Unitary spacetime translation operator Srednicki writes:  We can make this a little fancier by defining the unitary spacetime translation operator
$$  T(a) \equiv \exp(-iP^\mu a_\mu/ \hbar)  $$
Then we have 
$$ T(a)^{-1} \phi(x) T(a) = \phi(x-a)$$
How do we get the second equation from the first equation?
 A: It is unnecessary to use Hilbert space states to formally derive this result; it quickly follows from a useful result about the matrix exponential (which comes in very handy when one studies Lie algebras which, incidentally, is essentially what we're looking at here).  
Let $X$ be any $n$-by-$n$ complex matrix, then we define the linear operator $\mathrm{ad}_X$ on the vector space of such matrices by
$$
  \mathrm{ad}_X Y = [X,Y]
$$
for all $n$-by-$n$ complex matrices $Y$.  Here $[\cdot, \cdot]$ denotes the commutator often called the adjoint operator.  Then we have the following result:
$$
  e^XYe^{-X} = e^{\mathrm{ad}_X}Y
$$
Now, if we formally apply this result to linear operators on the Hilbert space of a quantum field theory, then we obtain
$$
  T(a)^{-1}\phi(x)T(a) = e^{ia_\mu P^\mu/\hbar}\phi(x)e^{-ia_\mu P^\mu/\hbar} = e^{\mathrm{ad}_{ia_\mu P^\mu/\hbar}}\phi(x) = \sum_{k=0}^\infty \frac{\mathrm{ad}^k_{ia_\mu P^\mu/\hbar}\phi(x)}{k!}
$$
Now we use the fact that for any field $\Phi$, we have
$$
  \mathrm{ad}_{ia_\mu P^\mu/\hbar}\phi(x)=\frac{ia_\mu}{\hbar} [P^\mu, \phi(x)] = \frac{ia_\mu}{\hbar}(i\hbar \partial^\mu)\phi(x) = -a_\mu \partial^\mu\phi(x)
$$
Applying this result $k$ times and inserting it into the series expansion for the exponential written above, we obtain
$$
  T(a)^{-1}\phi(x)T(a) = \sum_{k=0}^\infty\frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x)
$$
Now, we simply note that the right hand side is the Taylor expansion of $\phi(x-a)$.  Explictly
$$
  \phi(x-a) = \phi(x) -a_\mu\partial^\mu\phi +\frac{1}{2}(a_\mu\partial^\mu)^2\phi(x) + \cdots + \frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x)+\cdots
$$
and this gives the desired result.
A: [EDIT]
Supposing a state basis $|e_i\rangle $, we are going to use the following notation for a state: $|A(x)\rangle = \sum_i A_i(x)|e_i\rangle$
An operator $O$ appying on $A$ gives then: $O|A(x)\rangle = \sum_i OA_i(x)|e_i\rangle$
For instance, $\partial_{\mu}|A(x)\rangle = \sum\partial_{\mu}A_i(x)|e_i\rangle$
Now, instead of working with operators, I think it is simpler to work with states $\left|A(x)\right\rangle$ and $\left|B(x)\right\rangle$ such as: 
$$\left|B(x)\right\rangle  = \Phi(x) \left|A(x)\right\rangle  \tag{1}$$
This is true, of course, for $x-a$, that is: 
$$\left|B(x - a)\right\rangle  = \Phi(x- a) \left|A(x-a)\right\rangle \tag{2}$$
We know that: 
$$P_{\mu}\left|A(x)\right\rangle  = i \hbar \partial_{\mu}\left|A(x)\right\rangle.$$
So, we get:
\begin{align}
T(a)^{-1}\left|A(x)\right\rangle  
&=  e^{\large iP_\mu a^\mu/ \hbar}\left|A(x)\right\rangle\\  
&=e^{\large -a^{\mu}\partial_{\mu}} \left|A(x)\right\rangle\\  
&= \left|A(x - a)\right\rangle 
\end{align}
The last equality is simply the Taylor series of $\left|A(x - a)\right\rangle$ at $x$, that is:
$$\left|A(x - a)\right\rangle  = \left|A(x)\right\rangle  - a^{\mu}\partial_{\mu} \left|A(x)\right\rangle  + \frac{1}{2!} (a^{\mu}\partial_{\mu})^2\left|A(x)\right\rangle +\frac{(-1)^n}{n!}(a^{\mu}\partial_{\mu})^n\left|A(x)\right\rangle + \dots.$$
Now, applying $T(a)^{-1}$ to equation $(1)$, we get: 
$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)\left|A(x)\right\rangle.$$ 
That is:
$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)T(a)T(a)^{-1}\left|A(x)\right\rangle.$$
So, we get:
$$\left|B(x - a)\right\rangle =T(a)^{-1}\Phi (x)T(a)\left|A(x - a)\right\rangle.$$
Looking at equation $(2)$, we finally get: 
$$T(a)^{-1}\Phi (x)T(a) = \Phi (x-a)$$
