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This question pertains to Equation (2.33) in Peskin and Schroeder: $$ \hat{\vec P}=-\int d^3\!x\,\hat\pi(\vec x)\vec\nabla\hat\phi(\vec x)=\int d^3\!p\,\vec p\,\hat a_{\vec p}^\dagger\,\hat a_{\vec p} $$

I know there are several other questions on Stackexchange about this (for example), and my question exceeds all of them. Firstly, Eq (2.33) is wrong because $\hat\phi$ and $\hat\pi$ don't commute, so the first part of the above should be: $$ \hat{\vec P}=-\frac{1}{2}\int d^3\!x\,\left[\hat\pi(\vec x)\vec\nabla\hat\phi(\vec x)+\vec\nabla\hat\phi(\vec x)\hat\pi(\vec x)\right] $$

However, to develop my question I will follow Eq (2.33) as it appears in the book. Given \begin{align} \hat\phi(\vec x)&=\int \dfrac{d^3\!p}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_{\vec p}}}\left( \hat a_{\vec p}+\hat a^\dagger_{-\vec p} \right)e^{i\vec p\cdot\vec x}\\ \hat\pi(\vec x)&=-i\int \dfrac{d^3\!p}{(2\pi)^3}\sqrt{\dfrac{\omega_{\vec p}}{2}}\left( \hat a_{\vec p}-\hat a^\dagger_{-\vec p} \right)e^{i\vec p\cdot\vec x} \end{align}

we obtain \begin{align} \hat{\vec P}&=i\int d^3\!x \int \dfrac{d^3\!p'}{(2\pi)^3}\sqrt{\dfrac{\omega_{\vec p'}}{2}}\left( \hat a_{\vec p'}-\hat a^\dagger_{-\vec p'} \right)e^{i\vec p'\cdot\vec x}\int \dfrac{d^3\!p''}{(2\pi)^3}(i\vec p'')\dfrac{1}{\sqrt{2\omega_{\vec p''}}}\left( \hat a_{\vec p''}+\hat a^\dagger_{-\vec p''} \right)e^{i\vec p''\cdot\vec x}\\ &=\frac{-1}{2}\iint\dfrac{d^3\!p'\,d^3\!p''}{(2\pi)^3}\sqrt{\frac{\omega_{\vec p'}}{\omega_{\vec p''}}}\left( \hat a_{\vec p'}-\hat a^\dagger_{-\vec p'} \right) \left( \hat a_{\vec p''}+\hat a^\dagger_{-\vec p''} \right)\vec p''\int \dfrac{d^3\!x}{(2\pi)^3}e^{i\vec x\cdot(\vec p'+\vec p'')}\\ &=\frac{-1}{2}\iint\dfrac{d^3\!p'\,d^3\!p''}{(2\pi)^3}\sqrt{\frac{\omega_{\vec p'}}{\omega_{\vec p''}}}\left( \hat a_{\vec p'}-\hat a^\dagger_{-\vec p'} \right) \left( \hat a_{\vec p''}+\hat a^\dagger_{-\vec p''} \right)\vec p''\delta^{(3)}(\vec p'+\vec p'') \\ &=\frac{-1}{2}\int\dfrac{d^3\!p'}{(2\pi)^3} \left( \hat a_{\vec p'}-\hat a^\dagger_{-\vec p'} \right) \left( \hat a_{-\vec p'}+\hat a^\dagger_{\vec p'} \right)(-\vec p') \end{align}

Drop the prime for concision in notation to get: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\left( \hat a_{\vec p}-\hat a^\dagger_{-\vec p} \right) \left( \hat a_{-\vec p}+\hat a^\dagger_{\vec p} \right)\\ &=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\left( \hat a_{\vec p}\hat a_{-\vec p}+\hat a_{\vec p}\hat a^\dagger_{\vec p} -\hat a^\dagger_{-\vec p}\hat a_{-\vec p}-\hat a^\dagger_{-\vec p}\hat a^\dagger_{\vec p} \right) \\ \end{align}

The antisymmetric part of the integrand ($f(-x)=-f(x)$) vanishes leaving: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \hat a_{\vec p}\hat a^\dagger_{\vec p} -\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\,\hat a^\dagger_{-\vec p}\hat a_{-\vec p} \\ \end{align}

Introduce $\vec p'=-\vec p$ and $-d^3\!p'=d^3\!p$ in the second term to get: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \hat a_{\vec p}\hat a^\dagger_{\vec p} -\frac{1}{2}\int^{-\infty}_\infty\dfrac{-d^3\!p'}{(2\pi)^3}\, (-\vec p')\,\hat a^\dagger_{\vec p'}\hat a_{\vec p'} \\ &=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \hat a_{\vec p}\hat a^\dagger_{\vec p} +\frac{1}{2}\int_{-\infty}^\infty\dfrac{d^3\!p'}{(2\pi)^3}\, \vec p'\,\hat a^\dagger_{\vec p'}\hat a_{\vec p'} \end{align}

Rename the integration variable in the second term $\vec p'=\vec p$ to get: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \left( \hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p} \right) \end{align}

To obtain the answer in Peskin (2.33), I will add into the integrand $$ 0=\hat a^\dagger_{\vec p}\hat a_{\vec p} -\hat a^\dagger_{\vec p}\hat a_{\vec p} $$

to obtain: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \left( \hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p} +\underbrace{\hat a^\dagger_{\vec p}\hat a_{\vec p} -\hat a^\dagger_{\vec p}\hat a_{\vec p}}_{0}\right)\\ &=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \left( [\hat a_{\vec p},\hat a^\dagger_{\vec p}]+2\hat a^\dagger_{\vec p}\hat a_{\vec p} \right)\\ \end{align}

