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Positive definite Williamson normal form

Statement

Let $r \geq 1$ , $ℏ > 0$ , and $J = ⨁_{j = 1}^{r} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ . For every real symmetric positive definite $2 r \times 2 r$ matrix $V$ , there are an interleaved symplectic matrix $S$ ( $S^{T} J S = J$ ) and positive numbers $ν_{1}, \dots, ν_{r}$ such that $V = S D S^{T}, D = diag (ν_{1}, ν_{1}, \dots, ν_{r}, ν_{r}) .$ Moreover, $V + i ℏ J / 2 ⪰ 0 ⟺ ν_{j} \geq ℏ / 2 for every j .$ Here $A ⪰ 0$ means the Hermitian matrix $A$ has nonnegative quadratic form on every complex vector.

Facts & Assumptions

Given: $r \geq 1$ , $ℏ > 0$ , and a real symmetric positive definite $V \in M_{2 r} (R)$ .

[A1]

The dimension is finite and positive, the action scale satisfies $ℏ > 0$ , and $V = V^{T}$ obeys $x^{T} V x > 0$ for every nonzero real $x$ .

[F1]

In finite-dimensional Euclidean space, a real symmetric matrix has an orthonormal eigenbasis: maximize its Rayleigh quotient on the compact unit sphere to obtain an eigenvector, restrict to its invariant orthogonal complement, and induct on dimension. A positive definite matrix therefore has a real symmetric positive definite square root.

[F2]

The local convention uses interleaved quadrature order with $J = \oplus_{j} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ (Finite-mode quadratures and symplectic data); this item tests positivity of the Hermitian matrix $V + i ℏ J / 2$ defined in its Statement.

Proof

1.1givenA1F1algebra

First diagonalize $V$ by the Rayleigh-quotient induction in [F1]: a maximizer of $x^{T} V x$ on the unit sphere is an eigenvector, because the derivative at zero of the quotient at $e + t v$ vanishes for every $v ⊥ e$ ; the orthogonal complement is invariant under a symmetric matrix, so induction gives an orthonormal eigenbasis. Positivity makes each eigenvalue positive, and replacing it by its positive square root constructs the real symmetric positive definite $M = V^{1 / 2}$ . Now let $K = M J M$ . Then $K^{T} = - K$ , and $K$ is invertible because both $M$ and $J$ are. The matrix $B = - K^{2} = K^{T} K$ is real symmetric positive definite. Apply the same Rayleigh-quotient argument to $B$ to choose a unit eigenvector $e$ with eigenvalue $ν^{2} > 0$ and put $f = - K e / ν$ . Since $∥ K e ∥^{2} = e^{T} B e = ν^{2}$ , $f$ is unit; since $K$ is skew, $e^{T} f = 0$ . Also $K e = - ν f$ and $K f = - K^{2} e / ν = ν e$ . Thus in the orthonormal ordered pair $(e, f)$ , $(\begin{array}{cc} e^{T} K e & e^{T} K f \\ f^{T} K e & f^{T} K f \end{array}) = (\begin{array}{cc} 0 & ν \\ - ν & 0 \end{array}) .$ For any $x$ orthogonal to both $e, f$ , skew-symmetry gives $e^{T} K x = - (K e)^{T} x = ν f^{T} x = 0$ and $f^{T} K x = - (K f)^{T} x = - ν e^{T} x = 0$ , so their orthogonal complement is $K$ -invariant. Repeating the argument on this complement gives an orthogonal matrix $O$ and positive $ν_{j}$ with $O^{T} K O = J D$ , where $D = diag (ν_{1}, ν_{1}, \dots, ν_{r}, ν_{r})$ .

2.1F2step 1.1algebra

Set $S = M O D^{- 1 / 2}$ . Then $S D S^{T} = M O D^{- 1 / 2} D D^{- 1 / 2} O^{T} M = M^{2} = V,$ and $S^{T} J S = D^{- 1 / 2} O^{T} M J M O D^{- 1 / 2} = D^{- 1 / 2} (J D) D^{- 1 / 2} = J,$ because $D$ commutes with $J$ . This proves the Williamson factorization and symplecticity of $S$ .

3.1step 2.1algebra∎

For symplectic $S$ , congruence by $S^{- 1}$ maps $V + i ℏ J / 2$ to $D + i ℏ J / 2$ , since $S^{- 1} J S^{- T} = J$ . In each $(Q_{j}, P_{j})$ block the latter matrix is $(\begin{array}{cc} ν_{j} & i ℏ / 2 \\ - i ℏ / 2 & ν_{j} \end{array}),$ whose eigenvalues are $ν_{j} + ℏ / 2$ and $ν_{j} - ℏ / 2$ . Congruence by an invertible matrix preserves nonnegative quadratic forms, so the full matrix is positive semidefinite exactly when every $ν_{j} \geq ℏ / 2$ .

Depends on

Finite-mode quadratures and symplectic data

Positive definite Williamson normal form

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources