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In Steinmann's book, Perturbative Quantum Electrodynamics and Axiomatic Field Theory is stated that the commutator of two fields $\varphi$, which satisfy the Klein-Gordon equation $$ (\square-m^2)\varphi(x)=0, $$ is given by $$ [\varphi (x),\varphi (y)]=-i\hbar\Delta(x-y) $$ at arbitrary points. $\Delta (x)$ is the Pauli-Jordan function $$ \Delta(x)=\frac{1}{(2\pi)^3}\int d^4p\ \varepsilon(p_0)\delta(p^2-m^2)e^{-ipx} $$ with $\varepsilon (p_0)=\theta(p_0)-\theta(-p_0)$. From this is obtained the equal time commutators $$ [\varphi (t,\vec{x}),\varphi (t,\vec{y})]=0 \qquad [\dot{\varphi} (t,\vec{x}),\dot{\varphi} (t,\vec{y})]=0\\ [\dot{\varphi} (t,\vec{x}),\varphi (t,\vec{y})]=-i\hbar\delta^3(\vec{x}-\vec{y}) $$ where the dot over the fields denotes time derivative. How can I get this relations? I think I have to use the facts that $$ (\square-m^2)\Delta(x)=0 \qquad \Delta(0,\vec{x})=0 \qquad \frac{\partial}{\partial t}\Delta(t,\vec{x})|_{t=0}=\delta^3(\vec{x}) $$ but I do not know how.

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Let $x_0 = y_0$. Then you have

$\Delta(x-y) = \frac{1}{(2 \pi)^3} \int dp_0 \epsilon(p_0) \int d^3p \delta(p^2-m^2)e^{-i \vec{p} \vec{x}}$ (Vector arrows means: only 3 dimensions)

You can express the Delta Distribution as $\delta(p^2-m^2) = \delta(p_0-e_+)/|2e_+|+\delta(p_0-e_-)/|2e_-|$ where $e_{+,-} = \pm \sqrt{m^2+ \vec{p}^2}$. Evaluating Integration over $p_0$ and using $|e_+|=|e_-|$ you observe that the $\epsilon$ factor will cancel.

If you differentiate the $\Delta$ by $p_0$, you will also have an additional factor $-ip_0$ in your integral. But here will the Evaluation of the Delta function not cancel. You will have e.g. a term proportional to $\epsilon(p_0)p_0/|e_+|$ that is the same for positive and negative $p_0$.

All other Relations you obtain in a similar way.

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  • $\begingroup$ If i differentiate the commutator $[\varphi(x),\varphi(y)]$ by $x_0$, I have $$ \partial _0(\varphi(x),\varphi(y)-\varphi(y)\varphi(x))=-i\hbar\partial_0\Delta(x-y)$$ $$ (\partial _0\varphi(x)\varphi(y)+\varphi(x)\partial _0\varphi(y)-\partial _0\varphi(y)\varphi(x)-\varphi(y)\partial _0\varphi(x))=-i\hbar\partial_0\Delta(x-y)$$ $$ [\partial _0\varphi(x),\varphi(y)]+[\varphi(x),\partial _0\varphi(y)]=-i\hbar\partial_0\Delta(x-y)$$ letting $x_0=y_0$ $$ [\partial _0\varphi(ct,\vec{x}),\varphi(ct,\vec{y})]+[\varphi(ct,\vec{x}),\partial _0\varphi(ct,\vec{y})]=-i\hbar\delta(\vec{x}-\vec{y})$$ $\endgroup$
    – Daemonium
    Oct 19, 2018 at 12:53
  • $\begingroup$ Is that ok? I do not get the commutator required $$[\partial_0\varphi(ct,\vec{x}),\varphi(ct,\vec{y})]=-i\hbar\delta(\vec{x}-\vec{y})$$ or am I missing something? $\endgroup$
    – Daemonium
    Oct 19, 2018 at 12:55
  • $\begingroup$ As you had written (long long ago...), you are differentiating w.r.t. $x_0$ which is independent of $y_0$ so that before setting $x_0\overset{!}{=} y_0$, you have that the extra $\left[ \varphi(x), \partial_0 \varphi(y)\right]=0$ $\endgroup$
    – Noix07
    Jan 9, 2020 at 15:15

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