In A Brief Geometric Analysis of Harmonic Oscillators: Part 2 (Tensor Algebra), we generalized the methods used in A Brief Geometric Analysis of Harmonic Oscillators to analyze the behaviour of harmonic oscillators in phase space.
We began with Hooke’s law, \(\omega^{-1} \ddot{x} + \omega x = 0\) and explored the Hamiltonian flow described by it in the phase space \(S\) represented by \(\left( x, y \right)\) where \(y = \omega^{-1} \dot{x}\). We then showed using geometric notions that this flow resembles concentric circles in phase space, with the phase \(\theta \left( t \right)\) being proportional to \(t\) by a factor of \(\omega\), i.e. \(\theta \left( t \right) = \omega t\).
In this post, we will explicitly find the solution for the position vector \(\left( x, y \right)\) in phase space and derive the phase as a consequence of the same.
Hamilton’s equations
As found in the previously mentioned posts, in phase space, the equation of motion for a harmonic oscillator is ingrained in Hamilton’s equations of motion,
\[\dot{\pmb{u}} = \pmb{\omega} \left( \pmb{u} \right)\]where \(\pmb{\omega} = \omega \left( \partial_x \otimes \text{d} y - \partial_y \otimes \text{d} x \right)\) and \(\pmb{u} = x \partial_x + y \partial_y = x \partial_x + \dot{x} \partial_{\dot{x}}\).
In the matrix representation,
\[\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \omega \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\]Now, we will try to solve for \(\pmb{u} \left( t \right)\).
Time translation operator
Suppose we want to construct a time translation operator \(\mathcal{U}\) such that,
\[\mathcal{U}^t \pmb{u} \left( 0 \right) = \pmb{u} \left( t \right)\]The exponentiation of the time translation operator has to do with the way its compositions (products) add up in terms of their arguments,
\[\prod_{k} \mathcal{U}^{t_k} = \mathcal{U}^{\sum_k t_k}\]With this in mind, we have, using a series expansion of the time translation operator,
\[\begin{align} \mathcal{U}^t &= \sum_{n=0}^\infty \frac{1}{n!} t^n \frac{d^n}{dt^n} \mathcal{U}^t \Bigg|_{t=0} \\ & = \exp \left( t \frac{d}{dt} \right) \mathcal{U}^0 \\ & = \exp \left( t \frac{d}{dt} \right) \mathcal{I} \\ & = \exp \left( t \frac{d}{dt} \right) \end{align}\]where \(\mathcal{I}\) is the identity map \(\text{id}_S\).
Now, we perform a little trick. Since we imagine \(\mathcal{U}\) to act on \(\pmb{u} \left( t \right)\), the linear operator \(\frac{d}{dt}\) is the linear map \(\widehat{L} : \pmb{u} \to \dot{\pmb{u}}\). From Hamilton’s equations of motion, we know that this map is nothing but \(\pmb{\omega}\). Let us write \(\pmb{\omega} = \omega \pmb{J}\) where \(\pmb{J} = \partial_x \otimes \text{d} y - \partial_y \otimes \text{d} x\). Then, we have from the above equation,
\[\mathcal{U}^t = \exp \left( \omega t \pmb{J} \right)\]Therefore, from the definition of the time translation operator, we have,
\[\pmb{u} \left( t \right) = \exp \left( \omega t \pmb{J} \right) \: \pmb{u} \left( 0 \right)\]where,
\[\exp \left( \omega t \pmb{J} \right) = \sum_{n=0}^\infty \frac{1}{n!} \omega^n t^n \pmb{J}^n = \mathcal{U}^t\]In principle, this is indeed an explicit solution for \(\pmb{u} \left( t \right)\) in terms of initial conditions \(\pmb{u} \left( 0 \right)\). However, to extract the notion of periodicity from the exponential, we will need to derive Euler’s formula for matrix exponentials. 1
Euler’s formula for the time translation operator
To make sense of the solution we derived above for the position vector in phase space, we will do the following:
\[\mathcal{U}^t = \exp \left( \omega t \pmb{J} \right)\]Some useful facts to simplify the above expression are,
\[\begin{align} \pmb{J}^0 & = \pmb{\delta} \\ \pmb{J}^1 & = \pmb{J} \\ \pmb{J}^2 & = \left( \partial_x \otimes \text{d} y - \partial_y \otimes \text{d} x \right) \left( \partial_x \otimes \text{d} y - \partial_y \otimes \text{d} x \right) \\ & = - \partial_x \otimes \text{d} x - \partial_y \otimes \text{d} y \\ & = - \delta^i_{\phantom{i} j} \: \partial_i \otimes \text{d} x^j \\ & = - \pmb{\delta} \\ \pmb{J}^3 & = \pmb{J}^2 \pmb{J} \\ & = - \pmb{\delta} \pmb{J} \\ & = - \pmb{J} \\ \pmb{J}^4 & = \pmb{J}^3 \pmb{J} \\ & = - \pmb{J} \pmb{J} \\ & = - \pmb{J}^2 \\ & = \pmb{\delta} \\ & \vdots \\ \pmb{J}^n & = \pmb{J}^{n+4k} \: \forall \: n, k \in \mathbb{Z} \end{align}\]The above can be summarized in the following commutative diagram,
