$#*! I unlearned #2: causality and antiparticle

In the $#*! I unlearned #1, we saw that the propagator for a real Klein-Gordon scalar field is finite for two space-like separated points, which seems to violate causality. But not really. In the question of causality, we should not ask whether particles can propagate over space like intervals, but whether a measurement performed in one point can affect a measurement at another space-like separated point.

\[ \begin{aligned} [\phi(x), \phi(y)] &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 p^0}} \int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2 q^0}} \left[a_{\mathbf{p}} e^{-i p \cdot x} + a_{\mathbf{p}}^{\dagger} e^{i p \cdot x} ,\; a_{\mathbf{q}} e^{-i q \cdot y} + a_{\mathbf{q}}^{\dagger} e^{i q \cdot y} \right] \\ &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{2 p^0} \left( e^{-ip(x-y)} - e^{ip(x-y)} \right) = D(x-y) - D(y-x) \end{aligned} \]

When \( (x - y)^2 < 0 \), we can perform a Lorentz transformation from \( (x-y) \) to \( - (x-y) \)--going around the light cone--so that the two propagators are actually equal and therefore exactly cancel; causality if preserved. When \( (x - y)^2 > 0 \), there is no such Lorentz transformation. In this case, the amplitude is nonzero, and roughly \( ( e^{-imt} - c.c ) \) for the special case \( \mathbf{x} - \mathbf{y} = 0 \), which we also calculated. Therefore: no measurement in the Klein-Gordon theory can affect another measurement outside the light-cone.

But why do we have this magical cancellation? To answer we need to make the field complex with the action:

\[ \mathcal{L}[\phi, \phi^*; \partial_\mu \phi, \partial_\mu \phi^*] =\partial_{\mu} \phi^* \partial^{\mu} \phi - m^2 \phi^* \phi \,. \]

By definition, the conjugate momenta densities are:

\[ \pi = \frac{\partial\mathcal{L}}{\partial_0 \phi} = \partial^{0} \phi^{*} \,,\quad \pi^{*} = \frac{\partial\mathcal{L}}{\partial_0 \phi^{*}} = \partial^{0} \phi \,.\]

And the Hamiltonian operator in second quantization is:

\[ H = \int \left( \pi \partial_{0}\phi + \pi^{\dagger} \partial_{0}\phi^{\dagger} - \mathcal{L} \right) d^3x = \int \left(\pi^{\dagger} \pi + \nabla \phi^{\dagger} \cdot \nabla \phi + m^2 \phi^{\dagger} \phi \right) d^3x \,.\]

\( H \) is diagonalized in momentum space, so:

\[ H = \int \tilde{\pi}^{\dagger} (\mathbf{p}, t) \tilde{\pi} (\mathbf{p}, t) + E_{\mathbf{p}}^2 \tilde{\phi}^{\dagger} (\mathbf{p}, t) \tilde{\phi} (\mathbf{p}, t) \frac{d^3p}{(2\pi)^3} \,, \]

where, \( E_{\mathbf{p}}^2 = \mathbf{p}^2 + m^2 \). As usual, we should work out the commutation relation first:

\[ [ \tilde{\phi}(\mathbf{p},t), \tilde{\pi}(\mathbf{q},t) ] = \int [ \phi(\mathbf{x}, t), \pi(\mathbf{y}, t) ] e^{-i \mathbf{p}\cdot\mathbf{x}} e^{-i \mathbf{q} \cdot \mathbf{y}} \; d^3x\,d^3y = i (2\pi)^3 \delta^{(3)}(\mathbf{p} + \mathbf{q}) \,.\]

\( \phi \) is complex, so in general, its Fourier decomposition should be in term of two operators \( a_{\mathbf{p}} \) and \( b_{\mathbf{p}} \) like this:

\[ \begin{align} \phi(\mathbf{x},t) &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 E_{\mathbf{p}}}} \left[ a_{\mathbf{p}} e^{-i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + b_{\mathbf{p}}^{\dagger} e^{i E_{\mathbf{p}}t - i \mathbf{p} \cdot \mathbf{x}} \right] \\ \therefore \quad \tilde{\phi}(\mathbf{p}, t) &= \frac{1}{\sqrt{2E_{\mathbf{p}}}} \left( a_{\mathbf{p}} e^{-i E_{\mathbf{p}} t} + b_{-\mathbf{p}}^{\dagger} e^{i E_{\mathbf{p}} t}\right) \end{align} \]

