Gaussian coherence-breaking channels and coherence measures

Abstract

We give a characterization of arbitrary n-mode Gaussian coherence-breaking channels (GCBCs) and construct a kind of Gaussian coherence measure based on the topic of GCBCs. We show the measure can be calculated conveniently.

Appendix

Proof of Theorem 3.2.

The “if” part is obvious, let us check the “only if” part.

If an n-mode Gaussian channel $\Phi =\Phi (K, M, \bar{d})$ is coherence breaking, then $\Phi$ is an incoherent operation. By Theorem 1 in [28], an n-mode Gaussian channel $\Phi =\Phi (K, M, \bar{d})$ is incoherent (i.e., $\Phi ({\mathcal {I}}_C^G)\subseteq {\mathcal {I}}_C^G$) if and only if $\bar{d}=0$ and there exists a permutation $\pi$ on $\{1,2, \ldots , n\}$ such that

$$\begin{aligned} K= & {} (P_\pi \otimes I_2) (\oplus _{i=1}^n t_i {\mathcal {O}}_i), \end{aligned}$$

(6.1)

$$\begin{aligned} M= & {} \oplus _{i=1}^n \lambda _{\pi (i)}I_2, \end{aligned}$$

(6.2)

where $P_\pi$ is the $n\times n$ permutation matrix corresponding to $\pi$, ${\mathcal {O}}_i$s are some $2\times 2$ orthogonal matrices and $\lambda _j\ge \frac{1}{2}|t_j^2- 1|$ whenever ${{\mathcal {O}}}_i$ is symplectic or $\lambda _j\ge \frac{1}{2}|t_j^2+ 1|$ otherwise.

To complete the proof, let us show each $t_i=0$ for any $1\le i\le n$. Assume on the contrary $t_i\not =0$ for some i. We take the n-mode input state $\rho$ with its covariance matrix $\nu _\rho = I_2\oplus \cdots \oplus I_2 \oplus \nu _i \oplus I_2 \oplus \cdots \oplus I_2$, where $\nu _i=\left( \begin{array} {ccccccccc} a &{}\quad c \\ c &{} \quad b\\ \end{array}\right) ,$ where $a\ge 0, b\ge 0$, and $ab\ge c^2+\frac{1}{4}$. Recall that a 2$\times$2 real orthogonal matrix has one of the following forms:

$$\begin{aligned} \left( \begin{array} {ccccccccc} \cos \theta &{}\quad \sin \theta \\ -\sin \theta &{} \quad \cos \theta \\ \end{array}\right) ,\quad \left( \begin{array} {ccccccccc} \cos \theta &{}\quad \sin \theta \\ \sin \theta &{} \quad -\cos \theta \\ \end{array}\right) . \end{aligned}$$

(6.3)

Next we divide the proof into two cases.

If ${\mathcal {O}}_i$ has the first form in ( 6.3), then $t_i {\mathcal {O}}_i \nu _i {\mathcal {O}}_i^T + M_i$ has to be diagonal according to ( 6.1) and ( 6.2). It follows from a short computation that

$$\begin{aligned} t_i[-a \cos \theta \sin \theta - c \sin ^2 \theta +c \cos ^2 \theta + b\cos \theta \sin \theta ]=0 \end{aligned}$$

(6.4)

holds for all real numbers a, b, c with $a\ge 0, b\ge 0$ and $ab\ge c^2+\frac{1}{4}$. As $t_i\ne 0$ in Eq. (6.4), taking $c\ne 0$ and $a= b$ leads to $\cos ^2 \theta =\sin ^2 \theta$, that is, $\cos \theta =\pm \sin \theta$. Thus, $(b-a)\cos \theta \sin \theta$ is always zero in Eq  (6.4). Taking $a\ne b$, we get either $\cos \theta =0$ or $\sin \theta =0$. It follows from $\cos \theta =\pm \sin \theta$ that $\cos \theta =\sin \theta =0$, a contradiction. So $t_i=0$

Similarly, if ${\mathcal {O}}_i$ has the second form, we also have $t_i=0$.

