Plane wave shift in a differential operator Does anyone can help me to prove the following equation
\begin{equation}
e^{-i\vec{k}\cdot\vec{x}}f(\partial_{\mu})e^{i\vec{k}\cdot\vec{x}} = f(\partial_{\mu}+ik_{\mu})
\end{equation}
Where $\vec{k}\cdot\vec{x}=k^{\mu}x_{\mu}, \mu = 0,1,2,3$.
 A: Assume that $g$ is a function in the space of rapidly decreasing functions with all their derivatives of any order, the so-called Schwartz space ${\cal S}(\mathbb R^4)$. 
Define $$\hat{g}(k) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} e^{-i k_\mu x^\mu} g(x) d^4x\::$$ As a consequence, as is well known, $\hat{g} \in {\cal S}(\mathbb R^4)$ and
$$g(x) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:.$$ 
Moreover, by standard results on the interchange of integration and derivative symbols which are valid on the class of functions we are dealing with, we find 
$$\partial^{n_1}_{1}\cdots \partial^{n_4}_{4} g(x) = 
\frac{1}{(2\pi)^2} \int_{\mathbb R^4} \partial^{n_1}_{1}\cdots \partial^{n_4}_{4}e^{i k_\mu x^\mu} \hat{g}(k) d^4k$$
$$=\frac{1}{(2\pi)^2} \int_{\mathbb R^4} (ik)^{n_1}_{1}\cdots (ik)^{n_4}_{4} e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:.$$
and more generally assuming that the sum in the right-hand side encompasses a finite number of terms, where $a_{n_1\ldots n_4}$ are complex constants,
$$\sum_{n_1\ldots n_4}a_{n_1\ldots n_4}\partial^{n_1}_{1}\cdots \partial^{n_4}_{4} g(x) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} \sum_{n_1\ldots n_4}a_{n_1\ldots n_4}(ik)^{n_1}_{1}\cdots (ik)^{n_4}_{4} e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:.$$
In other words, if $p=p(\partial_{1},\cdots, \partial_{4})$ is a complex polynomial, where I assume $$c\partial^0_\alpha := cI$$ in particular,
$$\left(p(\partial_{1},\cdots, \partial_{4}) g\right)(x) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} p(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:.\tag{1}$$
There are several reasons to extend this identity to a class of functions $f$ larger than polynomials $p$, extending also the class of functions $g$ for instance to the space of Schwartz distributions if necessary. This is because the ring of polynomials if dense in larger spaces of functions with respect to relevant topologies.
This extension makes sense as soon as $p_n \to f$ implies 
$$\int_{\mathbb R^4} p_n(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k) d^4k \to 
\int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:.$$
If $$\int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k) d^4k$$
I henceforth assume 
$$\left(f(\partial_{1},\cdots, \partial_{4}) g\right)(x) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k) d^4k\:,\tag{2}$$
where $f$ is a polynomial or a more complicated function which can be suitably approximated by polynomials. This identity entails
$$e^{-ihx}\left(f(\partial_{1},\cdots, \partial_{4}) e^{ihx}g\right)(x) = \frac{e^{-ihx}}{(2\pi)^2} \int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \widehat{e^{ihx}g}(k) d^4k\:,$$
On the other hand 
$$\widehat{e^{ihx}g}(k) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{-i (k_\mu-h_\mu) x^\mu}  g(x)d^4x = \hat{g}(k-h)\:,$$
so that, using the translational invarience of the measure $d^4k$
$$e^{-ihx}\left(f(\partial_{1},\cdots, \partial_{4}) e^{ihx}g\right)(x) = 
\frac{e^{-ihx}}{(2\pi)^2} \int_{\mathbb R^4} f(ik_1,\cdots, ik_4) e^{i k_\mu x^\mu} \hat{g}(k-h) d^4k$$
$$=
\frac{e^{-ihx}}{(2\pi)^2} \int_{\mathbb R^4} f(i(k_1+h_1),\cdots, i(k_4+h_4))e^{i (k_\mu-h_\mu) x^\mu} \hat{g}(k) d^4k$$ $$ = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} f(i(k_1+h_1),\cdots, i(k_4+h_4)) e^{i k_\mu x^\mu} \hat{g}(k) d^4k$$
Comparing with (2), paying attention to the fact that $ih_k$ act as a multiplicative constant when $f$ is a polynomials and thus in general, so it is equivalent to $ih_k\partial_k^0 = ih_kI$ when acting on $g(x)$ and not $\hat{g}(k)$ we have 
$$e^{-ihx}\left(f(\partial_{1},\cdots, \partial_{4}) e^{ihx}g\right)(x)$$
$$= \frac{1}{(2\pi)^2} \int_{\mathbb R^4} f(i(k_1+h_1),\cdots, i(k_4+h_4)) e^{i k_\mu x^\mu} \hat{g}(k) d^4k$$
$$\left(f(\partial_{1}+ ih_1,\cdots, \partial_{4}+ih_4) g\right)(x)\:.$$
Since $g$ is generic,
$$e^{-ihx}f(\partial_{1},\cdots, \partial_{4}) e^{ihx} = f(\partial_{1}+ ih_1I,\cdots, \partial_{4}+ih_4I) \:.$$
A: PART 1 :  Taylor Expansion ( see  @Prahar comment ).
