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.

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.

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., $$ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p} \, a^\dagger_{p} a_p $$

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.

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):

$$ \mathbf{P} = -\int d^3x \, \pi(x) \nabla \phi(x) \\ \mathbf{P} = -\int d^3x \, \Bigg[ \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx} \Bigg]\Bigg[\int \frac{d^3p'}{(2\pi)^3} \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \, \nabla e^{ip'x} \Bigg] \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \int d^3x \, (-i^2) \, p' \, e^{i(p+p')x} \, \Bigg[\sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \Bigg]\Bigg[\frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \Bigg] \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} p' (2\pi)^3 \delta(p+p') \, \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} (-p) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{-p}}} (a_{-p} + a^\dagger_{p}) \\ $$ Since $w_{p} = w_{-p} = |p|^2 + m^2$, we get $$ \mathbf{P} = -\int \frac{d^3p}{2(2\pi)^3} (-p) \sqrt{\frac{w_p}{w_{p}}} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} \bigg[ a_p a_{-p} + a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} \bigg[ a_p a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] + \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} \bigg[ a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg] \\ $$

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 p. This line of reasoning gives us:

$$ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{p} a_{p}\bigg) = \int \frac{d^3p}{(2\pi)^3} \frac{p}{2} \, [ a_p,a^\dagger_{p}] \\ $$

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., $$ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} p \, a^\dagger_{p} a_p $$

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., $$ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p} \, a^\dagger_{p} a_p $$