# How does determinism manifest out of QFT?

Classical electrodynamics is deterministic. QED is indeterministic, or probabilistically random. Yet they agree with each other? What am I missing?

• an analogy, a random variable (e.g a dice) is random (indeterministic), yet the averages (e.g ratio of tails and heads in a given sequence) can be deterministic (or more correctly stationary) – Nikos M. Jun 23 '14 at 4:15
• @NikosM. Would the indeterminacy of QED manifest as an (extremely small) error in the calculations of classical electrodynamics? Just as an average of dice rolls will rarely be precisely the probabilistic average? – Ethan Reesor Jun 23 '14 at 5:33
• hmm, this is not the type of indeterminacy implied in QFT (or QM). The type of indeterminacy of finite length effects (approximation issues) can in principle be lifted by a better approximation (taking more time), however QFT indeterminacies cannot be lifted this way (like one cannot predict the next coin toss consistently) – Nikos M. Jun 23 '14 at 5:36

Classical electrodynamics and quantum electrodynamics don't agree with each other in general. They are distinct, inequivalent theories. Your observation that classical electrodynamics is deterministic, unlike QED, is one sufficient proof that they're inequivalent.

At most, the expectation values of some observables in quantum electrodynamics obey the same equations as the equations obeyed by the corresponding classical quantities in classical electrodynamics. Whenever this is true, it may be proven simply by inserting the Heisenberg equations of motion of the quantum theory to the expectation value brackets, $\langle \dots \rangle$.

But this doesn't even hold for the general operators because $$\langle X\rangle \langle Y \rangle \neq \langle XY \rangle$$ So if you first replace the quantum observables by their classical counterparts and then you multiply these classical quantities, you get a different result than if you multiply the quantum observables first and then you compute their expectation values! For this reason, most of the nonlinear equations of motion in a quantum theory disagree with the classical theory even at the level of the expectation values!

The linear equations, like Maxwell's equations in the vacuum, agree in the sense of the expectation values. But quantum theories are not just about expectation values. Quantum theories probabilistically predict lots of things that aren't included in the classical theory at all.

QED and classical electrodynamics also happen to agree in some formulae for various cross sections in simple problems etc. (when the QED loop processes are ignored). Those agreements are sort of coincidences arising from the simplicity of the theories and integrability (solvability) of these simple problems.

• Isn't this 'coincidence' actually due to the fact that classical electrodynamics (as well as any classical counterpart of a quantum system) is connected to QED by the limit $\hbar\rightarrow 0$? I mean you are alluding too to loop corrections, but then why saying coincidence – TwoBs Jun 23 '14 at 19:14
• No, TwoBs, the tree-level agreement in the formula for the elastic cross sections etc. isn't guaranteed for any theory. Note that "tree level" corresponds to neglecting corrections that scale like $e^2 \hbar$. But aside from these quantum loops in QED, quantum electrodynamics describing a point charge moving in an electromagnetic field also has lots of "quantum dynamics" not captured by the classical theory. So you may write down the corresponding theory for a different potential, like $1/r^3$, and ignore the QFT loops, but the agreement will disappear. – Luboš Motl Jun 24 '14 at 4:48
• Aside from cross sections, let me mention another example, the Bohr model of the hydrogen atom. One could have gotten the $-E_0/n^2$ energy spectrum even in a classical theory with classical trajectories quantized by a simple rule in the old Bohr model. It agrees with quantum mechanics but you surely agree it is a coincidence. None of the agreement survives for the helium atom or any other system. What the classical field theory doesn't really capture are the wave functions for the charged particles that it treats as point masses in classical mechanics! That's the problem. – Luboš Motl Jun 24 '14 at 4:51
• I see your point. Yet I am not sure I would call it coincidence (I could argue that wkb in the limit $\hbar$ vanishing...). Anyway, I am curious about the your comment on the $1/r^3$ potential, which arises e.g. by exchanging a massless pseudoscalar mediator (like a pion) or a $F_{\mu\nu}$ rather than $A_\mu$. What do you expect to be fundamentally different? Even if non-renormalizable at the field level in case of $F_{\mu\nu}$, or unbounded at the QM level, this potential is OK as an effective one, and at low energy and large distances I do not see problems at classical or quantum level. – TwoBs Jun 24 '14 at 5:37
• Dear TwoBs, whether one uses the word "coincidence" may perhaps be a matter of taste but what isn't a matter of taste is the fact that for generic potentials, the behavior of quantum particles in external fields where distances as short as the (would-be) quantum wavelength matter simply cannot be correctly modeled by the classical point-like particle and the potentials and problems where some results like the cross section as a function of energy or the spectrum agrees are extremely special, which is why I call it a coincidence. – Luboš Motl Jun 24 '14 at 5:51