A Commutation relations - field operators to ladder operators

1. Apr 26, 2016

spaghetti3451

I would like to show that the commutation relations $[a_{\vec{p}},a_{\vec{q}}]=[a_{\vec{p}}^{\dagger},a_{\vec{q}}^{\dagger}]=0$ and $[a_{\vec{p}},a_{\vec{q}}^{\dagger}]=(2\pi)^{3}\delta^{(3)}(\vec{p}-\vec{q})$ imply the commutation relations $[\phi(\vec{x}),\phi(\vec{y})]=[\pi(\vec{x}),\pi(\vec{y})]=0$ and $[\phi(\vec{x}),\pi(\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$.

Firstly,

$[\phi(\vec{x}),\phi(\vec{y})]$

$=\Big[ \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2\omega_{\vec{p}}}}(a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}+a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}}), \int \frac{d^{3}q}{(2\pi)^{3}} \frac{1}{\sqrt{2\omega_{\vec{q}}}}(a_{\vec{q}}e^{i\vec{q}\cdot{\vec{y}}}+a_{\vec{q}}^{\dagger}e^{-i\vec{q}\cdot{\vec{y}}}) \Big]$

$=\int \frac{d^{3}p\ d^{3}q}{(2\pi)^{6}}\frac{1}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big[ (a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}+a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}}), (a_{\vec{q}}e^{i\vec{q}\cdot{\vec{y}}}+a_{\vec{q}}^{\dagger}e^{-i\vec{q}\cdot{\vec{y}}}) \Big]$

$=\int \frac{d^{3}p\ d^{3}q}{(2\pi)^{6}}\frac{1}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big( [a_{\vec{p}},a_{\vec{q}}]e^{i(\vec{p}\cdot{\vec{x}}+\vec{q}\cdot{\vec{y}})}+[a_{\vec{p}}^{\dagger},a_{\vec{q}}]e^{i(-\vec{p}\cdot{\vec{x}}+\vec{q}\cdot{\vec{y}})}+[a_{\vec{p}},a_{\vec{q}}^{\dagger}]e^{i(\vec{p}\cdot{\vec{x}}-\vec{q}\cdot{\vec{y}})}+[a_{\vec{p}}^{\dagger},a_{\vec{q}}^{\dagger}]e^{i(-\vec{p}\cdot{\vec{x}}-\vec{q}\cdot{\vec{y}})} \Big)$

$=\int \frac{d^{3}p\ d^{3}q}{(2\pi)^{6}}\frac{1}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big( -[a_{\vec{q}},a_{\vec{p}}^{\dagger}]e^{i(-\vec{p}\cdot{\vec{x}}+\vec{q}\cdot{\vec{y}})}+[a_{\vec{p}},a_{\vec{q}}^{\dagger}]e^{i(\vec{p}\cdot{\vec{x}}-\vec{q}\cdot{\vec{y}})} \Big)$, where we used the relation $[a_{\vec{p}},a_{\vec{q}}]=[a_{\vec{p}}^{\dagger},a_{\vec{q}}^{\dagger}]=0$

$=\int \frac{d^{3}p\ d^{3}q}{(2\pi)^{6}}\frac{1}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big( -(2\pi)^{3}\delta^{(3)}(\vec{q}-\vec{p})e^{i(-\vec{p}\cdot{\vec{x}}+\vec{q}\cdot{\vec{y}})}+(2\pi)^{3}\delta^{(3)}(\vec{p}-\vec{q})e^{i(\vec{p}\cdot{\vec{x}}-\vec{q}\cdot{\vec{y}})} \Big)$

$=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2\omega_{\vec{p}}}\Big( -e^{i\vec{p}\cdot{(\vec{y}-\vec{x})}}+e^{i\vec{p}\cdot{(\vec{x}-\vec{y})}}\Big)$

$=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2\omega_{\vec{p}}}\Big( -e^{i\vec{p}\cdot{(\vec{x}-\vec{y})}}+e^{i\vec{p}\cdot{(\vec{x}-\vec{y})}}\Big)$, where I've flipped the limits of integration on the first term and used the relation $\omega_{\vec{p}}=\omega_{-\vec{p}}$

$=0$.

Is my working correct?

2. May 1, 2016