# Time Evolution of the Complex Scalar Field

1. Sep 26, 2014

### Xenosum

1. The problem statement, all variables and given/known data

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!

2. Relevant equations

3. The attempt at a solution

2. Sep 27, 2014

### Orodruin

Staff Emeritus
The Hamiltonian really should be the Hamiltonian at time $x^0 = y^0$ (note that you are integrating over the spatial components of $y$). As a result, $x$ and $y$ have space-like separation and the commutator $[\phi(x),\phi(y)]$ vanishes (they are even equal-time).

3. Sep 27, 2014

### Xenosum

Cool, thanks!