Recent content by samalkhaiat

I don’t know, I don’t use Zee’s books. May be he wants to introduce you to the adjoint representation, if he has not done that by this point. Note that “similar” thing can be done in the SU(n) group by expanding the traceless (1,1) tensor representation in terms of the hermitian generators in...

Weinberg constructs the field a_{\mu}(x) from the polarization “vector” e_{\mu}(\vec{p}, \sigma) and the annihilation and creation operators. Since the little group technics, he uses, is simpler to apply directly to the polarization vector, he derives the properties of the field a_{\mu} from...

The mathematical details is given on the pages 248 and 249 of Weinberg text. The math is easy if you are familiar with the technics of “little group”. If you are not, then observe the following problem: Consider the propagator of the massive vector field representation (1/2,1/2) (which is a true...

It is the point, A_{\mu}A^{\mu} is not physically meaningful quantity because it is neither gauge nor Lorentz invariant. I don’t know what you mean by those words. Can you translate your words to mathematical expressions so that I can understand them. I suggest you read Weinberg QFT Vol 1...

The potential with index up is not a contravariant Lorentz vector because it does not transform like A^{\mu} \to \Lambda^{\mu}{}_{\nu}A^{\nu}. And the potential with index down (i.e., the components of the connection 1-form) is not a covariant Lorentz vector because it does not transform like...

Gauge invariance must hold in all Lorentz frames, otherwise no physical meaning can be associated to a given quantity. Also, the gauge-invariant Maxwell equations are Lorentz covariant provided that the connection transforms as \bar{A}^{\mu}(\bar{x}) = \Lambda^{\mu}{}_{\nu}A^{\nu}(x) +...

The potential A_{\mu} is NOT a 4-vector. The fact that we can always choose A_{0} to be zero in ALL Lorentz frame show that A_{\mu} cannot be a 4-vector. Indeed, there exists no 4-vector representation (1/2,1/2) for massless particle of helicity \pm 1. Under the action of the Lorentz group, the...

Exercise: Consider the transformation of your tensor T under the rotation R \in SO(n): T \to \bar{T} = RTR^{-1}, Now consider the infinitesimal version of the transformation by writing R = 1 + \omega \cdot \mathcal{J}, where \omega \cdot \mathcal{J} = \omega^{a}\mathcal{J}_{a}, a = 1, \cdots \ ...

It is not clear what meaning one can associate with the product of operators on the left hand side. I you have an exact symmetry, then the Noether current is conserved, i.e., \partial_{\mu}j^{\mu}(x) = 0, and its associated charge, Q = \int d^{3}x \ j^{0}(x), generates the correct infinitesimal...

1)The Weyl Conformal Tensor: You really need to know where the Weyl tensor does come from. Weyl obtained his tensor C_{\rho\sigma\mu\nu} from the Riemann tensor R_{\rho\sigma\mu\nu} by subtracting all pieces that are not invariant under arbitrary rescaling of the metric (g_{\mu\nu}(x) \to...

I wrote an insight about it. I believe I called it: The classical limit of commutators, or something similar

This is not correct. Use g_{ad}\partial_{b}\chi^{d} = \partial_{b}\chi_{a} - \chi^{d}\partial_{b}g_{ad}. You made the same mistake in here. Using the following relations in the calculus of infinitesimals: \epsilon \ \bar{g}_{ab}(x) \approx \epsilon \ g_{ab}(x), \ \ \ \epsilon \...

Do you know how to solve differential equations by Green’s function methods? 1) K-G propagator: Recall that the FREE KG field satisfies \left( \partial^{2} + m^{2}\right) \phi (x) = 0 , \ \ \ \mbox{where} \ \partial^{2} = \partial_{\mu}\partial^{\mu}. \ \ (1) Now let us modify the above FREE KG...

Again, thanks for that elementary fact. You said: “…., if you measure the local field as a function of time, …” and I still don’t understand the meaning of “measure” in THAT sentence. So, what does it mean to “measure” say the KG field operator? And how you do it?

Thank you for that valuable piece of information :wink: If you cannot infer the value of \mathbf{j} from the reading of the ammeter, you can measure the electric field which, in many situations, allows you to determine the electric current by “Ohm’s law” \mathbf{j} = \sigma \mathbf{E}. Look, if...