# Commutator of conjugate momentum and field for complex field QFT

In Peskin & Schroeder's Introduction to QFT problem 2.2a), we are asked to find the equations of motion of the complex scalar field starting from the Lagrangian density. I want to show that: $$i\frac{\partial \pi^*}{\partial t}=\int d^3 x[\pi^*,H].$$ So plugging the Hamiltonian in, $$\int d^3 x[\pi^*,H]=\int d^3x [\pi^*, \pi^*\pi+\nabla \phi^*\nabla\phi+m^2\phi^*\phi].$$ The first term here is zero, and the second one can be manipulated by integration by parts and using the fact that the field vanishes at infinity: $$=\int d^3x [\pi^*, -\phi^*\nabla^2\phi+m^2\phi^*\phi]=\int d^3x [\pi^*, -\phi^*\nabla^2\phi]+\int d^3x [\pi^*,m^2\phi^*\phi]$$ Both commutators can be expanded using the fact that $$[A,BC]=[A,B]C+B[A,C]$$ $$\int d^3x (-[\pi^*, \phi^*]\nabla^2\phi)-\phi^*[\pi^*,\nabla^2\phi])+\int d^3x ([\pi^*,\phi^*]\phi+\phi^*[\pi^*,\phi])$$ Here, 3 out of the 4 commutators can be found using cannonical relations, i.e. $$[\phi^*(x),\pi^*(y) ]=i\delta^3(x-y)$$ I am unsure how to go about this one, however: $$[\pi^*,\nabla^2\phi]$$ I know that it must be zero, and that would be the case if one could simply take out the $$\nabla^2$$ operator from the commutator: $$=\nabla^2[\pi^*,\phi]=0$$ But I don't see why that would be allowed, since it would require $$\nabla^2\pi^*=0$$, which I don't see as obvious. What is the justification here?

• Hint: Think about what are the spacetime positions of the various operators. Commented Jan 15 at 16:29