I may have erred here since I have a scalar equal to an operator: $0=\hat a^\dagger\hat a-\hat a^\dagger\hat a $. If this is wrong, please explain. I think this ok, however, because I can just call that the zero operator: $\hat 0$. To continue, I will throw away the infinite term $[\hat a_{\vec p},\hat a^\dagger_{\vec p}]\propto\delta(0)$, and now I get: \begin{align} \hat{\vec P}&=\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \hat a^\dagger_{\vec p}\hat a_{\vec p} \\ \end{align}

This is the correct answer, per Peskin. However, I could have added the following into the integrand: $$ 0=\hat a_{\vec p}\hat a^\dagger_{\vec p} -\hat a_{\vec p}\hat a^\dagger_{\vec p} $$

to obtain: \begin{align} \hat{\vec P}&=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \left( \hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p}+\hat a_{\vec p}\hat a^\dagger_{\vec p} -\hat a_{\vec p}\hat a^\dagger_{\vec p} \right)\\ &=\frac{1}{2}\int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \left( 2\hat a_{\vec p}\hat a^\dagger_{\vec p}+[\hat a^\dagger_{\vec p},\hat a_{\vec p}]\right)\\ &\approx \int\dfrac{d^3\!p}{(2\pi)^3}\, \vec p\, \hat a_{\vec p}\hat a^\dagger_{\vec p} \end{align}

This is only equal to Peskin's Eq (2.33) if $\hat a_{\vec p}$ and $\hat a^\dagger_{\vec p}$ commute, but they do not. I assume the problem is the preferential ordering of the non-commuting $\hat\pi\nabla\hat\phi$ terms when I started and that a complete calculation will yield the correct answer. Before I do that, I'd like to get someone's comments. Why is one step at the end preferred over the other?

TLDR: Given: $$ 0=\hat a^\dagger_{\vec p}\hat a_{\vec p} -\hat a^\dagger_{\vec p}\hat a_{\vec p}\qquad\text{and}\qquad0=\hat a_{\vec p}\hat a^\dagger_{\vec p} -\hat a_{\vec p}\hat a^\dagger_{\vec p} $$

it follows that: \begin{align} \hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p}&=\hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p}\\ \underbrace{\hat a^\dagger_{\vec p}\hat a_{\vec p} -\hat a^\dagger_{\vec p}\hat a_{\vec p}}_{0} + \hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p} &=\hat a_{\vec p}\hat a^\dagger_{\vec p}+\hat a^\dagger_{\vec p}\hat a_{\vec p}+\underbrace{\hat a_{\vec p}\hat a^\dagger_{\vec p} -\hat a_{\vec p}\hat a^\dagger_{\vec p}}_{0}\\ [\hat a_{\vec p},\hat a^\dagger_{\vec p}]+2\hat a^\dagger_{\vec p}\hat a_{\vec p} &=2\hat a_{\vec p}\hat a^\dagger_{\vec p}+[\hat a^\dagger_{\vec p},\hat a_{\vec p}] \end{align}

These things don't seem equal to me, especially if we're throwing away the infinite part. What is going on here?

Qmechanic
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hodop smith
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    The answer to your literal question is that the difference between your two forms for $P$ is proportional to $\int d^3 p , \vec{p}$ which vanishes by symmetry. Peskin's form is preferred because it manifestly vanishes when it acts on the vacuum. – knzhou Sep 26 '23 at 04:07
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    However, your final remark makes me feel like your true confusion is about something much simpler. To keep the notation sane, let $A = \hat{a}{\vec{p}}$ and let $B = \hat{a}^\dagger\vec{p}$. Your final line is $[A, B] + 2 B A = 2 A B + [B, A]$ and you say these sides "don't seem equal to me". But they obviously are by the definition of the commutator. We have $[A, B] = - [B, A]$, so simplifying and dividing by $2$ gives $[A, B] = A B - B A $ which is literally just the definition of the commutator. – knzhou Sep 26 '23 at 04:09
  • When we drop the infinite part, the relationship that you're showing no longer holds. Other than that, I am aware that adding zero to both sides of an equation doesn't violate the equality. I think you have not read my question very closely. For example, "These things don't seem equal to me, especially if we're throwing away the infinite part.," makes reference to the condition you have ignored. – hodop smith Sep 26 '23 at 04:12
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    But you're asking two different questions and seem to be mixing them up. The thing in your "TLDR" is just an obviously true operator identity. It has nothing to do with the confusion in the rest of your question. – knzhou Sep 26 '23 at 04:16
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    When we "drop the infinite part", we're not changing the definition of the commutator or something silly like that. We are dropping a term out of an integrand because it doesn't contribute to the full integral, by symmetry. – knzhou Sep 26 '23 at 04:16
  • Also, the difference between my two forms for $P$ is $\int\dfrac{d^3!p}{(2\pi)^3}, \vec p(\hat a^\dagger_{\vec p}\hat a_{\vec p} -\hat a_{\vec p}\hat a^\dagger_{\vec p} )= \int d^3!p, \vec p\delta^{(3)}(0)$, which is, as you say, an odd function and must vanish. Thank you. – hodop smith Sep 26 '23 at 04:17
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    Yes, I have neglected to consider that $p$ times the commutator in question is an odd function. I thought we were dropping it because it was infinite, not because it integrates to zero. Very good and helpful. Thanks! – hodop smith Sep 26 '23 at 04:22

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