Derive the representation of the momentum acting on a field

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Homework Statement


consider the space-time transformation of translation

xμ → x'μ = xμ + aμ

where xμ is a point in space-time and aμis a constant 4-vector. Assuming translations are generated by the operator U=e-iPμaμ acting on fields Φ(x), derive the representation of Pμ 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.
 

Answers and Replies

  • #2
CAF123
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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)##.
 
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Thank you for the reply. I think I almost have it, so under the infinitesimal transformation x'μ=xμ+aμ, we have δΦ(x)=Φ'(x)-Φ(x). Rearranging the first equation gives xμ=x'μ-aμ. so we can rewrite δΦ=Φ'(x-a)-Φ(x), expanding this out gives δΦ=-aμμΦ(x).

Now we also have Φ'(x)=e-iPμaμΦ(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(-aμ)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.
 
  • #4
CAF123
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Thank you for the reply. I think I almost have it, so under the infinitesimal transformation x'μ=xμ+aμ, we have δΦ(x)=Φ'(x)-Φ(x). Rearranging the first equation gives xμ=x'μ-aμ. so we can rewrite δΦ=Φ'(x-a)-Φ(x), expanding this out gives δΦ=-aμμΦ(x).

Now we also have Φ'(x)=e-iPμaμΦ(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(-aμ)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)##.
 

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