# Linear momentum of the Klein Gordon field

Homework Statement:
I need to find an expression for the linear momentum of the Klein Gordon field in terms of creation and anihilation operators, but I fail to do it.
Relevant Equations:
##P_j = \int dx^3 \pi(x) \partial^j\phi(x)##
The correct answer is:
#P = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}} \big(a a^{\dagger} + a^{\dagger}a\big)#

But I get terms which are proportional to ##aa## and ##a^{\dagger}a^{\dagger}##

I hereunder display the procedure I followed:

First:
##\phi = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}}} \big(e^{-ipx}a(\vec{p})+e^{ipx}a^\dagger(\vec{p})\big)##

## \pi = \int \frac{dp^3}{(2\pi)^3}\frac{i}{2} \big(-e^{-ipx}a(\vec{p})+e^{ipx}a^\dagger(\vec{p}) \big)##

So that for the j-th component of the momentum we have

##P_j = \int dx^3 \pi(x) \partial^j\phi(x)=\int \frac{dp^3dp'^3dx^3}{(2\pi)^6}\frac{i}{4E_{\vec{p}}} \Big(-e^{-ipx}a(\vec{p}) + e^{ipx} a^{\dagger}(\vec{p}) \Big) \Big(-ip'_je^{-ip'x}a(\vec{p'}) +ip'_je^{ip'x} a^{\dagger}(\vec{p'}) \Big) = \int \frac{dp^3dp'^3dx^3}{(2\pi)^6}\frac{i}{4E_{\vec{p}}} \Big(- p_je^{-ix(p'+p)}a(\vec{p})a(\vec{p'}) - p_je^{ix(p+p')} a^{\dagger}(\vec{p})a^{\dagger}(\vec{p'})+ p_je^{ix(p-p')}(\vec{p})a^{\dagger}(\vec{p'}) + p_je^{ix(p-p')} a^{\dagger}(\vec{p})a(\vec{p'}) \Big) = \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})-p_ja^\dagger(\vec{p})a^\dagger(-\vec{p}) + p_j a(\vec{p})a^\dagger(\vec{p}) + pja^\dagger(\vec{p})a(\vec{p}) \Big)##

Note that there is also a factor 2 that disagrees with the desired solution.

Many thanks in advance.

## Answers and Replies

TSny
Homework Helper
Gold Member
##P_j = \int dx^3 \pi(x) \partial^j\phi(x)= ... = \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})-p_ja^\dagger(\vec{p})a^\dagger(-\vec{p}) + p_j a(\vec{p})a^\dagger(\vec{p}) + pja^\dagger(\vec{p})a(\vec{p}) \Big)##

Note that there is also a factor 2 that disagrees with the desired solution.
Consider ## \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})\Big)##

Show that the integrand is odd in ##\vec p##. That is changing ##\vec p## to ##-\vec p## just changes the overall sign of the integrand.

Your factor of 4 occurring in ##4E_{\vec p}## in the denominator looks correct to me. But I might be missing something.

You can write $$p_j a(\vec{p})a^\dagger(\vec{p}) + p_ja^\dagger(\vec{p})a(\vec{p})$$ as $$2 p_ja^\dagger(\vec{p})a(\vec{p}) + p_j[a(\vec{p}),a^\dagger(\vec{p})]$$ Integration over the 2nd term is trivial if you use the result for the commutator ##[a(\vec{p}),a^\dagger(\vec{p})]##. So, you are left with just an integral over the first term. This has a factor of 2 which can be used to change ##4E_{\vec p}## into ##2E_{\vec p}## in the denominator. You'll end up with a nice compact result for ##P_j##, but it still doesn't agree with the answer that was given to you as shown below:

The correct answer is:
$$P = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}}} \big(a a^{\dagger} + a^{\dagger}a\big)$$

This doesn't look right. Check to see if you typed this correctly.

Consider ## \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})\Big)##

Show that the integrand is odd in ##\vec p##. That is changing ##\vec p## to ##-\vec p## just changes the overall sign of the integrand.

Your factor of 4 occurring in ##4E_{\vec p}## in the denominator looks correct to me. But I might be missing something.

You can write $$p_j a(\vec{p})a^\dagger(\vec{p}) + p_ja^\dagger(\vec{p})a(\vec{p})$$ as $$2 p_ja^\dagger(\vec{p})a(\vec{p}) + p_j[a(\vec{p}),a^\dagger(\vec{p})]$$ Integration over the 2nd term is trivial if you use the result for the commutator ##[a(\vec{p}),a^\dagger(\vec{p})]##. So, you are left with just an integral over the first term. This has a factor of 2 which can be used to change ##4E_{\vec p}## into ##2E_{\vec p}## in the denominator. You'll end up with a nice compact result for ##P_j##, but it still doesn't agree with the answer that was given to you as shown below:

This doesn't look right. Check to see if you typed this correctly.
Yes. it is definitely not right. There's a two at the denominator missing and also the momentum p should multiply the whole expression. Thank you very much for your answer, what you told me about the expression being odd rescues everything.

TSny