I think one should be careful about the dependence on spacetime coordinates. I think what you mean by the first equation is actually: $$$$i\frac{\partial \pi^{\star}\left(y\right)}{\partial t} = \left[\pi^\star\left(y\right),H\right]$$$$ where, $$H = \int\mathrm{d}^3x\;\mathcal{H} = \int\mathrm{d}^3x \left[\pi^\star\pi+\nabla\phi^\star\nabla\phi + m^2\phi^\star\phi\right]$$. Now $$\pi$$ in the left part of the commutator should be regarded as a function of, say $$y$$ and the right part of the commutator is a function of $$x$$. Hence one can always move the Laplace operator out of the commutator, because it operates on $$x$$, not on $$y$$.
Remember that the Lagrangian density of a complex scalar field is given by $$\mathcal{L}= \dot{\phi^\ast} \dot{\phi}- \vec{\nabla}{\phi}^\ast \cdot \vec{\nabla}{\phi}-m^2 \phi^\ast \phi. \tag{1} \label{eq1}$$ As a consequence, we have $$\pi= \frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi^\ast}, \qquad \pi^\ast = \frac{\partial \mathcal{L}}{\partial \dot{\phi^\ast}}=\dot{\phi}. \tag{2} \label{eq2}$$ and the Hamilton density is given by $$\mathcal{H}(\phi, \phi^\ast \!, \pi, \pi^\ast)= \dot{\phi} \frac{\partial \mathcal{L}}{\partial \dot{\phi}} + \frac{\partial \mathcal{L}}{\partial \dot{\phi^\ast}} \dot{\phi^\ast}- \mathcal{L} = \pi^\ast \pi +\vec{\nabla}\phi^\ast \cdot \vec{\nabla} \phi +m^2 \phi^\ast \phi \tag{3} \label{eq3}$$
In QFT, the fields are promoted to field operators, satisfying the canonical equal-time commutation relations. For the conjugate pairs $$(\phi, \pi)$$ and $$(\phi^\dagger, \pi^\dagger)$$, respectively, we have $$[\phi(t,\vec{x}), \pi(t, \vec{y}) ] = i \delta^{(3)}(\vec{x}-\vec{y}), \qquad [\phi^\dagger (t, \vec{x}), \pi^\dagger(t, \vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y}) \tag{4} \label{eq4},$$ whereas the (equal-time) commutators of all other field combinations vanish, $$[\phi(t, \vec{x}), \phi(t, \vec{y}) ]= [\phi^\dagger(t, \vec{x}),\phi^\dagger(t, \vec{y})]= [\pi(t, \vec{x}), \pi(t, \vec{y})]= [\pi^\dagger(t, \vec{x}), \pi^\dagger(t,\vec{y}) ]=0, \tag{5} \label{eq5}$$ and also $$[\phi(t, \vec{x}),\phi^\dagger(t, \vec{y})]=[\pi(t, \vec{x}),\pi^\dagger(t, \vec{y})]=[\phi(t, \vec{x}),\pi^\dagger(t, \vec{y})] =[\phi^\dagger(t, \vec{x}), \pi(t, \vec{y})]=0. \tag{6} \label{eq6}$$ Note that eq. \eqref{eq6} solves already one of your problems. As the commutator $$[\phi(t, \vec{x}), \pi^\dagger(t, \vec{y})]$$ vanishes identically, we conclude that also $$\Delta_x [\phi(t,\vec{x}), \pi^\dagger(t, \vec{y}) ]=[\Delta_x \phi(t,\vec{x}), \pi^\dagger(t, \vec{y})]$$ has to vanish.
According to the Heisenberg equation of motion, the time evolution of $$\pi(t, \vec{x})$$ is determined by $$i \frac{\partial \pi(t, \vec{x})}{\partial t}= [ \pi(t,\vec{x}), H]. \tag{7} \label{eq7}$$ As the Hamilton operator $$H=\int d^3y \left(\pi^\dagger(t, \vec{y}) \pi(t, \vec{y}) + \vec{\nabla} \phi^\dagger(t, \vec{y})\cdot \vec{\nabla} \phi(t, \vec{y}) +m^2 \phi^\dagger(t, \vec{y}) \phi(t, \vec{y}) \right) \tag{8} \label{eq8}$$ is time independent, we are free to choose the same time $$t$$ in the Hamilton density of \eqref{eq8} as in the field $$\pi(t,\vec{x})$$, which allows us to apply the equal time commutation relations in the expression for the commutator $$[\pi(t, \vec{x}),H]$$. As already noted in a comment and in another answer, it is essential to distinguish the "free" vector $$\vec{x}$$ and the dummy (integration) variables $$\vec{y}$$ in the following calculation. It is now an easy task to compute the relevant field commutators using the formula $$[A,BC]=[A,B]C+B[A,C]$$ mentioned in your question. We find $$\left[\pi(t, \vec{x}), \pi^\dagger(t, \vec{y}) \pi(t,\vec{y})\right]=0 \tag{9} \label{eq9}$$ by using \eqref{eq5} and \eqref{eq6}, \begin{align}\left[\pi(t, \vec{x}), \vec{\nabla} \phi^\dagger(t, \vec{y}) \cdot \vec{\nabla}\phi(t,\vec{y})\right] &= \left(\frac{\partial}{\partial y_k} \phi^\dagger(t,\vec{y})\right) \left[\pi(t,\vec{x}), \frac{\partial}{\partial y_k}\phi(t,\vec{y})\right] \\ &= \left( \frac{\partial}{\partial y_k} \phi^\dagger(t, \vec{y}) \right) \frac{\partial}{\partial y_k}\left[\pi(t, \vec{x}), \phi(t,\vec{y}) \right] \\ &= -i \left(\frac{\partial}{\partial y_k} \phi^\dagger(t, \vec{y})\right) \frac{\partial}{\partial y_k}\delta^{(3)}(\vec{x}-\vec{y}) \end{align} \tag{10} \label{eq10} by using \eqref{eq4} and \eqref{eq6} and finally \begin{align} \left[ \pi(t, \vec{x}), \phi^\dagger(t, \vec{y}) \phi(t, \vec{y}) \right] &= \phi^\dagger(t, \vec{y}) \left[\pi(t, \vec{x}), \phi(t, \vec{y}) \right] \\ &=-i \phi^\dagger(t, \vec{y}) \delta^{(3)}(\vec{x}- \vec{y}),\end{align} \tag{11} \label{eq11} where we have again used \eqref{eq4} and \eqref{eq6}. Assembling all terms, we obtain the result \begin{align} [\pi(t, \vec{x}),H] &=-i\int d^3y \left[ \left(\frac{\partial}{\partial y_k} \phi^\dagger(t, \vec{y}) \right) \frac{\partial}{\partial y_k} \delta^{(3)}(\vec{x}-\vec{y})+ m^2 \phi^\dagger(t, \vec{y}) \delta^{(3)}(\vec{x}-\vec{y})\right] \\ &= i \left( \Delta -m^2 \right) \phi^\dagger(t, \vec{x}) \end{align} \tag{12} \label{eq12} and the Heisenberg equation $$\frac{\partial \pi(t, \vec{x})}{\partial t} = \left( \Delta -m^2 \right) \phi^\dagger(t, \vec{x}). \tag{13} \label{eq13}$$ As $$\pi(t,\vec{x})= \dot{\phi^\dagger}(t, \vec{x})$$ (c.f. eq. \eqref{eq2}), we recover indeed the Klein-Gordon equation $$\left(\partial^2/\partial t^2-\Delta+m^2 \right) \phi^\dagger(t,\vec{x})=0 \tag{14} \label{eq14}$$ for the field operator.