Commutation relations - field operators to ladder operators

[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?

 

[eljose79]

Yes, your working is correct. You have correctly used the commutation relations for the creation and annihilation operators to show that the commutation relations for the field operator and its conjugate momentum are satisfied. Well done!