\to \tilde{{ \mathcal M }}(\gamma \oplus {\gamma }_{\mathrm{anc}})$, where the measurement $\tilde{{ \mathcal M }}$ measures the last mode of γ, i.e.

  $$\begin{eqnarray*}\tilde{M}\left(\left(\begin{array}{ccc}A & C & 0\\ {C}^{T} & B & 0\\ 0 & 0 & {\gamma }_{\mathrm{anc}}\end{array}\right)\right)=(A\oplus {\gamma }_{\mathrm{anc}})-(C\oplus 0){(\pi B\pi )}^{{\rm{MP}}}{(C\oplus 0)}^{T}.\end{eqnarray*}$$

  Clearly, the resulting covariance matrices of the two sequences are the same and the costs are equivalent.

We can now easily prove the lemma. Let ${\gamma }_{0}\to \cdots \to \,{\gamma }_{n}$ be an arbitrary sequence with operations of type (O1)–(O5) or (O7). We can first move all measurements to the right of the sequence, i.e. we first perform all operations of type (O1)–(O5) and then all measurements. This is done using the observations above. Note also that this step is similar to the quantum circuit idea to 'perform all measurements last' (see [NC00], chapter 4).

Similarly, we can combine operations of type (O3) and (O4) and rearrange the other operations to obtain a new sequence as in equation (41) with at most the costs of the sequence ${\gamma }_{1}\to \cdots \,\to \,{\gamma }_{m}$ we started with.□

We can now slowly work towards theorem 5.1:

Lemma 5.3. Let $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$ be a covariance matrix, then

$$\begin{eqnarray}&&G(\gamma )=\inf \{F(S)| \gamma ={S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S,\,S\in {Sp}(2n),\,{\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}\geqslant {\mathbb{1}},\,{\gamma }_{\mathrm{noise}}\geqslant 0\}\mathrm{.}\end{eqnarray} \tag{ 42 }$$

Proof. First note that for any $\gamma \geqslant {\rm{i}}\sigma$, we can find $S\in {Sp}(2n)$, ${\gamma }_{0}\in {{\mathbb{R}}}^{2n\times 2n}$ with ${\gamma }_{0}\geqslant {\mathbb{1}}$ and ${\gamma }_{\mathrm{noise}}\in {{\mathbb{R}}}^{2n\times 2n}$ with ${\gamma }_{\mathrm{noise}}\geqslant 0$ such that $\gamma ={S}^{T}({\gamma }_{0}+{\gamma }_{\mathrm{noise}})S$ by using Williamson's theorem, hence the feasible set is never empty. The lemma is immediate by observing that for any $\gamma ={S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S$ since $({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})\geqslant {\mathbb{1}}$ we have $\gamma \geqslant {S}^{T}S$ and conversely, for any $\gamma \geqslant {S}^{T}S$, defining ${\gamma }_{0}:= {S}^{-T}\gamma {S}^{-1}\geqslant {\mathbb{1}}$, we have $\gamma ={S}^{T}{\gamma }_{0}S$. □

As an intermediate step we introduce the following notation:

$$\begin{eqnarray}&&\begin{array}{l}\tilde{G}(\gamma ):= \inf \{F(S)| \gamma ={ \mathcal M }({S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S),\,S\in {Sp}(2n),\\ \quad \qquad \quad {\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}\geqslant {{\mathbb{1}}}_{2n},\,{\gamma }_{\mathrm{noise}}\geqslant 0,{ \mathcal M }\,\mathrm{measurement}\}\mathrm{.}\end{array}\end{eqnarray} \tag{ 43 }$$

Then we have:

Lemma 5.4. For $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$ a covariance matrix, we have

$$\begin{eqnarray}&&\tilde{G}(\gamma )=\inf \{F({\hat{\gamma }}^{1/2})| \gamma ={ \mathcal M }(\tilde{\gamma }),\tilde{\gamma }\geqslant \hat{\gamma }\geqslant {\rm{i}}\sigma ,\,{ \mathcal M }\,\mathrm{measurement}\}\mathrm{.}\end{eqnarray} \tag{ 44 }$$

