How Do You Calculate the Structure Constant in Gell-Mann Matrices?

[Lapidus]

This is not a homework, only something embarrasing..

[T_8, T_4 + i T_5] = (3^(1/2) / 2) T_4 + i T_5

from http://phys.columbia.edu/~cyr/notes/QFT_3/lecture3.pdf"

I can't see how to get the structure constant (3^(1/2) / 2).

T_4 + i T_5 is a 3x3 matrix with a one at (2,3), the rest zeroes. I multiply T_8 with T_4 + i T_5, then T_4 + i T_5 with T_8, then substract.

I don't get (3^(1/2) / 2) times T_4 + i T_5.

I get (3/ 4x3^(1/2)) times T_4 + i T_5.thanks for any help

 

[jambaugh]

Hmmm...
[T_8, T_4 + i T_5] = \frac{1}{4}[\lambda^8,\lambda^4 +i\lambda^5] = \frac{1}{4}\left( [\lambda^8,\lambda^4] + i[\lambda^8,\lambda^5]\right)

I get [\lambda^8,\lambda^4] = \frac{3i}{\sqrt{3}}\lambda^5=i\sqrt{3}\lambda^5
and [\lambda^8,\lambda^5] = -i\sqrt{3}\lambda^4
[T_8, T_4 + i T_5] = \frac{\sqrt{3}}{4}\left(i\lambda^5 + \lambda^4\right)=\frac{\sqrt{3}}{2}\left(T_4 + i T_5\right)

It looks to me like you forgot to double when going back to the T form from the lambda form. Also note that \frac{a}{\sqrt{a}} = a^1 a^{-1/2} = a^{1-1/2} = a^{+1/2} = \sqrt{a}.

 

[Lapidus]

Ahh, of course! As I said, it is rather embarassing...

many thanks, Jambaugh

 

[jambaugh]

Ahh, of course! As I said, it is rather embarassing...

many thanks, Jambaugh

You shouldn't be embarrassed about making a mistake... (we all make them)... only about refusing to acknowledge your mistakes.