# How is the determinant of the correlations block a symplectic invariant in a covariance matrix

A well-known result is that every two-mode covariance matrix

$$\sigma_{AB} = \pmatrix{\alpha & \gamma \\ \gamma^T & \beta}$$

can be transformed via a symplectic transformation into a standard form with all diagonal 2 × 2 subblocks, $\alpha = diag(a,a)$, $\beta = diag(b,b)$, $\gamma = diag(c,d)$, where $a,b\geq1$, $c\geq|d|\geq0$. Four symplectic invariants can be identified, namely $\det(\alpha )$, $\det(\beta )$, $\det(\gamma )$, and $\det(\sigma_{AB})$.

The transformation can be done in steps by first performing symplectic diagonalization on the blocks $\alpha$ and $\beta$ (Williamson's theorem), leaving the correlations blocks still non-diagonal. The tranformation to standard form is completed using symplectic singular value decomposition of the correlations block; by now, the diagonal blocks are proportional to identity matrix and will not be affected by the orthogonal matrices diagonalizing $\gamma$ (see, e.g., section 2.3 in this reference).

However, aren't the singular values always positive, making $c,d\geq 0$? How can $\det(\gamma )$ be left invariant if we consider initially a $\sigma_{AB}$ with $\det(\gamma )<0$?

I must have misunderstood something about the transformation but I can't figure out what.

• I do not see what kind of transformation would allow you to change the sign of $\det(γ)$? To take a physics example, suppose you have $γ=\text{diag}(c,-c)$: To converting it to $γ=\text{diag}(c,c)$, you would need an operation which convert $P_A↦-P_A$, while keeping $Q_A$, $Q_B$ and $P_B$ constant. Such an operation is neither physical, nor symplectic. Jul 24 '17 at 15:41
• Isn't that transformation exactly the singular value decomposition?
– Kiro
Jul 25 '17 at 6:20
• No! Otherwise symplectic diagonalization would be named singular value decomposition. SVD uses orthogonal matrices (which preserve the identity), while symplectic diagonalization uses symplectic matrices (which preserve $Ω=\begin{bmatrix} 0,& 1\\-1,&0\end{bmatrix}^{⊗n}$. A squeezing operation is symplectic but not orthogonal. A rotation exchanging $Q_A$ and $Q_B$ without changing $P_A$ and $P_B$ is orthogonal, but not symplectic Jul 31 '17 at 15:03
• Got it. How would I go about finding the diagonal values of $\gamma$ then? I understand that Williamson's theorem applies to positive definite matrices, which $\gamma$ in general will not be, and Bloch-Messiah decomposition is for symplectic matrices which $\gamma$ in general will not be.
– Kiro
Aug 1 '17 at 5:46

$\DeclareMathOperator\diag{diag} \newcommand\Id{\mathbb I}$ Since I’m a physicist — and this question was originally migrated from physics.sx (Why? It’s a standard quantum optic question to me!) — I’ll answer to this question using physics techniques.

## I. Symplectic operations in quantum optics

Symplectic algebra offers a way to simply encode the Heisenberg uncertainty principle for multimode quantum optics: for a $n$ modes $2n×2n$ covariance matrix $σ$ where the quadatures are alternating ($Q₁, P₁, Q₂, P₂, …$) the Heisenberg uncertainty principle is \begin{align} σ + iΩ &≥0 & \text{with }& Ω=ω^{⊕n}= \begin{bmatrix}0&1\\-1&0\end{bmatrix}^{⊕n}\\ && &\text{in units where } \frac\hbar2≝1. \end{align} Physical transformations are the one which preserve this relations. Let us restrict ourselves to transformations $M$ linear in quadratures.The covariance matrix transforms as $σ↦M^TσM$. If we want $M$ to preserve Heisenberg uncertainty principle, it should preserve $Ω$: it is a symplectic transformations, defined by $M^TΩM=Ω.$

Some example of symplectic and non symplectic operations:

• Special orthogonal operations on a single mode, defined by rotation matrices $R⊕\mathbb I_{2(n-1)}$, corresponds to a dephasing in this mode. This operation is both orthogonal and symplectic.
• Squeezing operations, $M=\diag\left(s_1,\frac1{s_1},s_2,\frac1{s_2}, \dots\right)$, are symplectic, but not orthogonal.
• A beamsplitter between modes $1$ and $1$ is corresponds to the following symplectic and orthogonal operation $M=\begin{bmatrix}\Id_2\cosθ & -\Id_2\sinθ \\ \Id_2\sinθ & \Id_2\cosθ\end{bmatrix}⊕\Id_{2(n-2)}$.

The above operation generates the whole symplectic group. Note that it contains a non-orthogoal operations (squeezing operations) and that some orthogonal operations are excluded (like rotations involving $P_1$ and $Q_2$: e.g. $\begin{bmatrix} 1& 0& 0& 0\\ 0&0&1&0\\0&1&0&0\\0&0&0&1 \end{bmatrix}⊕\Id_{2(n-2)}$). Therefore, symplectic diagonalizations is not SVD, since the set of allowed transformations is different.

Another set of orthogonal but not symplectic operations are orthogonal operations of determinant $-1$ (reflections), which is essentially the answer to your question about why $c$ and $d$ cannot always be made positive.

## II. Local transformation of a 2-mode covariance matrix $σ_{AB}$ in its standard form

The task here is to transform a a two mode covariance matrix $σ_{AB}$ $$σ_{AB}=\begin{bmatrix} α & γ \\ γ^T & β\end{bmatrix}$$ in a standard form where the $2×2$ submatrices $α$, $β$ and $γ$ are diagonal and, if possible, $∝\Id_2$, using local operations, that is dephasing and squeezing.

Since $α$ is symmetric, it can be diagonalized into $\diag(a_1, a_2)$ using the local special orthogonal transformation $R_a⊕\Id_2$. This operation changes $γ$, but not $β$. Applying the squeezing operation $S_a=\diag\left(\sqrt{\frac{a_2}{a_1}}, \sqrt{\frac{a_1}{a_2}}, 1, 1\right)$ finally transforms $α$ into $a\Id_2$, with $a=\sqrt{a_1a_2}$, further changing $γ$, but not transforming $β$. The same method, using $\Id_2⊕S_bR_b$, transforms $β$ into $b\Id_2$ without changing $a\Id_2$.

Our covariance matrix is now under the form $\begin{bmatrix} a\Id_2 & γ' \\ γ'^T & b\Id_2\end{bmatrix}$.The SVD of $γ'$ gives us $c$,$d$ and the orthogonal operations $O_1$ and $O_2$ s.t. $\diag(|c|,|d|)=O_1^T γ' O_2$. Let us define the rotations $R_i≝\diag(1,\det(O_i))O_i$. We have then $\diag(c,d)=O_1^T γ' O_2$. Let us now apply $R_1⊕R_2$ to our covariance matrix:

• It diagonalizes $γ'$ and $γ'^T$, as expected.
• On the mode $A$, it has the effect of the rotation $R_1$, but since the covaraince matrix of this mode is $a\Id_2$, it is invariant by rotation, and it does not change.
• Similarly, the mode $B$ does not change.

The matrix is now under the standard form $\begin{bmatrix} a& &c\\ &a& &d\\c& &b\\ &d& &b\end{bmatrix}$. One could be tempted to change the balance beween $c$ and $d$ by squeezing operations, but it would destroy the balance between $a_1$ and $a_2$. All we can do is exhanging $c$ and $d$, and change both their signs simultaneously.

• Thanks a lot for taking the time to write this answer, this clarified the process for me. Cheers!
– Kiro
Aug 4 '17 at 9:03