When can we handle a quantum field like a classical field?

Question

I am curious that, are there any criterion to justify the use of a classical field to describe a fundamentally quantum field? To rephrase in another way, when we can take the classical limit of a quantum field (I am not asking how to take the classical limit)?

For instance, usually in the slow-roll inflation, the scalar field is taken as classical when it is far from the bottom of potential. Why can we do this? Are there any argument to show the quantum effects are negligible?

I think this is also related to the use of wave-packet in quantum mechanics.

Answer 1

A clean way to make the concept of a classical field precise it to phrase things in terms of a quantum effective action: given a generating function of connected and renormalised Green functions, $W(J)$, with $$ e^{W(J)} = \int \mathcal{D}\phi\,e^{-I[\phi]+\int d^dx\,J\phi} $$ the quantum effective action, $\Gamma[\varphi]$, is the Legendre transform (when it exists), $$ W(J) = -\Gamma[\varphi]+\int J\varphi, \quad {\rm where}\quad \varphi(J)\equiv \frac{\delta W(J)}{\delta J}. $$ It is assumed here that $\varphi(J)$ can be inverted (at least within perturbation theory) in the sense that one can derive from it an explicit expression for $J(\varphi)$. So we must assume that $J(\varphi)$ exists and is single-valued. Then the quantum effective action contains exact information about the full quantum theory.

Suppose now that $\bar{\varphi}$ solves the full quantum equations of motion of $\Gamma[\varphi]$, that is: $$ \frac{\delta \Gamma[\varphi]}{\delta \varphi}\Big|_{\varphi=\bar{\varphi}}=0. $$ It then follows from the Legendre transform above that this solution for the full quantum effective action is the one-point function: $$ \frac{\delta W(J)}{\delta J}\Big|_{J=0}=\bar{\varphi}. $$ Everything would be consistent if in our original path integral we were computing quantum fluctuations around the full quantum corrected background, i.e. if we expanded $\phi=\bar{\varphi}+\tilde{\phi}$ and integrated over $\tilde{\phi}$ in the defining path integral of $W(J)$. This is a non-trivial requirement, but it is understood how to do this.

So the quantity $\bar{\varphi}$ is an exact onshell field that minimises $\Gamma(\varphi)$ and $\varphi$ is the corresponding offshell field that appears in the generic action $\Gamma(\varphi)$. Now comes a crucial point: information about the full quantum theory is contained in the tree level diagrams of $\Gamma(\varphi)$, i.e. in the "classical equations of motion" of $\Gamma(\varphi)$. Furthermore, when you compute $\Gamma(\varphi)$ within perturbation theory you will find that you can rearrange it in a loop expansion (i.e. expansion in $\hbar$, forget about Wilsonian effective actions here), $$ \Gamma(\varphi) = \sum_{\ell=0}^{\infty}\Gamma_{\ell}(\varphi), $$ where $\ell$ denotes the loop order. The $\ell=0$ term is usually a manifestly local action and the higher order ($\ell>0$) terms are often highly non-local (especially in massless theories). (For massive theories the seemingly non-local terms can be written as an infinite superposition of local terms but this is not possible for massless theories. It is this latter expansion that makes the link to Wilsonian renormalisation, because here this superposition is organised in terms of energy scales. So all these concepts are very closely and intimately related.)

So finally we can answer your question: when the higher loop ($\ell>0$) terms in the expansion of $\Gamma(\varphi)$ are negligible the full dynamics is essentially captured by the classical equations of motion of the $\ell=0$ term, $\Gamma_0(\varphi)$. It is this quantity that one usually identifies with the dynamics of the classical field $\varphi$, and, e.g., in the context of inflation $\varphi$ (or, onshell $\bar{\varphi}$) would be the inflaton. To leading order it is often the case that $\Gamma_0(\varphi)$ coincides with the bare action $I(\phi)$ (in form) when the latter is treated classically. This is why we can consider classical fields and their classical dynamics, and this has its origins in the full quantum theory. It is often the case however that in the literature people do not distinguish between $\Gamma_0(\varphi)$ and $I(\phi)$, and this is what causes all the confusion that many people have. At least this is my understanding.

