# Detecting state of an electron

1. Jun 7, 2014

### RobikShrestha

I am referring to an experiment mentioned in Quantum Entanglement lecture:

(Starting from: 14:20)

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

Assume:
"up state" means 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"?

Last edited by a moderator: Sep 25, 2014
2. Jun 7, 2014

### Staff: Mentor

We also have conservation of energy; if the photon carries away some amount of energy then we know that the electron lost that amount. Thus, if we always get the same photon energy, we know that the electron only has two energy states, differing by the amount of energy in one photon.

3. Jun 7, 2014

### RobikShrestha

Is it possible that energy lost by electron does not depend on the angle it makes with the field?

4. Jun 7, 2014

### Staff: Mentor

Not only is it possible, that's how it is. The angle between the field we used for preparation and the angle we use for measurement does not affect the amount of energy released if the electron has to flip its spin to line up with the measuring field - it just affects the probability that the electron will have to flip its spin to line up with the measuring field. No flip, no photon; flip, photon.

There's some quantum weirdness going on here. If the preparation field is (for example) at an angle of 45 degrees from the vertical and the measuring field is vertical.... Then after the preparation we know the value of the spin around the 45-degree axis but, as far as the mathematical formalism of quantum mechanics is concerned, the spin around the vertical axis is completely undefined. It's not zero, it's not some value that we don't know, it just doesn't exist at all and it's meaningless to talk as if it does.

Only after we switch on the vertical measuring field and the electron interacts with it does it have a definite spin in the vertical direction (and then the previously known spin around the 45-degree axis becomes undefined).

5. Jun 7, 2014

### RobikShrestha

I thought we were trying to detect "previous state" with measurement. Like, trying to figure out what state electron was under 45-degree preparation field by measuring it with 90-degree field. I mean isn't that the whole point of measuring? Trying to figure out state of electron?

Seems like the only information we can get by measuring it in 90-degree field is whether electron's state had to be forced to align with the field or was it already aligned.
Isn't this some kind of limitation of experiment that we can't know the state without affecting the state? I mean does measurement under 90-degree field really answer the question of what state the electron was just before the field was turned on?

6. Jun 7, 2014

### Staff: Mentor

We're trying to detect the previous state; but the rules of quantum mechanics won't let us.

That's why we have to go through the preparation procedure to put the electron into a known state, namely spin up at 45 degrees. That state is the same as the state "probability X of finding the 90-degree spin to be up; probability Y of finding the 90-degree spin to be down" (where X and Y are values that we can calculate from the rules of QM).

7. Jun 7, 2014

### RobikShrestha

So, the state is just a superposition of all states with each state assigned a probability? That does make some sense.