# A Lorentz transformation of field with components

1. Nov 17, 2016

### ChrisVer

I am not looking for a solution, just a "starting point"/guidance for calculating the expression:
$[M^{\mu \nu} , \phi_a]$
with $M^{\mu \nu}$ being the angular-momentum operators and $\phi_a$ being the field's component, which happens to transform under Lorentz Transformations:
$x'^\mu = x^\mu + \delta \omega^{\rho \sigma} X ^\mu_{\rho \sigma}~~ X^{\mu}_{\rho \sigma} = \eta^{\mu}_\rho x_\sigma - \eta^\mu_\sigma x_\rho$
like:
$\phi_a'(x') = \phi_a(x) + \frac{1}{2} \delta \omega^{\mu \nu} (\Sigma_{\mu \nu})_a^b \phi_b(x)$

I have a tendency of writing $M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu )$ but I am not sure that this can help.

Last edited: Nov 17, 2016
2. Nov 17, 2016

### MathematicalPhysicist

In case you want a solution, I believe there's one in Atkinson's textbook, you should check it out.

3. Nov 17, 2016

### samalkhaiat

1) You have a factor of half missing in the Lorentz transformation of the coordinates: $$\bar{x}^{\tau} = x^{\tau} + \delta x^{\tau} ,$$ $$\delta x^{\tau} = \frac{1}{2}\omega_{\mu\nu} (\eta^{\tau \mu} x^{\nu} - \eta^{\tau \nu} x^{\mu}) . \ \ \ \ (1)$$
2) The commutator $[M , \varphi ]$ means that you are subjecting the field operator to an infinitesimal Lorentz transformation.
3) As an operator, $\varphi$ transforms by (infinite dimensional) unitary representation of the Lorentz group $$\bar{\varphi}_{a}(x) = e^{-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} \ \varphi_{a}(x) \ e^{\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} .$$ Expanding this to first order in $\omega$, you obtain $$\delta \varphi_{a}(x) = - \frac{i}{2}\omega_{\mu\nu}[M^{\mu\nu}, \varphi_{a}(x)] , \ \ \ \ \ \ \ \ \ (2)$$ $$\delta \varphi_{a}(x) \equiv \bar{\varphi}_{a}(x) - \varphi_{a}(x) .$$
4) Since $\varphi_{a}(x)$ is a finite-component field on Minkowski space, Lorentz group mixes its components by (finite-dimensional representation) matrix $D(\omega)$, and transform its argument by $\Lambda^{-1}$ $$\bar{\varphi}_{a}(x) = D_{a}{}^{b} (\omega) \varphi_{b}(\Lambda^{-1}x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ Infinitesimally, we may write $$D_{a}{}^{b} (\omega) = \delta_{a}{}^{b} - \frac{i}{2} \omega_{\mu\nu} \left( \Sigma^{\mu\nu}\right)_{a}{}^{b} ,$$ $$\Lambda^{-1}x = x - \delta x .$$ Using these expressions and keeping only the terms linear in $\omega$, Eq(3) becomes $$\delta \varphi_{a}(x) = - \delta x^{\tau} \partial_{\tau}\varphi_{a} - \frac{i}{2}\omega_{\mu\nu} \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .$$
Now, in this expression, substitute (1) and (2) to obtain $$[M^{\mu\nu} , \varphi_{a}(x)] = i \left(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}\right) \varphi_{a} + \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .$$
This expression has no meaning in field theory: on the left hand side, you have $M^{\mu\nu}$ which is an operator constructed out of $\varphi_{a}$ and its conjugate $\pi^{a}$, but the differential "operator" on the RHS, $x^{[\mu} \partial^{\nu]}$ is not an operator in field theory. If the generator $M^{\mu\nu}$ acts on the field $\varphi$ according to $$[M^{\mu\nu} , \varphi (x)] = i (x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}) \varphi (x) ,$$ this means that $\varphi$ is a scalar field, i.e., the spin matrix $\Sigma = 0$.