Proof. This follows from lemma 5.3:

$$\begin{eqnarray}\tilde{G}(\gamma ) & = & \inf \{F(S)| \gamma ={ \mathcal M }({S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S)\}\\ & = & \inf \{F(S)| \gamma ={ \mathcal M }(\tilde{\gamma }),\tilde{\gamma }={S}^{T}({\gamma }_{0}\oplus {\gamma }_{\mathrm{anc}}+{\gamma }_{\mathrm{noise}})S\geqslant {\rm{i}}\sigma \}\\ & \mathop{=}\limits^{\mathrm{Lemma}\,5.3} & \inf \{G(\tilde{\gamma })| \gamma ={ \mathcal M }(\tilde{\gamma }),\tilde{\gamma }\geqslant {\rm{i}}\sigma \}\\ & \mathop{=}\limits^{\mathrm{Prop}.\,4.2} & \inf \{F({\hat{\gamma }}^{1/2})| \gamma ={ \mathcal M }(\tilde{\gamma }),\tilde{\gamma }\geqslant \hat{\gamma }\geqslant {\rm{i}}\sigma \}\end{eqnarray} \tag{ 45 }$$

by taking the infimum over all measurements last. □

Note here, that equation (45) together with the following proposition 5.5 finishes the proof of proposition 4.6 via:

$$\begin{eqnarray}&&G(\gamma )=\inf \{G(\tilde{\gamma })| \gamma ={ \mathcal M }(\tilde{\gamma }),\tilde{\gamma }\geqslant {\rm{i}}\sigma \}\leqslant G(\gamma \oplus a{{\mathbb{1}}}_{{n}_{2}})\end{eqnarray} \tag{ 46 }$$

for $a\geqslant 1$, using that measuring the last modes we obtain ${ \mathcal M }(\gamma \oplus a{{\mathbb{1}}}_{{n}_{2}})=\gamma$ and therefore, $\gamma \oplus a{{\mathbb{1}}}_{{n}_{2}}$ is in the feasible set of $\tilde{G}(\gamma )=G(\gamma )$.

Proposition 5.5. For $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$ a covariance matrix we have

$$\begin{eqnarray*}&&\tilde{G}(\gamma )=G(\gamma ).\end{eqnarray*}$$

This proposition shows that G is operational if we exclude convex combinations (and therefore also non-selective measurements).

Proof. Using lemma 5.4, the proof of this proposition reduces to the question whether:

$$\begin{eqnarray}&&\inf \{F({\hat{\gamma }}^{1/2})| \tilde{\gamma }\geqslant \hat{\gamma }\geqslant {\rm{i}}\sigma ,{ \mathcal M }(\tilde{\gamma })=\gamma \}=\inf \{F({\overline{\gamma }}^{1/2})| \gamma \geqslant \overline{\gamma }\geqslant {\rm{i}}\sigma \}\mathrm{.}\end{eqnarray} \tag{ 47 }$$

Since we do not need to use measurements, ≤ is obvious.

Let $\tilde{\gamma }\geqslant \hat{\gamma }\geqslant {\rm{i}}\sigma$ for some ${ \mathcal M }(\tilde{\gamma })=\gamma$. Our first claim is that

$$\begin{eqnarray}&&\gamma \geqslant { \mathcal M }(\hat{\gamma })\geqslant {\rm{i}}\sigma \end{eqnarray} \tag{ 48 }$$

${ \mathcal M }(\hat{\gamma })\geqslant {\rm{i}}\sigma$ is clear from the fact that $\hat{\gamma }$ is a covariance matrix and a measurement takes states to states. $\gamma \geqslant { \mathcal M }(\hat{\gamma })$ is proved using Schur complements. Let ${ \mathcal M }$ be a Gaussian measurement as in equation (68) with ${\gamma }_{G}=\mathrm{diag}(d,1/d)$ with $d\in {{\mathbb{R}}}^{+}$. It is well-known that

$$\begin{eqnarray*}&&{ \mathcal M }(\gamma )={({\mathbb{1}}\oplus \mathrm{diag}(1/d,d)\gamma ({\mathbb{1}}\oplus \mathrm{diag}(1/d,d))+0\oplus {{\mathbb{1}}}_{2})}^{S},\end{eqnarray*}$$

where ^S denotes the Schur complement of the block in the lower-right corner of the matrix. For homodyne measurements, we take the limit $d\to \infty$. Since for any $\tilde{\gamma }\geqslant \hat{\gamma }\geqslant 0$, the Schur complements of the lower right block fulfil ${\tilde{\gamma }}^{S}\geqslant {\hat{\gamma }}^{S}\geqslant 0$ (see [Bha07], exercise 1.5.7), we have $\gamma \geqslant { \mathcal M }(\hat{\gamma })$ as claimed in equation (48).

