Characterization of Beams Separable in Cartesian Coordinates

Abstract

As we have discussed in Chap. 1, the characterization of 2D beams is a complex problem requiring the acquisition and processing of huge volume of data. Any hypothesis about the beam spatial structure, and specially those that can be experimentally checked, has to be used in order to simplify this task.

Appendices

Appendices

3.7 Demonstration of the Separability Test

The separability test affirms that a beam for which all its fractional power spectra satisfy

$$\begin{aligned} \left[ \int \mathrm {d}x~S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \right] \left[ \int \mathrm {d}y~S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \right] = S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \int \mathrm {d}\mathbf {r}^\prime ~S^{\gamma ,-\gamma }\left( \mathbf {r}^\prime \right) . \end{aligned}$$

(3.20)

must be separable in Cartesian coordinates. Introducing the FRFT kernel in both left- and right-hand sides of Eq. (3.20) we obtain that:

$$\begin{aligned} \mathrm {LHS}\left( \mathbf {r}\right)&\equiv \left[ \int \mathrm {d}x~S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \right] \left[ \int \mathrm {d}y~S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \right] \nonumber \\&= c_{\gamma }\int \mathrm {d}\mathbf {r}_1\mathrm {d}\mathbf {r}_2~\exp \left\{ \frac{\mathrm {i}\pi }{\sigma ^2\sin \gamma }\left[ \left( x_1^2 - x_2^2\right) \cos \gamma - 2\left( x_1 - x_2\right) x\right] \right\} \nonumber \\&\qquad \times \exp \left\{ -\frac{\mathrm {i}\pi }{\sigma ^2\sin \gamma }\left[ \left( y_1^2 - y_2^2\right) \cos \gamma - 2\left( y_1-y_2\right) y\right] \right\} \nonumber \\&\qquad \times \int \mathrm {d}x^\prime \Gamma \left( x^\prime ,y_1,x^\prime ,y_2\right) \int \mathrm {d}y^\prime ~\Gamma \left( x_1,y^\prime ,x_2,y^\prime \right) \end{aligned}$$

(3.21)

and

$$\begin{aligned} \mathrm {RHS}\left( \mathbf {r}\right)&\equiv S^{\gamma ,-\gamma }\left( \mathbf {r}\right) \int \mathrm {d}\mathbf {r}^\prime ~S^{\gamma ,-\gamma }\left( \mathbf {r}^\prime \right) \nonumber \\&= c_\gamma \int \mathrm {d}\mathbf {r}_1\mathrm {d}\mathbf {r}_2~\exp \left\{ \frac{\mathrm {i}\pi }{\sigma ^2\sin \gamma }\left[ \left( x_1^2 - x_2^2\right) \cos \gamma - 2\left( x_1 - x_2\right) x\right] \right\} \nonumber \\&\qquad \times \exp \left\{ -\frac{\mathrm {i}\pi }{\sigma ^2\sin \gamma }\left[ \left( y_1^2 - y_2^2\right) \cos \gamma - 2\left( y_1-y_2\right) y\right] \right\} \nonumber \\&\qquad \times \Gamma \left( x_1,y_1,x_2,y_2\right) \int \mathrm {d}\mathbf {r}^\prime ~\Gamma \left( \mathbf {r}^\prime , \mathbf {r}^\prime \right) , \end{aligned}$$

(3.22)

where $c_\gamma$ is a transformation constant that depends on $\gamma$ and the Parseval theorem has been applied to relate

$$\begin{aligned} \int \mathrm {d}\mathbf {r}^\prime ~S^{\gamma ,-\gamma }\left( \mathbf {r}\right) = \int \mathrm {d}\mathbf {r}^\prime ~S^{0,0}\left( \mathbf {r}^\prime \right) = \int \mathrm {d}\mathbf {r}^\prime ~\Gamma \left( \mathbf {r}^\prime ,\mathbf {r}^\prime \right) . \end{aligned}$$

(3.23)

As this condition must be held for all angles $\gamma$, it is equivalent to

$$\begin{aligned}&\int \mathrm {d}x^\prime \Gamma \left( x^\prime ,y_1,x^\prime ,y_2\right) \int \mathrm {d}y^\prime ~\Gamma \left( x_1,y^\prime ,x_2,y^\prime \right) \nonumber \\&\qquad \qquad = \Gamma \left( x_1,y_1,x_2,y_2\right) \int \mathrm {d}\mathbf {r}^\prime ~\Gamma \left( \mathbf {r}^\prime , \mathbf {r}^\prime \right) . \end{aligned}$$

(3.24)

The essence of the demonstration is to deduce that any beam that satisfies Eq. (3.24) must be separable.

Consider a beam described by the MI $\Gamma \left( \mathbf {r}_1,\mathbf {r}_2\right)$ which is not separable, but satisfies Eq. (3.24). Consider the following two functions obtained via

$$\begin{aligned} A\left( x_1,x_2\right)&= \int \mathrm {d}y^\prime ~\Gamma \left( x_1,y^\prime ,x_2,y^\prime \right) , \end{aligned}$$

(3.25)

$$\begin{aligned} B\left( y_1,y_2\right)&= \int \mathrm {d}x^\prime ~\Gamma \left( x^\prime ,y_1,x^\prime ,y_2\right) . \end{aligned}$$

(3.26)

Thus, Eq. (3.24) can be rewritten as

$$\begin{aligned} A\left( x_1,x_2\right) B\left( y_1,y_2\right)&= \Gamma \left( \mathbf {r}_1,\mathbf {r}_2\right) \int \mathrm {d}x^\prime A\left( x^\prime ,x^\prime \right) \nonumber \\&= \Gamma \left( \mathbf {r}_1,\mathbf {r}_2\right) \int \mathrm {d}y^\prime B\left( y^\prime ,y^\prime \right) . \end{aligned}$$

(3.27)

Notice that the power of the beam can be expressed as

$$\begin{aligned} P&= \int \mathrm {d}\mathbf {r}^\prime ~\Gamma \left( \mathbf {r}^\prime ,\mathbf {r}^\prime \right) = \int \mathrm {d}x^\prime ~A\left( x^\prime ,x^\prime \right) = \int \mathrm {d}y^\prime ~B\left( y^\prime ,y^\prime \right) \nonumber \\&= \sqrt{\int \mathrm {d}x^\prime ~A\left( x^\prime ,x^\prime \right) } \sqrt{\int \mathrm {d}y^\prime ~B\left( y^\prime ,y^\prime \right) }. \end{aligned}$$

(3.28)

Dividing both sides of Eq. (3.27) by the energy we obtain

$$\begin{aligned} \frac{A\left( x_1,x_2\right) }{\sqrt{\int \mathrm {d}x^\prime ~A\left( x^\prime ,x^\prime \right) }} \frac{B\left( y_1,y_2\right) }{\sqrt{\int \mathrm {d}y^\prime ~B\left( y^\prime ,y^\prime \right) }} = \Gamma \left( \mathbf {r}_1,\mathbf {r}_2\right) . \end{aligned}$$

(3.29)

This is the separability condition of our starting beam, which was considered not separable. Therefore, by reductio ad absurdum, a beam satisfying Eq. (3.24), or Eq. (3.20) for every $\gamma$, has to be separable.