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## more on the classical Harmonic oscillator Hamiltonian.

Posted by peeterjoot on October 3, 2009

[Click here for a PDF that incorporates this post with nicer formatting]

## Harmonic oscillator (change of variables.)

It was pointed out to me by Lut that the following rather strange looking change of variables has nice properties

\begin{aligned}P &= x \sqrt{\frac{k}{2}} + \frac{ p }{\sqrt{2 m}} \\ Q &= x \sqrt{\frac{k}{2}} - \frac{ p }{\sqrt{2 m}}\end{aligned} \quad\quad\quad(65)

In particular the Hamiltonian is then just

\begin{aligned}H = P^2 + Q^2\end{aligned} \quad\quad\quad(67)

Part of this change of variables, which rotates in phase space, as well as scales, looks like just a way of putting the system into natural units. We don’t however, need the rotation to do that. Suppose we write for just the scaling change of variables

\begin{aligned}p &= \sqrt{2m} P_s \\ x &= \sqrt{\frac{2}{k}} Q_s\end{aligned} \quad\quad\quad(68)

or

\begin{aligned}\begin{bmatrix}p \\ x \\ \end{bmatrix}=\begin{bmatrix}\sqrt{2m} & 0 \\ 0 & \sqrt{\frac{2}{k}} \end{bmatrix}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}\end{aligned} \quad\quad\quad(70)

This also gives the Hamiltonian 67, and the Hamiltonian equations are transformed to

\begin{aligned}\frac{d}{dt}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}&=\begin{bmatrix}1/\sqrt{2m} & 0 \\ 0 & \sqrt{\frac{k}{2}} \end{bmatrix}\begin{bmatrix}0 & - k \\ 1/m & 0 \\ \end{bmatrix}\begin{bmatrix}\sqrt{2m} & 0 \\ 0 & \sqrt{\frac{2}{k}} \end{bmatrix}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix} \\ &=\begin{bmatrix}0 & - \sqrt{\frac{k}{m}} \\ \sqrt{\frac{k}{m}} & 0 \\ \end{bmatrix}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}\end{aligned}

This first change of variables is nice since it groups the two factors $k$ and $m$ into a reciprocal pair. Since only the ratio is significant to the kinetics it is nice to have that explicit. Since $\sqrt{k/m}$ is in fact the angular frequency, we can define

\begin{aligned}\omega \equiv \sqrt{\frac{k}{m}}\end{aligned} \quad\quad\quad(71)

and our system is reduced to

\begin{aligned}\frac{d}{dt}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}&=\omega\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}\end{aligned} \quad\quad\quad(72)

Solution of this system now becomes particularily easy, especially if one notes that the matrix factor above can be expressed in terms of the $y$ axis Pauli matrix $\sigma_2$. That is

\begin{aligned}\sigma_2 = i\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\end{aligned} \quad\quad\quad(73)

Inverting this, and labelling this matrix $\mathcal{I}$ we can write

\begin{aligned}\mathcal{I} \equiv\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}=-i \sigma_2 \end{aligned} \quad\quad\quad(74)

Recalling that $\sigma_2^2 = I$, we then have $\mathcal{I}^2 = -I$, and see that this matrix behaves exactly like a unit imaginary. This reduces the Hamiltonian system to

\begin{aligned}\frac{d}{dt}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}&=\mathcal{I} \omega\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}\end{aligned} \quad\quad\quad(75)

We can now solve the system directly. Writing $\mathbf{z}_s = \bigl(\begin{smallmatrix} P_s \\ Q_s \\ \end{smallmatrix}\bigr)$, this is just

\begin{aligned}\mathbf{z}_s(t)=e^{\mathcal{I} \omega t} \mathbf{z}_s(0)&=\left( I \cos(\omega t) + \mathcal{I} \sin(\omega t) \right) \mathbf{z}_s(0)\end{aligned} \quad\quad\quad(76)

With just the scaling giving both the simple Hamiltonian, and a simple solution, what is the advantage of the further change of variables that mixes (rotates in phase space by 45 degrees with a factor of two scaling) the momentum and position coordinates? That second transformation is

\begin{aligned}P &= Q_s + P_s \\ Q &= Q_s - P_s \end{aligned} \quad\quad\quad(77)

Inverting this we have

\begin{aligned}\begin{bmatrix}P_s \\ Q_s \\ \end{bmatrix}=\frac{1}{{2}}\begin{bmatrix}1 & -1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix}P \\ Q \\ \end{bmatrix}\end{aligned} \quad\quad\quad(79)

The Hamiltonian after this change of variables is now

\begin{aligned}\frac{d}{dt}\begin{bmatrix}P \\ Q \\ \end{bmatrix}&=\frac{\omega}{2}\begin{bmatrix}1 & 1 \\ -1 & 1 \\ \end{bmatrix}\begin{bmatrix}0 & -1 \\ 1 & 0 \\ \end{bmatrix}\begin{bmatrix}1 & -1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix}P \\ Q \\ \end{bmatrix} \end{aligned} \quad\quad\quad(80)

But multiplying this out one finds that the equations of motion for the state space vector are unchanged by the rotation, and writing $\mathbf{z} = \bigl(\begin{smallmatrix} P \\ Q \\ \end{smallmatrix}\bigr)$ for the state vector, the Hamiltonian equations are

\begin{aligned}\mathbf{z}' = \mathcal{I} \omega \mathbf{z}\end{aligned} \quad\quad\quad(81)

This is just as we had before the rotation-like mixing of position and momentum coordinates. Now it looks like the rotational change of coordinates is related to the raising and lowering operators in the Quantum treatment of the Harmonic oscillator, but it is not clear to me what the advantage is in the classical context?