# 282: Two-particle Interference for Bosons and Fermions

A parametric down converter, PDC, transforms an incident photon into two lower energy photons. One photon takes the upper path and the other the lower path or vice versa. The principles of quantum mechanics require that the wave function for this event be written as the following entangled superposition.

Entangled superposition for bosons:

$\frac{1}{ \sqrt{2}}(U_1 D_2 + D_1 U_2)$

At the beam splitter, BS, the probability amplitude for transmission is $$\frac{1}{ \sqrt{2}}$$ and the probability amplitude for reflection is $$\frac{i}{ \sqrt{2}}$$. Therefore, for the four possible arrivals at the detectors we have,

$U_1 = \frac{1}{ \sqrt{2}} (i A_1 + B_1)$

$D_1 = \frac{1}{ \sqrt{2}} (A_1 + i B_1)$

$U_2 = \frac{1}{ \sqrt{2}} (i A_2 + B_2)$

$D_2 = \frac{1}{ \sqrt{2}} (A_2 + i B_2)$

Bosons are always observed at the same detector.

$\frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2} (i A_1 + B_1 \sqrt{2} (A_2 + iB_2 + \sqrt{2} (A_1 + i B_1 \sqrt{2} (i A_2 + B_2}\right] ~simplify~ \rightarrow \sqrt{2} \left( \frac{A_1 A_2}{2} + \frac{B_1 B_2}{2} \right) i$

Entangled superposition for fermions:

$\frac{1}{ \sqrt{2}} U_1 D_2 - D_1 U_2$

Fermions are never observed at the same detector.

$\frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2} (i A_1 + B_1 \sqrt{2} (A_2 + iB_2 - \sqrt{2} (A_1 + i B_1 \sqrt{2} (i A_2 + B_2}\right] ~simplify~ \rightarrow \frac{ \sqrt{2} A_2 B_1}{2} + \frac{ \sqrt{2} A_1 B_2}{2}$

In summary, the sociology of bosons and fermions can be briefly stated: bosons are gregarious and enjoy company; fermions are antisocial and prefer solitude.

Another way to do the calculation using Mathcad.

Bosons:

$\frac{1}{ \sqrt{2}} (U_1 D_2 + D_1 U_2)~ \begin{array}{|l} \text{substitute},~ U_1 = \frac{1}{ \sqrt{2}} (iA_1 + B_1) \\ \text{substitute},~ D_2 = \frac{1}{ \sqrt{2}} (A_2 + iB_2) \\ \text{substitute},~ D_1 = \frac{1}{ \sqrt{2}} (A_1 + iB_1) \\ \text{substitute},~ U_2 = \frac{1}{ \sqrt{2}} (iA_2 + B_2) \\ \text{simplify} \\ \end{array} \rightarrow \sqrt{2} \left( \frac{A_1 A_2}{2} + \frac{B_1 B_2}{2} \right) i$

Fermions:

$\frac{1}{ \sqrt{2}} (U_1 D_2 - D_1 U_2)~ \begin{array}{|l} \text{substitute},~ U_1 = \frac{1}{ \sqrt{2}} (iA_1 + B_1) \\ \text{substitute},~ D_2 = \frac{1}{ \sqrt{2}} (A_2 + iB_2) \\ \text{substitute},~ D_1 = \frac{1}{ \sqrt{2}} (A_1 + iB_1) \\ \text{substitute},~ U_2 = \frac{1}{ \sqrt{2}} (iA_2 + B_2) \\ \text{simplify} \\ \end{array} \rightarrow \frac{ \sqrt{2} A_2 B_1}{2} - \frac{ \sqrt{2} A_1 B_2}{2}$