Next, we claim

$$\begin{eqnarray}&&F({ \mathcal M }{(\hat{\gamma })}^{1/2})\leqslant F({\hat{\gamma }}^{1/2}).\end{eqnarray} \tag{ 49 }$$

To prove this claim, note that via the monotonicity of the exponential function on ${\mathbb{R}}$, it suffices to prove

$$\begin{eqnarray*}&&\displaystyle \prod _{j=1}^{m}{s}_{j}^{\downarrow }({ \mathcal M }(\hat{\gamma }))\leqslant \displaystyle \prod _{j=1}^{n}{s}_{j}^{\downarrow }(\hat{\gamma })\end{eqnarray*}$$

when we assume $\hat{\gamma }\in {{\mathbb{R}}}^{2n\times 2n}$ and ${ \mathcal M }(\hat{\gamma })\in {{\mathbb{R}}}^{2m\times 2m}$ with $m\leqslant n$. If we write

$$\begin{eqnarray*}\hat{\gamma }=\left(\begin{array}{cc}\hat{A} & \hat{C}\\ {\hat{C}}^{T} & \hat{B}\end{array}\right)\end{eqnarray*}$$

then the state after measurement is given by ${ \mathcal M }{(\hat{\gamma })=\hat{A}-\hat{C}(\hat{B}+\mathrm{diag}(d,1/d))}^{-1}{\hat{C}}^{T}$ or the limit $d\to \infty$ for homodyne measurements. In any case $\hat{C}{(\hat{B}+\mathrm{diag}(d,1/d))}^{-1}{\hat{C}}^{T}\geqslant 0$ and ${ \mathcal M }(\hat{\gamma })\leqslant \hat{A}$ and therefore, by Weyl's inequalities, also

$$\begin{eqnarray*}&&\displaystyle \prod _{j=1}^{m}{s}_{j}^{\downarrow }({ \mathcal M }(\hat{\gamma }))\leqslant \displaystyle \prod _{j=1}^{m}{s}_{j}^{\downarrow }(\hat{A}).\end{eqnarray*}$$

Now we use Cauchy's interlacing theorem (see [Bha96], corollary III.1.5): as $\hat{A}$ is a submatrix of $\hat{\gamma }$, we have ${\lambda }_{i}^{\downarrow }(\hat{A})\leqslant {\lambda }_{i}^{\downarrow }(\hat{\gamma })$ for all $i=1,\ldots ,2m$. Since at least m eigenvalues of $\hat{A}$ are bigger or equal one and at least n eigenvalues of $\hat{\gamma }$ are bigger or equal one, this implies

$$\begin{eqnarray}&&\displaystyle \prod _{j=1}^{m}{s}_{j}^{\downarrow }(\hat{A})=\displaystyle \prod _{j=1}^{m}{\lambda }_{j}^{\downarrow }(\hat{A})\leqslant \displaystyle \prod _{j=1}^{m}{\lambda }_{j}^{\downarrow }(\hat{\gamma })\leqslant \displaystyle \prod _{j=1}^{n}{\lambda }_{j}^{\downarrow }(\hat{\gamma })=\displaystyle \prod _{j=1}^{n}{s}_{j}^{\downarrow }(\hat{\gamma }).\end{eqnarray} \tag{ 50 }$$

In particular, this proves equation (49).