Answer 2

Adressing @AlQuemist's comment concerning a physical answer, I'd like to mention the concept of mean-field instability (where $\bar{\varphi}$ from @Wakabaloola's post would be the mean-field). As an illustration, consider the following (complex) model action:

$$ S = \int \mathrm{d}t\; \left\{\sum_{\alpha=1,2}\left[ \phi_{\alpha}^{*}i\partial_t\phi_{\alpha}^{} - \tfrac{U}{2} \left| \phi_{\alpha}^{}\right|^{4} \right] -J(\phi_{1}^{*} \phi_{2}^{} + \phi_{2}^{*} \phi_{1}^{}) \right\}. $$

This could for example describe cold atoms in a double-well potential, with a coherent coupling $J$ and a contact interaction $U$. Following the procedure sketched by @Wakabaloola, that is, calculating

$$ \left.\frac{\delta \Gamma[\phi_\alpha, \phi_\alpha^*]}{\delta\phi_\alpha^*}\right|_{\phi_\alpha=\Phi_\alpha,\; \phi_\alpha^*=\Phi_\alpha^*}=0, $$

one can derive two so-called Gross-Pitaevskii equations

\begin{align} \begin{split} i\partial_ t \Phi_{1}^{} &= J \Phi_{2}^{} + U \left| \Phi_{1}^{}\right|^{2} \Phi_{1}^{}, \\ i\partial_ t \Phi_{2}^{} &= J \Phi_{1}^{} + U \left| \Phi_{2}^{}\right|^{2} \Phi_{2}^{}, \end{split} \end{align} which are the "classical" equations of motion in the sense discussed above ($\bar{\varphi} \equiv \Phi_\alpha$). Setting the complex mean-fields to $\Phi_{\alpha}^{} = \sqrt[]{N_{\alpha}}e^{i\theta_{\alpha}}$, these are equivalent to

\begin{align} \begin{split} \dot{z} &= 2J\,\sqrt[]{1-z^2}\sin{\theta}, \\ \dot{\theta} &= NUz - 2J\frac{z }{\sqrt[]{1-z^2}}\cos{\theta}, \end{split} \end{align} where $ N = N_1+N_2 = \mathrm{const.}, z = (N_1 - N_2)/N, \theta = \theta_2 - \theta_1$. See also reference [1], where they give the analogy of a classical non-rigid pendulum described by these two equations.

Now we want to address the question of the stability of this mean-field by looking for the fixed points of these equations. Evidently, a set of fixed points is given by $(z^*,\theta^*) = (0, n\pi)$, where $n$ indicates multiples of $\pi$ and the asterisk signifies the fixed point ($z$ and $\theta$ are real). Let's restrict to the specific point $(z^*,\theta^*) = (0, \pi)$ [2]. The Jacobian of the two equations evaluated at this point is

\begin{align} \mathcal{J}(0, \pi) = \begin{pmatrix} 0 & 2J\\ NU - 2J & 0 \end{pmatrix}, \end{align} which has eigenvalues

\begin{align} \lambda = \begin{cases} \pm \, 2Ji\,\sqrt[]{|\Lambda-1|},\quad \Lambda < 1, \\ \pm \,2J\,\sqrt[]{\Lambda-1},\quad \Lambda > 1. \end{cases} \end{align}

We have defined $\Lambda = NU/2J$. For $\Lambda > 1$, the eigenvalues are evidently real, which means we have a so-called hyperbolic fixed point. The crucial point is now this: Whenever we are in the vicinity of an unstable fixed point ($\mathrm{Re}\,\lambda > 0$), the mean-field description is known to fail completely. One can prove this, for example, by including fluctuations into the description, which turn out to be large precisely under this condition. Connecting to @Wakabaloola's post, this is where one would have to take into account higher contributions in $\hbar$ with $l>0$.

What we have described is a dynamical instability for a system that has the classical analogy of a pendulum. The physical picture emerging is simple: the point $(0, \pi)$ is like the one where the pendulum is "upside down". Importantly, the non-linearity coming from the interaction is responsible for the existence of this fixed point. For "small" interactions ($\Lambda < 1$), the non-linearity stabilizes this point, and one gets non-trivial stable oscillations that are reasonably well described by the mean-field. When the interaction becomes larger ($\Lambda > 1$), the stability breaks down, and (quantum) fluctuations become relevant. Intuitively, this has to do with the fact that within the unstable regime, the slightest perturbation can make the pendulum topple over in an arbitrary direction ("spontaneous symmetry breaking"). In order to be able to predict this direction, one would have to know "everything" about the randomness entering via the quantum fluctuations. Whenever fluctuations are dominant, one is in the "non-perturbative" regime, where only highly sophisticated resummation techniques involving infinitely many diagrams can be employed (at best).

[1] Smerzi, A., Fantoni, S., Giovanazzi, S. and Shenoy, S.R., 1997. Quantum coherent atomic tunneling between two trapped Bose-Einstein condensates. Physical Review Letters, 79(25), p.4950.

[2] Vardi, A. and Anglin, J.R., 2001. Bose-Einstein condensates beyond mean field theory: Quantum backreaction as decoherence. Physical Review Letters, 86(4), p.568.