# Question about source terms in scalar quantum field theory

I'm having a bit of a mental block when trying to interpret the inhomogeneous Klein-Gordon equation $$(\Box +m^{2})\phi(x,t)=j(x,t)$$ In particular, how does one interpret the term on the right-hand side of the equation, $j(x,t)$ as a source term for the scalar field $\phi(x,t)$?

Is it analogous to the situation in classical electromagnetism? For example, the flux of an electric field through an closed surface, $S$ $\bigl(\oint_{S}(\mathbf{E}\cdot\mathbf{n})\,dS$, where $\mathbf{n}$ is the unit normal vector to the surface $S\bigr)$ is proportional to the total charge density $\rho$ in the enclosed volume $V$. This can be written in differential form as $$\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_{0}}$$ and so the charge density $\rho$ is interpreted as a source for a spatially varying electric field.

• That's pretty much it, at least in classical field theory. You can think of $j$ as a source in the sense that, for example, its presence means that $\phi = 0$ is not a solution. – Javier Feb 22 '16 at 13:33
• Ah ok, is it simply that the presence of $j$ means that there must necessarily be a non-trivial solution for $\phi$, i.e. a scalar field exists if $j$ is present? Also, is my understanding of the classical field theory case correct (referring to what I wrote in the original post)? In the classical case, the absence of a charge density in the volume $V$ enclosed by a surface $S$ would imply that the flux through $S$ is 0, and hence there is no electric field generated within the enclosed volume?! – Will Feb 22 '16 at 14:00
• Your first question is correct. As to the second, it's not so simple, since $\phi$ is not a vector field and its differential equation is more complicated. The KG equation is a kind of wave equation; if $j$ is concentrated at the origin and oscillates, you will get outgoing waves. – Javier Feb 22 '16 at 14:10
• @Javier you mean the trivial mode expansion of $\phi$ is not the solution ? (and not $\phi = 0$) – Bruce Lee Feb 22 '16 at 14:14
• @BruceLee: $\phi=0$ is a solution to the sourceless equation, but it is not a solution to the equation with sources. And I was talking about classical field theory. – Javier Feb 22 '16 at 14:31

The source terms have different interpretations depending on context.

Within classical field theory, they are called sources because they affect the solutions of the equation; in particular, they force the field to be nonzero. Take the equations of electrostatics, for example:

$$\nabla \cdot \mathbf{E} = \rho \qquad \nabla \times \mathbf{E} = 0$$

If $\rho = 0$, then $\mathbf{E} = 0$ is a possible solution. Of course there are others, but if, for example, we fix as boundary conditions that the field go to zero at infinity, then that's the only one. On the other hand, if $\rho$ is nonzero, $\mathbf{E} = 0$ is not a solution. In this case the interpretation of the "sources" is pretty easy, because we know what charges and currents are, and it is intuitive to say that they are the source of electromagnetic fields.

The Klein-Gordon equation doesn't correspond to a familiar classical situation, but we deal with sources in the same way. The free equation $(\partial^2 + m^2)\phi = 0$ is a kind of wave equation; if we solve by Fourier transform, the equation says that the wave vectors satisfy the dispersion relation $\omega^2 = \mathbf{k}^2 + m^2$. But again, if $j$ is nonzero the $\phi = 0$ solution disappears. The physical intuition is trickier because I can't give one unless I know what $\phi$ represents, but we can take a special case: if $m=0$ we recover the usual wave equation, and we know that the electromagnetic potentials satisfy that one. So we could interpret $j$ as being the "charge" for a potential $\phi$ that satisfies a slightly weird dispersion relation. You can try plugging in simple functions in place of $j$ (for example, one that varies sinusoidally in time) and see what you get.

In quantum field theory, there are two ways of thinking about the sources that I know of. One is to interpret them as background fields. This is an approach taken, for example, by Peskin and Schroeder when discussing electron scattering of a proton. If we make the approximation that the proton is stationary during scattering, we can treat its field as classical, and only quantize the electron field. This is done by adding to the free electron Lagrangian a term $\mathcal{L}_{\text{int}} = e \bar{\psi} \gamma_\mu \psi A^\mu_\text{cl}$, where $A^\mu_\text{cl}$ is a fixed classical field. This adds certain kinds of vertices to the Feynman diagrams, similar to the ones in fully quantum QED. One can derive Rutherford's formula with this formalism. This is covered, for example, in exercise 4.4 of Peskin and Schroeder, or section 6.4 of Weinberg (warning: if you read Weinberg you might end up more confused than when you started!).