We can then complete the proof: let $\tilde{\gamma }\geqslant \hat{\gamma }\geqslant {\rm{i}}\sigma$ for some ${ \mathcal M }(\tilde{\gamma })=\gamma$ in equation (47). We have just seen that this implies $\gamma \geqslant { \mathcal M }(\hat{\gamma })\geqslant {\rm{i}}\sigma$ via equation (48) and furthermore that $F{({\hat{\gamma }}^{1/2})\geqslant F({ \mathcal M }(\hat{\gamma })}^{1/2})$ via equation (49). But this means that we have found $\overline{\gamma }:= { \mathcal M }(\hat{\gamma })$ such that $\gamma \geqslant \overline{\gamma }\geqslant {\rm{i}}\sigma$. Hence $\overline{\gamma }$ is in the feasible set of the right-hand side of (47) and $F({\tilde{\gamma }}^{1/2})\geqslant F({\overline{\gamma }}^{1/2})$, which implies ≥ in equation (47).□

Finally, we can prove theorem 5.1 by also covering convex combinations:

Proof. Let $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$ be a covariance matrix. First consider only sequences $\vec{\gamma }$: we replace any sequence by the special type of sequence of lemma 5.3. For these sequences, we have seen that the minimum cost is given by $G(\gamma )$ in proposition 5.5.

However, we explicitly excluded convex combinations (O6) by considering only sequences and not trees $\hat{\gamma }$: consider a tree of operations (O1)–(O7) which has γ at its root and ${\gamma }_{0}={\mathbb{1}}$ as leaves. Let us consider any node closest to the leaves. At such a node, we start with two covariance matrices ${\gamma }_{1}$ and ${\gamma }_{2}$ that were previously constructed without using convex combinations and with costs $G({\gamma }_{1})$ and $G({\gamma }_{2})$. The combined matrix would be $\tilde{\gamma }:= \lambda {\gamma }_{1}+(1-\lambda ){\gamma }_{2}$ for some $\lambda \in (0,1)$ and the costs would be $\lambda G({\gamma }_{1})+(1-\lambda )G({\gamma }_{2})$.

By convexity of G (see theorem 4.3):

$$\begin{eqnarray*}&&G(\lambda {\gamma }_{1}+(1-\lambda ){\gamma }_{2})\leqslant \lambda G({\gamma }_{1})+(1-\lambda )G({\gamma }_{2})\end{eqnarray*}$$

which means that we can find a sequence (without any convex combinations) producing $\lambda {\gamma }_{1}+(1-\lambda ){\gamma }_{2}$ which is cheaper than first producing ${\gamma }_{1}$ and ${\gamma }_{2}$ and then taking a convex combination. Iteratively, this means we can eliminate every node from the tree and replace the tree by a sequence of operations (O1)–(O5) and (O7), which is cheaper than the tree and trees do not matter.□

We have now seen that the measure G can be interpreted as a measure of the amount of single-mode squeezing needed to create a state ρ. Let us now take a different perspective, which is the analogue of the entanglement of formation for squeezing: consider covariance matrices of the form

$$\begin{eqnarray}{\gamma }_{s}:= \left(\begin{array}{cc}s & 0\\ 0 & {s}^{-1}\end{array}\right).\end{eqnarray} \tag{ 51 }$$

These are single-mode squeezed states with squeezing parameter $s\geqslant 1$. We will now allow these states as resources and ask the question: given a (Gaussian) state ρ with covariance matrix γ, what is the minimal amount of these resources needed to construct γ, if we can freely transform the state by the same operations as before excluding squeezing ((O1)–(O7) excluding (O4)).

The corresponding measure is once again G:

Theorem 5.6. Let $\rho$ be an $n$-mode state with covariance matrix $\gamma \in {{\mathbb{R}}}^{2n\times 2n}$. Then

$$\begin{eqnarray}&&G(\gamma )=\inf \left\{\displaystyle \sum _{i=1}^{m}\displaystyle \frac{1}{2}\mathrm{log}({s}_{m})| \gamma ={ \mathcal T }\left.\,(\underset{i=1}{\overset{m}{\displaystyle \bigoplus }}{\gamma }_{{s}_{i}}\right)\right\},\end{eqnarray} \tag{ 52 }$$

where ${ \mathcal T }\,:{{\mathbb{R}}}^{2m\times 2m}\to {{\mathbb{R}}}^{2n\times 2n}$ is a combination of the operations (1)–(6) above.

