Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

If I define T_{ij}= a^{+}_{i}a_{j}, then

C_{2}= T_{11}T_{11}+ T_{12}T_{21}+ T_{21}T_{12}+ T_{22}T_{22}is a second order casimir operator.

For SU(2), it's [tex]\frac{N}{2}[/tex] ([tex]\frac{N}{2}[/tex] + 1)

But as I calculate it directly,

C_{2}= a^{+}_{1}a_{1}a^{+}_{1}a_{1}+ a^{+}_{1}a_{2}a^{+}_{2}a_{1}+ a^{+}_{2}a_{1}a^{+}_{1}a_{1}+ a^{+}_{2}a_{2}a^{+}_{2}a_{2}=

a^{+}_{1}a_{1}a^{+}_{1}a_{1}+ a^{+}_{1}(a^{+}_{2}a_{2}+ 1)a_{1}+ a^{+}_{2}(a^{+}_{1}a_{1}+ 1)a_{2}+ a^{+}_{2}a_{2}a^{+}_{2}a_{2}=

N_{1}N_{1}+ N_{1}(N_{2}+ 1) + N_{2}(N_{1}+ 1) + N_{2}N_{2}= (N_{1}+ N_{2})^{2}+ N_{1}+ N_{2}= N(N + 1)

which is different from above. Can you let me know what is wrong with my argument? Thank you very much!

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

Dismiss Notice

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!

# Casimir operator in su(2)

Loading...

Similar Threads for Casimir operator | Date |
---|---|

Why the Casimir operator is proportional to the unit matrix ? | Feb 22, 2013 |

Casimir operators and rest mass | Oct 31, 2011 |

Casimir operators of QCD | Jun 2, 2011 |

Quadratic Casimir Operator of SU(3) | Jan 31, 2009 |

Casimir operator | Feb 13, 2007 |

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