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A Commutation relations - field operators to ladder operators

  1. Apr 26, 2016 #1
    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. jcsd
  3. May 1, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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