Deriving the Lorentz Commutator and Factor of 2

[spookyfish]

I am trying to derive the algebra and I get a factor of 2 wrong...
Consider the Lorentz group elements near the identity
$$\Lambda_1^\mu\,_\nu = \delta^\mu\,_\nu + \omega_1^\mu\,_\nu, \quad \Lambda_2^\mu\,_\nu = \delta^\mu\,_\nu + \omega_2^\mu\,_\nu$$
and write a representation as
$$U(\Lambda)=U(1 +\omega)=1-\frac{1}{2}i\omega^{\mu \nu} M_{\mu \nu}$$
where $M$ is a generator. Now the term $\Lambda \equiv (\Lambda_2^{-1}\Lambda_1^{-1}\Lambda_2 \Lambda_1)$ belongs to the group and up to 2nd order is $1+[\omega_2,\omega_1]$.
So its representation is
$$U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$
I know this is wrong and I am supposed to get
$$1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$

 

[samalkhaiat]

I am trying to derive the algebra

What algebra? Is it Lorentz algebra?

 

[spookyfish]

What algebra? Is it Lorentz algebra?

yes.

 

[MathematicalPhysicist]

I am trying to derive the algebra and I get a factor of 2 wrong...
Consider the Lorentz group elements near the identity
$$\Lambda_1^\mu\,_\nu = \delta^\mu\,_\nu + \omega_1^\mu\,_\nu, \quad \Lambda_2^\mu\,_\nu = \delta^\mu\,_\nu + \omega_2^\mu\,_\nu$$
and write a representation as
$$U(\Lambda)=U(1 +\omega)=1-\frac{1}{2}i\omega^{\mu \nu} M_{\mu \nu}$$
where $M$ is a generator. Now the term $\Lambda \equiv (\Lambda_2^{-1}\Lambda_1^{-1}\Lambda_2 \Lambda_1)$ belongs to the group and up to 2nd order is $1+[\omega_2,\omega_1]$.
So its representation is
$$U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$
I know this is wrong and I am supposed to get
$$1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$

I am not planning to make the calculation, but did you use the fact that for $$\Lambda_{\mu}^{\nu} = \delta_{\mu}^{\nu} + \omega_{\mu}^{\nu}$$
then its inverse matrix is: $$(\Lambda^{-1})_{\mu}^{\nu} = \delta_{\mu}^{\nu} - \omega_{\mu}^{\nu} +O(\omega^2)$$?

 

[spookyfish]

I am not planning to make the calculation, but did you use the fact that for $$\Lambda_{\mu}^{\nu} = \delta_{\mu}^{\nu} + \omega_{\mu}^{\nu}$$
then its inverse matrix is: $$(\Lambda^{-1})_{\mu}^{\nu} = \delta_{\mu}^{\nu} - \omega_{\mu}^{\nu} +O(\omega^2)$$?

Yes. I used it, and I am pretty sure it leads to $\Lambda = 1+ [\omega_2,\omega_1]$. I am not sure then how it would lead to
$$U(1+[\omega_2,\omega_1])=1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$
(there is a missing factor 1/2 according to my convention)

 

[MathematicalPhysicist]

Yes, you are, your calculation is right.

Do you use a specific reference where this equation is given there?

I must say that I myself find QFT quite hard as you can follow my posts in the forum in the last month or so.

 

[spookyfish]

Yes, you are, your calculation is right.

Do you use a specific reference where this equation is given there?

I must say that I myself find QFT quite hard as you can follow my posts in the forum in the last month or so.

Yes. Sorry, I am using the following reference:
http://www.damtp.cam.ac.uk/user/ho/GNotes.pdf
and trying to fill in the gaps. This is at the end of page 41, and on page 42. My problem is the top of eq. (4.30) given the definition (4.29).

 

[samalkhaiat]

So its representation is
$$U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$
I know this is wrong

No, in fact that is the correct form.

and I am supposed to get
$$1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}$$

This is wrong. If you use it, you will have a factor (2) multiplying the RHS of the Lorentz algebra.

 

[spookyfish]

Thank you. you are right, it works out! I stopped when I didn't understand this line, which is a typo, and should have continued in the first place

 

[samalkhaiat]

Thank you. you are right, it works out!

The point of having (1/2) in the $U( \Lambda )$ is to get "mormalized" algebra. There is nothing wrong in choosing $U = 1 + i \omega^{ \mu \nu } M_{ \mu \nu }$, but your algebra, in this case becomes $[M , M] = (1/2) ( g M + \cdots )$. So, when you mormalize $M \rightarrow (1/2) M$ you get the "nice" algebra $[M , M] = g M + \cdots$.
The point to remember is this, once you made a choice, you have to use it for all group elements: $( \Lambda ), ( \Lambda_{1} \Lambda_{2} ) , \cdots , ( \Lambda_{1} \cdots \Lambda_{n} )$.

I stopped when I didn't understand this line, which is a typo, and should have continued in the first place

I think it is lazyness not a "typo", because he repeats the same mistake when he "derives" the Poincare' algebra.

Sam