What Does Feynman Mean When He Says Amplitude and Probabilities?

Question

In Feynman lectures on gravitation section 1.4, he tries to debate over whether one should quantize the gravitation or not. He provides a two-slit diffraction experiment with a gravity detector, which is assumed to be classical. The position of electron is described by amplitude, and it cannot be that a particle described by an amplitude has an interaction (gravitation here) described by probabilities. He then claims that if the gravity interacts through a field, it follows that the gravity field must have an amplitude corresponding to gravity field of an electron pass through upper slit and the other. It is confusing to use 'amplitude' and 'probabilities' to separate the concept of 'quantum' and 'classical'. As far as I'm concerned, Feynman's 'amplitude' means the wave function of the particle, on the other hand, the probabilities means that the observer does several measurements for independent experiments, and record the probabilities of an electron passing through the upper slit and passing through the lower slit.

I'm not sure about my opinion on Feynman's terms is correct? And he points out in page 15, that the amplitudes would become probabilities for very complex objects, which then confuses me, can anybody explain clearly what he means here?

Answer 1

I think the confusion starts from an interpretation of quantum mechanics I have seen many times applied to atomic orbitals or Young's double-slit experiment (YDSE).

In the former case the electron has a wave function $\psi_{nlm}\left(\vec r\right)$, while the latter at a fixed position on the screen, as two complex amplitude from the path's through each slit, $A_L$ and $A_U$.

The atomic orbital is often described to beginners as follows: well, it describes a probability cloud and if you were to, at some time $t_i$, measure its position, measuring $\vec r_i$, and then repeat this many many times, you would find the position distributed as $$ P(\vec r_i) = \left|\psi_{nlm}\left(\vec r\right)\right|^2. $$ This is a true statement (in theory: I don't think there is a way to do the measurement, though I am accepting input from atomic physicists).

People (especially those making chemistry videos on YouTube) then get the impression that an atom has a point electron randomly zipping around a stationary proton with aforementioned probability distribution.

This is just wrong. First: remember $\vec r $ is in the reduced mass coordinates—both the electron and the proton are in these weird orbitals, but ignoring, we can consider just the electron zipping around: it's too classical, and classical models just don't describe quantum—and Feynman is accurate when he uses "probability" to mean classical. It is classical.

It's wrong for several reasons. The ground state of hydrogen, e.g., is a stationary state. The time evolution is an unobservable global phase: $$\psi_{nlm}\left(\vec r, t\right) = \psi_{nlm}\left(\vec r\right)e^{-iE_{nlm}t/\hbar}. $$ With QM, you leave it at that. In a (qualitative) QFT interpretation, the electron is not a point particle, it's a non-localized quantum field with an amplitude, which as Feynman says: it's a quantum thing.

So your interpretation of the YDSE probability distribution is very similar, that a series of experiment would detect which slit the electron traverses.

That's just not how it works. If you know which slit it went through, there is no interference pattern. But there is interference, and the electron went through both slits, mostly with equal $|A_i|^2$ (at a point on the screen, there is a slow spatial variation from the diffraction pattern of each slit).

Then Feynman's question is about the gravity field of said electron—is it from a distributed source? What does that look like? If not, and you can determine the slit, what happens to the interference? IDK.

Answer 2

I believe your interpretation of Feynman is fully correct here. He is drawing a line between which objects in the theory need to have wavefunctions (amplitudes) and which can be described by ordinary probability distributions. Where the quantum things are the ones where ordinary probability distributions don't work and wavefunctions are needed.

Re-saying what I understand him to mean in a more maths-ish way, if gravitational fields are a classical thing then the density matrix for them will always be diagonal, but if they are quantum then that density matrix will (sometimes) need off-diagonal elements.