Hi guys, why does the following mean B is unitary?(adsbygoogle = window.adsbygoogle || []).push({});

if we have two rotations such that;

b1= B_{11}a1+ B_{12}a2

b2= B_{21}a1+ B_{22}a2

and the following commutator results are;

[b1,b1(dagger)] = |B_{11}|^2 + |B_{12}|^2 --> 1

[b2,b2(dagger)] = |B_{21}|^2 + |B_{22}|^2 --> 1

[b1,b2(dagger)] = [B_{11}B*_{21}] + B_{12}B*_{22}--> 0

thus B is unitary.

I'm assuming it's something to do with the probabilities adding to 1, but i'm hoping for a more 'visual' understanding.

Thanks in advance.

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Beam Splitter - Commutation relations

Loading...

Similar Threads - Beam Splitter Commutation | Date |
---|---|

I How to create the operator of unknown beamsplitter | Feb 27, 2018 |

I Beam splitter in an entanglement swapping experiment | Feb 22, 2018 |

B Beam splitter, one photon, one detector? | Nov 21, 2017 |

I Hong-Ou-Mandel with two beam splitters | Mar 16, 2017 |

A Quantum entanglement by the means of beam splitters | Jan 16, 2016 |

**Physics Forums - The Fusion of Science and Community**