<s>[INST] Derivation of Total Momentum Operator for Klein-Gordon Field Quantization I am studying the second chapter of Peskin and Schroeder's QFT text. In equation 2.27 and 2.28, the book defines the field operators:
$$
\phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 w_p}} (a_p + a^\dagger_{-p}) \, e^{ipx} \\
\pi(x) = \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx}
$$
The momentum operator is then calculated in equation 2.33. However, my own derivation gives a different answer. I am reproducing my steps hoping that someone will be able to find where I went wrong. Starting with the definition of the momentum (conserved charge of spatial translations):
$$
\mathbf{P} = -\int d^3x \, \pi(x) \nabla \phi(x) \\
\mathbf{P} = -\int d^3x \, \Bigg[ \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx} \Bigg]\Bigg[\int \frac{d^3p'}{(2\pi)^3} \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \, \nabla e^{ip'x} \Bigg] \\
\mathbf{P} = -\int \int  \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \int d^3x \, (-i^2) \, \mathbf{p}' \, e^{i(p+p')x}  \, \Bigg[\sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \Bigg]\Bigg[\frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \Bigg] \\
\mathbf{P} = -\int \int  \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \mathbf{p}' (2\pi)^3 \delta(p+p')  \, \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \\
\mathbf{P} = -\int \int  \frac{d^3p}{(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{-p}}} (a_{-p} + a^\dagger_{p}) \\
$$
Since $w_{p} = w_{-p} = |p|^2 + m^2$, we get
$$
\mathbf{P} = -\int \frac{d^3p}{2(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{w_{p}}} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} + a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] + \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg] \\
$$
The first integral is odd with respect to p, and vanishes. For the second term, we can formally prove that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$, but we can also argue that from noting that this operator pair creates a particle but then destroys it, with any possible constants only depending on the magnitude of \mathbf{p}. This line of reasoning gives us:
$$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{p} a_{p}\bigg) = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \, [ a_p,a^\dagger_{p}] \\
$$
The commutator here is proportional to the delta function, and hence this expression doesn't match what Peskin & Schroeder, and other QFT books have, i.e.,
$$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p} \, a^\dagger_{p} a_p
$$
UPDATE:
I realized later that my assumption that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$ was wrong. When I was trying to prove this using the expansion of the ladder operators in terms of $\phi(x)$ and $\pi(x)$ I was making an algebra error.
 [/INST] From
$$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg)$$
you actually have  $$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{\mathbf{p}}{2}a^\dagger_{-p} a_{-p}\right) = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{-\mathbf{p}}{2}a^\dagger_{p} a_{p}\right)
=   \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a^\dagger_{p} a_p + \frac{\mathbf{p}}{2}a^\dagger_{p} a_{p}\right) + \mbox{renormalization term.}\\
$$
Dropping the infinite renormalization term due to $\delta({\bf 0})$
 $$
\mathbf{P} 
=   \int \frac{d^3p}{(2\pi)^3} \mathbf{p}\: a^\dagger_{p} a_p \:.
$$
</s>