# Proving that the 3-current density corresponding to the global phase invariance vanishes at infinity

The components $j^i$ of the 3-current density $\textbf{j}$ corresponding to the global phase invariance of the action of a complex scalar field $\phi$ i.e., $\phi\to e^{-iq\theta}\phi$ is given by $$j^i=iq[(\partial^i\phi)\phi^*-(\partial^i\phi^*)\phi].\tag{1}$$

Each component of the 3-current density must go to zero at spatial infinity for the corresponding Noether's charge $$Q=\int j^0 d^3\textbf{x}$$ to be conserved.

So my questions are:

1. Is there a way to prove that $j^i$ as given in equation (1) vanishes at spatial infinity?

2. Even if we assume $j^i$ to be zero at spatial infinity what would it mean physically?

• From what givens do you want to "prove" this? Also, is it not obvious that the current vanishing at infinity means no charge is leaking "out of the universe"? Is there something more you want to hear when you ask what it would "mean physically"? – ACuriousMind Dec 25 '17 at 16:25
• @ACuriousMind It can be shown that $Q=\int d^3x j^0=q\int d^3p[a^*_pa_p-b^*_pb_p]$ . Taking $d/dt$ of both sides immediately implies $Q$ is conserved, classically. So it must be true that $j^i$ vanishes at the surface. – SRS Dec 25 '17 at 16:29

You may have already seen this, but Schwartz says

$$\partial_t Q = \int d^3 x \partial_t J_0 = \int d^3 x \vec{\nabla}\cdot \vec J = 0.\tag{3.28}$$

"In the last step we have assumed $\vec J$ vanishes at the spatial boundary, since, by assumption, nothing is leaving our experiment."

Without some assumption about $\vec J$ at the spatial boundary, it is not true in general.

Edit: To clarify this assumption, Schwartz states in section 3.2 "We will always make the physical assumption that our fields vanish on these asymptotic boundaries" (referring to spatial and temporal infinity).

• But somehow it works. Right? Therefore, one must be able to show this or at least justify it. – SRS Dec 25 '17 at 16:01
• Could you explain what you want to justify or show? Our assumption about $\vec J$ together with Stokes' Theorem gives us (3.28). – Dwagg Dec 25 '17 at 16:03
• In this case, you take the Noether charge to be conserved and therefore assume the 3-current vanishes at infinity. How do you know your assumption is valid? In this case, you have the explicit Lagrangian, explicit symmetry transformation to check whether your assumption is valid. @Dwagg – SRS Dec 25 '17 at 16:08
• In field theory, we always make the physical assumption that our fields vanish at spatial and temporal infinity. You immediately see that your current from Eq. (1) vanishes at infinity. Note I never assumed the Noether charge is conserved; it is derived from the above physical assumption. – Dwagg Dec 25 '17 at 16:23
• It is a standard assumption that fields vanish at spatial infinity. They do not in general vanish at timelike infinity. – Prahar Dec 25 '17 at 17:03

Existence and uniqueness theorem for KG equation works assuming the field and its temporal derivative are smooth and have compact support on a 3-surface at constant time. The unique corresponding solution has support included in the causal future and caysal past of the support of the initial conditions. This implies that, on every constant time surface (also referred to a different Minkowskian coordinate system), the solution has compact support. In turn, this fact implies that every boundary term as the one you consider can be neglected trivially and the charge is well-defined and independent from the considered constant time 3-surface. This result is also valid in generic globally hyperbolic spacetimes using spacelike smooth Cauchy surfaces. In Minkowski spacetime it is also possible to deal with weaker hypotheses assuming that initial conditions are rapidly vanishing at infinity on the initial 3-space (they belong to Schwartz' space). This feature propagates into the associated solutions and the same result as above arises.

The requirement that the field goes to zero at infinity is a more general requirement, not related to the conservation of the Noether charge. It follows from the requirement that the field must have a finite energy: $$\int |\phi|^2 d^3x < \infty .$$ The finite energy requirement, can only be satisfied if the field goes to zero at infinity.

• Why did you get $|\phi^2|$ from? @flippiefanus – SRS Dec 27 '17 at 19:03
• Not sure I understand what you are asking. The modulus square of the scalar field is obtained by multiplying the field with its complex conjugate: $|\phi|^2=\phi \phi^*$. – flippiefanus Dec 29 '17 at 5:16
• But that is not the Hamiltonian density – SRS Dec 29 '17 at 5:22
• Unless there is a mass term $m^2 |\phi|^2$. But that is not the issue here. The field can be interpreted as a wavefunction and its modulus square must integrate to one to conserve probability. It is by implication a finite energy function. – flippiefanus Dec 29 '17 at 5:26