# Heisenberg evolution equation for $\hat{\phi}$

Consider quantum Hamiltonian of free massive scalar particle: $$\hat{H} = \int d^3x \left[\frac{1}{2} \hat{\pi}^2 (t, \vec{x}) + \frac{1}{2} \partial_i \hat{\phi}(t, \vec{x}) \partial_i \hat{\phi}(t, \vec{x}) + \frac{m_0^2}{2} \hat{\phi}^2(t, \vec{x}) \right].$$ I'd like to show that Heisenberg evolution equation for $\hat{\phi}$ holds, i.e.: $$\partial_t \hat{\phi} (t, \vec{x}) = i [\hat{H}, \hat{\phi}(t, \vec{x})].$$ My problem is to compute the following term: $$\int d^3 y \left[\partial_i \hat{\phi}(t, \vec{y}) \partial_i \hat{\phi}(t, \vec{y}), \hat{\phi}(t, \vec{x})\right].$$ I would appreciate some help on how should I rearrange this commutator to see it vanishes.

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The derivative on the $\hat \phi$'s are taken with respect to $y$ in your formula, so using $$[A,B^2]=[A,B]B+B[A,B]$$ and $$\bigl[ \frac{\partial \hat \phi(t,\vec y)}{\partial y^i},\hat \phi(t,\vec x)\bigr] = \frac{\partial}{\partial y^i}\bigl[\hat \phi(t,\vec y),\hat \phi(t,\vec x)\bigr]=0$$ you get your result.
Edit: let me comment on why does the above identity holds (the last equality). We know that for spacelike separated points $x$ and $y$, the commutator $\bigl[\hat \phi(y),\hat \phi(x)\bigr]$ vanishes. This happens in particular if $x^0=y^0=t$ and $\vec x \not = \vec y$, showing the result in this case. Moreover, when $\vec x = \vec y$, then the result holds as well since the commutator of an operator with itself is always zero, completing the proof.
Thanks, it's clear now. I missed the fact that differentiation is with respect to $y$. –  qoqosz Mar 31 '13 at 9:23
Use the rule for commutators, $$[fg,h]=f[g,h]+[f,h]g$$ which reduces the problem to showing that $[\partial_{i}\phi(t,y),\phi(t,x)]=0$. Now use the analogy with point mechanics, $$q^{i}(t)=q(t,i)\rightarrow q(t,x) \rightarrow \phi(t,x)$$ so that the commutator is $[q^{i}-q^{j},q^{k}]$ when the labels $i,j,k$ are analogous to the three points $y+\delta y, y,x$ in the field theory and the bracket is zero because $[q^{i},q^{j}]=0$.