2
$\begingroup$

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?

$\endgroup$
1
  • 3
    $\begingroup$ Hint: Think about what are the spacetime positions of the various operators. $\endgroup$
    – Qmechanic
    Jan 15 at 16:29

2 Answers 2

1
$\begingroup$

I think one should be careful about the dependence on spacetime coordinates. I think what you mean by the first equation is actually: \begin{equation} i\frac{\partial \pi^{\star}\left(y\right)}{\partial t} = \left[\pi^\star\left(y\right),H\right] \end{equation} 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 $y$.

$\endgroup$
1
  • $\begingroup$ That makes a lot of sense, thank you! $\endgroup$ Jan 15 at 16:56
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.