In summary, $t_i=0$ for all i. We complete the proof.$\square$

Proof of Theorem 4.1

It is obvious that $C_{CB}^G$ is always non-negative. A fact is that a Gaussian state $\rho$ is incoherent iff it is a thermal state (its covariance matrix is the multiple of the identity and displacement vector is zero [28]. Next we show $C_{CB}^G(\rho )=0$ iff $\rho$ is incoherent. On the one hand, if $\rho$ is incoherent, then $\nu _\rho =\omega _\rho I_2$ and $\bar{d}=0$. It follows that $\Vert \nu _\rho -\phi _0(\nu _\rho )\Vert +\Vert \bar{d}\Vert _2=0$ with the GCBC $\phi _0$ satisfying $\phi _0(\nu _\rho )=\omega _\rho I_2$. So $C_{CB}^G(\rho )=0$. On the other hand, if $C_{CB}^G(\rho )=0$, then $\bar{d}=0$ and $\mathrm{min}_{\phi \in \Omega _{CB}^G}\Vert \nu _\rho -\phi (\nu _\rho )\Vert =0$. It follows that there is $\phi _0\in \Omega _{CB}^G$ such that $\Vert \nu _\rho -\phi _0(\nu _\rho )\Vert =0$. Without loss of generality, assume that $\phi _0(\nu _\rho )=\lambda _\rho I_2$. So $\nu _\rho =\lambda _\rho I_2$. It follows that $\rho$ is incoherent. We complete the proof.$\square$

Proof of Theorem 4.2

By (4.2), it is enough to show $\Vert S_U\nu _\rho S_U^T-\omega I_2\Vert _F=\Vert \nu _\rho -\omega I_2\Vert _F$ for an arbitrary $2\times 2$ real symplectic matrix $S_U$. Indeed,

$$\begin{aligned} \begin{array}{llll}\Vert S_U\nu _\rho S_U^T-\omega I_2\Vert _F&{}=\Vert S_U(\nu _\rho -\omega I_2) S_U^T\Vert _F\\ &{}=\sqrt{\mathrm{tr}(S_U(\nu _\rho -\omega I_2)S_U^TS_U(\nu _\rho -\omega I_2)^TS_U^T)}\\ &{}=\sqrt{\mathrm{tr}(S_U(\nu _\rho -\omega I_2)^2S_U^T)}\\ &{}=\sqrt{\mathrm{tr}((\nu _\rho -\omega I_2)^2)}\\ &{}=\Vert \nu _\rho -\omega I_2\Vert _F.\end{array} \end{aligned}$$

We complete the proof.$\square$

Proof of Theorem 4.3

Now for an arbitrary incoherent Gaussian channel $\Psi [t,O,N,\bar{d}]$ with $t\le 1$,

$$\begin{aligned} \begin{array}{ll} C_{CB}^G(\Psi (\rho ))&{} = \mathrm{min}_{\psi \in \Omega _{CB}^G}\Vert \psi (\nu _\rho )-\phi (\psi (\nu _\rho ))\Vert _F+\Vert {{\pm \sqrt{t}\mathcal O}}\bar{d}\Vert _2 \\ &{} \le \mathrm{min}_{\omega ^\prime \in \Omega _T}\Vert t {\mathcal {O}}^T \nu _\rho {\mathcal {O}} + \omega I_2-\omega ^\prime I_2\Vert _F+\Vert \bar{d}\Vert _2 \ \ \ \ \ \ \ (\phi (\psi (\nu _\rho ))=\omega ^\prime I_2)\\ &{} \le \mathrm{min}_{\omega ^\prime \in \Omega _T}\Vert {\mathcal {O}}^T (t\nu _\rho + \omega I_2-\omega ^\prime I_2) {\mathcal {O}}\Vert _F+\Vert \bar{d}\Vert _2\\ &{} \le \mathrm{min}_{\omega ^\prime }\Vert t\nu _\rho + \omega I_2-(\omega ^\prime +\omega ) I_2\Vert _F+\Vert \bar{d}\Vert _2\\ &{}= \sqrt{2}|tc|+\sqrt{d_1^2+d_2^2} \\ &{} \le \sqrt{2}|c|+\sqrt{d_1^2+d_2^2}\\ &{} =C_{CB}^G(\rho ),\end{array} \end{aligned}$$

where $\nu _\rho =\left( \begin{array}{cc} a &{} \quad c \\ c &{} \quad a \\ \end{array}\right)$. We complete the proof.$\square$