Proof. ≤: Note that for any feasible $S\in {Sp}(2n)$ in $G(\gamma )$, i.e. any S with ${S}^{T}S\leqslant \gamma$, we can find $O\in {Sp}(2n)\cap O(2n)$ and $D={\bigoplus }_{i=1}^{n}{\gamma }_{{s}_{i}}$ with ${S}^{T}S={O}^{T}{DO}$ via the Euler decomposition. Using that the Euler decomposition minimises F, we have $F(S)=\tfrac{1}{2}F(D)={\sum }_{i=1}^{n}\tfrac{1}{2}\mathrm{log}({s}_{i})$. But then, since we can find ${\gamma }_{\mathrm{noise}}\geqslant 0$ such that $\gamma ={O}^{T}{\bigoplus }_{i=1}^{n}{\gamma }_{{s}_{i}}O+{\gamma }_{\mathrm{noise}}$, we have that D is a feasible resource state to produce γ. This implies ${G}^{\mathrm{resource}}(\gamma )\leqslant G(\gamma )$.

≥: For the other direction, the proof proceeds exactly as the proof of theorem 5.1. First, we exclude convex combinations. Then, we realise that we can change the order of the different operations (even if we include adding resource states during any stage of the preparation process) according to lemma 5.2, making sure that any preparation procedure can be implemented via:

$$\begin{eqnarray*}&&\gamma ={ \mathcal M }\left(O\left(\underset{i=1}{\overset{m}{\displaystyle \bigoplus }}{\gamma }_{{s}_{i}}\oplus {{\mathbb{1}}}_{2{m}^{\prime }}+{\gamma }_{\mathrm{noise}}\right){O}^{T}\right),\end{eqnarray*}$$

where $O\in {Sp}(2m+2{m}^{\prime })\cap O(2m+2{m}^{\prime })$, ${\gamma }_{\mathrm{noise}}\in {{\mathbb{R}}}^{2m+2{m}^{\prime }\times 2m+2{m}^{\prime }}$ with ${\gamma }_{\mathrm{noise}}\geqslant 0$ and ${ \mathcal M }$ a measurement. Now the only difference to proof of 5.1 is that we had the vacuum ${\mathbb{1}}$ instead of ${\bigoplus }_{i=1}^{m}{\gamma }_{{s}_{i}}\oplus {{\mathbb{1}}}_{2{m}^{\prime }}$ and an arbitrary symplectic matrix S instead of O, but the two ways of writing the maps are completely interchangeable, so that the proof proceeds as in theorem 5.1.□

We could call this measure the '(Gaussian) squeezing of formation', as it is the analogue to the Gaussian entanglement of formation. Note also that the measure is similar to the Gaussian entanglement of formation as defined in [Wol+04]. One natural further question would be whether 'distillation of squeezing' is possible with Gaussian operations. It is impossible in some sense for the minimal eigenvalue via [Kra+03], while it is possible and has been investigated for non-Gaussian states in many papers (see [Fil13, Hee+06] and references therein). In our case, it is not immediately clear whether extraction of single-mode squeezed states with less squeezing is possible or not. This could be investigated in future work.

Proposition 6.1. Let $n=1$, then $G({\rm{\Gamma }})=-\tfrac{1}{2}\,{\min }_{i}\,\mathrm{log}({\lambda }_{i}({\rm{\Gamma }}))$ for all ${\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}$.

Proof. Note that this is the lower bound in proposition 4.9, hence $-\tfrac{1}{2}\,{\min }_{i}\,\mathrm{log}({\lambda }_{i}({\rm{\Gamma }}))\leqslant G({\rm{\Gamma }})$. Now consider the diagonalisation ${\rm{\Gamma }}=O\,\mathrm{diag}({\lambda }_{1},{\lambda }_{2}){O}^{T}$ with $O\in {SO}(2)$ and assume ${\lambda }_{1}\geqslant {\lambda }_{2}$. Then, ${\lambda }_{2}^{-1}\leqslant {\lambda }_{1}$ since otherwise, ${\rm{\Gamma }}\,\not\hspace{-2pt}{\geqslant }\,{\rm{i}}{J}$.

