# Conjugate momentum for constant scalar field

I am reading Witten's Why Does Quantum Field Theory in Curved Spacetime Make Sense?, and I am caught up on what appears to be a straightforward computation. The discussion (on page six) centers around zero modes of the massless scalar field theory with Lagrangian density

$$\mathcal{L}(x) = -\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi$$

with signature $$(-++...+)$$. Witten points out that the constant mode $$\phi_0$$ is such a mode. I assume here he means $$\phi_0$$ truly does not depend on spacetime whatsoever. What I am stuck on is the subsequent computation of the conjugate momentum field $$\pi$$.

Using the expression

$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{\phi}}$$

with $$\dot{\phi} \equiv \partial_0 \phi$$, I find that

$$\pi = -g^{0\mu} \partial_\mu \phi.$$

Witten writes that the canonical conjugate to $$\phi_0$$, denoted $$\pi_0$$ should also be a constant mode. However, plugging $$\phi = \phi_0$$ into the above expression, I find $$\pi = 0$$. Of course this is constant, but this is not what I expected, since I cannot see how to sensibly impose canonical conjugation relations.

Where is the flaw in my reasoning? What is the correct way to carry out this analysis?

I haven't gotten around to reading that paper yet but zero modes are not pure numbers. Before we quantize a free scalar field in a finite box, we write it as $$$$\phi(\textbf{x}, t) = \sum_{\textbf{k}} \phi_{\textbf{k}}(t) e^{i\textbf{k} \cdot \textbf{x}}.$$$$ In this discrete sum (the $$k_i$$ might be $$2\pi n_i / L$$ for instance), there is a single term with no spatial dependence called the zero mode $$\phi_0(t)$$. But crucially it does have time dependence. Therefore, upon quantization, it will behave as a decoupled oscillator just like the other modes.

• In the case of a massless field though, wouldn't $\mathbf{k} = 0$ imply that $\omega = 0$, so really there is no time dependence for this particular Fourier mode? Commented Aug 15, 2022 at 11:46

I would like to try and answer my own question, which I was able to do taking hint from Connor's answer. The essential point is that the differential equation $$f''(x) = 0$$ has a qualitatively distinct solutions from $$f''(x) = -\lambda^2 f(x)$$ for $$\lambda > 0$$. It is, in fact, a special case that we need to treat separately.

We will look for zero modes of the Lagrangian in the OP. We assume a globally hyperbolic spacetime with Cauchy surface $$S$$ to serve as a setting for an initial value problem. We also assume a timelike Killing vector $$K$$ which serves as our time coordinate, with respect to which the spatial coordinates are orthogonal. This implies the metric satisfies

$$\partial_0 g_{\mu\nu} = 0, \quad g_{0i} = 0.$$

where latin indices $$i, j$$ imply spatial coordinates. The Euler-Lagrange equation for $$\mathcal{L}$$ gives

$$\square\phi \equiv g^{\mu\nu} \nabla_\mu \nabla_\nu \phi = 0$$

where $$\square$$ is the D'Alembertian. Following the procedure from Sean Carroll's Spacetime and Geometry (Sec. 9.4, pg 400), we can incorporate our conditions on $$g_{\mu\nu}$$ above to write

$$\partial_0^2 \phi = L_{\mathbf{x}}(\phi)$$

where $$L_{\mathbf{x}}$$ is a 2nd-order differential operator with respect to spatial coordinates. Precisely because of the massless setting, (and no coupling with the curvature), $$L_{\mathbf{x}} (\phi) = 0$$ if $$\phi$$ has no spatial dependence. As we often do, we will seek a separable solution $$\phi(t, \mathbf{x}) = T(t) X(\mathbf{x})$$, where in particular the time portion satisfies

$$T''(t) = -\omega^2 T.$$

Setting $$\omega = 0$$ defines what we mean by searching for a zero mode, in this context. The spatial part of the zero mode satisfies $$L_{\mathbf{x}} (X) = 0$$, which can be satisfied by assuming $$X = \text{const}$$. Solving the simple equation for the time part $$T$$, we find that the zero mode $$\phi_{\rm ZM}$$ satisfies

$$\phi_{\rm ZM} (t) = \dot{\phi}_0 t + \phi_0$$

for initial conditions $$\dot{\phi}_0$$ and $$\phi_0$$.

The major point, as Connor pointed out, is that $$\dot{\phi}_{\rm ZM}$$ is not zero. Hence, neither is the conjugate momentum $$\pi_{\rm ZM}$$, though both $$\phi_{\rm ZM}$$ and $$\pi_{\rm ZM}$$ lack spatial dependence. Upon quantization, the only natural choice for equal-time commutation relations is

$$[\phi_{\rm ZM}(t), \pi_{\rm ZM}(t)] = i.$$

Note the lack of spatial dirac delta $$\delta(\mathbf{x} - \mathbf{x'})$$, which makes sense because if the classical fields have no spatial dependence, neither should the quantized versions. Therefore, this zero mode provides a representation of a single harmonic oscillator.