Is measuring energy of photon released by electron enough to detect its previous state?

Question

I am referring to an experiment mentioned in Quantum Entanglement lecture: https://www.youtube.com/watch?v=VtBRKw1Ab7E (Starting from: 14:20)

In that lecture, as far as I understand, the description of the experiment is as follows:

Assume: "UP state" means electron's (assumed as a bar magnet) north pole is vertically upwards

1. Preparing the state: Prepare electron to be in certain state (an arbitrary angle) by placing it in a magnetic field at certain angle.

2. Turn off that magnetic field.

3. Detecting the state: Turn on another magnetic field to detect the state of the electron. The north pole of external magnet is vertically downwards and the south pole is vertically upwards. Detect the state of the electron's state (up or down) by detecting if the electron releases any photon or not. If the electron releases photon, then it is "concluded" that previously, the electron had its north pole downwards i.e. electron was in "down" state. If electron does not emit any photon, then it is assumed that previously it was in "up" state. That "photon" released always has same energy (which is equal to energy released when transitioning from "down" to "up"). This is true even in cases where electron has been prepared in some other angle (i.e. neither down nor up but something in between).

The argument made here is that, no matter what "state" (angle) the electron is prepared for (in step 1), while "detecting the state" we find that either electron releases no energy or some constant energy. So "detecting the state" results in either "up" or "down", but not any other state.

But just because electron emits constant energy (equal to transition from down to up), why should it be "concluded" that it was previously "down"? May be the photon has quantized energy, so no matter what state it previously was in, it always releases that particular amount of energy to get to state "up".

My question is: is it correct to assume electron was previously "down" just because it releases that particular amount of energy (equal to transitioning from down to up) to go to state "up"?

Answer 1

First, this is clearly only a thought experiment. The electron's magnetic moment is only about 60 μeV/T, so even for an enormous 10 T measurement field you'd have a very far-infrared 1.2 meV photon from a spin flip. Non-detection of one such photon would usually mean that you missed it, not necessarily that it wasn't there.

That said, the point of this exercise is to contrast the behavior of the electron with the behavior of a classical dipole. A classical dipole $\mu$ making an angle $\theta$ to a field $B$ has energy $-\mu B \cos\theta$; if you were to prepare three such dipoles parallel to, normal to, and antiparallel to the field, they'd release energy 0, $\mu B$, $2\mu B$ as they rotated into the favored position. But a single electron would only ever release $2\mu B$ — not $\mu B$, or $1.7\mu B$, or any other quantity.

It's not that the photon energy is quantized — if you want a particular photon energy, you just set the field strength $B$ to whatever you like. It's that the angle an electron makes with any particular axis is quantized, and can only take on one of exactly two values.