The proper interpretation of this is that \( \phi \) annihilates a particle with positive energy and creates an antiparticle with negative energy. \( \phi^{\dagger} \) does the opposite. Actually, we haven't shown that \( a_{\mathbf{p}} \) and \( b_{\mathbf{p}} \) are annihilation operators of particles and antiparticles, but we'll denote them as such. Now the conjugate momenta, by definition:

\[ \begin{align} \pi(\mathbf{x}, t) = \frac{\partial \phi^{\dagger}(\mathbf{x},t)}{\partial t} &= \int \frac{d^3p}{(2\pi)^3} i \sqrt{\frac{E_{\mathbf{p}}}{2}} \left[ a_{\mathbf{p}}^{\dagger} e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}} - b_{\mathbf{p}} e^{-i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x})} \right] \\ \therefore \quad \tilde{\pi}(-\mathbf{p}, t) &= i \sqrt{\frac{E_{\mathbf{p}}}{2}} \left( a_{\mathbf{p}}^{\dagger} e^{i E_{\mathbf{p}} t} - b_{-\mathbf{p}} e^{-i E_{\mathbf{p}} t} \right) \,. \end{align} \]

We can invert this to get:

\[ \begin{align} a_{\mathbf{p}} e^{-i E_{\mathbf{p}} t} = \sqrt{\frac{E_{\mathbf{p}}}{2}} \tilde{\phi}(\mathbf{p},t) + i \sqrt{\frac{1}{2E_{\mathbf{p}}}} \tilde{\pi}^{\dagger}(-\mathbf{p},t)\,, \quad b_{-\mathbf{p}} e^{-i E_{\mathbf{p}} t} = \sqrt{\frac{E_{\mathbf{p}}}{2}} \tilde{\phi}^{\dagger}(\mathbf{p},t) + i \sqrt{\frac{1}{2E_{\mathbf{p}}}} \tilde{\pi}(-\mathbf{p},t) \end{align} \]

This gives the Hamiltonian:

\[ H = \int E_{\mathbf{p}} \left( a_{\mathbf{p}}^{\dagger} a_{\mathbf{p}} + b_{\mathbf{p}}^{\dagger} b_{\mathbf{p}} \right) \frac{d^3p}{(2\pi)^3} + {\rm const} \,, \]

since,

\[ [a_{\mathbf{p}}, a_{\mathbf{q}}^{\dagger} ] = [b_{\mathbf{p}}, b_{\mathbf{q}}^{\dagger} ] = (2\pi)^3 \delta^{(3)} (\mathbf{p} - \mathbf{q} ) \,. \]

So the complex scalar field theory describes two excitations of the same mass.

Now let us reconsider the commutator of field. We want to create a particle at space-time point \( x \) then destroy it at \( y \):

\[ \begin{aligned} [ \phi(x), \phi^{\dagger}(y) ] &= \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{q}}} \left[ a_{\mathbf{p}} e^{-i p \cdot x} + b_{\mathbf{p}}^{\dagger} e^{i p \cdot x}, a_{\mathbf{q}}^{\dagger} e^{i q \cdot y} + b_{\mathbf{q}} e^{-i q \cdot y} \right] \\ &= D(x-y) - D(y-x) \,. \end{aligned} \]

So we get the same two propagators as before, but now the interpretation is a lot more clear: one comes from the particle operator \( a_{\mathbf{p}} \), the other comes from the antiparticle operator \( b_{\mathbf{p}} \). Therefore causality is preserved by the exact cancellation of these two fields outside the lightcone. Also, inside the lightcone, the two propagators are equal up to a phase. So it is equivalent to propagate the field \( \phi \) from \( x \) to \( y \) as to propagate the field \( \phi^* \) backward from \( y \) to \( x \). \( \phi \) and \( \phi^* \) have the same mass but opposite quantum numbers. So they are in fact particle and antiparticle pair. And a real scalar field is its own antiparticle.