- #1

Markus Kahn

- 112

- 14

- 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?

$$\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?