Derivation of the full generator of the Lorentz transformations

Let us study the subgroup of the Poincare group that leaves the point $x=0$ invariant, that is the Lorentz group. The action of an infinitesimal Lorentz transformation on a field $\Phi(0)$ is $L_{\mu \nu}\Phi(0) = S_{\mu \nu}\Phi(0)$. By use of the commutation relations of the Poincare group, we translate the generator $L_{\mu \nu}$ to a nonzero value of $x$: $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} e^{-ix^{\sigma}P_{\sigma}} = S_{\mu \nu} - x_{\mu}P_{\nu} + x_{\nu}P_{\mu}\,\,\,\,\,\,\,\,\,(1),$$ where the RHS is computed using the Baker-Campbell-Hausdorff formula to the second term. Then we can write the action of the generators $$P_{\mu} \Phi(x) = -i\partial_{\mu}\Phi(x) \,\,\,\,\text{and}\,\,\,\,L_{\mu \nu}\Phi(x) = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})\Phi(x) + S_{\mu \nu}\Phi(x)\,\,\,\,\,(2)$$

Could someone explain the contents of Eqn (1) and how (2) is deduced from it? I think there is an unitary transformation on the LHS of Eqn (1), but I don't understand what the equation is doing and why the contents are what they are in the exponents of the exp terms.

• What is exactly your question (asking about the content of an equation is very broad in my opinion)? Do you want to understand where equation (1) is coming from? Also, what is $S_{\mu \nu}$ (you don't seem to have defined it)? – Hunter Jun 17 '14 at 12:10
• Hi Hunter, thanks for your comment. Yes, I am just trying to understand the motivation for introducing the quantity on the LHS and what it represents. $S_{\mu \nu}$ is the spin matrix of the field $\Phi$. – CAF Jun 17 '14 at 12:13

The LHS describes the transformation of $L_{\mu \nu}$ under "space-time translations". This is the natural generalization of how time evolution works in the Heisenberg picture. So instead of acting by just the generator of time translation $e^{-\frac{i H t}{\hbar}}$, you also have the $p_i$ in the exponent, which are the generators of space translations. You start with the action of the subgroup which leaves $x=0$ invariant, this is just the spin part of the angular momentum. Your total angular momentum operator is the sum of the orbital angular momentum and spin angular momentum which is evident from your equation 2.

You get equation 2 by just Taylor expanding $\Phi$ about 0, and then comparing with equation 1. (Here you are comparing the Heisenberg and Schrödinger pictures)

$$L_{\mu \nu}\omega^{\mu \nu}\Phi(x)=S_{\mu \nu} \omega^{\mu \nu} \Phi(0)+ x_{\mu} \omega^{\mu \nu}\partial_{\nu}\Phi(0)$$

where $\omega_{\mu \nu}$ are the boost and rotation paramaters, and I have considered the infinitesimal lorentz transformation $x^{\rho}=0 + \omega^{\rho}_{\nu} x^{\nu}$ and noting $\omega_{\mu \nu}$ is antisymmetric you get the correct actions.

THe infinitesimal version of equation 1 is just the commutator of the field with $L_{\mu \nu}$. This is a common feature of actions in QFT, the infinitesimal lie algebra version acts through the commutator which on exponentiating gives the conjugation action defined above.

• Thanks ramanujan_dirac! My expansion is like so $$\Phi(0+x) \approx \Phi(0) + x^{\mu}\partial_{\mu}\Phi(x)|_{x=0} = \Phi(0) + \omega^{\mu \sigma} x_{\sigma} \partial_{\mu}\Phi(0)$$ using the Lorentz transformation $x^{\mu} = \omega^{\mu}_{\,\,\,\nu}x^{\nu}$ What did you do from here to obtain the equation you listed in your answer? – CAF Jun 17 '14 at 14:11
• I got that by considering an infinitesimal lorentz transformation of the fields. So $\Phi^{\prime}(x^{\prime})=(1+iL_{\mu \nu}\omega ^{\mu \nu})\Phi(0-\omega^{\rho}_{\nu}x^{\nu})$, to first order the second term is the linear change in $x^{\rho}$ multiplied by 1. (The identity operator.) I have used the fact that scalar fields transform like $\Phi^{\prime}(x^{\prime})=\Phi(x)$ – user7757 Jun 17 '14 at 14:40
• If I expand that to first order then $$\Phi'(x') = \Phi(0) - ix^{\rho}\partial_{\rho}\Phi(0) + iL_{\mu \nu}w^{\mu \nu} \Phi(0) + L_{\mu \nu}w^{\mu \nu} x^{\rho} \partial_{\rho} \Phi(0)$$ but I do not seem to recover the terms you wrote down. – CAF Jun 17 '14 at 15:06
• Is using the fact that scalar fields transform trivially not considering a special case though? Given $\Phi'(x') = \Phi(x)$, I could then multiply the LHS of my last equation such that it matches yours but then that introduces a $L_{\mu \nu}\omega^{\mu \nu}$ term on the $x^{\rho}\partial_{\rho}$ term above. – CAF Jun 17 '14 at 15:30
• @CAF: Your second term should contain a factor $\omega$, and your last term $\omega^2$ after expanding $\Phi$ in $\Phi^{\prime}(x^{\prime})=(1+iL_{\mu \nu}\omega ^{\mu \nu})\Phi(0-\omega^{\rho}_{\nu}x^{\nu})$, you ignore the last term as it is second order in $\omega$, and you use antisymmetry of $\omega$ to get the final expression. Yes, we are doing this for the special case of scalar fields which is what you mention in your question. For a general field that transforms under a representation $L_A^B$ you will have an extra term corresponding to the generators of L. – user7757 Jun 18 '14 at 5:14

I am still not entirely sure what your question is, but I will try to explain the left-hand side of equation (1) in general. More information can be found in "Lie algebras in particle physics" by Georgi.

A representation of matrix Lie group can in be written as: $$D(g) = e^{i \alpha_a X_a}$$ where $X_a$ denotes a generator and $\alpha_a$ denotes the group parameter. The representation acts on some vector space: $$|i \rangle \to |i' \rangle = e^{i \alpha_a X_a} |i \rangle$$ Now, let $O$ denote an operator giving the ket $O|i\rangle$. Clearly, this will transform as: $$O|i \rangle \to O' | i' \rangle = e^{i \alpha_a X_a} O |i \rangle = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a} e^{i \alpha_a X_a} |i \rangle = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a} |i' \rangle$$ Thus, we see from above that the operator transforms as: $$O' = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a}$$ which corresponds to the left-hand side of equation (1) in your question. In other words, the left-hand side shows how the operator $L_{\mu \nu}$ transforms under the 4-momentum generator (i.e. space and time translations).

• Thanks Hunter! That is what I was alluding to when I said unitary transformation, but it is nice to have it spelled out. – CAF Jun 17 '14 at 14:12