# Time evolution of a quantum field via classical field theory

How do quantum fields evolve in time? (Heisenberg Picture)

How does time evolution relate to the (E-L) equations of motion?

I’ve had this understanding that there is a duality between classical and quantum fields:

Take the fourier transform of a classical field, multiply it by creation operators, and then fourier transform back to get a corresponding quantum field.

Now here is my picture of time-evolution (*): We evolve a quantum field by going to its corresponding classical field, evolve the classical field (via the equations of motion), and then find the corresponding quantum field. Keep in mind that this is all using the creation operators of the free field.

However, in many QFT textbooks, the correspondence between quantum and classical fields is a function of time. This is because the creation operators are a function of time. So "My picture of time evolution (*)" is wrong in some way! What is the correct picture? I’ve had this idea of how to remedy the situation:

If $\hat a_{\mathbf{k}}^\dagger$ is the quantum field corresponding to the classical field $\varphi(\mathbf{x},0)=e^{i \mathbf{k} \cdot \mathbf{x}}$,

then, $\hat a_{\mathbf{k}}^\dagger(t)$ is the quantum field corresponding to the classical field $\varphi(\mathbf{x},t)$, where $\varphi(\mathbf{x},t)$ is obtained by applying the equations of motion (E-L) to the classical field $\varphi(\mathbf{x},0)$.

That is, $\hat a_{\mathbf{k}}^\dagger(t)$ is just $\hat a_{\mathbf{k}}^\dagger$ evolved in time in the sense according to “My picture of time evolution (*)". Because of the superposition principle, this would validate my picture of time evolution, because all fields can be represented by their fourier transforms, and time evolution is linear.

If this is also wrong, then how else can the equations of a motion of a classical field dictate the evolution of a quantum field?

• What "time evolution" is depends on whether you want to look at it in the Schrödinger, Heisenberg or interaction picture. Which are you looking at? (The fact that you want to evolve the fields excludes the Schrödinger picture, but the other two remain open) – ACuriousMind Oct 5 '14 at 20:52
• I would say the Heisenberg picture. – Dave Oct 5 '14 at 20:59
• Quantum "time evolution" is "unitary", meaning that the time (duration) parameter merely changes the phase of the quantum states (Schrodiger picture) or the phase of observables (Heisenberg picture). Actual "evolution" (meaning actual change of state) happens on a quantum measurement (measurement problem) – Nikos M. Oct 5 '14 at 21:27
• @NikosM.: So the Higgs doesn't really show up in the collisions of the LHC beams, it only happens because ATLAS and CMS are surrounding the intersection points? If I turn the power off on either of my detectors, would there be any Higgs created? – CuriousOne Oct 5 '14 at 21:50
• @NikosM.: Careful with these statements about phases - even in the Schrödinger picture, only Hamiltonian eigenstates evolve only with a phase, others are actually changing. And in the Heisenberg picture, the action of $\mathrm{e}^{\mathrm{i}Ht}$ from the left and its conjugate from the right on the observable is far from being just a phase in the general setting. – ACuriousMind Oct 5 '14 at 21:57

There is but one truth about time evolution in quantum mechanics:

The Hamiltonian is the generator of time translations.

In the Heisenberg picture, this means we directly borrow the quantized1 version of the evolution of observables on phase space, i.e.

$$\frac{\mathrm{d}}{\mathrm{d}t}A(\vec x,t) = \mathrm{i}[H,A(\vec x, t)]$$

holds for all operators/operator-valued distributions on the Hilbert space of our QFT. Now, solving this equation yields (see also BCH-formula)

$$A(\vec x,t) = \mathrm{e}^{\mathrm{i}H(t-t_0)} A(\vec x,t_0) \mathrm{e}^{-\mathrm{i}H(t-t_0)}$$

From this, it follows that the fields, which are operator-valued distributions, also must evolve like this. Since we implemented the quantized version of the classical time evolution, it follows that the classical equations of motion implied by it also hold for the quantum fields as an operator equation.

But nowhere here did we ever need classical pictures. We don't even care if $H$ comes from quantizing a classical system or if we just made it up. Quantum physics - and especially QFT - works without any recourse to the classical world. And that is good, because the classical world is supposed to emerge in certain limits from the quantum theory, which is more fundamental and has a broader range of phenomena.

1Here, quantization means replacing the Poisson bracket $\{\dot{},\dot{}\}$ by the commutator $\mathrm{i}[\dot{},\dot{}]$.

• I suppose it might be worth pointing out that you've set $\hbar=1$ here without explicitly stating it. – Kyle Kanos Oct 6 '14 at 2:46

Simply, the classical evolution does not dictate quantum evolution.

In a suitable sense (I have no time of explaining it here in details) you can obtain the classical dynamics from the quantum dynamics (even in a rigorous mathematical fashion in some cases) in the limit $\hslash\to 0$.

So it is really the contrary of what you say. This is actually natural: it is common folklore that classical dynamics is just an approximation of quantum dynamics when the quantum effects are negligible and not vice versa.

Properties of the classical dynamics (such as symmetries, dispersive effects...) may have a quantum counterpart, and thus their study is important. Nevertheless, having a well-defined classical dynamics is not sufficient to define a corresponding quantum dynamics.

• But the equations of motion of a quantum field are obtained via classical field theory (i.e. E-L equation)! Also, just the mere fact that we discuss "scalar" fields means that the dynamics are of $\phi:\mathbb{R}^4\to\mathbb{R}$, not some operator valued distribution. – Dave Oct 5 '14 at 20:58
• Context may help here: I'm trying to follow my book's derivation of the LSZ formula. The book discusses classical field theory first, and then writes Lagrangians for quantum systems and expects us to infer that everything just transfers over. – Dave Oct 5 '14 at 21:03
• The equation of motion of a quantum field are not obtained via the classical theory. Why do you think it is the case? In most of the realistic cases, QFT hamiltonians are not mathematically well defined. And that is a problem indeed. Nevertheless (at least when it is well-defined, or in theories with cut offs), it is the quantum Hamiltonian that generates the quantum dynamics (even of quantum fields obviously). – yuggib Oct 5 '14 at 21:03
• At least with the free-quantum field, the E-L equations are satisfied. (i.e. $(\square+m^2)\phi=0$). So the principle of least action holds at least in this case. – Dave Oct 5 '14 at 21:04
• "Why do I think it is the case"? Because we obtain the equations of motion via the E-L equation. At least in the free theory. – Dave Oct 5 '14 at 21:16