Momentum Operator for the real scalar field

[Markus Kahn]

Homework Statement

Show that the momentum for the real scalar field reduces to
$$\vec{P} = \int \frac{dp^3}{(2\pi)^3 2e(p)} \vec{p} a_p^\dagger a_p.$$

Relevant Equations

$$\vec{P}=-\int dx^3 \pi \nabla \phi$$
$$\phi(\vec x )= \int \frac{dp^3}{(2\pi)^3 2e(p)} (a_pe^{ipx} + a^\dagger_p e^{-ipx})$$
$$\pi(\vec x )= \int \frac{dp^3}{(2\pi)^3 }\frac{i}{2} (a_p^\dagger e^{-ipx} - a_p e^{ipx})$$

I think the solution to this problem is a straightforward calculation and I think I was able to make reasonable progress, but I'm not sure how to finish this...

$$\begin{align*} \vec{P}&=-\int dx^3 \pi \nabla \phi\\
&= -\int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{i}{2} \left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\nabla \left(a_pe^{ipx} + a^\dagger_p e^{-ipx}\right)\\
&= -\int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{i}{2} (ip)\left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\left(a_pe^{ipx} - a^\dagger_p e^{-ipx}\right)\\
&= \int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{p}{2} \left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\left(a_pe^{ipx} - a^\dagger_p e^{-ipx}\right)\\
&= \int\int\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3}\times \\& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{p}{2} \int dx^3 \left(a_u^\dagger a_p e^{i(p-u)x} +a_ua_p^\dagger e^{i(u-p)x} -a_u^\dagger a_p^\dagger e^{-i(u+p)x} - a_ua_p e^{i(p+u)x}\right)
\end{align*}$$

I know applied to every term of the sum the following formula:
$$\delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ip(x-\alpha )}\ dp,$$
which resulted in

$$\begin{align*}\vec{P}&= \int\int\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3}\times \\& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{p}{2}(2\pi)^3 \left(a_u^\dagger a_p \delta(p-u) +a_ua_p^\dagger \delta(u-p) -a_u^\dagger a_p^\dagger \delta(u+p) - a_ua_p \delta(p+u)\right)\\
&= \int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \int du^3 \left(a_u^\dagger a_p \delta(u-p) +a_ua_p^\dagger \delta(u-p) -a_u^\dagger a_p^\dagger \delta(u+p) - a_ua_p \delta(u+p)\right)
\end{align*}$$
I'm not too sure about this step, but I basically used
$$\int f(x)\,\delta(x-a)\,dx =\int f(x)\,\delta(a-x)\,dx.$$
If I continue from here on I arrive at
$$\begin{align*}\vec P &= \int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \int du^3 \left((a_u^\dagger a_p+a_ua_p^\dagger )\delta(u-p) -(a_u^\dagger a_p^\dagger + a_ua_p) \delta(u+p)\right)\\
&=\int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times (a_p^\dagger a_p+a_pa_p^\dagger )-(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)\\
&=\int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times (2a_p^\dagger a_p+1 )-(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)\\
&=\int \frac{dp^3}{(2\pi)^3 2e(p)} p (a_p^\dagger a_p+\frac{1}{2} ) -(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)
\end{align*}$$
The term corresponding to the $1/2$ in parenthesis can be resolved with renormalization (as far as I understand), so we don't really need to bother with it. But what about the $(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)$? How do I get rid of that?

 

[mitochan]

Hi.
Say E(p)=E(-p), the integrand of
-\frac{1}{4}\int \frac{dp^3}{(2\pi)^3E(p)}p(a^\dagger_{-p} a^\dagger_p+a_{-p}a_p)
is antisymmetric for inverse of p . So the integral vanishes.

 

[Ribeiro]

Hello Markus Kanh!
So, I examined your calculation and I have feeling that something is odd in your definition of conjugate momentum of the field. In the books that I have learning I never see yet that definition. I often see the following definition $$\pi(\vec x)=-i\int \frac {d^{3}p} {(2\pi)^{3} } {\sqrt{E_{p}/ 2}}(a_{p}e^{ipx}-a_{p}e^{-ipx})$$ One example is in the Peskin's book ''An introduction to QFT''. Anyway... this is just one comment.
About your question, we can show that the second term in your last equation is zero$\int \frac p {(2\pi)^{3}2E_p} (a_{-p}^\dagger a_{p}^\dagger+a_{-p}a_{p})\,d^{3}p=^{!}0$.
Consider the following integral $$I=\int \frac p {(2\pi)^{3}2E_p} a_{-p}^\dagger a_{p}^\dagger\,d^{3}p$$ we can change the variable $p\to-p$ That change don't change the value of integral because that integral is evaluated in all p-space. $$I=-\int \frac p {(2\pi)^{3}2E_p} a_{p}^\dagger a_{-p}^\dagger\,d^{3}p$$ summing these two pieces above, we have $$2I=\int \frac p {(2\pi)^{3}2E_p} (a_{-p}^\dagger a_{p}^\dagger-a_{p}^\dagger a_{-p}^\dagger)\,d^{3}p$$ that is $$2I=\int \frac p {(2\pi)^{3}2E_p}[a_{-p}^\dagger,a_{p}^\dagger]\,d^{3}p$$ note that $[a_{-p}^\dagger,a_{p}^\dagger]=0$ in this way we can saying that $$I=0$$ Similarly, we can show that the second term in the your last equation is also zero.