- #1

- 90

- 12

- 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.

#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.