Consider $\mathrm{diag}({\lambda }_{1},{\lambda }_{2})\geqslant {O}^{-T}{S}^{T}{{SO}}^{-1}$ for some $S\in {Sp}(2)$ with eigenvalues $s\geqslant 1$ and ${s}^{-1}$. Since $\mathrm{diag}({\lambda }_{1},{\lambda }_{2})\geqslant {O}^{-T}{S}^{T}{{SO}}^{-1}$, this implies in particular that ${s}^{-1}\leqslant {\lambda }_{2}$ by Weyl's inequality. Since $F({S}^{T}S)=\mathrm{log}s$, in order to minimise F(S) over ${S}^{T}S\leqslant {\rm{\Gamma }}$, we need to maximize ${s}^{-1}$. Setting ${s}^{-1}={\lambda }_{2}$ we obtain $s={\lambda }_{2}^{-1}\leqslant {\lambda }_{1}$ and $\mathrm{diag}({\lambda }_{1},{\lambda }_{2})\geqslant \mathrm{diag}(s,{s}^{-1})$. Since ${SO}(2)=K(1)$, ${S}^{T}S:= {O}^{T}\,\mathrm{diag}({\lambda }_{1},{\lambda }_{2})O\leqslant {\rm{\Gamma }}$ is the minimising matrix in G and $G({\rm{\Gamma }})=F(S)=\tfrac{1}{2}\mathrm{log}{\lambda }_{2}^{-1}$.□

Proposition 6.2. Let $\rho$ be a pure, Gaussian state with covariance matrix ${\rm{\Gamma }}\in {{\mathbb{R}}}^{2n\times 2n}$. Then $G({\rm{\Gamma }})=F({{\rm{\Gamma }}}^{1/2})$.

Proof. From proposition 2.2, we know that $\det ({\rm{\Gamma }})=1$ in particular. Therefore, the bounds in proposition 4.11 are tight and $G({\rm{\Gamma }})=F({{\rm{\Gamma }}}^{1/2})$. □

The crucial observation to numerically find the optimal squeezing measure is given in lemma 4.4: if we use G in the form of equation (25), we know that the function to be minimised is convex on ${ \mathcal H }$. In general, convex optimisation with convex constraints is efficiently implementable and there is a huge literature on the topic (see [BV04] for an overview).

In our case, a certain number of problems occur when performing convex optimisation:

• (1)  
  The function f in equation (28) is highly nonlinear. It is also not differentiable at eigenvalue crossings of A + iB or $H\in { \mathcal H }$. In particular, it is not differentiable when one of the eigenvalues becomes zero, which is to be expected at the minimum.
• (2)  
  While the constraints ${{ \mathcal C }}^{-1}(\gamma )\geqslant H$ and ${\mathbb{1}}\gt H\gt -{\mathbb{1}}$ are linear in matrices, they are nonlinear in simple parameterisations of matrices.
• (3)  
  For γ on the boundary of the set of allowed density operators, the set of feasible solutions might not have an inner point.

The first and second problem imply that most optimisation methods are unsuitable, as they are either gradient-based or need more problem structure. It also means that there is no guarantee for good stability of the solutions. The third problem implies that interior point methods become unsuitable on the boundary, which limits applications. For instance, our example of the next section (see equation (53)) lies on the boundary. As a proof of principle implementation, we used the Matlab-based solver SolvOpt (for details see the manual [KK97]). We believe our implementation could be made more efficient and more stable, but it seems to work well in most cases for less than ten modes. More information on the programme is provided in appendix E.

Let us now work with a particular example that has been studied in the quantum information literature. In [MK08], Mišta Jr and Korolkova define the following three-parameter group of three-mode states where the modes are labelled $A,B,C$:

$$\begin{eqnarray}&&\gamma ={\gamma }_{{AB}}\oplus {{\mathbb{1}}}_{C}+x({q}_{1}{q}_{1}^{T}+{q}_{2}{q}_{2}^{T})\end{eqnarray} \tag{ 53 }$$