Proof of Theorem 4.4

Now for an arbitrary n-mode incoherent Gaussian channel $\Psi [t_i, O_i, M, \bar{d}]$ with each $t_i\le 1$,

$$\begin{aligned} C_{CB}^G(\Psi (\rho ))= & {} \mathrm{min}_{\psi \in \Omega _{CB}^G}\Vert \psi (\nu _\rho )\\&-\phi (\psi (\nu _\rho ))\Vert _F+\Vert {{{K}}}\bar{d}\Vert _2\\\le & {} \mathrm{min}_{\omega ^\prime \in \Omega _T^n}\Vert (P_\pi \otimes I_2)(\oplus _{i=1}^n t_i {\mathcal {O}}_i) \nu _\rho (\oplus _{i=1}^n t_i {\mathcal {O}}_i^T)(P_\pi \otimes I_2)\\&+ \oplus _i N_i-\oplus _{i=1}^n\omega ^\prime _i I_2\Vert _F+\Vert \bar{d}\Vert _2 \\= & {} \mathrm{min}_{\omega ^\prime \in \Omega _T^n}\Vert (P_\pi \otimes I_2)(\oplus _{i=1}^n t_i {\mathcal {O}}_i)( \nu _\rho \\&+ \oplus _i N_i-\oplus _{i=1}^n\omega ^\prime _i I_2)(\oplus _{i=1}^n t_i {\mathcal {O}}_i^T)(P_\pi \otimes I_2)\Vert _F+\Vert \bar{d}\Vert _2 \\= & {} \mathrm{min}_{\omega ^\prime \in \Omega _T^n}\Vert (\oplus _{i=1}^n {\mathcal {O}}_i) (\oplus _{i=1}^n t_i I_2)( \nu _\rho \\&+ \oplus _i N_i-\oplus _{i=1}^n\omega ^\prime _i I_2)(\oplus _{i=1}^n t_i I_2)(\oplus _{i=1}^n {\mathcal {O}}_i^T)\Vert _F+\Vert \bar{d}\Vert _2\\= & {} \mathrm{min}_{\omega ^\prime \in \Omega _T^n}\Vert (\oplus _{i=1}^n t_i I_2)( \nu _\rho \\&+ \oplus _i N_i-\oplus _{i=1}^n\omega ^\prime _i I_2)(\oplus _{i=1}^n t_i I_2)\Vert _F+\Vert \bar{d}\Vert _2\\= & {} \mathrm{min}_{\omega ^\prime \in \Omega _T^n}\Vert \nu _\rho \\&+ \oplus _i N_i-\oplus _{i=1}^n\omega ^\prime _i I_2\Vert _F+\Vert \bar{d}\Vert _2\\= & {} \sqrt{2}\sum _{i\ne j}^{2n}|t_it_jv_{ij}|+\sqrt{\sum _{j=1}^{2n}d_j^2}\\\le & {} \sqrt{2}\sum _{i\ne j}^{2n}|v_{ij}|+\sqrt{\sum _{j=1}^{2n}d_j^2}\\= & {} C_{CB}^G(\rho ), \end{aligned}$$

where $\nu _\rho$ is an n-mode Gaussian state $\rho$ with the covariance matrix $\nu _\rho =(v_{ij})_{2n\times 2n}$ satisfying $v_{2i-1, 2i-1}=v_{2i,2i}=a_i\ge 1$ and displacement vector $\bar{d}= (d_1, d_2, \ldots , d_{2n})$. We complete the proof.$\square$