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The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$

Example

Let $n = 1$ , $U = R^{2}$ with coordinates $(q, p)$ , and $H (q, p) = (p^{2} + q^{2}) / 2$ . Then:

$X_{H} = (p, - q)$ and its flow is $Φ_{t} (q, p) = (q \cos t + p \sin t, - q \sin t + p \cos t)$ , defined for all $t \in R$ ;
the derivative $Y (t) = D Φ_{t} (q, p) = (\begin{array}{cc} \cos t & \sin t \\ - \sin t & \cos t \end{array})$ satisfies $Y^{'} (t) = J H e s s H Y (t)$ , $Y (0) = I_{2}$ , $Y (t)^{T} J Y (t) = J$ and $\det Y (t) = 1$ for all $t$ , in accordance with The derivative of the Hamiltonian flow satisfies $Y^{T} J Y = J$ and $\det Y = 1$ .

Facts & Assumptions

Given: The Hamiltonian $H (q, p) = (p^{2} + q^{2}) / 2$ on $R^{2}$ and the matrix $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ .

[F1]

Hamiltonian field (The Hamiltonian field of a $C^{2}$ function on a phase domain is $J \nabla H$ , and the Poisson bracket is $\sum_{i} (\partial_{q^{i}} f \partial_{p_{i}} g - \partial_{p_{i}} f \partial_{q^{i}} g)$ ): $H_{q} = q$ , $H_{p} = p$ , so $X_{H} = (H_{p}, - H_{q}) = (p, - q)$ ; hence the flow solves $q^{'} = p$ , $p^{'} = - q$ .

[F2]

Trigonometric calculus (The derivatives of sine and cosine are cosine and minus sine, The addition formulas for sine and cosine, Sine and cosine defined by their real power series): $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , and $Φ_{t}$ as displayed is the solution of the linear system with $Φ_{0} = i d$ , since differentiating the displayed expressions gives $q^{'} = - q \sin t + p \cos t = p (t)$ and $p^{'} = - q \cos t - p \sin t = - q (t)$ .

[F3]

The matrix identity (The derivative of the Hamiltonian flow satisfies $Y^{T} J Y = J$ and $\det Y = 1$ , The standard two-form of a phase domain is closed and nondegenerate, with contraction $ι_{v} ω = \sum_{i} a_{i} d p_{i} - \sum_{i} b_{i} d q^{i}$ , The Hessian matrix and critical points of a scalar field): for a $C^{2}$ Hamiltonian the flow derivative satisfies $Y^{'} = J H e s s H Y$ , $Y^{T} J Y = J$ and $\det Y = 1$ , the Hessian being the matrix of second partial derivatives; here $H e s s H = I_{2}$ .

Verification

technique · direct; verify the displayed flow by differentiation and multiply the two-by-two matrices

1.1F1F2algebra

Clause 1: by [F1] the field is $X_{H} = (p, - q)$ and the flow solves the linear system; the displayed $Φ_{t}$ satisfies $Φ_{0} = i d$ and, by [F2], its derivatives are $q^{'} = p (t)$ and $p^{'} = - q (t)$ , so by uniqueness it is the flow, defined for all real $t$ .

1.2F3algebra

Clause 2: differentiating $Y (t)$ entrywise gives $Y^{'} (t) = (\begin{array}{cc} - \sin t & \cos t \\ - \cos t & - \sin t \end{array})$ , while $J H e s s H Y = J Y = (\begin{array}{cc} - \sin t & \cos t \\ - \cos t & - \sin t \end{array})$ ; thus $Y^{'} = J Y$ with $Y (0) = I_{2}$ .

2.1F3step 1.2algebra∎

The orthogonality of $Y (t)$ gives $Y^{T} J Y = J$ by direct multiplication, and $\det Y = \cos^{2} t + \sin^{2} t = 1$ ; hence both conclusions of [F3] hold explicitly for this flow.

The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

The isotropic oscillator: the explicit flow, YTJY=JYTJY=J and det⁡Y=1detY=1

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$