with

$$\begin{eqnarray*}\begin{array}{rcl}{\gamma }_{{AB}} & = & \left(\begin{array}{cccc}{e}^{2d}a & 0 & -{e}^{2d}c & 0\\ 0 & {e}^{-2d}a & 0 & {e}^{-2d}c\\ -{e}^{2d}c & 0 & {e}^{2d}a & 0\\ 0 & {e}^{-2d}c & 0 & {e}^{-2d}a\end{array}\right),\\ {q}_{1} & = & {(0,\sin \phi ,0,-\sin \phi ,\sqrt{2},\sqrt{2})}^{T},\\ {q}_{2} & = & {(\cos \phi ,0,\cos \phi ,0\sqrt{2},\sqrt{2})}^{T},\end{array}\end{eqnarray*}$$

where $a=\cosh (2r)$, $c=\sinh (2r)$, $\tan \phi ={e}^{-2r}\sinh (2d)+\sqrt{1+{e}^{-4r}{\sinh }^{2}(2d)}$. The remaining parameters are $d\geqslant r\gt 0$ and $x\geqslant 0$. For

$$\begin{eqnarray*}&&x={x}_{\mathrm{sep}}\geqslant \displaystyle \frac{2\sinh (2r)}{{e}^{2d}{\sin }^{2}\phi +{e}^{-2d}{\cos }^{2}\phi }\end{eqnarray*}$$

the state becomes fully separable [MK08]. The state as such is a special case of a bigger family described in [Gie+01]. In [MK08], it was used to entangle two systems at distant locations using fully separable mediating ancillas (here the system labelled C). Therefore, Mišta Jr and Korolkova considered also an LOCC procedure to prepare the state characterised by (53). For our purposes, this is less relevant and we allow for arbitrary preparations of the state. This was also done in [MK08] by first preparing modes A and B each in a pure squeezed-state with position quadratures ${e}^{2(d-r)}$ and ${e}^{2(d+r)}$. A vacuum mode in C was added and $x({q}_{1}{q}_{1}^{T}+{q}_{2}{q}_{2}^{T})$ was added as random noise. Therefore, the squeezing needed to produce this state in this protocol is given by

$$\begin{eqnarray}&&c=\displaystyle \frac{1}{2}\mathrm{log}({e}^{2(d-r)}\cdot {e}^{2(d+r)})=2d.\end{eqnarray} \tag{ 54 }$$

We numerically approximated the squeezing measure for ${\gamma }_{{ABC}}$, choosing $x={x}_{\mathrm{sep}}$, which leaves a two-parameter family of states. We chose parameters d and r according to

$$\begin{eqnarray}&&r=0.1+j\cdot 0.05,\qquad d=r+i\cdot 0.03\end{eqnarray} \tag{ 55 }$$

with $i,j\in \{1,\ldots ,30\}$ for a total of 900 data points. Since the algorithm is not an interior point algorithm as described above, to check the result, we reprepared the state in the following way:

• (1)  
  Let S be the symplectic matrix at the value optimum found by SolvOpt for a covariance matrix ${\gamma }_{{ABC}}$.
• (2)  
  Calculate ${S}^{-T}{\gamma }_{{ABC}}{S}^{-1}$ and calculate its lowest eigenvalue ${\lambda }_{2n}$.
• (3)  
  Define $\tilde{\gamma }:= {S}^{-T}{\gamma }_{{ABC}}{S}^{-1}+(1-\min \{1,{\lambda }_{2n}\}){\mathbb{1}}\geqslant {\mathbb{1}}$. Calculate the largest singular value of ${S}^{T}\tilde{\gamma }S-\gamma$.

If S was a feasible point, then ${S}^{T}\tilde{\gamma }S=\gamma$. Since it is obvious how to prepare $\tilde{\gamma }$ with operations specified in section 5, the largest singular value of ${S}^{T}\tilde{\gamma }S-\gamma$ is an indicator of how well we can approximate the state we want to prepare by a state with comparably low squeezing costs.

The results of the numerical computation are shown in figure 1. We computed the minimum both with the help of numerical and analytical subgradients and took the value with a better approximation error. At rare occasions, one algorithm failed to obtain a minimum. Possible reasons for this are discussed in appendix E. The optimal values computed by the algorithm are close to the lower bound and a lot better than the upper bound and the costs obtained by equation (54). One can easily see that ${\gamma }_{{ABC}}$ cannot achieve the spectral lower bound as the assumptions of lemma 4.10 are not met.

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