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I want to derive eq. (2.4.3) in S. Weinberg, The Quantum Theory of fields, Vol. 1. The derivations start from expanding inhomogenous Lorentz transforms near identity

$$\Lambda^{\mu}_{\nu} ~=~ \delta^{\mu}_{\gamma}+\omega^{\mu}_{\nu}, \qquad a^{\mu}~=~\epsilon^{\mu}.$$ $\Lambda^{\mu}_{\nu}= \delta^{\mu}_{\gamma}$ and $a^{\mu}=0$ at identity.

Then the Unitary operator is expanded as follows:

$$U(1+\omega,\epsilon) ~=~ 1 + \frac{1}{2} i \omega_{\rho \sigma} J^{\rho \sigma}- i \epsilon_{\rho}P^{\rho}\ldots $$

I was wondering how this equation was derived. I know that near the identity, the Unitary operator can be expanded as

$$ U ~=~ 1+i \epsilon t.$$

Not able to see how to extend this to above equation.

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It's just one derivative in each independent direction--- the number of direction is the number of generators. –  Ron Maimon Apr 25 '12 at 4:16

1 Answer 1

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Here is one possible derivation.

I) The unitary operator $U=U(\Lambda,a)$ depends on a Lorentz transformation $\Lambda$ and a translation $a$.

II) It is assumed that $$U(\Lambda={\bf 1},a=0)~=~{\bf 1}.$$

III) Define $$ \tag{2.4.1} \Lambda^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}+\omega^{\mu}{}_{\nu}. $$

IV) Lower the indices with the Minkowski metric $\eta_{\mu\nu}$,

$$ \omega_{\mu\nu}~=~\sum_{\lambda}\eta_{\mu\lambda} \omega^{\lambda}{}_{\nu} , \qquad a_{\mu}~=~ \sum_{\nu}\eta_{\mu\nu} a^{\nu}. $$

V) Prove that

$$\tag{2.4.2}\omega_{\mu\nu}~=~-\omega_{\nu\mu}$$

is an antisymmetric matrix if $\omega^{\mu}{}_{\nu}$ is infinitesimal.

VI) Assume e.g. that the entries $\omega_{\mu\nu}$ above the diagonal $\mu<\nu$ are the independent d.o.f. of the $\omega$ matrix. (The entries $\omega_{\mu\nu}$ below the diagonal $\mu>\nu$ are then fully determined as the opposite values.)

VII) For $\mu<\nu$, define angular momentum

$$J^{\mu\nu}~=~ -i \left.\frac{\partial U(\Lambda={\bf 1}+\omega,a=0)}{\partial\omega_{\mu\nu}}\right|_{\omega=0}.$$ Extend $J^{\mu\nu}$ to an antisymmetric matrix $J^{\mu\nu}=-J^{\nu\mu}$.

VIII) Similarly, define $4$-momentum

$$P^{\mu}~=~ i \left.\frac{\partial U(\Lambda={\bf 1},a)}{\partial a_{\mu}}\right|_{a=0}.$$

IX) Taylor expand to first order

$$ U({\bf 1}+\omega,a) ~=~ {\bf 1} + i\sum_{\mu<\nu} \omega_{\mu\nu} J^{\mu\nu} - i \sum_{\mu}a_{\mu}P^{\mu}+\ldots $$ $$\tag{2.4.3} ~=~ {\bf 1} + \frac{i}{2}\sum_{\mu\nu} \omega_{\mu\nu} J^{\mu\nu} - i \sum_{\mu}a_{\mu}P^{\mu}+\ldots.$$

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Thanks for the response. In the equation 2.4.3, is –  qmfan Apr 24 '12 at 21:53
    
What is $U(1+\omega,a)$ in terms of exp ? –  qmfan Apr 24 '12 at 21:59

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