# What does the optical Hamiltonian mean?

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.

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.

• Could you maybe elaborate on how you get the relations between momenta? I know almost nothing about relativity. The question came up while answering a question in my classical mechanics course. Thank you. Commented Jun 23 at 16:20

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

1. We start by identifying the optical length with a square root action$$^1$$ \begin{align}S_0[{\bf r}]~=~&\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L_0, \qquad L_0~:=~n({\bf r})|\dot{\bf r}|, \cr |\dot{\bf r}|~:=~&\sqrt{\dot{\bf r}^2}, \qquad \dot{\bf r}~:=~\frac{d{\bf r}}{d\lambda}, \qquad {\bf r}~=~\begin{pmatrix}r^1\cr r^2\cr r^3\end{pmatrix}. \end{align}\tag{1} The Euler-Lagrange (EL) equation is the ray equation. Note that the action (1) has a worldline (WL) reparametrizations gauge symmetry $$\lambda\quad\longrightarrow\quad \lambda^{\prime}~=~f(\lambda).\tag{2}$$

2. To get rid of the square root, we introduce an auxiliary einbein field $$e(\lambda)>0$$: $$S[{\bf r},e]~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L, \qquad L~:=~\frac{\dot{\bf r}^2}{2e}+\frac{e}{2}n({\bf r})^2. \tag{3}$$ The EL equation for the einbein field is$$^2$$ $$e~\stackrel{(3)}{\approx}~\frac{|\dot{\bf r}|}{n({\bf r})}.\tag{4}$$ If we eliminate/integrate out the einbein field $$e$$, the action (3) reduces to the original action (1).

3. The momentum is $${\bf p}~=~\frac{\partial L}{\partial \dot{\bf r}}~\stackrel{(3)}{=}~\frac{\dot{\bf r}}{e}.\tag{5}$$

4. The Hamiltonian action becomes $$S_H[{\bf r},{\bf p},e]~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda ~ L_H, \qquad L_H~:=~{\bf p}\cdot\dot{\bf r}-\underbrace{\frac{e}{2}({\bf p}^2-n({\bf r})^2)}_{\text{Hamiltonian}}.\tag{6}$$ If we eliminate/integrate out the momentum $${\bf p}$$, the Hamiltonian action (6) reduces to the Lagrangian action (3).

5. The einbein $$e$$ in the Hamiltonian action (6) becomes a Lagrange multiplier that imposes the constraint $$|{\bf p}|~\stackrel{(6)}{\approx}~ n({\bf r}).\tag{7}$$ It follows that minus OP's square root Hamiltonian is $$|p_3|$$, cf. LPZ's answer.

6. Axial gauge $$r^3=\lambda$$. If we eliminate/integrate out $$p_3$$ and $$e$$, 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{8}

References:

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

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$$^1$$ This is a modification of the action for a relativistic point particle, cf. e.g. this and this Phys.SE posts.

$$^2$$ The $$\approx$$ symbol means equality modulo eqs. of motion.