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## 1D SHO phase space

Posted by peeterjoot on February 2, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Let’s review the 1D SHO to get a better feel for the ideas of phase space. Given a spring and mass system

\begin{aligned}F = - k x = -(\boldsymbol{\nabla} \phi)_x\end{aligned} \hspace{\stretch{1}}(1.0.1)

our potential is

\begin{aligned}\phi = \frac{1}{{2}} k x^2,\end{aligned} \hspace{\stretch{1}}(1.0.2)

So, our Hamiltonian is

\begin{aligned}H = \frac{1}{{2m}}{p^2} + \frac{1}{{2}} k x^2.\end{aligned} \hspace{\stretch{1}}(1.0.3)

Hamilton’s equations follow from $H = p \dot{x} - \mathcal{L}$

\begin{aligned}\frac{\partial {H}}{\partial {p}} = \dot{x}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\frac{\partial {H}}{\partial {x}} = -\dot{p}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

For the SHO this is

\begin{aligned}\frac{d{{}}}{dt}\begin{bmatrix}p \\ x\end{bmatrix}=\begin{bmatrix}-\frac{\partial {H}}{\partial {x}} \\ \frac{\partial {H}}{\partial {p}} \end{bmatrix}=\begin{bmatrix} - k x \\ p/m\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

It’s convenient to non-dimensionalize this. Using $\omega = \sqrt{k/m}$, which has dimensions of $1/T$, we form

\begin{aligned}\frac{d{{}}}{dt}\begin{bmatrix}p/m \\ \omega x\end{bmatrix}=\begin{bmatrix} - (k/m) x \\ (\omega) p/m\end{bmatrix}=\omega\begin{bmatrix} - \omega x \\ p/m\end{bmatrix}=\omega\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}p/m \\ \omega x\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.6)

With definitions

\begin{aligned}i = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}\mathbf{x} =\begin{bmatrix}p/m \\ \omega x\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.7b)

the SHO Hamilton’s equations are just

\begin{aligned}\boxed{\mathbf{x}' = i \omega \mathbf{x}.}\end{aligned} \hspace{\stretch{1}}(1.0.8)

The solution follows immediately

\begin{aligned}\mathbf{x} = e^{i \omega t} \mathbf{x}_0.\end{aligned} \hspace{\stretch{1}}(1.0.9)

We expect matrix exponential to have the structure of a rotation matrix, so let’s write it out explicitly to see its structure

\begin{aligned}e^{i \omega t} = I \cos(\omega t)+ i \sin(\omega t)=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\cos(\omega t)+ \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\sin(\omega t)=\begin{bmatrix}\cos(\omega t) & - \sin(\omega t) \\ \sin(\omega t) & \cos(\omega t)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.10)

In this non-dimensionalized phase space, with $p/m$ on the horizontal axis and $\omega x$ on the vertical axis, this is a counterclockwise rotation. The (squared) radius of the rotation is

\begin{aligned}(p_0/m)^2+(\omega x_0)^2 = \frac{2}{m}\left( {\frac{p_0^2}{m}+ \frac{1}{{2}} \omega^2 m x_0^2} \right)= \frac{2}{m}\left( {\frac{p_0^2}{2m}+ \frac{1}{{2}} k x_0^2} \right)=\frac{2 E}{m}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

It makes sense to put the initial position in phase space in polar form too. We can write

\begin{aligned}\begin{bmatrix}p_0/m \\ \omega x_0\end{bmatrix}=\sqrt{\frac{2 E}{m}}e^{i \theta}\begin{bmatrix}1 \\ 0\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.12)

where

\begin{aligned}\theta = \arctan \left( { \omega m \frac{x_0}{p_0} } \right).\end{aligned} \hspace{\stretch{1}}(1.0.13)

Now the non-dimensionalized phase space solution takes the particularly simple form

\begin{aligned}\boxed{\mathbf{x} = \sqrt{ \frac{2 E}{m} } e^{i (\omega t + \theta)} \begin{bmatrix}1 \\ 0\end{bmatrix}.}\end{aligned} \hspace{\stretch{1}}(1.0.14)

Removing the non-dimensionalization

Written explicitly, our momentum and position trace out, elliptical trajectories

\begin{aligned}p = p_0 \cos(\omega t) - \omega m x_0 \sin(\omega t) \end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}x = \frac{p_0}{m \omega} \sin(\omega t) + x_0 \cos(\omega t).\end{aligned} \hspace{\stretch{1}}(1.0.15b)

With the initial phase space point specified as a rotation from the momentum axis as in 1.0.13, this is just

\begin{aligned}p = \sqrt{ \frac{2 E}{m} } m \cos(\omega t + \theta) = \sqrt{ 2 m E } \cos(\omega t + \theta) \end{aligned} \hspace{\stretch{1}}(1.0.16a)

\begin{aligned}x = \sqrt{ \frac{2 E}{m} } \frac{1}{{\omega}} \sin(\omega t + \theta) = \sqrt{ \frac{2 E}{k} } \sin(\omega t + \theta) \end{aligned} \hspace{\stretch{1}}(1.0.16b)

This is plotted in (Fig 1)

Fig 1: A SHO phase space trajectory

Observe that the rotation angle $\theta$ doesn’t specify a geometric rotation of the ellipse. Instead, it is a function of the starting point of the elliptical trajectory through phase space.

Aside. Complex representation of phase space points

It’s interesting to note that we can also work in a complex representation of phase space, instead of a matrix picture (for this 1D SHO problem).

\begin{aligned}\frac{d{{}}}{dt} \left( {\frac{p}{m} + i \omega x} \right)=\omega \left( {- \omega x + i \frac{p}{m}} \right)=i \omega \left( {\frac{p}{m} + i \omega x} \right).\end{aligned} \hspace{\stretch{1}}(1.0.17)

Writing

\begin{aligned}z = \frac{p}{m} + i \omega x, \end{aligned} \hspace{\stretch{1}}(1.0.18)

Hamilton’s equations take the form

\begin{aligned}z' = i \omega z.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Again, we can read off the solution by inspection

\begin{aligned}z = e^{i \omega t} z_0.\end{aligned} \hspace{\stretch{1}}(1.0.20)