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How can I see that for a scalar field

$$\phi(x)=e^{i\hat{p}\cdot x}\phi(0)e^{-i\hat{p}\cdot x}$$

if we have translation invariance?

Qmechanic
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    If $U(\Lambda,a)$ is a Poincaré transformation acting on the Hilbert space a scalar field obeys, by definition, $$U(\Lambda,a)\phi(x)U(\Lambda,a)^{-1} = \phi(\Lambda x+a).$$ Your equation is a particular case with $\Lambda=1$ and $a=x$, and where we apply the transformation to the operator $\phi(0)$. – Gold Apr 22 '22 at 21:54
  • related: https://physics.stackexchange.com/q/532254/226902 , https://physics.stackexchange.com/q/566963/226902 , – Quillo Apr 23 '22 at 14:20

1 Answers1

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This is a straightforward application of Lagrange's shift operator, namely the formal rewriting of Taylor's expansion around 0. ($\phi$ being translationally invariant means its form does not change under translations--only its argument.) However, you must be very careful with the labelling of your double-duty variables, since the momentum operator acts to shift the implicit "x" variable of the field, starting from 0, by an amount x. To avoid confusion, and mistakes, we rename that shift to y. Work in one dimension, as extension to more dimensions is self-evident.

Now, when you are working in the x-representation, the arguments of the field, you have $\hat p=-i\hbar \partial_x$, hence Lagrange's/Taylor's shift is just $$ e^{y \cdot \partial_x }\phi(x)= \phi(x+y) e^{y \cdot \partial_x } ~~~\leadsto \\ e^{y \cdot \partial_x }\phi(x) e^{-y \cdot \partial_x } = \phi(x+ y) $$ for an arbitrary differentiable function $\phi(x)$.

Now, take x=0, $$ e^{iy \cdot \hat p/\hbar }\phi(0) e^{-iy \cdot \hat p /\hbar } = \phi( y),$$ and rename $ y\to x$, to get your dimensionally consistent expression. (Instead, you appear to have set $\hbar=1$.)

Cosmas Zachos
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