Quantum operator calculations 
We define the quantum operator 
  $$
P^\mu=\int{\frac{d^3p}{(2\pi)^3}}p^\mu a_p^\dagger a_p
$$
  Now how can I calculate
  $$
\langle p_2|P^\mu|p_1\rangle~?
$$

My attempt:
$$
\langle p_2|P^\mu|p_1\rangle =\int{\frac{d^3p}{(2\pi)^3}}\langle 0|a_{p_2} p^\mu a_p^\dagger a_p a_{p_1}^\dagger|0\rangle.
$$
Now we know that $\langle0|a_p a_q^\dagger|0\rangle =\delta^{(3)}(p-q)$ but I'm not quite sure how it works with multiple states in the bra-ket.
 A: Use the canonical commutation relation $[a_p, a^{\dagger}_q] = (2\pi)^3\delta^3(p-q)$:
We have that $a_{p_2}p^{\mu}a^{\dagger}_pa_pa^{\dagger}_{p_1} =p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}a_p + p^{\mu}a_{p_2}a^{\dagger}_p[a_p, a^{\dagger}_{p_1}]$
The first term is ignored, because when considering 
$<0|p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}a_p|0>$
We have an annihilation operator hitting the vacuum state, so this term in the integrand must vanish. Plugging in the commutator, your integral is equal to
$\int\frac{d^3p}{(2\pi)^3}p^{\mu}(2\pi)^3\delta^3(p-p_1)<0|a_{p_2}a^{\dagger}_p|0> $
You can now integrate out the delta function:
$p_1^{\mu}<0|a_{p_2}a^{\dagger}_{p_1}|0> $
Repeating the same trick with commuting the operators and using $<0|0>=1$, we have 
$p_1^{\mu}(2\pi)^3\delta^3(p_1-p_2)$
Pretty much as expected. The much easier route is if you recognize that states $|p_1>$ are eigenstates of the operator $P^\mu$ with eigenvalue $p_1^\mu$. Using this,
$<p_2|P^\mu|p_1> = p_1^\mu<p_2|p_1>$
and using the normalization of singly excited states,
$=p_1^\mu(2\pi)^3\delta^3(p_1-p_2)$
A: $\newcommand{ket}[1]{\left|#1\right>}$Marcus has given the mathematical answer using the commutation rules of ladder operators $a_p$ and $a_q^\dagger$. I want to give a bit the intuition behind the operator $P^\mu$ and how you can almost guess the answer. Note that this intuition is solidified by the calculation that Marcus gives and in no way replaces it. 
Recall that $a_p^\dagger$ creates a state with momentum quanta $p$ and $a_p$ kills a state unless the state has $p$ momentum quanta. In more mathematical terms, if we have a superposition state $\ket\psi = \int d^3q/(2 \pi)^3 f(q) a_q^\dagger\ket0 $ for an arbitrary function $f(q)$, then 
$$ a_p \ket \psi  \neq 0 \iff f(p) \neq 0$$
In fact $a_p$ "picks out" the $p$ momentum quanta i.e.
$$ a_p \ket \psi = f(p) \ket0$$
Hence, the operator $a_p^\dagger a_p$ first picks out the $p$ momentum quanta by $a_p$ and then replenishes it back by $a_p^\dagger$. You can even show that the operator $\int d^3q/(2 \pi)^3 a_q^\dagger a_q$ is the identity because at each value of $q$ we undo what we have done. From this intuition it should be clear that 
$$P^\mu = \int \frac{d^3q}{(2 \pi)^3}  q^\mu a_q^\dagger a_q$$
is the operator that picks the $q$ momentum quanta weights it with the momentum $q^\mu$ and then sums over all possible $q$'s. Since the states $\ket{p_1}$ by definition have one $p_1$ momentum quanta we expect 
$$ P^\mu \ket{p_1} = p_1^\mu \ket{p_1}$$
From which the result follows.