The other use of sources is inside a path integral. In this formalism, the time-ordered correlation function of a number of fields is given by Feynman's path integral:

$$\langle 0 | T \phi(x_1) \cdots \phi(x_n) | 0\rangle = \int \mathcal{D}\phi\ e^{i\int d^4x \mathcal{L}} \phi(x_1) \cdots \phi(x_n)$$

The way to calculate this is to add to the Lagrangian a $J\phi$ term, with $J$ some fixed function (we don't actually care what it is) called a classical source or just a source. If we define the generating functional

$$Z[J] = \int \mathcal{D}\phi\ e^{i\int d^4x (\mathcal{L} + J\phi)}$$

we can insert factors of $\phi$ in the integral by taking functional derivatives with respect to $J$ and then setting $J=0$ (this is why we don't care what $J$ is). For example, take the 1-point function:

$$\langle 0 | \phi(x_1) | 0 \rangle = \int \mathcal{D}\phi\ e^{i\int d^4x \mathcal{L}} \phi(x_1) = \frac{1}{Z[0]} \frac{1}{i} \frac{\delta}{\delta J(x_1)} \int \mathcal{D}\phi\ e^{i\int d^4x (\mathcal{L} + J\phi)}$$

If you know $Z[J]$, you can calculate any correlation function. This approach is used extensively in any QFT book that uses path integrals. See for example Srednicki, in particular sections 6 to 9.

• Brilliant answer! If one considers $\phi$ to be a scalar quantum field can one interpret $j$ as "sourcing" a propagating wave through the quantum field? (This kind of makes sense to me as this is how the propagator is derived for a scalar field in QFT) – Will Feb 23 '16 at 18:21
• @Will: Yes, you can, though the background field interpretation is not used very often. There are some exercises in P&S where they ask you to calculate the expected number of particles after a source is turned on and off. – Javier Feb 23 '16 at 18:34
• Ah ok, thanks for your help. I think it's becoming a bit clearer now. – Will Feb 23 '16 at 18:41
• Also, just to clarify in the electrostatics example that you gave, is the reason why the only solution for the case in which the electric field goes to zero at infinity is $\mathbf{E}=0$ because the solution to $\nabla\cdot\mathbf{E}=0$ is a constant vector, and hence the only way this can vanish at infinity is if this constant vector is the zero vector?! – Will Feb 23 '16 at 20:14
• @Will: The solution to $\nabla \cdot \mathbf{E} = 0$ is not necessarily constant. If $\rho = 0$ then the electrostatic potential satisfies $\nabla^2 \phi = 0$. This has many solutions, but there is a theorem that states that given boundary conditions the solution is unique. In this case the boundary condition is that $\phi \to 0$ at infinity, and $\phi = 0$ is the unique solution that satisfies it. – Javier Feb 23 '16 at 20:41

Generally, source terms are what stands in the rhs of the differential equation $$\hat{\Theta}_x f(x) = j(x),$$ where $\hat{\Theta}_x$ is some linear differential operator (in your case $\hat{\Theta}_x = \Box_x + m^2$).

A generic solution of this equation is a sum of any solution of the homogeneous equation $\hat{\Theta}_x f(x) = 0$ (which usually is a superposition of plane waves) and any particular solution of the nonhomogeneous equation. Moreover, this particular solution is given by the Green's function:

$$f(x) = f_0 (x) + \int d^4 y \, G(x, y) j(y)$$

where $G(x, y)$ is the Green's function of the differential operator $\hat{\Theta}_x$ (which means that it satisfies $\hat{\Theta}_x G(x, y) = \delta^{(4)}(x - y)$ as a distribution), and $f_0 (x)$ is any solution of the homogeneous equation (a superposition of plane waves).

• Thanks for the detailed answer, but I really want to understand why it is referred to as a "source term"? Is it simply because it "sources" a solution for $f$ (i.e. it affects the form of the solution)? – Will Feb 23 '16 at 12:52
• @Will exactly. My point was that $f(x)$ depends on $j(x)$ linearly, so $j$ is kind of a source for nonzero $f$. You shouldn't expect the terminology to be 100% reasonable though. – Prof. Legolasov Feb 23 '16 at 21:06