How to interpret the results of translation operator acting on a momentum state? I am reading the "Quantum Field Theory for the Gifted Amateur" by Lancaster & Bundell and coming cross a problem on Page 81 in section 9.1 "Translations in spacetime".
It is shown that the translation operator $\hat{U}(a)$ acting on a momentum state $|q\rangle$, the calculation is:
$$\hat{U}(a)|q\rangle=e^{-iq\cdot a}|q\rangle$$
When projecting along the coordinate direction gives us a translated wave function:
$$\langle x|\hat{U}(a)|q\rangle=\langle x|q\rangle e^{-iq\cdot a}=\frac{1}{\sqrt{V}}e^{iq\cdot (x-a)}$$
In fact, i can finish these deductions from the mathematical view. However, i cannot interpret these results. My question is :Why the spacetime translation operator $\hat{U}(a)$ will have impact on the momentum state?
 A: It has no physical effect on a momentum eigenstate, only an overall phase difference:
$$\def\bra#1{\left\langle#1\right|}\def\ket#1{\left|#1\right\rangle}\hat{U}(a)\ket{q}=e^{-iqa}\ket{q}$$
Note that both $q$ and $a$ are constants. Momentum eigenstates are invariant under translation, up to a phase difference.
A: Momentum is intimately tied to translation because in position space the momentum operator is a derivative. The derivative can be used to construct ('generate') a translation operator.
For $\epsilon$ infinitessimal we have
\begin{align}
f(x+\epsilon)&\approx f(x)+\epsilon f'(x)\\
&=\left(1+\epsilon\frac{d}{dx}\right)f(x)\\
&=(1+i\epsilon\hat p)f(x)
\end{align}
A finite translation can be constructed by composing $n$ of these infinitessimal translations and taking the limit $n\rightarrow\infty$. In this limit keep $n\epsilon=1$
\begin{align}
f(x+a)&=f(x+n\epsilon a)\\
&=\left(1+\frac{i\hat p a}{n}\right)^nf(x)\\
&\equiv e^{i\hat pa }f(x)\\
&=(1+i\hat pa+\frac{(i\hat p a)^2}{2!}+\dots)f(x)
\end{align}
Note the translation operator has a minus sign before $a$ i.e. $\hat U_af(x)=f(x-a)$. The momentum state$|q\rangle$, being an eigenstate of momentum, behaves simply under translation.
\begin{align}
\hat U_a e^{iqx}&=(1-i\hat pa+\frac{(-i\hat p a)^2}{2!}+\dots)e^{iqx}\\
&=(1-iqa+\frac{(-iq a)^2}{2!}+\dots)e^{iqx}\\
&=e^{-iqa}e^{iqx}
\end{align}
So the translation operator just acts on momentum eigenstates by multiplying them by a phase factor. So this leaves the wavefunction invariant since global phase factors don't change the physical wave function. If you construct a linear combination of eigenstates like in the Fourier transform
it will act to shift the entire wavefunction. You can easily check this.
$$\psi(x)=\int dk\,\tilde \psi(k)e^{ikx}\\\hat U_a\psi(x)=\ ?$$
Final note: throughout this answer I used the position representation to make things easier i.e. $f(x)$ instead of $\langle x|f\rangle$. With some effort you could translate this back to Bra-Ket notation. For example $\hat U_ae^{iqx}=\langle x|\hat U_a|q\rangle$. But I won't do it here because the general idea still holds.
