Xenosum
- 18
- 2
Homework Statement
Consider the Lagrangian, L, given by
L = \partial_{\mu}\phi^{*}(x)\partial^{\mu}\phi(x) - m^2\phi^{*}(x)\phi(x) .
The conjugate momenta to \phi(x) and \phi^{*}(x) are denoted, respectively, by \pi(x) and \pi^{*}(x). Thus,
\pi(x) = \frac{\partial L}{\partial(\partial_{0}\phi(x))} = \partial_0\phi^{*}(x)
\pi^{*}(x) = \frac{\partial L}{\partial(\partial_{0}\phi^{*}(x))} = \partial_0\phi(x) .
Upon quantizing the system, \phi(x) and \phi^{*}(x) are promoted to operators which satisfy the equal-time commutation relations:
[ \phi(x) , \pi(y) ] = i\delta^{(3)}(\vec{x} - \vec{y})
[ \phi^{*}(x) , \pi^{*}(y) ] = i\delta^{(3)}(\vec{x} - \vec{y})
(all others zero). In the Heisenberg regime, the time evolution of the operator \phi(x), i \partial_0 \phi(x), is given by
i \partial_0 \phi(x) = \left[ \phi(x) , H(y) \right].
The Hamiltonian may be derived from the Lagrangian, and we find that
i\frac{\partial \phi(x)}{\partial t} = \int d^{3}y \left( \left[ \phi(x) , \pi(y)\pi^{*}(y) \right] + \left[ \phi(x) , \nabla\phi^{*}(y) \cdot \nabla\phi(y) \right] + m^2 \left[ \phi(x) , \phi^{*}(y)\phi(y) \right] \right).
Now here's my question. When we evaluate the commutators both my professor and a solution manual to Peskin and Schroeder claim that only the first commutator survives, because \phi(x) commutes with everything except for the its conjugate momentum (by the canonical commutation relations). I don't see why. The canonical commutation relations only give us a relationship between \phi(x) and \pi(y), not e.g. \phi(x) and \phi(y). The point is pressed by the fact that one can only show that the commutator \left[ \phi(x) , \phi(y) \right] vanishes for space-like separation between the points x and y (this is the condition which preserves causality).
I guess it would be resolved if the commutator were instead \left[ \phi(x) , H(x) \right], but this doesn't seem to be how it's done.
Thanks for any help!