Armed with this knowledge,let us collect even and odd terms in the power series for \(\mathcal{U}^t\),
\[\begin{align} \mathcal{U}^t & = \sum_{n=0}^\infty \frac{1}{n!} \omega^n t^n \pmb{J}^n \\ & = \sum_{n=0}^\infty \frac{1}{\left( 2n \right)!} \omega^{2n} t^{2n} \pmb{J}^{2n} + \sum_{n=0}^\infty \frac{1}{\left( 2n+1 \right)!} \omega^{2n+1} t^{2n+1} \pmb{J}^{2n+1} \\ & = \sum_{n=0}^\infty \frac{\left( -1 \right)^n}{\left( 2n \right)!} \omega^{2n} t^{2n} \pmb{\delta}^{2n} + \sum_{n=0}^\infty \frac{\left( -1 \right)^n}{\left( 2n+1 \right)!} \omega^{2n+1} t^{2n+1} \pmb{J} \\ & = \cos \left( \omega t \right) \pmb{\delta} + \sin \left( \omega t \right) \pmb{J} \end{align}\]Thus,
\[\begin{align} \pmb{u} \left( t \right) & = \mathcal{U}^t \pmb{u} \left( 0 \right) \\ & = \left[ \cos \left( \omega t \right) \pmb{\delta} + \sin \left( \omega t \right) \pmb{J} \right] \pmb{u} \left( 0 \right) \\ & = \cos \left( \omega t \right) \pmb{u} \left( 0 \right) + \sin \left( \omega t \right) \pmb{J} \pmb{u} \left( 0 \right) \\ & = \cos \left( \omega t \right) \left( x \left( 0 \right) \partial_x + y \left( 0 \right) \partial_y \right) + \sin \left( \omega t \right) \left( \partial_x \otimes \text{d} y - \partial_y \otimes \text{d} x \right) \left( x \left( 0 \right) \partial_x + y \left( 0 \right) \partial_y \right) \\ & = \cos \left( \omega t \right) \left( x \left( 0 \right) \partial_x + y \left( 0 \right) \partial_y \right) + \sin \left( \omega t \right) \left( y \left( 0 \right) \partial_x - x \left( 0 \right) \partial_y \right) \\ & = \left( \cos \left( \omega t \right) x \left( 0 \right) + \sin \left( \omega t \right) y \left( 0 \right) \right) \partial_x + \left( \cos \left( \omega t \right) y \left( 0 \right) - \sin \left( \omega t \right) x \left( 0 \right) \right) \partial_y \end{align}\]Once again, in the matrix representation,
\[\begin{align} \mathcal{U}^t & = \cos \left( \omega t \right) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sin \left( \omega t \right) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\ & = \begin{pmatrix} \cos \left( \omega t \right) & \sin \left( \omega t \right) \\ - \sin \left( \omega t \right) & \cos \left( \omega t \right) \end{pmatrix} \\ \pmb{u} \left( t \right) & = \mathcal{U}^t \pmb{u} \left( 0 \right) \\ & = \begin{pmatrix} \cos \left( \omega t \right) & \sin \left( \omega t \right) \\ - \sin \left( \omega t \right) & \cos \left( \omega t \right) \end{pmatrix} \begin{pmatrix} x \left( 0 \right) \\ y \left( 0 \right) \end{pmatrix} \\ & = \begin{pmatrix} \cos \left( \omega t \right) x \left( 0 \right) + \sin \left( \omega t \right) y \left( 0 \right) \\ - \sin \left( \omega t \right) x \left( 0 \right) + \cos \left( \omega t \right) y \left( 0 \right) \end{pmatrix} \end{align}\]From this, we can concretely conclude that \(\pmb{u} \left( t \right)\) is a periodic function with a period of \(T = \frac{2 \pi}{\omega}\). Furthermore, the argument \(\omega t\) assumes the role of an angle parameter \(\theta \left( t \right)\) in characterizing \(\pmb{u} \left( t \right)\).
Conclusion
Through a series of posts, we have managed to demonstrate the periodic evolution of harmonic oscillators, following as a geometric consequence of the structure of their equation of motion. The basis for this argument was the Hamiltonian flow of a harmonic oscillator in phase space, which was then explored using a range of mathematical toolkits, namely: vector calculus, tensor algebra and finally, matrix exponentials.
This analysis hopefully lays the groundwork for building theories of more complicated systems resembling harmonic oscillators, such as Klein-Gordon fields and quantum harmonic oscillators — topics we will want to explore on this blog soon.
As always, thanks so much for reading and feel free to leave suggestions or any other comments below :)
Pedantically, matrices are only representations of underlying tensors. What we will really be doing is deriving Euler’s formula for antisymmetric tensors. ↩