Derive the representation of the momentum acting on a field

[D_Cross]

Homework Statement

consider the space-time transformation of translation

x^μ → x'^μ = x^μ + a^μ

where x^μ is a point in space-time and a^μis a constant 4-vector. Assuming translations are generated by the operator U=e^-iP_μaμ acting on fields Φ(x), derive the representation of P^μ on the field Φ(x).

Homework Equations

The Attempt at a Solution

From looking at it I would assume it would look something like i∂_μΦ
Though if this is the case I'm not entirely sure why.

 

[CAF123]

An abstract element of the translation group looks like $\exp(-ia^{\mu}P_{\mu})$. When this acts on fields, $P_{\mu}$ indeed manifests itself as a derivative and we say the representation of the generator of translations on space time fields is a differential operator. To see this, start by shifting the space time point $x'^{\mu} \rightarrow x^{\mu} + a^{\mu}$, where $a^{\mu}$ is an infinitesimal shift and look at the corresponding transformation of the field from $\phi(x)$ to $\phi'(x)$.

 

[D_Cross]

Thank you for the reply. I think I almost have it, so under the infinitesimal transformation x'^μ=x^μ+a^μ, we have δΦ(x)=Φ'(x)-Φ(x). Rearranging the first equation gives x^μ=x'^μ-a^μ. so we can rewrite δΦ=Φ'(x-a)-Φ(x), expanding this out gives δΦ=-a^μ∂_μΦ(x).

Now we also have Φ'(x)=e^-iPμa_μΦ(x), but I don't know what to do next. I've seen in a book that it should be Φ'(x'-a)=e^-iPμ(-a_μ)Φ'(x'), but I don't see how to get to this.

I'm sorry if I'm actually going in completely the wrong direction.

 

[CAF123]

Thank you for the reply. I think I almost have it, so under the infinitesimal transformation x'^μ=x^μ+a^μ, we have δΦ(x)=Φ'(x)-Φ(x). Rearranging the first equation gives x^μ=x'^μ-a^μ. so we can rewrite δΦ=Φ'(x-a)-Φ(x), expanding this out gives δΦ=-a^μ∂_μΦ(x).

Now we also have Φ'(x)=e^-iPμa_μΦ(x), but I don't know what to do next. I've seen in a book that it should be Φ'(x'-a)=e^-iPμ(-a_μ)Φ'(x'), but I don't see how to get to this.

I'm sorry if I'm actually going in completely the wrong direction.

It's correct - given $\phi'(x) = \exp(-i P_{\mu} a^{\mu}) \phi(x)$, expand the rhs for an infinitesimal shift $a^{\mu}$ and compare with your derivation of $\delta \phi = \phi'(x)-\phi(x)$.