What does the optical Hamiltonian mean?

Question

So I was trying to demonstrate Snell's law with Hamilton's equations, and when I got the Hamiltonian: $$H = -\sqrt{n^2-p_{1}^2-p_{2}^2}.$$

I had a question about what this Hamiltonian indicates. I mean, we know that, for example, under certain circumstances $H=T+U$, but in this case, what is the physical content of it?

Could someone maybe explain to me what the physical content of this optical Hamiltonian is?

Reference: This is from the book "Lagrangian Optics" by Vasudevan Lakshminarayanan et al., pages 93-95.

Answer 1

You can view it as $H = p_3$, the momentum component in the third direction. It is analogous to relativity where you view energy as the time comment of 4-momentum. Similarly, the components are not independent and are subject to the constraint, which follows from your equation: $$ p_3 = -\sqrt{n^2-p_1^2-p_2^2} \\ p_1^2+p_2^2+p_3^2 = n^2 $$ just like in relativity. Extending the analogy with relativity, in a non constant metric, you do not have the clean separation of the Hamiltonian of $H=T+U$, they are jumbled together. However, just as the separation occurs in the non relativistic limit, in the paraxial limit you do recover a similar separation in optics: $$ H = \frac1{2n}(p_1^2+p_2^2)-n $$ where you can identify the two terms as kinetic and potential "energy" respectively. Just as in classical mechanics, if $n$ is independent of $x_3$, then $H$ is conserved and it is precisely the conserved quantity in Snell-Descartes's law if the diopter boundary is in a $13$ plane.

Physically, you can also relate $H=p_3$ to the wavevector component in wave optics.

Answer 2

OP's square root Hamiltonian taken from Ref. 1 can more systematically be derived as follows:

1. We start with the Hamiltonian action $$ \begin{align} S_H[{\bf r},{\bf p},e]~=~&\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L_H, \cr L_H~:=~&{\bf p}\cdot\dot{\bf r}-\underbrace{\frac{e}{2}({\bf p}^2-n({\bf r})^2)}_{\text{Hamiltonian}},\end{align}\tag{1} $$ cf. my Phys.SE answer here.

2. The einbein $e$ in the Hamiltonian action (1) becomes a Lagrange multiplier that imposes the constraint $$ |{\bf p}|~\stackrel{(1)}{\approx}~ n({\bf r}).\tag{2}$$ It follows that minus OP's square root Hamiltonian is $|p_3|$, cf. LPZ's answer.

3. Axial gauge $r^3=\lambda$. If we eliminate/integrate out $p_3$ and $e$ in the Hamiltonian action (1), we get OP's square root Hamiltonian $$\begin{align} \left. L_H\right|_{r^3=\lambda} \quad\stackrel{p_3}{\longrightarrow}&\quad p_1\dot{r}^1+p_2\dot{r}^2 +\underbrace{\left(\frac{1}{2e}+\frac{e}{2}(n({\bf r})^2-p_1^2-p_2^2)\right)}_{\text{minus Hamiltonian}}\cr\cr \quad\stackrel{e}{\longrightarrow}&\quad p_1\dot{r}^1+p_2\dot{r}^2 +\underbrace{\sqrt{n({\bf r})^2-p_1^2-p_2^2}}_{\text{minus Hamiltonian}} .\end{align}\tag{3}$$

References:

1. V. Lakshminarayanan, A. Ghatak & K. Thyagarajan, Lagrangian Optics, 2002; p. 9-11 and p. 93-95.