In the following :
$$
\mathbf{x}=x^{\mu},  \quad  \mathbf{k}=k_{\mu},  \quad  \partial_{\mu}=\dfrac{\partial}{\partial x^{\mu}} , \quad  \mu=0,1,2,3
\tag{1-01}
$$
$$
\mathbf{k}\cdot\mathbf{x}=k_{\mu}x^{\mu}=k^{\mu}x_{\mu}, \quad  \text{Einstein's convention on}\: \mu
\tag{1-02}
$$
Suppose now that the function $\:f\left(z\right)\:$ could be expanded in Taylor series, see  https://en.wikipedia.org/wiki/Taylor_series. 
$$
f\left(z\right)=\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:z^{n}
\tag{1-03}
$$
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(z\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:e^{-i\mathbf{k}\cdot\mathbf{x}}z^{n}e^{i\mathbf{k}\cdot\mathbf{x}}
\tag{1-04}
$$
For $z=\partial_{\mu}$ equation (1-04) gives
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(\partial_{\mu}\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}
\tag{1-05}
$$
In PART 2  we prove by induction the following equation :
$$
\bbox[#FFFF88,12px]{e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}}
\tag{1-06}
$$
so from (1-05)
$$
e^{-i\mathbf{k}\cdot\mathbf{x}} f\left(\partial_{\mu}\right)e^{i\mathbf{k}\cdot\mathbf{x}} =\sum_{n=0}^{n=\infty}\dfrac{f^{\left(n\right)}\left(0\right)}{n!}\:\left( \partial_{\mu} + ik_{\mu}\right)^{n}=f\left( \partial_{\mu} + ik_{\mu}\right)
\tag{1-07}
$$
QED.

PART 2 :  Proof  by induction of equation (1-06) : $\:e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}$
Let a function $\:g\left(\mathbf{x}\right)\:$  infinitely differentiable. Then
(1). For $\:n=1\:$
\begin{align}
\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]\:g\left(\mathbf{x}\right) & =e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:\left[e^{i\mathbf{k}\cdot\mathbf{x}}\:g\left(\mathbf{x}\right)\right]\\
& =e^{-i\mathbf{k}\cdot\mathbf{x}}\:\left[e^{i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:g\left(\mathbf{x}\right)+ik_{\mu}e^{i\mathbf{k}\cdot\mathbf{x}}g\left(\mathbf{x}\right)\right]
\tag{2-01}\\
& = \left( \partial_{\mu}+ ik_{\mu} \right)g\left(\mathbf{x}\right) 
\end{align}
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}\:e^{i\mathbf{k}\cdot\mathbf{x}}= \left( \partial_{\mu}+ ik_{\mu} \right)
\tag{2-02}
$$ 
(2) Suppose that for a positive integer  $\:n\:$ 
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}
\tag{2-03}
$$
then
(3) for $\:n+1\:$ we have
\begin{align}
\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n+1)}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]\:g\left(\mathbf{x}\right) & = e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\left(\partial_{\mu}\left[ e^{i\mathbf{k}\cdot\mathbf{x}}\:g\left(\mathbf{x}\right)\right] \right)  \\
& = e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\left[ e^{i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}g\left(\mathbf{x}\right)+ik_{\mu}e^{i\mathbf{k}\cdot\mathbf{x}}g\left(\mathbf{x}\right) \right]\\
& =\underbrace{\left[e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n)}\:e^{i\mathbf{k}\cdot\mathbf{x}}\right]}_{\left( \partial_{\mu} + ik_{\mu}\right)^{n}}\left( \partial_{\mu}+ ik_{\mu} \right)g\left(\mathbf{x}\right)\\
&= \left( \partial_{\mu}+ ik_{\mu} \right)^{n+1} g\left(\mathbf{x}\right)
\end{align}
so
$$
e^{-i\mathbf{k}\cdot\mathbf{x}}\:\partial_{\mu}^{(n+1)}\:e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu}+ ik_{\mu} \right)^{n+1} 
\tag{2-04}
$$
proving finally (1-06)
$$
\bbox[#FFFF88,12px]{e^{-i\mathbf{k}\cdot\mathbf{x}}\partial_{\mu}^{\left(n\right)}e^{i\mathbf{k}\cdot\mathbf{x}}=\left( \partial_{\mu} + ik_{\mu}\right)^{n}}
\tag{1-06}
$$
